Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion...

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Research Article Approximation by -Transformation of Double Walsh-Fourier Series to Multivariable Functions Yi Zhao 1 and Dansheng Yu 2 1 Department of Mathematics, Hangzhou Dianzi University, Hangzhou, Zhejiang 310018, China 2 Department of Mathematics, Hangzhou Normal University, Hangzhou, Zhejiang 310036, China Correspondence should be addressed to Yi Zhao; [email protected] Received 20 January 2014; Accepted 9 March 2014; Published 22 April 2014 Academic Editors: D. G. Costa, C. Guti´ errez, and D.-X. Zhou Copyright © 2014 Y. Zhao and D. Yu. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We study the Walsh series expansion of multivariate functions in (1 ≤ ≤ ∞) and, in particular, in Lip(, ). e rate of uniform approximation by T-transformation of rectangular partial sums of double Walsh to these functions is investigated. By extending the concepts of rest (head) bounded variation series, which was introduced by Leindler (2004), we generalize the related results of M´ oricz and Rhoades (1996), Nagy (2012). Our results can be applied to many summability methods, including the N¨ orlund summability and weighted summability. 1. Introduction Let 0 () be the function defined on := 0 = [0, 1) by 0 () = { { { { { { { 1, ∈ [0, 1 2 ), −1, ∈[ 1 2 , 1) , 0 ( + 1) = 0 () . (1) e Rademacher system is defined by () = 0 (2 ) . (2) Let 0 () = 1, () = ∞ ∏ =0 ( ()) (3) be the Walsh functions, where ∈ [0, 1), ∈ N with the dyadic expansion =∑ ∞ =0 2 −−1 and =∑ ∞ =0 2 , respectively; here , ∈ {0, 1} for , = 0,1,.... We also write ⌈⌉ := {max , ̸ = 0}. e idea of using products of Rademacher’s functions to construct the Walsh system originated from Paley [1]. As an important orthonormal bases, Walsh functions have most of the properties of Fourier series but are more suited to nonlinear studies. If a zero-memory nonlinear transformation is applied to a Walsh series, the output series can be derived by simple algebraic processes. Corrington [2] proved that nonlinear differential and integral equations can be solved by Walsh series. Meanwhile, the Walsh functions are of great practical interest. ey have many applications in signal processing [3], dynamic systems, identification, control [4, 5], and so on. In the above-mentioned issues, Walsh series expansion of certain function and its convergence to that function play very important roles. In this paper, we are interested in the Walsh expansion of multivariate functions and discuss the convergence of its T-transformation to these functions (we mention here that, in order to avoid notational difficulties, we restrict ourselves to the case of bivariate functions). Furthermore, the results are applied to some summability methods. e Walsh-Dirichlet kernels and Walsh-Fej´ er kernels are defined by () = −1 ∑ =0 () , 0 () := 0, (4) () = 1 ∑ =1 () , ( ∈ N), 0 () = 0. (5) Hindawi Publishing Corporation ISRN Mathematical Analysis Volume 2014, Article ID 713175, 14 pages http://dx.doi.org/10.1155/2014/713175

Transcript of Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion...

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Research ArticleApproximation by 119879-Transformation of Double Walsh-FourierSeries to Multivariable Functions

Yi Zhao1 and Dansheng Yu2

1 Department of Mathematics Hangzhou Dianzi University Hangzhou Zhejiang 310018 China2Department of Mathematics Hangzhou Normal University Hangzhou Zhejiang 310036 China

Correspondence should be addressed to Yi Zhao mathyizhaogmailcom

Received 20 January 2014 Accepted 9 March 2014 Published 22 April 2014

Academic Editors D G Costa C Gutierrez and D-X Zhou

Copyright copy 2014 Y Zhao and D Yu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study the Walsh series expansion of multivariate functions in 119871119901 (1 le 119901 le infin) and in particular in Lip(120572 119901) The rate ofuniform approximation by T-transformation of rectangular partial sums of double Walsh to these functions is investigated Byextending the concepts of rest (head) bounded variation series which was introduced by Leindler (2004) we generalize the relatedresults ofMoricz andRhoades (1996)Nagy (2012) Our results can be applied tomany summabilitymethods including theNorlundsummability and weighted summability

1 Introduction

Let 1199030(119909) be the function defined on 119868 = 1198680 = [0 1) by

1199030 (119909) =

1 119909 isin [01

2)

minus1 119909 isin [1

2 1)

1199030 (119909 + 1) = 1199030 (119909) (1)

The Rademacher system is defined by

119903119899 (119909) = 1199030 (2119899119909) (2)

Let

1199080 (119909) = 1 119908119899 (119909) =

infin

prod

119896=0

(119903119896 (119909))119899119896 (3)

be the Walsh functions where 119909 isin [0 1) 119899 isin N withthe dyadic expansion 119909 = sum

infin

119894=01199091198942minus119894minus1 and 119899 = sum

infin

119895=01198991198952119895

respectively here 119909119894 119899119895 isin 0 1 for 119894 119895 = 0 1 We alsowrite lceil119899rceil = max 119895 119899119895 = 0 The idea of using productsof Rademacherrsquos functions to construct the Walsh systemoriginated from Paley [1]

As an important orthonormal bases Walsh functionshave most of the properties of Fourier series but are moresuited to nonlinear studies If a zero-memory nonlinear

transformation is applied to a Walsh series the output seriescan be derived by simple algebraic processes Corrington [2]proved that nonlinear differential and integral equations canbe solved by Walsh series Meanwhile the Walsh functionsare of great practical interestThey have many applications insignal processing [3] dynamic systems identification control[4 5] and so on

In the above-mentioned issues Walsh series expansionof certain function and its convergence to that function playvery important roles In this paper we are interested in theWalsh expansion of multivariate functions and discuss theconvergence of its T-transformation to these functions (wemention here that in order to avoid notational difficultieswe restrict ourselves to the case of bivariate functions)Furthermore the results are applied to some summabilitymethods

The Walsh-Dirichlet kernels and Walsh-Fejer kernels aredefined by

119863119899 (119909) =

119899minus1

sum

119896=0

119908119896 (119909) 1198630 (119909) = 0 (4)

119870119899 (119909) =1

119899

119899

sum

119896=1

119863119896 (119909) (119899 isin N) 1198700 (119909) = 0 (5)

Hindawi Publishing CorporationISRN Mathematical AnalysisVolume 2014 Article ID 713175 14 pageshttpdxdoiorg1011552014713175

2 ISRNMathematical Analysis

It is known that [6]

1198632119899 (119909) = 2119899 119909 isin 119868119899

0 119909 isin 119868 119868119899(6)

where 119868119899 denotes the dyadic interval in [0 1) defined by 119868119899 =[(1198942119899) ((119894 + 1)2

119899)] and 119894 = 0 1 2119899 minus 1 119899 isin N

In addition we point out that the standard representa-tions for the Walsh-Dirichlet kernel [7 8] are

1198632119899+119895 (119909) = 1198632119899 (119909) + 119903119899 (119909)119863119895 (119909) (7)

1198632119895+119894 (119909) minus 1198632119895+1 (119909) = minus1199082119895+1minus1 (119909)1198632119895minus119894 (119909) (8)

For the Walsh-Fejer kernel 119870119894 let 119894 ge 1 Yano [9] provedthat

100381710038171003817100381711987011989410038171003817100381710038171le 2 (9)

Let 119879 = 119905119898119899119895119896 be a doubly infinite matrix It is saidto be doubly triangular if 119905119898119899119895119896 = 0 for 119895 gt 119898 or 119896 gt 119899In the recent research [10] the authors established necessaryconditions for a general inclusion theorem involving a pair ofdoubly triangular matrices

Given a double sequence 119904119895119896 119895 119896 = 0 1 of complexnumbers the 119898119899th term of the 119879-transformation of 119904119895119896 isdefined by

119905119898119899 =

119898

sum

120583=0

119899

sum

]=0119905119898119899120583]119904120583] (10)

If sum119898120583=0

sum119899

]=0 119905119898119899120583] = 1 then we say that 119879 is normalLet 119875 = 119901119895119896 119895 119896 = 0 1 be a double sequence of

nonnegative numbers 11990100 gt 0 Taking

119905119898119899119895119896 =

119901119898minus119895119899minus119896

119875119898119899

0 le 119895 le 119898 0 le 119896 le 119899

119905119898119899119895119896 = 0 if 119895 gt 119898 or 119896 gt 119899(11)

where 119875119898119899 = sum119898

119895=0sum119899

119896=0119901119895119896 Then the corresponding 119905119898119899 is

known as the Norlund means The Cesaro summability oforders 120574 120575 gt minus1 denoted by (119862 120574 120575) is a special case of theNorlund summability with

119901119895119896 = 119860120574minus1

119895119860120575minus1

119896 119895 119896 = 0 1 2

119860120574

119897= (

120574 + 119897

119897) =

(120574 + 1) (120574 + 2) sdot sdot sdot (120574 + 119897)

119897

(12)

for 119897 = 1 2 and 1198601205740= 1 In this case

119875119898119899 = 119860120574

119898119860120575

119899 119898 119899 = 0 1 2 (13)

If we take

119905119898119899119895119896 =

119901119895119896

119875119898119899

0 le 119895 le 119898 0 le 119896 le 119899

119905119898119899119895119896 = 0 if 119895 gt 119898 or 119896 gt 119899(14)

the corresponding 119905119898119899 is the well-knownRieszmeans of 119904119895119896Let 119871119901 (1 le 119901 le infin) denote the Lebesgue function spaces

on the torus 1198682 that is119891 isin 119871119901(1198682)The doubleWalsh (Walsh-

Fourier) series of such function is defined by

119891 (119909 119910) sim 119878 (119909 119910) =

infin

sum

119894=0

infin

sum

119895=0

119891 (119894 119895) 119908119894 (119909)119908119895 (119910) (15)

and the 119906Vth rectangular partial sum of 119878 is

119878119906V (119909 119910) =

119906minus1

sum

119894=0

Vminus1

sum

119895=0

119891 (119894 119895) 119908119894 (119909) 119908119895 (119910) (16)

where

119891 (119894 119895) = ∬

1

0

119891 (119904 119905) 119908119894 (119904) 119908119895 (119905) 119889119904 119889119905 (17)

The119898119899th T-transformation of 119878119906V is defined by

119879119898119899 = 119879119898119899 (119891 119909 119910) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119878119906V (119909 119910) (18)

By (16) we have

119879119898119899 (119891 119909 119910) = ∬

1

0

119891 (119909 + 119904 119910 + 119905)119870119898119899 (119904 119905) 119889119904 119889119905 (19)

where

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905) 119898 119899 = 0 1

(20)

119863119906(119904) and 119863V(119905) are the Walsh-Dirichlet kernels in terms of119906 and V respectively

For any function 119891 isin 119871119901(1198682) when 119879 is normal

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

times (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

= ∬

1

0

119870119898119899 (119904 119905) (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(21)

Recall that the modulus of continuity of the function119891(119909 119910) isin 119871119901(1198682) 1-periodic in each variable is defined by

120596119901(119891 120575) = sup

|119905|le120575

1003817100381710038171003817119891 (119909 + 119906 119910 + V) minus 119891 (119909 119910)1003817100381710038171003817119901

|119905| = radic1199062 + V2 120575 ge 0

(22)

For each 120572 gt 0 the Lipschitz classes in 119871119901 are defined by

119871119894119901 (120572 119901) = 119891 isin 119871119901 120596119901(119891 120575) = O (120575

120572) (23)

ISRNMathematical Analysis 3

The (total) modulus of continuity of function 119891(119909 119910) isin119871119901(1198682) 1-periodic in each variable is defined by

120596119901

12(119891 1205751 1205752) = sup 1003817100381710038171003817119891 (119909 + 119906 119910 + V) minus 119891 (119909 + 119906 119910)

minus119891 (119909 119910 + V) + 119891 (119909 119910)1003817100381710038171003817119901

|119906| le 1205751 |V| le 1205752

(24)

It is easy to verify that there is a constantC gt 0 such that

120596119901

12(119891 1205751 1205752) le C (120596

119901(119891 1205751) + 120596

119901(119891 1205752)) (25)

Moricz and Siddiqi [11] studied the rate of uniformapproximation by Norlund means of Walsh (Walsh-Fourier)series of 119891 isin 119871119901[0 1) Later Moricz and Rhoades [12] stud-ied the corresponding approximation problem by weightedmeans of Walsh-Fourier series Their main results in [12] canbe read as follows

Theorem A Let 119891 isin 119871119901 1 le 119901 le infin 119899 = 2119898+ 119896 1 le 119896 le

2119898 119898 ge 1(i) If 119901119896 is nondecreasing and satisfies the condition

119899119901119899

119875119899

= O (1) (26)

then1003817100381710038171003817119905119899 (119891) minus 119891

1003817100381710038171003817119901

le3

119875119899

119898minus1

sum

119895=0

21198951199012119895+1minus1120596119901 (119891 2

minus119895) + O (120596119901 (119891 2

minus119898))

(27)

(ii) If 119901119896 is nonincreasing then

1003817100381710038171003817119905119899 (119891) minus 1198911003817100381710038171003817119901le

3

119875119899

119898minus1

sum

119895=0

21198951199012119895120596119901 (119891 2

minus119895) + O (120596119901 (119891 2

minus119898))

(28)

Theorem B Let 119891 isin 119871119894119901(120572 119901) for 120572 gt 0 and 1 le 119901 le infinIf 119901119896 is nondecreasing then one has the following estimates

1003817100381710038171003817119905119899 (119891) minus 1198911003817100381710038171003817119871119901

=

O (119899minus120572) 0 lt 120572 lt 1

O (119899minus1 log 119899) 120572 = 1

O (119899minus1) 120572 gt 1

(29)

For any fourfold sequence 119886119898119899119895119896 write

Δ 11119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895+1119896 minus 119886119898119899119895119896+1 + 119886119898119899119895+1119896+1

Δ 01119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895119896+1

Δ 10119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895+1119896

(30)

The sequence 119886119898119899119895119896 is called nondecreasing if it isnondecreasing in both 119895 and 119896 that is Δ 01119886119898119899119895119896 le 0 andΔ 10119886119898119899119895119896 le 0 for every 119895 119896 = 0 1 The nonincreasing caseis defined analogously

Recently Nagy [13] did some research on the approxima-tion by Norlund means of double Walsh-Fourier series forLipschitz functions and generalizedTheorems A and B to thefunctions of two variablesWe present one of themain resultsin [13] here

Theorem C Let 119891 isin 119871119894119901(120572 119901) for some 120572 gt 0 and 1 le 119901 le

infin let 119902119895119896 be a double sequence of nonnegative numbers suchthat it is nondecreasing Δ 11119902119895119896 is of fixed sign and satisfies theregularity condition

(119898 + 1) (119899 + 1) 119902119898119899

119876119898119899

= O (1)

119876119898119899 =

119898

sum

119894=0

119898

sum

119895=0

119902119894119895 (119898 119899 = 0 1 )

(31)

then1003817100381710038171003817119882119898119899 (119891) minus 119891

1003817100381710038171003817119871119901

le

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(32)

where

119882119898119899 (119891 119909 119910) =1

119876119898119899

119898

sum

119894=0

119898

sum

119895=0

119902119898119899119898minus119894119899minus119895119878119894119895 (119909 119910) (33)

and 119878119894119895(119909 119910) is defined as in (16)

We know that in the theory of Fourier series it is ofmain interest how to approximate the function from thepartial sums of its Fourier series The purpose of the presentpaper is to get the rate of uniform approximation by 119879-transformation with doubly triangular We give the outlineof the paper In Section 2 we state the main results Someauxiliary lemmas are given in Section 3 and the proofs ofthe main theorems are presented in Section 4 Our newresults can be applied tomany classical summability methodssuch as Norlund summability and Riesz summability As animportant application we will apply them to the Norlundsummability and weighted means in Section 5 We will seethat not only Theorems A B and C are corollaries ofour results but also some other new types of estimates arepresented in this paper

2 The Main Results

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numbers tending tozero is called rest bounded variation or briefly 120572119899 isin RBVS ifthere is a constant119870(120572119899) only depending on 120572119899 such that

infin

sum

119896=119898

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (34)

holds for all natural numbers119898

4 ISRNMathematical Analysis

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numberstending to zero is called head bounded variation or briefly120572119899 isin HBVS if there is a constant 119870(120572119899) only depending on120572119899 such that

119898

sum

119896=0

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (35)

for all natural numbers 119898 or only for all 119898 le 119873 if thesequence 120572119899 has only finite nonzero terms and the lastnonzero term is 119886119899119873

Remark AThe definitions of RBVS andHBVS are introducedby Leindler [14] to generalize the monotonicity conditionson sequences In fact RBVS and HBVS generalized mono-tone nonincreasing sequences and monotone nondecreasingsequences respectively

Remark B Since it involves a sequence119870(120572119899) there should bea constant119870 such that 0 lt 119870(120572119899) le 119870

Now we extend the concepts of RBVS and HBVS to thedouble sequences as follows

Definition 1 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDRBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

infin

sum

120583=119895

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

infin

sum

]=119896

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

infin

sum

120583=119895

infin

sum

]=119896

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870 (120572119898119899) 119886119898119899119895119896 (36)

Definition 2 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDHBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

119895

sum

120583=0

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

119896

sum

]=0

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

119895

sum

120583=0

119896

sum

]=0

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870(120572119898119899) 119886119898119899119895119896 (37)

It is also required that the sequence 119870(120572119898119899) is boundedthat is there is a positive constant 119870 such that 0 lt 119870(120572119898119899) le

119870We state our main theorems as follows

Theorem 3 Let 119891 isin 119871119901 119905119898119899119906V be a a doubly normaltriangular matrix of nonnegative numbers 119905119898119899119906V isin DRBVSand let 119864 = lceil119898rceil 119865 = lceil119899rceil Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(38)

Also if 119891 isin 119871119901 119905119898119899119906V is doubly normal triangular ofnonnegative numbers 119905119898119899119906V isin DHBVS then one has theconclusion as follows

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

times (120596119901(119891 2minus119896) +C120596

119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(39)

Theorem 4 Suppose that 119891 isin 119871119894119901(119901 120572) for 120572 gt 0 1 le 119901 le

+infin Let 119905119898119899119906V be nonnegative doubly normal triangular and119905119898119899119906V isin DHBVS satisfying 119898119899119905119898119899119898119899 = O(1) Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(40)

While for 119891 isin 119871119894119901(119901 120572) and for 120572 gt 0 1 le 119901 le

+infin let 119905119898119899119906V be nonnegative normal doubly triangular and119905119898119899119906V isin DRBVS one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901= O (1)

119864minus1

sum

119896=0

119865minus1

sum

119897=0

(2119896(1minus120572)+119897

+ 2119896+119897(1minus120572)

) 11990511989811989921198962119897

(41)

ISRNMathematical Analysis 5

3 Lemmas and Proofs

Lemma 1 Suppose that 119905119898119899119906V is doubly normal triangularand 119863119906 is defined as (4) write 119864 = lceil119898rceil 119865 = lceil119899rceil then onehas

1198601 =

2119864minus1

sum

119906=0

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904) 1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905)

times (21198961198702119896 (119904)

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+1 (119904) 1199082119897+1minus1 (119905)

times (21198971198702119897 (119905)

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

+

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

(42)

Proof We can rewrite 1198601 as

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+119894 (119904)1198632119897+119895 (119905) (43)

and keep dividing the above into 4 parts

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895) (1198632119896+119894 (119904) minus 1198632119896+1 (119904))

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119897+1 (119905)

+

119864minus1

sum

119896=0

119865minus1

sum

l=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))1198632119896+1 (119904)

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

= 11986011 + 11986012 + 11986013 + 11986014

(44)

By using (8) we have

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1199082119896+1minus1 (119904)

times 1198632119896minus119894 (119904) 1199082119897+1minus1 (119905) 1198632119897minus119895 (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896

sum

119894=1

2119897

sum

119895=1

119905119898119899(2119896+1minus119894)(2119897+1minus119895)1199082119896+1minus1 (119904)

times 119863119894 (119904) 1199082119897+1minus1 (119905) 119863119895 (119905)

(45)

Applying double Abelrsquos transformation

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (

2119896

sum

1198941=1

2119897

sum

1198951=1

1198631198941(119904)1198631198951

(119905) 11990511989811989921198962119897

+

2119897

sum

1198951=1

1198631198951(119905)

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)2119897

6 ISRNMathematical Analysis

times

119894

sum

1198941=1

1198631198941(119904)

+

2119896

sum

1198941=1

1198631198941(119904)

2119897minus1

sum

119895=1

Δ 011199051198981198992119896(2119897+1minus119895)

times

119895

sum

1198951=1

1198631198951(119905)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times

119894

sum

1198941=1

1198631198941(119904)

119895

sum

1198951=1

1198631198951(119905))

(46)

Next (5) leads to

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904)1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(47)

Using (8) (5) and Abelrsquos transformation similar to theestimate of 11986011 (more easily actually)

11986012 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905) 21198961198702119896 (119904)

times

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1

times

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904)

(48)

Analogously

11986013 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 1198632119896+1 (119904) 21198971198702119897 (119905)

times

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+11199082119896+1minus1 (119904)

times

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 10119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904)

(49)

Combining the above estimates of 1198601119894 (119894 = 1 2 3 4) weobtain the conclusion of Lemma 1

Lemma 2 Let 119905119898119899119906V be the same matrix as in Lemma 1 thenone has

1198602 =

2119864minus1

sum

119906=0

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865(119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times119894119895119870119894 (119904) 119870119895 (119905))

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

2 ISRNMathematical Analysis

It is known that [6]

1198632119899 (119909) = 2119899 119909 isin 119868119899

0 119909 isin 119868 119868119899(6)

where 119868119899 denotes the dyadic interval in [0 1) defined by 119868119899 =[(1198942119899) ((119894 + 1)2

119899)] and 119894 = 0 1 2119899 minus 1 119899 isin N

In addition we point out that the standard representa-tions for the Walsh-Dirichlet kernel [7 8] are

1198632119899+119895 (119909) = 1198632119899 (119909) + 119903119899 (119909)119863119895 (119909) (7)

1198632119895+119894 (119909) minus 1198632119895+1 (119909) = minus1199082119895+1minus1 (119909)1198632119895minus119894 (119909) (8)

For the Walsh-Fejer kernel 119870119894 let 119894 ge 1 Yano [9] provedthat

100381710038171003817100381711987011989410038171003817100381710038171le 2 (9)

Let 119879 = 119905119898119899119895119896 be a doubly infinite matrix It is saidto be doubly triangular if 119905119898119899119895119896 = 0 for 119895 gt 119898 or 119896 gt 119899In the recent research [10] the authors established necessaryconditions for a general inclusion theorem involving a pair ofdoubly triangular matrices

Given a double sequence 119904119895119896 119895 119896 = 0 1 of complexnumbers the 119898119899th term of the 119879-transformation of 119904119895119896 isdefined by

119905119898119899 =

119898

sum

120583=0

119899

sum

]=0119905119898119899120583]119904120583] (10)

If sum119898120583=0

sum119899

]=0 119905119898119899120583] = 1 then we say that 119879 is normalLet 119875 = 119901119895119896 119895 119896 = 0 1 be a double sequence of

nonnegative numbers 11990100 gt 0 Taking

119905119898119899119895119896 =

119901119898minus119895119899minus119896

119875119898119899

0 le 119895 le 119898 0 le 119896 le 119899

119905119898119899119895119896 = 0 if 119895 gt 119898 or 119896 gt 119899(11)

where 119875119898119899 = sum119898

119895=0sum119899

119896=0119901119895119896 Then the corresponding 119905119898119899 is

known as the Norlund means The Cesaro summability oforders 120574 120575 gt minus1 denoted by (119862 120574 120575) is a special case of theNorlund summability with

119901119895119896 = 119860120574minus1

119895119860120575minus1

119896 119895 119896 = 0 1 2

119860120574

119897= (

120574 + 119897

119897) =

(120574 + 1) (120574 + 2) sdot sdot sdot (120574 + 119897)

119897

(12)

for 119897 = 1 2 and 1198601205740= 1 In this case

119875119898119899 = 119860120574

119898119860120575

119899 119898 119899 = 0 1 2 (13)

If we take

119905119898119899119895119896 =

119901119895119896

119875119898119899

0 le 119895 le 119898 0 le 119896 le 119899

119905119898119899119895119896 = 0 if 119895 gt 119898 or 119896 gt 119899(14)

the corresponding 119905119898119899 is the well-knownRieszmeans of 119904119895119896Let 119871119901 (1 le 119901 le infin) denote the Lebesgue function spaces

on the torus 1198682 that is119891 isin 119871119901(1198682)The doubleWalsh (Walsh-

Fourier) series of such function is defined by

119891 (119909 119910) sim 119878 (119909 119910) =

infin

sum

119894=0

infin

sum

119895=0

119891 (119894 119895) 119908119894 (119909)119908119895 (119910) (15)

and the 119906Vth rectangular partial sum of 119878 is

119878119906V (119909 119910) =

119906minus1

sum

119894=0

Vminus1

sum

119895=0

119891 (119894 119895) 119908119894 (119909) 119908119895 (119910) (16)

where

119891 (119894 119895) = ∬

1

0

119891 (119904 119905) 119908119894 (119904) 119908119895 (119905) 119889119904 119889119905 (17)

The119898119899th T-transformation of 119878119906V is defined by

119879119898119899 = 119879119898119899 (119891 119909 119910) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119878119906V (119909 119910) (18)

By (16) we have

119879119898119899 (119891 119909 119910) = ∬

1

0

119891 (119909 + 119904 119910 + 119905)119870119898119899 (119904 119905) 119889119904 119889119905 (19)

where

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905) 119898 119899 = 0 1

(20)

119863119906(119904) and 119863V(119905) are the Walsh-Dirichlet kernels in terms of119906 and V respectively

For any function 119891 isin 119871119901(1198682) when 119879 is normal

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

times (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

= ∬

1

0

119870119898119899 (119904 119905) (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(21)

Recall that the modulus of continuity of the function119891(119909 119910) isin 119871119901(1198682) 1-periodic in each variable is defined by

120596119901(119891 120575) = sup

|119905|le120575

1003817100381710038171003817119891 (119909 + 119906 119910 + V) minus 119891 (119909 119910)1003817100381710038171003817119901

|119905| = radic1199062 + V2 120575 ge 0

(22)

For each 120572 gt 0 the Lipschitz classes in 119871119901 are defined by

119871119894119901 (120572 119901) = 119891 isin 119871119901 120596119901(119891 120575) = O (120575

120572) (23)

ISRNMathematical Analysis 3

The (total) modulus of continuity of function 119891(119909 119910) isin119871119901(1198682) 1-periodic in each variable is defined by

120596119901

12(119891 1205751 1205752) = sup 1003817100381710038171003817119891 (119909 + 119906 119910 + V) minus 119891 (119909 + 119906 119910)

minus119891 (119909 119910 + V) + 119891 (119909 119910)1003817100381710038171003817119901

|119906| le 1205751 |V| le 1205752

(24)

It is easy to verify that there is a constantC gt 0 such that

120596119901

12(119891 1205751 1205752) le C (120596

119901(119891 1205751) + 120596

119901(119891 1205752)) (25)

Moricz and Siddiqi [11] studied the rate of uniformapproximation by Norlund means of Walsh (Walsh-Fourier)series of 119891 isin 119871119901[0 1) Later Moricz and Rhoades [12] stud-ied the corresponding approximation problem by weightedmeans of Walsh-Fourier series Their main results in [12] canbe read as follows

Theorem A Let 119891 isin 119871119901 1 le 119901 le infin 119899 = 2119898+ 119896 1 le 119896 le

2119898 119898 ge 1(i) If 119901119896 is nondecreasing and satisfies the condition

119899119901119899

119875119899

= O (1) (26)

then1003817100381710038171003817119905119899 (119891) minus 119891

1003817100381710038171003817119901

le3

119875119899

119898minus1

sum

119895=0

21198951199012119895+1minus1120596119901 (119891 2

minus119895) + O (120596119901 (119891 2

minus119898))

(27)

(ii) If 119901119896 is nonincreasing then

1003817100381710038171003817119905119899 (119891) minus 1198911003817100381710038171003817119901le

3

119875119899

119898minus1

sum

119895=0

21198951199012119895120596119901 (119891 2

minus119895) + O (120596119901 (119891 2

minus119898))

(28)

Theorem B Let 119891 isin 119871119894119901(120572 119901) for 120572 gt 0 and 1 le 119901 le infinIf 119901119896 is nondecreasing then one has the following estimates

1003817100381710038171003817119905119899 (119891) minus 1198911003817100381710038171003817119871119901

=

O (119899minus120572) 0 lt 120572 lt 1

O (119899minus1 log 119899) 120572 = 1

O (119899minus1) 120572 gt 1

(29)

For any fourfold sequence 119886119898119899119895119896 write

Δ 11119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895+1119896 minus 119886119898119899119895119896+1 + 119886119898119899119895+1119896+1

Δ 01119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895119896+1

Δ 10119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895+1119896

(30)

The sequence 119886119898119899119895119896 is called nondecreasing if it isnondecreasing in both 119895 and 119896 that is Δ 01119886119898119899119895119896 le 0 andΔ 10119886119898119899119895119896 le 0 for every 119895 119896 = 0 1 The nonincreasing caseis defined analogously

Recently Nagy [13] did some research on the approxima-tion by Norlund means of double Walsh-Fourier series forLipschitz functions and generalizedTheorems A and B to thefunctions of two variablesWe present one of themain resultsin [13] here

Theorem C Let 119891 isin 119871119894119901(120572 119901) for some 120572 gt 0 and 1 le 119901 le

infin let 119902119895119896 be a double sequence of nonnegative numbers suchthat it is nondecreasing Δ 11119902119895119896 is of fixed sign and satisfies theregularity condition

(119898 + 1) (119899 + 1) 119902119898119899

119876119898119899

= O (1)

119876119898119899 =

119898

sum

119894=0

119898

sum

119895=0

119902119894119895 (119898 119899 = 0 1 )

(31)

then1003817100381710038171003817119882119898119899 (119891) minus 119891

1003817100381710038171003817119871119901

le

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(32)

where

119882119898119899 (119891 119909 119910) =1

119876119898119899

119898

sum

119894=0

119898

sum

119895=0

119902119898119899119898minus119894119899minus119895119878119894119895 (119909 119910) (33)

and 119878119894119895(119909 119910) is defined as in (16)

We know that in the theory of Fourier series it is ofmain interest how to approximate the function from thepartial sums of its Fourier series The purpose of the presentpaper is to get the rate of uniform approximation by 119879-transformation with doubly triangular We give the outlineof the paper In Section 2 we state the main results Someauxiliary lemmas are given in Section 3 and the proofs ofthe main theorems are presented in Section 4 Our newresults can be applied tomany classical summability methodssuch as Norlund summability and Riesz summability As animportant application we will apply them to the Norlundsummability and weighted means in Section 5 We will seethat not only Theorems A B and C are corollaries ofour results but also some other new types of estimates arepresented in this paper

2 The Main Results

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numbers tending tozero is called rest bounded variation or briefly 120572119899 isin RBVS ifthere is a constant119870(120572119899) only depending on 120572119899 such that

infin

sum

119896=119898

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (34)

holds for all natural numbers119898

4 ISRNMathematical Analysis

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numberstending to zero is called head bounded variation or briefly120572119899 isin HBVS if there is a constant 119870(120572119899) only depending on120572119899 such that

119898

sum

119896=0

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (35)

for all natural numbers 119898 or only for all 119898 le 119873 if thesequence 120572119899 has only finite nonzero terms and the lastnonzero term is 119886119899119873

Remark AThe definitions of RBVS andHBVS are introducedby Leindler [14] to generalize the monotonicity conditionson sequences In fact RBVS and HBVS generalized mono-tone nonincreasing sequences and monotone nondecreasingsequences respectively

Remark B Since it involves a sequence119870(120572119899) there should bea constant119870 such that 0 lt 119870(120572119899) le 119870

Now we extend the concepts of RBVS and HBVS to thedouble sequences as follows

Definition 1 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDRBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

infin

sum

120583=119895

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

infin

sum

]=119896

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

infin

sum

120583=119895

infin

sum

]=119896

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870 (120572119898119899) 119886119898119899119895119896 (36)

Definition 2 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDHBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

119895

sum

120583=0

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

119896

sum

]=0

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

119895

sum

120583=0

119896

sum

]=0

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870(120572119898119899) 119886119898119899119895119896 (37)

It is also required that the sequence 119870(120572119898119899) is boundedthat is there is a positive constant 119870 such that 0 lt 119870(120572119898119899) le

119870We state our main theorems as follows

Theorem 3 Let 119891 isin 119871119901 119905119898119899119906V be a a doubly normaltriangular matrix of nonnegative numbers 119905119898119899119906V isin DRBVSand let 119864 = lceil119898rceil 119865 = lceil119899rceil Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(38)

Also if 119891 isin 119871119901 119905119898119899119906V is doubly normal triangular ofnonnegative numbers 119905119898119899119906V isin DHBVS then one has theconclusion as follows

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

times (120596119901(119891 2minus119896) +C120596

119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(39)

Theorem 4 Suppose that 119891 isin 119871119894119901(119901 120572) for 120572 gt 0 1 le 119901 le

+infin Let 119905119898119899119906V be nonnegative doubly normal triangular and119905119898119899119906V isin DHBVS satisfying 119898119899119905119898119899119898119899 = O(1) Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(40)

While for 119891 isin 119871119894119901(119901 120572) and for 120572 gt 0 1 le 119901 le

+infin let 119905119898119899119906V be nonnegative normal doubly triangular and119905119898119899119906V isin DRBVS one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901= O (1)

119864minus1

sum

119896=0

119865minus1

sum

119897=0

(2119896(1minus120572)+119897

+ 2119896+119897(1minus120572)

) 11990511989811989921198962119897

(41)

ISRNMathematical Analysis 5

3 Lemmas and Proofs

Lemma 1 Suppose that 119905119898119899119906V is doubly normal triangularand 119863119906 is defined as (4) write 119864 = lceil119898rceil 119865 = lceil119899rceil then onehas

1198601 =

2119864minus1

sum

119906=0

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904) 1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905)

times (21198961198702119896 (119904)

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+1 (119904) 1199082119897+1minus1 (119905)

times (21198971198702119897 (119905)

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

+

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

(42)

Proof We can rewrite 1198601 as

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+119894 (119904)1198632119897+119895 (119905) (43)

and keep dividing the above into 4 parts

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895) (1198632119896+119894 (119904) minus 1198632119896+1 (119904))

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119897+1 (119905)

+

119864minus1

sum

119896=0

119865minus1

sum

l=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))1198632119896+1 (119904)

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

= 11986011 + 11986012 + 11986013 + 11986014

(44)

By using (8) we have

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1199082119896+1minus1 (119904)

times 1198632119896minus119894 (119904) 1199082119897+1minus1 (119905) 1198632119897minus119895 (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896

sum

119894=1

2119897

sum

119895=1

119905119898119899(2119896+1minus119894)(2119897+1minus119895)1199082119896+1minus1 (119904)

times 119863119894 (119904) 1199082119897+1minus1 (119905) 119863119895 (119905)

(45)

Applying double Abelrsquos transformation

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (

2119896

sum

1198941=1

2119897

sum

1198951=1

1198631198941(119904)1198631198951

(119905) 11990511989811989921198962119897

+

2119897

sum

1198951=1

1198631198951(119905)

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)2119897

6 ISRNMathematical Analysis

times

119894

sum

1198941=1

1198631198941(119904)

+

2119896

sum

1198941=1

1198631198941(119904)

2119897minus1

sum

119895=1

Δ 011199051198981198992119896(2119897+1minus119895)

times

119895

sum

1198951=1

1198631198951(119905)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times

119894

sum

1198941=1

1198631198941(119904)

119895

sum

1198951=1

1198631198951(119905))

(46)

Next (5) leads to

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904)1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(47)

Using (8) (5) and Abelrsquos transformation similar to theestimate of 11986011 (more easily actually)

11986012 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905) 21198961198702119896 (119904)

times

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1

times

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904)

(48)

Analogously

11986013 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 1198632119896+1 (119904) 21198971198702119897 (119905)

times

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+11199082119896+1minus1 (119904)

times

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 10119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904)

(49)

Combining the above estimates of 1198601119894 (119894 = 1 2 3 4) weobtain the conclusion of Lemma 1

Lemma 2 Let 119905119898119899119906V be the same matrix as in Lemma 1 thenone has

1198602 =

2119864minus1

sum

119906=0

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865(119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times119894119895119870119894 (119904) 119870119895 (119905))

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Stochastic AnalysisInternational Journal of

Page 3: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

ISRNMathematical Analysis 3

The (total) modulus of continuity of function 119891(119909 119910) isin119871119901(1198682) 1-periodic in each variable is defined by

120596119901

12(119891 1205751 1205752) = sup 1003817100381710038171003817119891 (119909 + 119906 119910 + V) minus 119891 (119909 + 119906 119910)

minus119891 (119909 119910 + V) + 119891 (119909 119910)1003817100381710038171003817119901

|119906| le 1205751 |V| le 1205752

(24)

It is easy to verify that there is a constantC gt 0 such that

120596119901

12(119891 1205751 1205752) le C (120596

119901(119891 1205751) + 120596

119901(119891 1205752)) (25)

Moricz and Siddiqi [11] studied the rate of uniformapproximation by Norlund means of Walsh (Walsh-Fourier)series of 119891 isin 119871119901[0 1) Later Moricz and Rhoades [12] stud-ied the corresponding approximation problem by weightedmeans of Walsh-Fourier series Their main results in [12] canbe read as follows

Theorem A Let 119891 isin 119871119901 1 le 119901 le infin 119899 = 2119898+ 119896 1 le 119896 le

2119898 119898 ge 1(i) If 119901119896 is nondecreasing and satisfies the condition

119899119901119899

119875119899

= O (1) (26)

then1003817100381710038171003817119905119899 (119891) minus 119891

1003817100381710038171003817119901

le3

119875119899

119898minus1

sum

119895=0

21198951199012119895+1minus1120596119901 (119891 2

minus119895) + O (120596119901 (119891 2

minus119898))

(27)

(ii) If 119901119896 is nonincreasing then

1003817100381710038171003817119905119899 (119891) minus 1198911003817100381710038171003817119901le

3

119875119899

119898minus1

sum

119895=0

21198951199012119895120596119901 (119891 2

minus119895) + O (120596119901 (119891 2

minus119898))

(28)

Theorem B Let 119891 isin 119871119894119901(120572 119901) for 120572 gt 0 and 1 le 119901 le infinIf 119901119896 is nondecreasing then one has the following estimates

1003817100381710038171003817119905119899 (119891) minus 1198911003817100381710038171003817119871119901

=

O (119899minus120572) 0 lt 120572 lt 1

O (119899minus1 log 119899) 120572 = 1

O (119899minus1) 120572 gt 1

(29)

For any fourfold sequence 119886119898119899119895119896 write

Δ 11119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895+1119896 minus 119886119898119899119895119896+1 + 119886119898119899119895+1119896+1

Δ 01119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895119896+1

Δ 10119886119898119899119895119896 = 119886119898119899119895119896 minus 119886119898119899119895+1119896

(30)

The sequence 119886119898119899119895119896 is called nondecreasing if it isnondecreasing in both 119895 and 119896 that is Δ 01119886119898119899119895119896 le 0 andΔ 10119886119898119899119895119896 le 0 for every 119895 119896 = 0 1 The nonincreasing caseis defined analogously

Recently Nagy [13] did some research on the approxima-tion by Norlund means of double Walsh-Fourier series forLipschitz functions and generalizedTheorems A and B to thefunctions of two variablesWe present one of themain resultsin [13] here

Theorem C Let 119891 isin 119871119894119901(120572 119901) for some 120572 gt 0 and 1 le 119901 le

infin let 119902119895119896 be a double sequence of nonnegative numbers suchthat it is nondecreasing Δ 11119902119895119896 is of fixed sign and satisfies theregularity condition

(119898 + 1) (119899 + 1) 119902119898119899

119876119898119899

= O (1)

119876119898119899 =

119898

sum

119894=0

119898

sum

119895=0

119902119894119895 (119898 119899 = 0 1 )

(31)

then1003817100381710038171003817119882119898119899 (119891) minus 119891

1003817100381710038171003817119871119901

le

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(32)

where

119882119898119899 (119891 119909 119910) =1

119876119898119899

119898

sum

119894=0

119898

sum

119895=0

119902119898119899119898minus119894119899minus119895119878119894119895 (119909 119910) (33)

and 119878119894119895(119909 119910) is defined as in (16)

We know that in the theory of Fourier series it is ofmain interest how to approximate the function from thepartial sums of its Fourier series The purpose of the presentpaper is to get the rate of uniform approximation by 119879-transformation with doubly triangular We give the outlineof the paper In Section 2 we state the main results Someauxiliary lemmas are given in Section 3 and the proofs ofthe main theorems are presented in Section 4 Our newresults can be applied tomany classical summability methodssuch as Norlund summability and Riesz summability As animportant application we will apply them to the Norlundsummability and weighted means in Section 5 We will seethat not only Theorems A B and C are corollaries ofour results but also some other new types of estimates arepresented in this paper

2 The Main Results

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numbers tending tozero is called rest bounded variation or briefly 120572119899 isin RBVS ifthere is a constant119870(120572119899) only depending on 120572119899 such that

infin

sum

119896=119898

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (34)

holds for all natural numbers119898

4 ISRNMathematical Analysis

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numberstending to zero is called head bounded variation or briefly120572119899 isin HBVS if there is a constant 119870(120572119899) only depending on120572119899 such that

119898

sum

119896=0

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (35)

for all natural numbers 119898 or only for all 119898 le 119873 if thesequence 120572119899 has only finite nonzero terms and the lastnonzero term is 119886119899119873

Remark AThe definitions of RBVS andHBVS are introducedby Leindler [14] to generalize the monotonicity conditionson sequences In fact RBVS and HBVS generalized mono-tone nonincreasing sequences and monotone nondecreasingsequences respectively

Remark B Since it involves a sequence119870(120572119899) there should bea constant119870 such that 0 lt 119870(120572119899) le 119870

Now we extend the concepts of RBVS and HBVS to thedouble sequences as follows

Definition 1 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDRBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

infin

sum

120583=119895

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

infin

sum

]=119896

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

infin

sum

120583=119895

infin

sum

]=119896

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870 (120572119898119899) 119886119898119899119895119896 (36)

Definition 2 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDHBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

119895

sum

120583=0

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

119896

sum

]=0

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

119895

sum

120583=0

119896

sum

]=0

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870(120572119898119899) 119886119898119899119895119896 (37)

It is also required that the sequence 119870(120572119898119899) is boundedthat is there is a positive constant 119870 such that 0 lt 119870(120572119898119899) le

119870We state our main theorems as follows

Theorem 3 Let 119891 isin 119871119901 119905119898119899119906V be a a doubly normaltriangular matrix of nonnegative numbers 119905119898119899119906V isin DRBVSand let 119864 = lceil119898rceil 119865 = lceil119899rceil Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(38)

Also if 119891 isin 119871119901 119905119898119899119906V is doubly normal triangular ofnonnegative numbers 119905119898119899119906V isin DHBVS then one has theconclusion as follows

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

times (120596119901(119891 2minus119896) +C120596

119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(39)

Theorem 4 Suppose that 119891 isin 119871119894119901(119901 120572) for 120572 gt 0 1 le 119901 le

+infin Let 119905119898119899119906V be nonnegative doubly normal triangular and119905119898119899119906V isin DHBVS satisfying 119898119899119905119898119899119898119899 = O(1) Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(40)

While for 119891 isin 119871119894119901(119901 120572) and for 120572 gt 0 1 le 119901 le

+infin let 119905119898119899119906V be nonnegative normal doubly triangular and119905119898119899119906V isin DRBVS one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901= O (1)

119864minus1

sum

119896=0

119865minus1

sum

119897=0

(2119896(1minus120572)+119897

+ 2119896+119897(1minus120572)

) 11990511989811989921198962119897

(41)

ISRNMathematical Analysis 5

3 Lemmas and Proofs

Lemma 1 Suppose that 119905119898119899119906V is doubly normal triangularand 119863119906 is defined as (4) write 119864 = lceil119898rceil 119865 = lceil119899rceil then onehas

1198601 =

2119864minus1

sum

119906=0

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904) 1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905)

times (21198961198702119896 (119904)

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+1 (119904) 1199082119897+1minus1 (119905)

times (21198971198702119897 (119905)

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

+

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

(42)

Proof We can rewrite 1198601 as

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+119894 (119904)1198632119897+119895 (119905) (43)

and keep dividing the above into 4 parts

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895) (1198632119896+119894 (119904) minus 1198632119896+1 (119904))

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119897+1 (119905)

+

119864minus1

sum

119896=0

119865minus1

sum

l=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))1198632119896+1 (119904)

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

= 11986011 + 11986012 + 11986013 + 11986014

(44)

By using (8) we have

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1199082119896+1minus1 (119904)

times 1198632119896minus119894 (119904) 1199082119897+1minus1 (119905) 1198632119897minus119895 (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896

sum

119894=1

2119897

sum

119895=1

119905119898119899(2119896+1minus119894)(2119897+1minus119895)1199082119896+1minus1 (119904)

times 119863119894 (119904) 1199082119897+1minus1 (119905) 119863119895 (119905)

(45)

Applying double Abelrsquos transformation

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (

2119896

sum

1198941=1

2119897

sum

1198951=1

1198631198941(119904)1198631198951

(119905) 11990511989811989921198962119897

+

2119897

sum

1198951=1

1198631198951(119905)

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)2119897

6 ISRNMathematical Analysis

times

119894

sum

1198941=1

1198631198941(119904)

+

2119896

sum

1198941=1

1198631198941(119904)

2119897minus1

sum

119895=1

Δ 011199051198981198992119896(2119897+1minus119895)

times

119895

sum

1198951=1

1198631198951(119905)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times

119894

sum

1198941=1

1198631198941(119904)

119895

sum

1198951=1

1198631198951(119905))

(46)

Next (5) leads to

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904)1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(47)

Using (8) (5) and Abelrsquos transformation similar to theestimate of 11986011 (more easily actually)

11986012 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905) 21198961198702119896 (119904)

times

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1

times

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904)

(48)

Analogously

11986013 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 1198632119896+1 (119904) 21198971198702119897 (119905)

times

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+11199082119896+1minus1 (119904)

times

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 10119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904)

(49)

Combining the above estimates of 1198601119894 (119894 = 1 2 3 4) weobtain the conclusion of Lemma 1

Lemma 2 Let 119905119898119899119906V be the same matrix as in Lemma 1 thenone has

1198602 =

2119864minus1

sum

119906=0

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865(119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times119894119895119870119894 (119904) 119870119895 (119905))

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

4 ISRNMathematical Analysis

For a fixed 119899 120572119899 = 119886119899119896 of nonnegative numberstending to zero is called head bounded variation or briefly120572119899 isin HBVS if there is a constant 119870(120572119899) only depending on120572119899 such that

119898

sum

119896=0

1003816100381610038161003816119886119899119896 minus 119886119899119896+11003816100381610038161003816 le 119870 (120572119899) 119886119899119898 (35)

for all natural numbers 119898 or only for all 119898 le 119873 if thesequence 120572119899 has only finite nonzero terms and the lastnonzero term is 119886119899119873

Remark AThe definitions of RBVS andHBVS are introducedby Leindler [14] to generalize the monotonicity conditionson sequences In fact RBVS and HBVS generalized mono-tone nonincreasing sequences and monotone nondecreasingsequences respectively

Remark B Since it involves a sequence119870(120572119899) there should bea constant119870 such that 0 lt 119870(120572119899) le 119870

Now we extend the concepts of RBVS and HBVS to thedouble sequences as follows

Definition 1 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDRBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

infin

sum

120583=119895

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

infin

sum

]=119896

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

infin

sum

120583=119895

infin

sum

]=119896

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870 (120572119898119899) 119886119898119899119895119896 (36)

Definition 2 A double sequence 120572119898119899 = 119886119898119899119895119896 is calledDHBVS if there is a constant 119870(120572119898119899) such that for 119895 119896 =

0 1 2

119895

sum

120583=0

10038161003816100381610038161003816Δ 10119886119898119899120583119896

10038161003816100381610038161003816

119896

sum

]=0

10038161003816100381610038161003816Δ 01119886119898119899119895]

10038161003816100381610038161003816

119895

sum

120583=0

119896

sum

]=0

10038161003816100381610038161003816Δ 11119886119898119899119895]

10038161003816100381610038161003816

le 119870(120572119898119899) 119886119898119899119895119896 (37)

It is also required that the sequence 119870(120572119898119899) is boundedthat is there is a positive constant 119870 such that 0 lt 119870(120572119898119899) le

119870We state our main theorems as follows

Theorem 3 Let 119891 isin 119871119901 119905119898119899119906V be a a doubly normaltriangular matrix of nonnegative numbers 119905119898119899119906V isin DRBVSand let 119864 = lceil119898rceil 119865 = lceil119899rceil Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(38)

Also if 119891 isin 119871119901 119905119898119899119906V is doubly normal triangular ofnonnegative numbers 119905119898119899119906V isin DHBVS then one has theconclusion as follows

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

times (120596119901(119891 2minus119896) +C120596

119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(39)

Theorem 4 Suppose that 119891 isin 119871119894119901(119901 120572) for 120572 gt 0 1 le 119901 le

+infin Let 119905119898119899119906V be nonnegative doubly normal triangular and119905119898119899119906V isin DHBVS satisfying 119898119899119905119898119899119898119899 = O(1) Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(40)

While for 119891 isin 119871119894119901(119901 120572) and for 120572 gt 0 1 le 119901 le

+infin let 119905119898119899119906V be nonnegative normal doubly triangular and119905119898119899119906V isin DRBVS one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901= O (1)

119864minus1

sum

119896=0

119865minus1

sum

119897=0

(2119896(1minus120572)+119897

+ 2119896+119897(1minus120572)

) 11990511989811989921198962119897

(41)

ISRNMathematical Analysis 5

3 Lemmas and Proofs

Lemma 1 Suppose that 119905119898119899119906V is doubly normal triangularand 119863119906 is defined as (4) write 119864 = lceil119898rceil 119865 = lceil119899rceil then onehas

1198601 =

2119864minus1

sum

119906=0

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904) 1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905)

times (21198961198702119896 (119904)

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+1 (119904) 1199082119897+1minus1 (119905)

times (21198971198702119897 (119905)

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

+

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

(42)

Proof We can rewrite 1198601 as

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+119894 (119904)1198632119897+119895 (119905) (43)

and keep dividing the above into 4 parts

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895) (1198632119896+119894 (119904) minus 1198632119896+1 (119904))

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119897+1 (119905)

+

119864minus1

sum

119896=0

119865minus1

sum

l=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))1198632119896+1 (119904)

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

= 11986011 + 11986012 + 11986013 + 11986014

(44)

By using (8) we have

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1199082119896+1minus1 (119904)

times 1198632119896minus119894 (119904) 1199082119897+1minus1 (119905) 1198632119897minus119895 (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896

sum

119894=1

2119897

sum

119895=1

119905119898119899(2119896+1minus119894)(2119897+1minus119895)1199082119896+1minus1 (119904)

times 119863119894 (119904) 1199082119897+1minus1 (119905) 119863119895 (119905)

(45)

Applying double Abelrsquos transformation

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (

2119896

sum

1198941=1

2119897

sum

1198951=1

1198631198941(119904)1198631198951

(119905) 11990511989811989921198962119897

+

2119897

sum

1198951=1

1198631198951(119905)

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)2119897

6 ISRNMathematical Analysis

times

119894

sum

1198941=1

1198631198941(119904)

+

2119896

sum

1198941=1

1198631198941(119904)

2119897minus1

sum

119895=1

Δ 011199051198981198992119896(2119897+1minus119895)

times

119895

sum

1198951=1

1198631198951(119905)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times

119894

sum

1198941=1

1198631198941(119904)

119895

sum

1198951=1

1198631198951(119905))

(46)

Next (5) leads to

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904)1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(47)

Using (8) (5) and Abelrsquos transformation similar to theestimate of 11986011 (more easily actually)

11986012 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905) 21198961198702119896 (119904)

times

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1

times

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904)

(48)

Analogously

11986013 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 1198632119896+1 (119904) 21198971198702119897 (119905)

times

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+11199082119896+1minus1 (119904)

times

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 10119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904)

(49)

Combining the above estimates of 1198601119894 (119894 = 1 2 3 4) weobtain the conclusion of Lemma 1

Lemma 2 Let 119905119898119899119906V be the same matrix as in Lemma 1 thenone has

1198602 =

2119864minus1

sum

119906=0

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865(119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times119894119895119870119894 (119904) 119870119895 (119905))

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

ISRNMathematical Analysis 5

3 Lemmas and Proofs

Lemma 1 Suppose that 119905119898119899119906V is doubly normal triangularand 119863119906 is defined as (4) write 119864 = lceil119898rceil 119865 = lceil119899rceil then onehas

1198601 =

2119864minus1

sum

119906=0

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904) 1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905)

times (21198961198702119896 (119904)

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+1 (119904) 1199082119897+1minus1 (119905)

times (21198971198702119897 (119905)

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

+

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

(42)

Proof We can rewrite 1198601 as

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+119894 (119904)1198632119897+119895 (119905) (43)

and keep dividing the above into 4 parts

1198601 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895) (1198632119896+119894 (119904) minus 1198632119896+1 (119904))

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119897+1 (119905)

+

119864minus1

sum

119896=0

119865minus1

sum

l=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)

times (1198632119897+119895 (119905) minus 1198632119897+1 (119905))1198632119896+1 (119904)

+

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1198632119896+1 (119904)1198632119897+1 (119905)

= 11986011 + 11986012 + 11986013 + 11986014

(44)

By using (8) we have

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=0

2119897minus1

sum

119895=0

119905119898119899(2119896+119894)(2119897+119895)1199082119896+1minus1 (119904)

times 1198632119896minus119894 (119904) 1199082119897+1minus1 (119905) 1198632119897minus119895 (119905)

=

119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896

sum

119894=1

2119897

sum

119895=1

119905119898119899(2119896+1minus119894)(2119897+1minus119895)1199082119896+1minus1 (119904)

times 119863119894 (119904) 1199082119897+1minus1 (119905) 119863119895 (119905)

(45)

Applying double Abelrsquos transformation

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (

2119896

sum

1198941=1

2119897

sum

1198951=1

1198631198941(119904)1198631198951

(119905) 11990511989811989921198962119897

+

2119897

sum

1198951=1

1198631198951(119905)

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)2119897

6 ISRNMathematical Analysis

times

119894

sum

1198941=1

1198631198941(119904)

+

2119896

sum

1198941=1

1198631198941(119904)

2119897minus1

sum

119895=1

Δ 011199051198981198992119896(2119897+1minus119895)

times

119895

sum

1198951=1

1198631198951(119905)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times

119894

sum

1198941=1

1198631198941(119904)

119895

sum

1198951=1

1198631198951(119905))

(46)

Next (5) leads to

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904)1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(47)

Using (8) (5) and Abelrsquos transformation similar to theestimate of 11986011 (more easily actually)

11986012 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905) 21198961198702119896 (119904)

times

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1

times

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904)

(48)

Analogously

11986013 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 1198632119896+1 (119904) 21198971198702119897 (119905)

times

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+11199082119896+1minus1 (119904)

times

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 10119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904)

(49)

Combining the above estimates of 1198601119894 (119894 = 1 2 3 4) weobtain the conclusion of Lemma 1

Lemma 2 Let 119905119898119899119906V be the same matrix as in Lemma 1 thenone has

1198602 =

2119864minus1

sum

119906=0

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865(119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times119894119895119870119894 (119904) 119870119895 (119905))

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Differential EquationsInternational Journal of

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International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

6 ISRNMathematical Analysis

times

119894

sum

1198941=1

1198631198941(119904)

+

2119896

sum

1198941=1

1198631198941(119904)

2119897minus1

sum

119895=1

Δ 011199051198981198992119896(2119897+1minus119895)

times

119895

sum

1198951=1

1198631198951(119905)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times

119894

sum

1198941=1

1198631198941(119904)

119895

sum

1198951=1

1198631198951(119905))

(46)

Next (5) leads to

11986011 =

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904) 1199082119897+1minus1 (119905)

times (2119896+119897119905119898119899211989621198971198702119896 (119904)1198702119897 (119905)

+

2119896minus1

sum

119894=1

1198942119897Δ 10119905119898119899(2119896+1minus119894)2119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

1198952119896Δ 011199051198981198992119896(2119897+1minus119895)119870119895 (119905) 1198702119896 (119904)

+

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(47)

Using (8) (5) and Abelrsquos transformation similar to theestimate of 11986011 (more easily actually)

11986012 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1 (119905) 21198961198702119896 (119904)

times

2119897minus1

sum

119895=0

1199051198981198992119896(2119897+119895)

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119896+1minus1 (119904)1198632119897+1

times

2119896minus1

sum

119894=1

2119897minus1

sum

119895=0

Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)119894119870119894 (119904)

(48)

Analogously

11986013 = minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 1198632119896+1 (119904) 21198971198702119897 (119905)

times

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)2119897

minus

119864minus1

sum

119896=0

119865minus1

sum

119897=0

1198632119896+11199082119896+1minus1 (119904)

times

2119896minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 10119905119898119899(2119896+119894)(2119897+1minus119895)119895119870119895 (119904)

(49)

Combining the above estimates of 1198601119894 (119894 = 1 2 3 4) weobtain the conclusion of Lemma 1

Lemma 2 Let 119905119898119899119906V be the same matrix as in Lemma 1 thenone has

1198602 =

2119864minus1

sum

119906=0

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865(119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times119894119895119870119894 (119904) 119870119895 (119905))

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

ISRNMathematical Analysis 7

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865 (119905)

minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(50)

Meanwhile

1198603 =

119898

sum

119906=2119864

2119865minus1

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

= minus

119865minus1

sum

119897=0

119898minus2119864

sum

119894=0

1199082119897+1minus1 (119905) 1198632119864(119904)

times (21198971198702119897 (119905) 119905119898119899(2119864+119894)2119897

+

2119897minus1

sum

119895=1

Δ 01119905119898119899(2119864+119894)(2119897+1minus119895)119895119870119895 (119905))

minus

119865minus1

sum

119897=0

1199082119897+1minus1 (119905) 119903119864 (119904)

times (11990511989811989911989821198972119897(119898 minus 2

119864)1198702119897 (119905) 119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)2119897 1198942119897119870119894 (119904) 1198702119897 (119905)

+

2119897minus1

sum

119895=1

Δ 01119905119898119899119898(2119897+1minus119895) (119898 minus 2119864)

times 119895119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

2119897minus1

sum

119895=1

Δ 11119905119898119899(2119864+119894)(2119897+1minus119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

minus

119865minus1

sum

119897=0

2119897minus1

sum

119895=0

119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)(2119897+119895)1198632119864 (119904)1198632119897+1 (119905)

minus

119865minus1

sum

119897=0

119903119864 (119904)1198632119897+1 (119905)

times (

2119897minus1

sum

119895=0

119905119898119899119898(2119897+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

2119897minus1

sum

119895=0

119898minus2119860minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119897+119895)119894119870119894 (119904))

(51)

Proof Rewrite 1198602 by some different decomposition methodcombining the decomposition we used in Lemma 1 it followsthat

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899

sum

V=2119865119905119898119899(2119896+119894)V1198632119896+119894 (119904)119863V (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+119894 (119904)1198632119865+119895 (119905)

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (1198632119896+119894 (119904) minus 1198632119896+1 (119904))1198632119865+119895 (119905)

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)1198632119865+119895 (119905)

(52)

Furthermore by properties (7) and (8) of119863119899 we have

1198602 =

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895) (minus1199082119896+1minus1 (119904)1198632119896minus119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

+

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)1198632119896+1 (119904)

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119864minus1

sum

119896=0

2119896minus1

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119896+119894)(2119865+119895)

times (minus1198632119865 (119905) 1199082119896+1minus1 (119904)1198632119896minus119894 (119904)

minus 1199082119896+1minus1 (119904)1198632119896minus119894 (119904) 119903119865 (119905) 119863119895 (119905)

+ 1198632119896+1 (119904)1198632119865 (119905)

+1198632119896+1 (119904) 119903119865 (119905) 119863119895 (119905))

= 11986021 + 11986022 + 11986023 + 11986024

(53)

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

8 ISRNMathematical Analysis

By changing 1198632119891minus119894 to 119863119894 and applying Abelrsquos transforma-tion and (5)

11986021 = minus

119864minus1

sum

119896=0

2119896

sum

119894=1

119899minus2119865

sum

119895=0

119905119898119899(2119896+1minus119894)(2119865+119895)1198632119865 (119905) 1199082119896+1minus1 (119904)119863119894 (119904)

= minus

119864minus1

sum

119896=0

119899minus2119865

sum

119895=0

1199082119896+1minus1 (119904)1198632119865 (119905)

times (21198961198702119896 (119904) 1199051198981198992119896(2119865+119895)

+

2119896minus1

sum

119894=1

Δ 10119905119898119899(2119896+1minus119894)(2119865+119895)119894119870119894 (119904))

(54)

Similarly

11986022 = minus

119864minus1

sum

119896=0

1199082119896+1minus1119903119865 (119905)

times (11990511989811989921198961198992119896(119899 minus 2

119865)1198702119896 (119904) 119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

Δ 10119905119898119899(2119896+1minus119894)119899119894 (119899 minus 2119865)

times 119870119894 (119904) 119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 011199051198981198992119896(2119865+119895)21198961198951198702119896 (119904) 119870119895 (119905)

+

2119896minus1

sum

119894=1

119899minus2119865minus1

sum

119895=0

Δ 11119905119898119899(2119896+1minus119894)(2119865+119895)

times 119894119895119870119894 (119904) 119870119895 (119905))

(55)

We also obtain the estimate for

11986024 = minus

119864minus1

sum

119896=0

1198632119896+1 (119904) 119903119865 (119905)

times (

2119896minus1

sum

119894=0

119905119898119899(2119896+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

2119896minus1

sum

119894=0

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119896+119894)(2119865+119895)119895119870119895 (119905))

(56)

Combining the estimates of 1198602119894 (119894 = 1 2 3 4) we havethe conclusion for 1198602 in Lemma 2 and the discussion for1198603

is similar (as for the decomposition of 1198602 we denote 1198603 =sum4

119894=11198603119894) This completes the proof of Lemma 2

Lemma 3 Let 119905119898119899119906V be the same matrix as in Lemma 1Then one has

1198604 =

119898

sum

119906=2119864

119899

sum

V=2119865119905119898119899119906V119863119906 (119904)119863V (119905)

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

1198632119864 (119904) 119903119865 (119905)

times (119905119898119899(2119864+119894)119899 (119899 minus 2119865)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899(2119864+119894)(2119865+119895)119895119870119895 (119905))

+

119899minus2119865

sum

119895=0

119903119864 (119904)1198632119865 (119905)

times (119905119898119899119898(2119865+119895) (119898 minus 2119864)119870119898minus2119864 (119904)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)(2119865+119895)119894119870119894 (119904))

+ 119903119864 (119904) 119903119865 (119905)

times (119905119898119899119898119899 (119898 minus 2119864) (119899 minus 2

119865)119870119898minus2119864 (119904) 119870119899minus2119865 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 10119905119898119899(2119864+119894)119899119894 (119899 minus 2119865)119870119894 (119904)119870119899minus2119865 (119905)

+

119899minus2119865minus1

sum

119895=0

Δ 01119905119898119899119898(2119865+119895)119895 (119898 minus 2119864)119870119898minus2119864 (119904) 119870119895 (119905)

+

119898minus2119864minus1

sum

119894=0

Δ 11119905119898119899(2119864+119894)(2119865+119895)119894119895119870119894 (119904) 119870119895 (119905))

(57)

Proof Decompose 1198604 into 4 parts

1198604 =

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864+119894 (119904)1198632119865+119895 (119905)

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Stochastic AnalysisInternational Journal of

Page 9: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

ISRNMathematical Analysis 9

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895) (1198632119864 (119904) + 119903119864 (119904)119863119894 (119904))

times (1198632119865 (119905) + 119903119865 (119905) 119863119895 (119905))

=

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)1198632119864 (119904) 119903119865 (119905) 119863119895 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904)119863119894 (119904)1198632119865 (119905)

+

119898minus2119864

sum

119894=0

119899minus2119865

sum

119895=0

119905119898119899(2119864+119894)(2119865+119895)119903119864 (119904) 119903119865 (119905) 119863119894 (119904)119863119895 (119905)

= 11986041 + 11986042 + 11986043 + 11986044

(58)

Using the techniques as in Lemmas 1 and 2 we can easilyget the result of Lemma 3

By Lemmas 1 2 and 3 and (20) 119870119898119899(119904 119905) can be writtenas

119870119898119899 (119904 119905) =

119898

sum

119906=0

119899

sum

V=0119905119898119899119906V119863119906 (119904)119863V (119905)

=

4

sum

119894=1

119860 119894 =

4

sum

119894=1

4

sum

119895=1

119860 119894119895

(59)

Therefore by (21) we have

119879119898119899 (119909 119910) minus 119891 (119909 119910)

= ∬

1

0

4

sum

119894=1

4

sum

119895=1

119860 119894119895 (119891 (119909 + 119904 119910 + 119905) minus 119891 (119909 119910)) 119889119904 119889119905

(60)

We denote by P119899 the set of Walsh polynomials of orderless than 119899 that is

P119899 = 119875 (119909) 119875 (119909) =119899minus1

sum

119894=0

119888119894119908119894 (119909) (61)

where 119899 ge 1 and 119888119894 denotes the real or complex numbersOn the torus 119868

2 we define the two-dimensional Walshpolynomials of order less than (119898 119899) as

P119898119899 = 1198751 (119904) times 1198752 (119905) 1198751 isin P119898 1198752 isin P119899 119904 119905 isin 119868 (62)

Lemma 4 Consider

Θ =

1003817100381710038171003817100381710038171003817int11986821198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

1003817100381710038171003817100381710038171003817119901

le C (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(63)

Proof By (6) and Holder inequality we have

Θ le

100381710038171003817100381710038171003817100381710038171003817

int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905) (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le [int1198682(int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot))119901119889119904 119889119905) 119889119909 119889119910]

1119901

(64)

Furthermore using the generalized Minkowski inequal-ity we have

Θ le int119868119896times119868119897

1198632119896 (119904)1198632119897 (119905)

times (int1198682(119891 (sdot + 119904 sdot + 119905) minus 119891(sdot sdot))

119901119889119909 119889119910)

1119901

119889119904 119889119905

le 120596119901(119891 2minus119896 2minus119897) le C (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(65)

This completes the proof of Lemma 4

Lemma 5 Let 119875 isin P2119896 2119897 119876 isin P2119896 1 le 119901 le infin 119896 119897 isin Nthen one has

(i)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904) 119903119897 (119905) 119875 (119904 119905) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198751120596119901

12(119891 2minus119896 2minus119897)

(66)

(ii)100381710038171003817100381710038171003817100381710038171003817

∬

1

0

(119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119903119896 (119904)1198632119897 (119905) 119876 (119904) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

le 1198621198761120596119901

12(119891 2minus119896 2minus119897)

(67)

Proof Note that 119875 isin P2119896 2119897 and119876 isin P2119896 are constants on thesets 119868119896 times 119868119897 and 119868119896 respectively By Lemma 4 it is not difficultto prove Lemma 5 We can also find the conclusions in [13]

Lemma 6 Suppose that

(i) 119905119898119899119906V isin 119863119877119861119881119878 and it is doubly triangular then forany 1198941 ge 1198942 or 1198951 ge 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)(ii) 119905119898119899119906V isin 119863119867119861119881119878 and it is doubly triangular then for

any 1198941 le 1198942 or 1198951 le 1198952 one has 1199051198981198991198941119895 = O(1199051198981198991198942119895) or1199051198981198991198941198951

= O(1199051198981198991198941198952)

Proof (i) Since 119905119898119899119906V is triangular

1199051198981198991198941119895le

119898

sum

119906=1198941

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816le

119898

sum

119906=1198942

10038161003816100381610038161003816Δ 10119905119898119899119906119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (68)

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

10 ISRNMathematical Analysis

Similarly we have

1199051198981198991198941198951le

119898

sum

V=1198951

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 le

119898

sum

V=1198952

1003816100381610038161003816Δ 10119905119898119899119894V1003816100381610038161003816 = O (1199051198981198991198941198952) (69)

(ii) Since

100381610038161003816100381610038161199051198981198991198941119895

minus 1199051198981198991198942119895

10038161003816100381610038161003816le

1198942

sum

119894=1198941

10038161003816100381610038161003816Δ 10119905119898119899119894119895

10038161003816100381610038161003816= O (1199051198981198991198942119895

) (70)

it implies that

1199051198981198991198941119895= O (1199051198981198991198942119895

) (71)

Similarly we have

1199051198981198991198941198951= O (1199051198981198991198941198952

) (72)

This completes the proof of Lemma 6

4 Proofs of Theorems 3 and 4

We denote the norm of∬10119860 119894119895(119891(119909 + 119904 119910 + 119905) minus 119891(119909 119910))119889119904 119889119905

by

Φ119894119895 =

100381710038171003817100381710038171003817100381710038171003817

∬

1

0

119860 119894119895 (119891 (sdot + 119904 sdot + 119905) minus 119891 (sdot sdot)) 119889119904 119889119905

100381710038171003817100381710038171003817100381710038171003817119901

(73)

Then by (60) and Minkowski inequality we have

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901le

4

sum

119894=1

4

sum

119895=1

Φ119894119895 (74)

Proof of Theorem 3 (I) When 119905119898119899119894119895 isin DRBVS we deduceΦ119894119895 separately by (9) and Lemma 5(i)

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816120596119901

12(119891 2minus119896 2minus119897)

+C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816

times 119894119895120596119901

12(119891 2minus119896 2minus119897)

(75)

Since 119905119898119899119894119895 isin DRBVS by the definition of DRBVS we have

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+11989711990511989811989921198962119897

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+11989711990511989811989921198962119897

(76)

Substituting the above intoΦ11 yields

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897120596

119901

12(119891 2minus119896 2minus119897) (77)

Now by (9) and Lemma 5(ii)

Φ12 le C119864minus1

sum

119896=0

(

119865minus1

sum

119897=0

2119897minus1

sum

119895=1

21198961199051198981198992119896(2119897+119895)

+

119865minus1

sum

119897=0

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

11989410038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)(2119897+119895)

10038161003816100381610038161003816)

times 120596119901(119891 2minus119896)

(78)

Applying Lemma 6(i) and still the definition of DRBVS wehave

Φ12 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896120596119901(119891 2minus119896)

times (

2119897minus1

sum

119895=1

1199051198981198992119896(2119897+119895) + 11990511989811989921198962119897)

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119896) 11990511989811989921198962119897

(79)

Analogously

Φ13 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897120596119901(119891 2minus119897) 11990511989811989921198962119897 (80)

By Lemmas 4 and 6(i) it is easy to verify that

Φ14 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897)) (81)

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

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Stochastic AnalysisInternational Journal of

Page 11: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

ISRNMathematical Analysis 11

Similar discussion deduces that (by using Lemmas 4 and 5 (i)and (ii) separately)

Φ21 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865120596

119901(119891 2minus119896)

Φ22 le C(

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899

+

119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865)

times 120596119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 11990511989811989921198962119865 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 1199051198981198992119896119899120596

119901(119891 2minus119865)

+C119864minus1

sum

119896=0

(119899 minus 2119865) 11990511989811989921198962119865120596

119901(119891 2minus119865)

(82)

Note the definition of 119864 and 119865 clearly 119898 + 1 le 2119864+1 and

119899 + 1 le 2119865+1 thus we have

119898 minus 2119864le 2119864minus 1 =

119864minus1

sum

119896=0

2119896

119899 minus 2119865le 2119865minus 1 =

119865minus1

sum

119897=0

2119897

(83)

Applying inequality (83) and Lemma 6 to (82) also usingthe monotonicity of continuous modulus it is not difficult todeduce that

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(84)

Φ3119894 (119894 = 1 2 3 4) goes analogously as Φ2119894 (119894 = 1 2 3 4)thus we have

4

sum

119894=1

Φ3119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(85)

Next we estimate Φ4119894 (119894 = 1 2 3 4) Applying Lemmas4 and 6 we easily get

Φ41 le C (119898 minus 2119864) (119899 minus 2

119865) 11990511989811989921198642119865

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

Φ42 le C120596119901(119891 2minus119865)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198992119864119899 + (119899 minus 2

119865) 11990511989811989921198642119865)

Φ43 le C120596119901(119891 2minus119864)

times ((119898 minus 2119864) (119899 minus 2

119865) 1199051198981198991198982119865 + (119898 minus 2

119864) 11990511989811989921198642119865)

(86)

and the last term

Φ44 le C120596119901(119891 2minus119864 2minus119865) (119898 minus 2

119864) (119899 minus 2

119865)

times (119905119898119899119898119899 + 1199051198981198992119864119899 + 1199051198981198991198982119865 + 11990511989811989921198642119865)

(87)

Applying Lemma 6(i) and (83) to (86) and (87) we have4

sum

119894=1

Φ4119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(88)

Combining all the estimates (77)ndash(81) (84) (85) and(88) ofΦ119894119895 yields that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119871119901

le

4

sum

119894=1

4

sum

119895=1

Φ119894119895

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+11989711990511989811989921198962119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(89)

(II) When 119905119898119899119894119895 isin DHBVS the proof goes similarly Wemainly point out the differences in this case and prove someterms of Φ119894119895 as examples

Since 119905119898119899119894119895 isin DHBVS by the definition of DHBVS wehave

2119896minus1

sum

119894=1

1198942119897 10038161003816100381610038161003816Δ 10119905119898119899(2119896+1minus119894)2119897

10038161003816100381610038161003816le 2119896+119897119905119898119899(2119896+1minus1)2119897

2119897minus1

sum

119895=1

1198952119896 10038161003816100381610038161003816Δ 011199051198981198992119896(2119897+1minus119895)

10038161003816100381610038161003816le 2119896+1198971199051198981198992119896(2119897+1minus1)

2119896minus1

sum

119894=1

2119897minus1

sum

119895=1

10038161003816100381610038161003816Δ 11119905119898119899(2119896+1minus119894)(2119897+1minus119895)

10038161003816100381610038161003816119894119895 le 2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)

(90)

Thus by Lemma 5(i) (9) (71) and (72) we have

Φ11 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (91)

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

12 ISRNMathematical Analysis

Lemma 5(ii) (9) (71) (72) and (83) yield that

Φ1119894 le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119905119898119899(2119896+1minus1)(2119897+1minus1)120596

119901

12(119891 2minus119896 2minus119897) (92)

Now the definition of DHBVS Lemmas 5 and 6 and (9)deduce that

Φ21 le C119864minus1

sum

119896=0

2119896((119899 minus 2

119865) 1199051198981198992119896119899 + 119905119898119899(2119896+1minus1)119899) 120596

119901(119891 2minus119896)

le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896)

Φ22 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119896 2minus119865)

Φ23 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

Φ24 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899120596

119901(119891 2minus119865)

(93)

While in this case we keep 119905119898119899(2119896+1minus1)119899 in the sum so wehave

4

sum

119894=1

Φ2119894 le C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119905119898119899(2119896+1minus1)119899

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

(94)

Analogously

4

sum

119894=1

Φ3119894 le C119865minus1

sum

119897=0

(119898 minus 2119864) 2119897119905119898119899119898(2119897+1minus1)

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119897))

(95)

4

sum

119894=1

Φ4119894 le C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(96)

Combining all the estimates of (92)ndash(96) we obtain theconclusion of the case of 119905119898119899119894119895 isin DHBVS inTheorem 3

Remark C Actually in the case of 119905119898119899119894119895 isin DHBVS (96) canbe represented in amore complicated form if we calculate theterm with the same manner For instance

4

sum

119894=1

Φ4119894 le C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C119898minus2119864

sum

119894=0

119905119898119899(2119864+119894)119899 (119899 minus 2119865) 120596119901(119891 2minus119865)

+C (119898 minus 2119864) (119899 minus 2

119865) 119905119898119899119898119899

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(97)

while we give a simpler form in Theorem 3 of the presentpaper

Proof of Theorem 4 Since 119905119898119899119906V isin DHBVS and 119898119899119905119898119899119898119899 =O(1) in the case of DHBVS it is easy to calculate that

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (2minus119864120572

+ 2minus119865120572

) 0 lt 120572 lt 1

O (1198642minus119864+ 1198652minus119865) 120572 = 1

O (2minus119864+ 2minus119865) 120572 gt 1

(98)

Note that 119898 sim 2119864 and 119899 sim 2119865 the above estimate isequivalent to

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(99)

While for 119905119898119899119906V isin DRBVS the conclusion is obviousfromTheorem 3 and the supposed 119891 isin Lip(120572 119901)

Comparing Theorems 3 and 4 with Theorems A B andC we find that the former are the generalizations of the latterfrom the sense of monotonicity on one hand On the otherhand 119879-transformation is the generalization of some meansof series such as Norlund means Cesaro means and Rieszmeans In the next section we give applications of our resultsto some summability methods

5 Applications to Noumlrlund Means andWeighted Means

Corollary 5 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin 119863119867119861119881119878 for 119891 isin 119871119901(119868

2) Then one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(100)

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

ISRNMathematical Analysis 13

For 119901119895119896 isin 119863119877119861119881119878 one has

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896+1+1119899minus2119897+1+1

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119897))

+C119864minus1

sum

119896=0

2119896(119899 minus 2

119865) 119901119898minus2119896+1+10

times (120596119901(119891 2minus119896) + 120596119901(119891 2minus119865))

+C119865minus1

sum

119897=0

2119897(119898 minus 2

119864) 1199010119899minus2119897+1+1

times (120596119901(119891 2minus119864) + 120596119901(119891 2minusl))

+C (119898 minus 2119864) (119899 minus 2

119865) 11990100

times (120596119901(119891 2minus119864) + 120596119901(119891 2minus119865))

(101)

Proof Note that in the case when 119901119895119896 isin DHBVS we have119905119898119899119895119896 isin DRBVS By the first part of Theorem 3 it is easy todeduce that

1198751198981198991003817100381710038171003817119879119898119899 minus 119891

1003817100381710038171003817119901

le C119864minus1

sum

119896=0

119865minus1

sum

119897=0

2119896+119897119901119898minus2119896 119899minus2119897 (120596

119901(119891 2minus119896) + 120596119901(119891 2minus119897))

(102)

When 119901119895119896 isin DRBVS (corresponding to the case119905119898119899119895119896 isin DHBV) we get the conclusion from the second partof Theorem 3 immediately

Remark D When 119901119895119896 is nondecreasing and Δ 11119901119895119896 is offixed sign it is obvious that 119901119895119896 isin DHBVS ThereforeTheorem 3 of [13] can be deduced directly from the firstinequality of Corollary 5 Furthermore we exclude somepreconditions in this case For the nonincreasing case ofTheorem in [13] as we figure out in Remark C the secondinequality of Corollary 5 can be seen as the generalization ofthis case as long as it takes the more complicated form ofΦ4119894119894 = 1 2 3 4

The following is a Corollary of Theorem 4 for the caseof Norlund means Since the nondecreasing of 119901119895119896 andfixed sign of Δ 11119901119895119896 imply that 119901119895119896 isin DHBVS it is thegeneralization of Theorem C

Corollary 6 Let 119879 be the matrix defined by (11) Suppose that119901119895119896 isin DHBVS satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (103)

Then for 119891 isin 119871119894119901(120572 119901) one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(104)

As far as the weighted means is concerned we have thefollowing corollary which generalizes Theorems A and B

Corollary 7 Let 119879 be the matrix defined by (14) and 119891 isin

119871119901(1198682) Suppose that 119901119895119896 isin 119863119867119861119881119878 Then one has (101) For

119901119895119896 isin 119863119867119861119881119878 one has (100)Suppose that 119891 isin 119871119894119901(120572 119901) 119901119895119896 isin 119863119867119861119881119878 satisfies

(119898 + 1) (119899 + 1) 119901119898119899

119875119898119899

= O (1) (105)

Then one has

1003817100381710038171003817119879119898119899 minus 1198911003817100381710038171003817119901=

O (119898minus120572+ 119899minus120572) 0 lt 120572 lt 1

O (119898minus1 log119898 + 119899

minus1 log 119899) 120572 = 1

O (119898minus1+ 119899minus1) 120572 gt 1

(106)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] A Paley ldquoA remarkable series of orthogonal functionsrdquoProceed-ings of the London Mathematical Society vol 34 no 1 pp 241ndash279 1932

[2] M S Corrington ldquoSolution of differential and integral equa-tions with Walsh functionsrdquo IEEE Transactions on CircuitTheory vol 20 no 5 pp 470ndash476 1973

[3] R R Coifman and M V Wickerhauser ldquoEntropy-based algo-rithms for best basis selectionrdquo IEEE Transactions on Informa-tion Theory vol 38 no 2 pp 713ndash718 1992

[4] P N Paraskevopoulos ldquoChebyshev series approach to systemidentification analysis and optimal controlrdquo Journal of theFranklin Institute vol 316 no 2 pp 135ndash157 1983

[5] G B Mahapatra ldquoSolution of optimal control problem of lineardiffusion equations via Walsh functionsrdquo IEEE Transactions onAutomatic Control vol 25 no 2 pp 319ndash321 1980

[6] F Schipp W R Wade P Simon and J Pal Walsh Series AnIntroduction toDyadicHarmonic Analysis AdamHilger BristolUK 1990

[7] B I Golubov A V Efimov and V A Skvortsov Ryady i Pre-obrazovaniya Uolsha Teoriya i Primeneniya Nauka MoscowRussia 1987 English Translation Walsh Series and TransformsTheTheory and Applications Kluwer Academic DordrechtTheNetherlands 1991

[8] U Goginava ldquoApproximation properties of (119862 120572) means ofdouble Walsh-Fourier seriesrdquo Analysis in Theory and Applica-tions vol 20 no 1 pp 77ndash98 2004

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

14 ISRNMathematical Analysis

[9] S Yano ldquoOn Walsh-Fourier seriesrdquo The Tohoku MathematicalJournal vol 3 pp 223ndash242 1951

[10] E Savas and B E Rhoades ldquoNecessary conditions for inclusionrelations for double absolute summabilityrdquo Proceedings of theEstonian Academy of Sciences vol 60 no 3 pp 158ndash164 2011

[11] FMoricz and A H Siddiqi ldquoApproximation byNorlundmeansof Walsh-Fourier seriesrdquo Journal of Approximation Theory vol70 no 3 pp 375ndash389 1992

[12] F Moricz and B E Rhoades ldquoApproximation by weightedmeans of Walsh-Fourier seriesrdquo International Journal of Mathe-matics amp Mathematical Sciences vol 19 no 1 pp 1ndash8 1996

[13] K Nagy ldquoApproximation by Norlund means of double Walsh-Fourier series for Lipschitz functionsrdquo Mathematical Inequali-ties amp Applications vol 15 no 2 pp 301ndash322 2012

[14] L Leindler ldquoOn the degree of approximation of continuousfunctionsrdquo Acta Mathematica Hungarica vol 104 no 1-2 pp105ndash113 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article Approximation by -Transformation of Double ...We study the Walsh series expansion of multivariate functions in (1) and, in particular, in Lip (,) .erateof uniform

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Methoden der Psychologie Multivariate Analysemethoden Günter Meinhardt Johannes Gutenberg Universität Mainz Multivariate Distanz – Multivariate Normalverteilung.

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Telephone Operation Guide For LIP-9030 & LIP-9040

Telephone Operation Guide For LIP-9030 & LIP-9040

Lip Fusion |Lip Fusion Reviews |Lip Fusion - Lip Plumper

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LIP-8050V User Guide - VKTEK USERGUIDES/LIP-8050V...warranted to be true in all cases. ... LIP-8050V User Guide Issue 1.0 ... 1.5 LIP-8050V Phone Installation

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Marianne E. Walsh and Michael R. Walsh Charles A. Ramsey

Web view08.01.2016 · Lip Sync Battle. Thanks to our fabulous teacher performers!!!! Author: its Created Date: 01/11/2016 06:09:00 Last modified by: WALSH, TALEESE

Web view08.01.2016 · Lip Sync Battle. Thanks to our fabulous teacher performers!!!! Author: its Created Date: 01/11/2016 06:09:00 Last modified by: WALSH, TALEESE

New Walsh & Associates - Professional Safety Servicesprofessionalsafetyservices.ca/photos/custom/walsh... · 2016. 1. 5. · Walsh & Associates Professional Safety Services staff

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Pg. 60-62 - Cloud Object Storage | Store & Retrieve … Proof Fixture Replacement ORDERING INFORMATION: Lenses Lip Style/Latch Locations ROLLED LIP ROLLED LIP ROLLED LIP ROLLED LIP

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Photo Story of Diane Walsh and Lana Walsh

Introduction to Multivariate Analysis and Multivariate Distances

Introduction to Multivariate Analysis and Multivariate Distances

Care Drinks · 2020-06-08 · Lip balm Lip balm basic Fairtrade lip balm Lip butter alu Lip balm eco Lip balm recycled Lip balm specials Nail files Tooth picks Dental floss Tooth

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MULTIVARIATE SPATIAL STATISTICS - sct.uab.catsct.uab.cat/estadistica/sites/sct.uab.cat.estadistica/files/Dmayer... · MULTIVARIATE SPATIAL STATISTICS ... using multivariate geostatistics

MULTIVARIATE SPATIAL STATISTICS - sct.uab.catsct.uab.cat/estadistica/sites/sct.uab.cat.estadistica/files/Dmayer... · MULTIVARIATE SPATIAL STATISTICS ... using multivariate geostatistics

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Final Judgment Of Walsh v Walsh Divorce

Multivariate polynomial approximation and convex bodies · 2018. 6. 6. · Outline 1 Part I: Approximation by holomorphic polynomials in Cd; Oka-Weil and Bernstein-Walsh 2 Part 2:

Multivariate polynomial approximation and convex bodies · 2018. 6. 6. · Outline 1 Part I: Approximation by holomorphic polynomials in Cd; Oka-Weil and Bernstein-Walsh 2 Part 2:

Multivariate L1 Statistical Methods: The Package MNM1. Introduction Classical multivariate statistical inference methods (Hotelling’s T2, multivariate analysis of variance, multivariate

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