Research Article Pointwise Analog of the Ste Ikin ...W Bodzimierz A enski Faculty of Mathematics,...

Hindawi Publishing CorporationJournal of MathematicsVolume 2013, Article ID 426347, 8 pageshttp://dx.doi.org/10.1155/2013/426347

Research ArticlePointwise Analog of the SteIkin Approximation Theorem

WBodzimierz Aenski

Faculty of Mathematics, Computer Science and Econometrics, University of Zielona Góra,Ulica Szafrana 4a, 65-516 Zielona Góra, Poland

Correspondence should be addressed to Włodzimierz Łenski; [email protected]

Received 11 November 2012; Accepted 20 January 2013

Academic Editor: Roberto Renò

Copyright © 2013 Włodzimierz Łenski.This is an open access article distributed under theCreative CommonsAttribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We show the pointwise version of the Stečkin theorem on approximation by de la Vallée-Poussin means. The result on normapproximation is also derived.

1. Introduction

Let 𝐿𝑝 (1 ≤ 𝑝 < ∞) [𝐶] be the class of all 2𝜋-periodic real-valued functions integrable in the Lebesgue sense with 𝑝thpower (continuous) over 𝑄 = [−𝜋, 𝜋], and let 𝑋𝑝 = 𝐿𝑝 when1 ≤ 𝑝 < ∞ or 𝑋𝑝 = 𝐶 when 𝑝 = ∞.

Let us define the norms of 𝑓 ∈ 𝑋𝑝 as

𝑓 =

𝑓𝑋𝑝

=𝑓 (⋅)

𝑋𝑝

:=

{{{

{{{

{

{∫𝑄

𝑓 (𝑥)

𝑝𝑑𝑥}

1/𝑝

when 1 ≤ 𝑝 < ∞,

sup𝑥∈𝑄

𝑓 (𝑥) when 𝑝 = ∞,

𝑓𝑥,𝛿

=𝑓

𝑋𝑝,𝑥,𝛿=

𝑓 (⋅)𝑋𝑝 ,𝑥,𝛿

:= sup0

2 Journal of Mathematics

=

{{{{{

{{{{{

{

sup0

Journal of Mathematics 3

𝜎𝑛,𝑚𝑓 (𝑥) − 𝑓 (𝑥) ≤ 𝜋2𝐸∘

𝑛−𝑚(𝑓, 𝑥,

𝜋

2𝑛 − 𝑚 + 1)𝑋𝑝

+ 6𝐹∘

𝑛−𝑚,𝑚(𝑓, 𝑥)

𝑋𝑝

+ ∫

𝜋/(𝑚+1)

𝜋/(2𝑛−𝑚+1)

𝐸∘

𝑛−𝑚(𝑓, 𝑥, 𝑡)

𝑋𝑝

𝑡𝑑𝑡

+ 𝐸∘

𝑛−𝑚(𝑓, 𝑥; 0)

𝑋𝑝,

𝜎𝑛,𝑚𝑓 (𝑥) − 𝑓 (𝑥) ≤ (6 + 𝜋

2) 𝐹𝑛−𝑚,𝑚(𝑓, 𝑥)𝑋𝑝

× [1 + ln 𝑛 + 1𝑚 + 1

]

+ 𝐸𝑛−𝑚(𝑓, 𝑥; 0)𝑋𝑝.

(18)

Now, we can present the main result on pointwiseapproximation.

Theorem 2. If 𝑓 ∈ 𝑋𝑝, then, for any positive integer 𝑚 ≤ 𝑛and all real 𝑥,

𝜎𝑛,𝑚𝑓 (𝑥) − 𝑓 (𝑥)

≤ 𝐾

𝑛

∑

𝜈=0

𝐹𝑛−𝑚+𝜈,𝑚(𝑓, 𝑥)𝑋𝑝+ 𝐹𝑛−𝑚+𝜈,𝜈(𝑓, 𝑥)𝑋𝑝

𝑚 + 𝜈 + 1

+ 𝐸2𝑛(𝑓, 𝑥; 0)𝑋𝑝.

(19)

Remark 3. Theorem 2, in case 𝑓 ∈ 𝐶, immediately yields theresult of Stečkin [2]. The result of Stečkin type also holds ifinstead of 𝐶 we consider the spaces 𝑋𝑝 with 1 < 𝑝 < ∞.Then, in the proof, we need the Hardy-Littlewood estimate ofthe maximal function.

At every 𝑋𝑝-point 𝑥 of 𝑓

Ω𝑥𝑓(𝛾)𝑋𝑝= 𝑜𝑥 (1) as 𝛾 → 0+ (20)

and thus fromTheorem 1, we obtain the corollarywhich statesthe result of the Tanović-Miller type [3].

Corollary 4. If 𝑓 ∈ 𝑋𝑝, then, for any positive integer 𝑚 ≤ 𝑛at every 𝑋𝑝-point 𝑥 of 𝑓,

𝜎𝑛,𝑚𝑓 (𝑥) − 𝑓 (𝑥) = 𝑜𝑥 (1) [1 + ln

𝑛 + 1

𝑚 + 1] as 𝑛 → ∞.

(21)

3. Auxiliary Results

In order to prove our theorems, we require some lemmas.

Lemma 5. If 𝑇𝑛 is the trigonometric polynomial of the degreeat most 𝑛 of the best approximation of 𝑓 ∈ 𝑋𝑝 with respect tothe norm ‖ ⋅ ‖𝑋𝑝 , then it is also the trigonometric polynomial ofthe degree at most 𝑛 of the best approximation of 𝑓 ∈ 𝑋𝑝 withrespect to the norm ‖ ⋅ ‖𝑋𝑝 ,𝑥,𝛿 for any 𝛿 ∈ [0, 𝜋].

Proof. From the inequalities𝐸𝑛(𝑓, ⋅, 𝛿)𝑋𝑝

𝑋𝑝≥

𝐸∘

𝑛(𝑓, ⋅, 𝛿)

𝑋𝑝𝑋𝑝

=

𝑓 − 𝑇𝑛,𝛿

∘

𝑋𝑝 ,⋅,𝛿

𝑋𝑝=

𝑓 − 𝑇𝑛,𝛿𝑋𝑝

≥𝑓 − 𝑇𝑛

𝑋𝑝= 𝐸𝑛(𝑓)𝑋𝑝

,

𝐸∘

𝑛(𝑓, ⋅, 𝛿)

𝑋𝑝𝑋𝑝

≤

𝑓 − 𝑇𝑛

∘

𝑋𝑝,⋅,𝛿

𝑋𝑝

=𝑓 − 𝑇𝑛

𝑋𝑝= 𝐸𝑛(𝑓)𝑋𝑝

,

(22)

where 𝑇𝑛,𝛿 and 𝑇𝑛 are the trigonometric polynomials of thedegree at most 𝑛 of the best approximation of 𝑓 ∈ 𝑋𝑝 withrespect to the norms ‖ ⋅ ‖∘

𝑋𝑝 ,𝑥,𝛿and ‖ ⋅ ‖𝑋𝑝 , respectively, we

obtain relation𝑓 − 𝑇𝑛,𝛿

𝑋𝑝=

𝑓 − 𝑇𝑛𝑋𝑝

= 𝐸𝑛(𝑓)𝑋𝑝, (23)

whence 𝑇𝑛,𝛿 = 𝑇𝑛 for any 𝛿 ∈ [0, 𝜋] by uniqueness ofthe trigonometric polynomial of the degree at most 𝑛 of thebest approximation of 𝑓 ∈ 𝑋𝑝 with respect to the norm‖ ⋅ ‖𝑋𝑝 (e.g., see [5] p. 96). We can also observe that for such𝑇𝑛 and any ℎ ∈ [0, 𝛿],

𝑓 − 𝑇𝑛

∘

𝑋𝑝 ,𝑥,ℎ= 𝐸∘

𝑛(𝑓, 𝑥, ℎ)

𝑋𝑝≤ 𝐸𝑛(𝑓, 𝑥, 𝛿)𝑋𝑝

≤𝑓 − 𝑇𝑛

𝑋𝑝,𝑥,𝛿.

(24)

Hence,

𝐸𝑛(𝑓, 𝑥, 𝛿)𝑋𝑝=

𝑓 − 𝑇𝑛𝑋𝑝 ,𝑥,𝛿 (25)

and our proof is complete.

Lemma 6. If 𝑛 ∈ N0 and 𝛿 > 0, then 𝐸𝑛(𝑓, 𝑥; 𝛿)𝑋𝑝 isnonincreasing function of 𝑛 and nondecreasing function of 𝛿.These imply that for 𝑚, 𝑛 ∈ N the function 𝐹𝑛,𝑚(𝑓, 𝑥)𝑋𝑝 isnonincreasing function of 𝑛 and 𝑚 simultaneously.

Proof. The first part of our statement follows from theproperty of the norm ‖ ⋅ ‖𝑥,𝛿 and supremum.The second partis a consequence of the calculation

𝐹𝑛,𝑚+1(𝑓, 𝑥)𝑋𝑝

𝐹𝑛,𝑚(𝑓, 𝑥)𝑋𝑝

=𝑚 + 1

𝑚 + 2(1 +

𝐸𝑛(𝑓, 𝑥; 𝜋/ (𝑚 + 2))𝑋𝑝

∑𝑚

𝑘=0𝐸𝑛(𝑓, 𝑥; 𝜋/ (𝑘 + 1))𝑋𝑝

)

≤𝑚 + 1

𝑚 + 2(1 +

𝐸𝑛(𝑓, 𝑥; 𝜋/ (𝑚 + 1))𝑋𝑝

∑𝑚

𝑘=0𝐸𝑛(𝑓, 𝑥; 𝜋/ (𝑚 + 1))𝑋𝑝

)

=𝑚 + 1

𝑚 + 2(1 +

1

𝑚 + 1) = 1.

(26)

Lemma 7. Let 𝑚, 𝑛, 𝑞 ∈ N0 such that 𝑚 ≤ 𝑛 and 𝑞 ≥ 𝑚 + 1. If𝑓 ∈ 𝑋

𝑝, then

𝜎𝑛+𝑞,𝑚𝑓 (𝑥) − 𝜎𝑛,𝑚𝑓 (𝑥)

≤ 𝐾𝐹𝑛−𝑚,𝑚(𝑓, 𝑥)𝑋𝑝

𝑞−1

∑

𝜈=0

1

𝑚 + 𝜈 + 1.

(27)


Proof. It is clear that

𝜎𝑛,𝑚𝑓 (𝑥) =1

𝑚 + 1

𝑛

∑

𝑘=𝑛−𝑚

1

𝜋∫

𝜋

−𝜋

𝑓 (𝑥 + 𝑡)𝐷𝑘 (𝑡) 𝑑𝑡

=1

𝜋∫

𝜋

−𝜋

𝑓 (𝑥 + 𝑡) 𝑉𝑛,𝑚 (𝑡) 𝑑𝑡,

(28)

where

𝑉𝑛,𝑚 (𝑡) =1

𝑚 + 1

𝑛

∑

𝑘=𝑛−𝑚

𝐷𝑘 (𝑡) ,

𝐷𝑘 (𝑡) =sin ((2𝑘 + 1) 𝑡/2)

2 sin (𝑡/2).

(29)

Hence, by orthogonality of the trigonometric system,

𝜎𝑛+𝑞,𝑚𝑓 (𝑥) − 𝜎𝑛,𝑚𝑓 (𝑥)

=1

𝜋∫

𝜋

−𝜋

[𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)]

× (𝑉𝑛+𝑞,𝑚 (𝑡) − 𝑉𝑛,𝑚 (𝑡)) 𝑑𝑡

=1

𝜋 (𝑚 + 1)

𝑛

∑

𝑘=𝑛−𝑚

∫

𝜋

−𝜋

[𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)]

× (𝐷𝑘+𝑞 (𝑡) − 𝐷𝑘 (𝑡)) 𝑑𝑡

=1

𝜋 (𝑚 + 1)

×

𝑛

∑

𝑘=𝑛−𝑚

∫

𝜋

−𝜋

[𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)]

×sin ((2𝑘 + 2𝑞 +1) 𝑡/2) − sin ((2𝑘 + 1) 𝑡/2)

2 sin (𝑡/2)𝑑𝑡

=1

𝜋 (𝑚 + 1)

𝑛

∑

𝑘=𝑛−𝑚

∫

𝜋

−𝜋

[𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)]

×sin (𝑞𝑡/2) cos ((2𝑘 + 𝑞1) 𝑡/2)

sin (𝑡/2)𝑑𝑡,

(30)

with trigonometric polynomial𝑇𝑛−𝑚 of the degree at most 𝑛−𝑚 of the best approximation of 𝑓.

Using the notations

𝐼1 = [−𝜋

𝑞,𝜋

𝑞] , 𝐼2 = [−

𝜋

𝑚 + 1, −

𝜋

𝑞] ∪ [

𝜋

𝑞,

𝜋

𝑚 + 1] ,

𝐼3 = [−𝜋, −𝜋

𝑚 + 1] ∪ [

𝜋

𝑚 + 1, 𝜋] ,

(31)

we get

∑ =1

𝜋 (𝑚 + 1)

×

𝑛

∑

𝑘=𝑛−𝑚

(∫𝐼1

+∫𝐼2

+∫𝐼3

)

× [𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)]

×sin (𝑞𝑡/2) cos ((2𝑘 + 𝑞 + 1) 𝑡/2)

sin (𝑡/2)𝑑𝑡

= ∑

1

+∑

2

+∑

3

,

∑

1

≤1

𝜋 (𝑚 + 1)

𝑛

∑

𝑘=𝑛−𝑚

∫𝐼1

𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡) 𝑞𝑑𝑡

=𝑞

𝜋∫𝐼1

𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡) 𝑑𝑡

≤ 2𝐸𝑛−𝑚(𝑓, 𝑥;𝜋

𝑞)

𝑋𝑝

.

(32)

We next evaluate the sums ∑2 and ∑3 using the partialintegrating and Lemma 5. Thus,

∑

2

≤ ∫𝐼2

𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)

𝑡𝑑𝑡

= 2[1

2𝑡∫

𝑡

−𝑡

𝑓 (𝑥 + 𝑢) − 𝑇𝑛−𝑚 (𝑥 + 𝑢) 𝑑𝑢]

𝑡=𝜋/(𝑚+1)

𝑡=𝜋/𝑞

+ 2∫

𝜋/(𝑚+1)

𝜋/𝑞

1

𝑡[

1

2𝑡∫

𝑡

−𝑡

𝑓 (𝑥 + 𝑢) − 𝑇𝑛−𝑚 (𝑥 + 𝑢) 𝑑𝑢]𝑑𝑡


𝑚 + 1)𝑋𝑝

+ 2∫

𝑡

−𝑡

1

𝑡𝐸𝑛−𝑚(𝑓, 𝑥; 𝑡)𝑋𝑝

𝑑𝑡


𝑚 + 1)𝑋𝑝

[1 + ln𝑞

𝑚 + 1]


𝑚 + 1)𝑋𝑝

[1 +

𝑞−1

∑

𝜈=0

1

𝑚 + 𝜈 + 1] ,

∑

3

≤1

𝑚 + 1

× ∫𝐼3

𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)

𝑡

×

𝑛

∑

𝑘=𝑛−𝑚

cos(𝑘𝑡 +𝑞 + 1

2𝑡)

𝑑𝑡


≤1

𝑚 + 1

× ∫𝐼3

𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)

𝑡

×

2 sin((𝑛 + 1) 𝑡/2) cos ((2𝑛 − 𝑚 + 𝑞 + 1) 𝑡/2)2 sin (𝑡/2)

𝑑𝑡

≤𝜋

𝑚 + 1∫𝐼3

𝑓 (𝑥 + 𝑡) − 𝑇𝑛−𝑚 (𝑥 + 𝑡)

𝑡2𝑑𝑡

=𝜋

𝑚 + 1{2 [

1

2𝑡∫

𝑡

−𝑡

𝑓 (𝑥 + 𝑢)

−𝑇𝑛−𝑚 (𝑥 + 𝑢) 𝑑𝑢]

𝑡=𝜋

𝑡=𝜋/(𝑚+1)

+ 4∫

𝜋

𝜋/(𝑚+1)

1

𝑡2[

1

2𝑡∫

𝑡

−𝑡

𝑓 (𝑥 + 𝑢)

− 𝑇𝑛−𝑚 (𝑥 + 𝑢) 𝑑𝑢] 𝑑𝑡}

≤𝜋

𝑚 + 1{2𝐸𝑛−𝑚(𝑓, 𝑥; 𝜋)𝑋𝑝

+ 4∫

𝜋

𝜋/(𝑚+1)

1

𝑡2𝐸𝑛−𝑚(𝑓, 𝑥; 𝑡)𝑋𝑝

𝑑𝑡}

=2𝜋

𝑚 + 1{𝐸𝑛−𝑚(𝑓, 𝑥; 𝜋)𝑋𝑝

+2∫

𝑚+1

1

𝐸𝑛−𝑚(𝑓, 𝑥; 𝜋/𝑢)𝑋𝑝

𝜋2/𝑢2

𝜋𝑑𝑢

𝑢2}

=2𝜋


+2

𝜋

𝑚−1

∑

𝑘=0

∫

𝑘+2

𝑘+1

𝐸𝑛−𝑚(𝑓, 𝑥;𝜋

𝑢)𝑋𝑝

𝑑𝑢}

=2𝜋


+2

𝜋

𝑚−1

∑

𝑘=0


𝑘 + 1)𝑋𝑝

}

≤2𝜋 + 2/𝜋

𝑚 + 1

𝑚−1

∑

𝑘=0


𝑘 + 1)𝑋𝑝

,

(33)

which proves Lemma 7.

Before formulating the next lemmas, we define a newdifference. Let 𝑚, 𝑛 ∈ N0 and 𝑚 ≤ 𝑛. Denote that

𝜏𝑛,𝑚𝑓 (𝑥) := (𝑚 + 1) {𝜎𝑛+𝑚+1,𝑚𝑓 (𝑥) − 𝜎𝑛,𝑚𝑓 (𝑥)} . (34)

Lemma 8. Let 𝑚, 𝑛, 𝜇 ∈ N0 such that 2𝜇 ≤ 𝑚 ≤ 𝑛. If 𝑓 ∈ 𝑋𝑝,then

𝜏𝑛,𝑚𝑓 (𝑥) − 𝜏𝑛−𝜇,𝑚−𝜇𝑓 (𝑥)

≤ 𝐾𝜇𝐹𝑛−𝜇+1,𝜇−1(𝑓, 𝑥)𝑋𝑝

ln 𝑚𝜇

.

(35)

Proof. The proof follows by the method of Leindler [6].Namely,


= (

𝑛+𝑚+1

∑

𝑘=𝑛+𝑚−2𝜇+2

− 2

𝑛

∑

𝑘=𝑛−𝜇+1

)[𝑆𝑘𝑓 (𝑥) − 𝑓 (𝑥)] ,


≤

(

𝑛+𝑚−𝜇+1

∑

𝑘=𝑛+𝑚−2𝜇+2

−

𝑛

∑

𝑘=𝑛−𝜇+1

)[𝑆𝑘𝑓 (𝑥) − 𝑓 (𝑥)]

+

(

𝑛+𝑚+1

∑

𝑘=𝑛+𝑚−𝜇+2

−

𝑛

∑

𝑘=𝑛−𝜇+1

)[𝑆𝑘𝑓 (𝑥) − 𝑓 (𝑥)]

= 𝜇𝜎𝑛+𝑚−𝜇+1,𝜇−1𝑓 (𝑥) − 𝜎𝑛,𝜇−1𝑓 (𝑥)

+ 𝜇𝜎𝑛+𝑚+1,𝜇−1𝑓 (𝑥) − 𝜎𝑛,𝜇−1𝑓 (𝑥)

.

(36)

By Lemma 6, for 2𝜇 ≤ 𝑚,𝜏𝑛,𝑚𝑓 (𝑥) − 𝜏𝑛−𝜇,𝑚−𝜇𝑓 (𝑥)

≤ 𝐾𝜇𝐹𝑛−𝜇+1,𝜇−1(𝑓, 𝑥)𝑋𝑝[1 + ln

(𝑛 − 𝜇 + 1) + 𝜇 − 1

𝜇]

+ 𝐾𝜇𝐹𝑛−𝜇+1,𝜇−1(𝑓, 𝑥)𝑋𝑝[1 + ln

𝑚 + 𝜇 − 1

𝜇]

≤ 𝐾𝜇𝐹𝑛−𝜇+1,𝜇−1(𝑓, 𝑥)𝑋𝑝[1 + ln 𝑚

𝜇] ,

(37)

and our proof is complete.

Lemma 9. Let 𝑚, 𝑛 ∈ N0 and 𝑚 ≤ 𝑛. If 𝑓 ∈ 𝑋𝑝, then

𝜏𝑛,𝑚𝑓 (𝑥) ≤ 𝐾

𝑛

∑

𝑘=𝑛−𝑚

𝐹𝑘,𝑘−𝑛+𝑚(𝑓, 𝑥)𝑋𝑝. (38)

Proof. Our proof runs parallel with the proof ofTheorem 1 in[2].

If 𝑚 = 0, then𝜏𝑛,0𝑓 (𝑥)

=𝜎𝑛+1,0𝑓 (𝑥) − 𝜎𝑛,0𝑓 (𝑥)

≤ 𝐾𝐹𝑛,0(𝑓, 𝑥)𝑋𝑝, (39)

and if 𝑚 = 1, then𝜏𝑛,1𝑓 (𝑥)

≤ 2𝜎𝑛+2,1𝑓 (𝑥) − 𝜎𝑛,1𝑓 (𝑥)

≤ 𝐾𝐹𝑛−1,1(𝑓, 𝑥)𝑋𝑝


≤ 𝐾 [𝐹𝑛−1,1(𝑓, 𝑥)𝑋𝑝+ 𝐹𝑛−1,1(𝑓, 𝑥)𝑋𝑝

]

≤ 𝐾 [𝐹𝑛−1,1(𝑓, 𝑥)𝑋𝑝+ 𝐹𝑛,1(𝑓, 𝑥)𝑋𝑝

]

(40)

by Lemmas 6 and 7.Next, we construct the same decreasing sequence (𝑚𝑠) of

integers that was given by Stečkin. Let

𝑚0 = 𝑚, 𝑚𝑠 = 𝑚𝑠−1 − [𝑚𝑠−1

2] (𝑠 = 1, 2, . . .) , (41)

where [𝑦] denotes the integral part of 𝑦. It is clear that thereexists smallest index 𝑡 ≥ 1 such that 𝑚𝑡 = 1 and

𝑚 = 𝑚0 > 𝑚1 > ⋅ ⋅ ⋅ > 𝑚𝑡 = 1. (42)

By the definition of the numbers 𝑚𝑠, we have

𝑚𝑠 ≥𝑚𝑠−1

2,

𝑚𝑠−1 − 𝑚𝑠 = [𝑚𝑠−1

2] ≥ [

𝑚𝑠−1

3] (𝑠 = 1, 2, . . . , 𝑡)

(43)

whence

𝑚𝑡−1 = 2, 𝑚𝑡−1 − 𝑚𝑡 = 1,

𝑚𝑠−1 − 𝑚𝑠 ≤ 𝑚𝑠 ≤ 3 (𝑚𝑠 − 𝑚𝑠+1) (𝑠 = 1, 2, . . . , 𝑡 − 1)

(44)

follow.Under these notations, we get the following equality:

𝜏𝑛,𝑚𝑓 (𝑥) =

𝑡

∑

𝑠=1

(𝜏𝑛−𝑚+𝑚𝑠−1,𝑚𝑠−1𝑓 (𝑥) − 𝜏𝑛−𝑚+𝑚𝑠 ,𝑚𝑠

𝑓 (𝑥))

+ 𝜏𝑛−𝑚+𝑚𝑡 ,𝑚𝑡𝑓 (𝑥) ,

(45)

whence, by 𝑚𝑡 = 1,

𝜏𝑛,𝑚𝑓 (𝑥) ≤

𝑡

∑

𝑠=1

𝜏𝑛−𝑚+𝑚𝑠−1,𝑚𝑠−1

𝑓 (𝑥) − 𝜏𝑛−𝑚+𝑚𝑠 ,𝑚𝑠𝑓 (𝑥)

+𝜏𝑛−𝑚+𝑚𝑡 ,𝑚𝑡

𝑓 (𝑥)

(46)

follows.It is easy to see that the terms in the sum ∑𝑡

𝑠=1, by

Lemma 8, with 𝜇 = 𝑚𝑠−1 − 𝑚𝑠 and 𝑚 = 𝑚𝑠−1 do not exceed

𝐾(𝑚𝑠−1 − 𝑚𝑠) 𝐹𝑛−𝑚+𝑚𝑠+1,𝑚𝑠−1−𝑚𝑠(𝑓, 𝑥)

𝑋𝑝ln

𝑚𝑠−1

𝑚𝑠−1 − 𝑚𝑠

,

where (𝑠 = 1, 2, . . . , 𝑡 − 1) ,(47)

and by Lemma 7, we get𝜏𝑛−𝑚+1,1𝑓 (𝑥)

≤ 2𝜎𝑛−𝑚+2,1𝑓 (𝑥) − 𝜎𝑛−𝑚+1,1𝑓 (𝑥)

≤ 𝐾𝐹𝑛−𝑚,1(𝑓, 𝑥)𝑋𝑝.

(48)

Thus,

𝜏𝑛,𝑚𝑓 (𝑥) ≤ 𝐾

𝑡−1

∑

𝑠=1

3 (𝑚𝑠 − 𝑚𝑠+1) 𝐹𝑛−𝑚+𝑚𝑠+1,𝑚𝑠(𝑓, 𝑥)

𝑋𝑝ln 3

+ 𝐾𝐹𝑛−𝑚+2,𝑚−2(𝑓, 𝑥)𝑋𝑝+ 𝐾𝐹𝑛−𝑚,1(𝑓, 𝑥)𝑋𝑝

,

(49)

whence, by the monotonicity of 𝐹𝜈,𝜇(𝑓, 𝑥)𝑋𝑝 ,𝜏𝑛,𝑚𝑓 (𝑥)

≤ 𝐾(

𝑡−1

∑

𝑠=1

𝑚𝑠

∑

𝜈=𝑚𝑠+1+1

𝐹𝑛−𝑚+𝜈+1,𝜈(𝑓, 𝑥)𝑋𝑝

+

2

∑

𝜈=0

𝐹𝑛−𝑚+𝜈,𝑚−𝜈−1(𝑓, 𝑥)𝑋𝑝)

+ 𝐾𝐹𝑛−𝑚,1(𝑓, 𝑥)𝑋𝑝

≤ 𝐾

𝑚1+1

∑

𝜈=0

𝐹𝑛−𝑚+𝜈,𝜈(𝑓, 𝑥)𝑋𝑝+ 𝐾𝐹𝑛−𝑚,1(𝑓, 𝑥)𝑋𝑝

≤ 𝐾

𝑚

∑

𝜈=0

𝐹𝑛−𝑚+𝜈,𝜈(𝑓, 𝑥)𝑋𝑝+ 𝐾𝐹𝑛−𝑚,1(𝑓, 𝑥)𝑋𝑝

≤ 𝐾

𝑛

∑

𝑘=𝑛−𝑚

𝐹𝑘,𝑘−𝑛+𝑚(𝑓, 𝑥)𝑋𝑝+ 𝐾𝐹𝑛−𝑚,1(𝑓, 𝑥)𝑋𝑝

.

(50)

4. Proofs of the Results

Proof of Theorem 2. The proof follows the lines of the proofsof Theorem 4 in [2] and Theorem in [6]. Therefore, let 𝑛 > 0and 𝑚 ≤ 𝑛 be fixed. Let us define an increasing sequence(𝑛𝑠 : 𝑠 = 0, 1, . . . , 𝑡) of indices introduced by Stečkin inthe following way. Set 𝑛0 = 𝑛. Assuming that the numbers𝑛0, . . . , 𝑛𝑠 are already defined and 𝑛𝑠 < 2𝑛, we define 𝑛𝑠+1 asfollows. Let 𝜈𝑠 denote the smallest natural number such that

𝐹𝑛𝑠−𝑚+𝜈𝑠 ,𝜈(𝑓, 𝑥)

𝑋𝑝≤

1

2𝐹𝑛𝑠−𝑚,𝜈

(𝑓, 𝑥)𝑋𝑝

(𝜈 = 0, 1, . . . , 𝑛) .

(51)

According to the magnitude of 𝜈𝑠, we define

𝑛𝑠+1 =

{{

{{

{

𝑛𝑠 − 𝑚 + 1 for 𝜈𝑠 ≤ 𝑚,𝑛𝑠 + 𝜈𝑠 for 𝑚 + 1 ≤ 𝜈𝑠 < 2𝑛 + 𝑚 − 𝑛𝑠,2𝑛 + 𝑚 for 𝜈𝑠 ≥ 2𝑛 + 𝑚 − 𝑛𝑠.

(52)

If 𝑛𝑠+1 < 2𝑛, we continue the procedure, and if once 𝑛𝑠+1 ≥ 2𝑛,then we stop the construction and define 𝑡 := 𝑠 + 1.

By the previous definition of (𝑛𝑠), we have the followingobvious properties:

𝑡 ≥ 1, 𝑛 = 𝑛0 < 𝑛1 < ⋅ ⋅ ⋅ < 𝑛𝑡, 2𝑛 ≤ 𝑛𝑡 ≤ 2𝑛 + 𝑚,

𝑛𝑠+1 − 𝑛𝑠 ≥ 𝑚 + 1 (𝑠 = 0, 1, . . . , 𝑡 − 1) ,

(53)


and relations

𝐹𝑛𝑠+1−𝑚,𝜈(𝑓, 𝑥)

𝑋𝑝≤

1



for 𝑠 = 0, 1, . . . , 𝑡 − 2,

1



≤ 𝐹𝑛𝑠+1−𝑚−1,𝜈(𝑓, 𝑥)

𝑋𝑝

for 𝑠 = 0, 1, . . . 𝑡 − 1,

(54)

whenever 𝑛𝑠+1 − 𝑛𝑠 > 𝑚 + 1.Let us start with

𝜎𝑛,𝑚𝑓 (𝑥) − 𝑓 (𝑥)

=

𝑡−1

∑

𝑠=0

[𝜎𝑛𝑠 ,𝑚

𝑓 (𝑥) − 𝑓 (𝑥)−

𝜎𝑛𝑠+1 ,𝑚

𝑓 (𝑥) − 𝑓 (𝑥)]

+𝜎𝑛𝑡 ,𝑚

𝑓 (𝑥) − 𝑓 (𝑥)

≤

𝑡−1

∑

𝑠=0

𝜎𝑛𝑠+1,𝑚

𝑓 (𝑥) − 𝜎𝑛𝑠 ,𝑚𝑓 (𝑥)

+

𝜎𝑛𝑡 ,𝑚

𝑓 (𝑥) − 𝑓 (𝑥)

=

𝑡−1

∑

𝑠=0

1

𝑚 + 1𝜏𝑛𝑠,𝑚

𝑓 (𝑥)

+

𝜎𝑛𝑡 ,𝑚

𝑓 (𝑥) − 𝑓 (𝑥).

(55)

UsingTheorem 1 and that 2𝑛 ≤ 𝑛𝑡 ≤ 2𝑛 + 𝑚, we get

𝜎𝑛𝑡 ,𝑚

𝑓 (𝑥) − 𝑓 (𝑥)

≤ 𝐾𝐹𝑛𝑡−𝑚,𝑚(𝑓, 𝑥)

𝑋𝑝[1 + ln

𝑛𝑡 + 1

𝑚 + 1]

+𝑓 (𝑥) − 𝑇𝑛𝑡−𝑚

(𝑥)

≤ 𝐾

𝑛

∑

𝜈=0

𝐹𝑛−𝑚+𝜈,𝑚(𝑓, 𝑥)𝑋𝑝

𝑚 + 𝜈 + 1+

𝑓 (𝑥) − 𝑇𝑛𝑡−𝑚

(𝑥)

≤ 𝐾

𝑛

∑

𝜈=0


𝑚 + 𝜈 + 1

+𝑓 (𝑥) − 𝑇𝑛𝑡−𝑚

(𝑥)

≤ 𝐾

𝑛

∑

𝜈=0


𝑚 + 𝜈 + 1

+ 𝐸𝑛𝑡−𝑚(𝑓, 𝑥; 0)

𝑋𝑝

≤ 𝐾

𝑛

∑

𝜈=0


𝑚 + 𝜈 + 1

+ 𝐸2𝑛(𝑓, 𝑥; 0)𝑋𝑝.

(56)

The estimate of the sum in the right hand side of (55) wederive from the following one

1


𝑓 (𝑥)

≤ 𝐾

𝑛𝑠+1−𝑛𝑠−1

∑

𝜈=0

𝐹𝑛𝑠−𝑚+𝜈,𝑚(𝑓, 𝑥)

𝑋𝑝+ 𝐹𝑛𝑠−𝑚+𝜈,𝑚


𝑚 + 𝜈 + 1.

(57)

We split the proof of this inequality in two parts. If 𝑛𝑠+1−𝑛𝑠 =𝑚 + 1, then by Lemma 9,

1

𝑚 + 1𝜏𝑛𝑠 ,𝑚

𝑓 (𝑥)

≤ 𝐾

1

𝑚 + 1

𝑛𝑠

∑

𝑘=𝑛𝑠−𝑚

𝐹𝑘,𝑘−𝑛𝑠+𝑚(𝑓, 𝑥)

𝑋𝑝

≤ 𝐾


∑

𝜈=0

𝐹𝑛𝑠−𝑚+𝜈,𝜈(𝑓, 𝑥)

𝑋𝑝

𝑚 + 𝜈 + 1.

(58)

If 𝑛𝑠+1 − 𝑛𝑠 > 𝑚 + 1, then, by Lemma 7,

1


𝑓 (𝑥)

≤ 𝐾𝐹𝑛𝑠−𝑚,𝑚(𝑓, 𝑥)

𝑋𝑝


∑

𝜈=0

1

𝑚 + 𝜈 + 1,

(59)

and since (1/2)𝐹𝑛𝑠−𝑚,𝑚(𝑓, 𝑥)𝑋𝑝 ≤ 𝐹𝑛𝑠+1−𝑚−1,𝑚(𝑓, 𝑥)𝑋𝑝 , we have

1


𝑓 (𝑥)

≤ 2𝐾𝐹𝑛𝑠+1−𝑚−1,𝑚(𝑓, 𝑥)

𝑋𝑝


∑

𝜈=0

1

𝑚 + 𝜈 + 1

≤ 2𝐾


∑

𝜈=0


𝑋𝑝

𝑚 + 𝜈 + 1.

(60)

Consequently,𝑡−1

∑

𝑠=0

1


𝑓 (𝑥)

≤ 2𝐾

𝑡−1

∑

𝑠=0


∑

𝜈=0


𝑋𝑝+ 𝐹𝑛𝑠−𝑚+𝜈,𝜈


𝑚 + 𝜈 + 1.

(61)

Since 𝑛𝑠+1 − 𝑛𝑠 ≤ 2𝑛 + 𝑚 − 𝑛 − 1 = 𝑛 + 𝑚 − 1 for all 𝑠 ≤ 𝑡 − 1,changing the order of summation, we get𝑡−1

∑

𝑠=0

1


𝑓 (𝑥)

≤ 2𝐾

𝑛+𝑚−1

∑

𝜈=0

1

𝑚 + 𝜈 + 1

× ∑

𝑠:𝑛𝑠+1−𝑛𝑠>𝜈

[𝐹𝑛𝑠−𝑚+𝜈,𝑚(𝑓, 𝑥)



] .

(62)


Using the inequality

𝐹𝑛𝑠+1−𝑚,𝜈(𝑓, 𝑥)

𝑋𝑝≤

1



for { 𝜈 = 0, 1, 2, . . . , 𝑛𝑠+1 − 𝑛𝑠 − 1,𝑠 = 0, 1, 2, . . . , 𝑡 − 2,

(63)

we obtain

∑

𝑠:𝑛𝑠+1−𝑛𝑠>𝜈

[𝐹𝑛𝑠−𝑚+𝜈,𝑚(𝑓, 𝑥)



]

= 𝐹𝑛𝑝−𝑚+𝜈,𝑚(𝑓, 𝑥)

𝑋𝑝+ 𝐹𝑛𝑝−𝑚+𝜈,𝜈


+ ∑

𝑠≥𝑝+1:𝑛𝑠+1−𝑛𝑠>𝜈

[𝐹𝑛𝑠+1−𝑚+𝜈,𝑚(𝑓, 𝑥)

𝑋𝑝

+𝐹𝑛𝑠+1−𝑚+𝜈,𝜈(𝑓, 𝑥)

𝑋𝑝]

≤ 𝐹𝑛𝑝−𝑚+𝜈,𝑚(𝑓, 𝑥)



+ ∑

𝑠:𝑠≥𝑝+1

𝐹𝑛𝑠−𝑚,𝑚(𝑓, 𝑥)

𝑋𝑝[𝐹𝑛𝑠−𝑚,𝑚


+𝐹𝑛𝑠−𝑚,𝜈(𝑓, 𝑥)

𝑋𝑝]

≤ 𝐹𝑛𝑝−𝑚+𝜈,𝑚(𝑓, 𝑥)



+ 2 [𝐹𝑛𝑝+1−𝑚,𝑚(𝑓, 𝑥)

𝑋𝑝+ 𝐹𝑛𝑝+1−𝑚,𝜈


]

≤ 3 [𝐹𝑛𝑝−𝑚+𝜈,𝑚(𝑓, 𝑥)



] ,

(64)

where 𝑝 denotes the smallest index 𝑠 having the property𝑛𝑠+1 − 𝑛𝑠 > 𝜈. Hence,

𝑡−1

∑

𝑠=0

1


𝑓 (𝑥)

≤ 𝐾

𝑛+𝑚−1

∑

𝜈=0


𝑚 + 𝜈 + 1

≤ 𝐾

𝑛

∑

𝜈=0


𝑚 + 𝜈 + 1,

(65)

and our proof follows.

References

[1] S. Aljančić, R. Bojanić, andM. Tomić, “On the degree of conver-gence of Fejéer-Lebesgue sums,” L’EnseignementMathématique.Revue Internationale. IIe Série, vol. 15, pp. 21–28, 1969.

[2] S. B. Stečkin, “On the approximation of periodic functions byde la Vallée Poussin sums,” Analysis Mathematica, vol. 4, no. 1,pp. 61–74, 1978.

[3] N. Tanović-Miller, “On some generalizations of the Fejér-Lebesgue theorem,” Unione Matematica Italiana. Bollettino. B.Serie 6, vol. 1, no. 3, pp. 1217–1233, 1982.

[4] W. Łenski, “Pointwise approximation by de la Vallée-Poussinmeans,” East Journal on Approximations, vol. 14, no. 2, pp. 131–136, 2008.

[5] P. L. Butzer and R. J. Nessel, Fourier Analysis and Approxima-tion, Academic Press, New York, NY, USA, 1971.

[6] L. Leindler, “Sharpening of Stečkin’s theorem to strong approx-imation,” Analysis Mathematica, vol. 16, no. 1, pp. 27–38, 1990.

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