Hilbert Pachpatte Inequalities

8/4/2019 Hilbert Pachpatte Inequalities

1/28

Hilbert-Pachpattes Inequalities

Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 1 of 28

Go Back

Full Screen

Close

SOME NEW HILBERT-PACHPATTESINEQUALITIES

WENGUI YANGSchool of Mathematics and Computational ScienceXiangtan University, Xiangtan, 411105Hunan, P.R. China

EMail: [email protected]

Received: 05 September, 2008

Accepted: 2 January, 2009

Communicated by: W.S. Cheung

2000 AMS Sub. Class.: 26D15.

Key words: Hilbert-Pachpattes inequality; Hlders inequality; Jensens inequality; Nonneg-ative sequences.

Abstract: Some new Hilbert-Pachpatte discrete inequalities and their integral analogues areestablished in this paper. Other inequalities are also given in remarks.
mailto:[email protected]:[email protected]://jipam.vu.edu.au/http://close/http://fullscreen/http://goback/http://prevpage/http://lastpage/mailto:[email protected]://jipam.vu.edu.au/


2/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 2 of 28

Go Back

Full Screen

Close

Contents

1 Introduction 3

2 Main Results 6

3 Integral Analogues 17
http://jipam.vu.edu.au/http://close/http://fullscreen/http://goback/http://prevpage/http://lastpage/mailto:[email protected]://jipam.vu.edu.au/


3/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 3 of 28

Go Back

Full Screen

Close

1. Introduction

Let p 1, q 1 and {am} and {bn} be two nonnegative sequences of real numbersdefined for m = 1, 2, . . . , k and n = 1, 2, . . . , r, where k and r are natural numbersand define Am =

ms=1 as and Bn =

nt=1 bt. Then

(1.1)k

m=1

rn=1

ApmBqnm + n

C(p,q,k,r)

km=1

(k m + 1)(Ap1m am)2 1

2

rn=1

(r n + 1)(Bq1n bn)21

2

,

unless {am} or {bn} is null, where C(p,q,k,r) = 12pq

kr.

An integral analogue of (1.1) is given in the following result.Let p 1, q 1 and f() 0, g() 0 for (0, x), (0, y), where x, yare positive real numbers and define F(s) =

s0 f()d and G(t) =

t0 g()d, for

s (0, x), t (0, y). Then

(1.2)x0

y0

Fp(s)Gq(t)dsdt

s + t D(p,q,x,y)

x0

(x s)(Fp1(s)f(s))2ds 1

2

y0

(y t)(Gq1(t)g(t))2dt12

,

unless f() 0 or g() 0, where D(p,q,x,y) = 12pq

xy.Inequalities (1.1) and (1.2) are the well known Hilbert-Pachpatte inequalities [1],

which gave new estimates on Hilbert type inequalities [2]. It is well known that theHilbert-Pachpatte inequalities play a dominant role in analysis, so the literature onsuch inequalities and their applications is vast [3] [8].


4/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 4 of 28

Go Back

Full Screen

Close

Young-Ho Kim [9] gave new inequalities similar to the Hilbert-Pachpatte inequal-ities as follows.

Let p 1, q 1, > 0, and {am} and {bn} be two nonnegative sequences ofreal numbers defined for m = 1, 2, . . . , k and n = 1, 2, . . . , r, where k and r arenatural numbers and define Am =

ms=1 as and Bn =

nt=1 bt. Then

(1.3)k

m=1

rn=1

ApmBqn

(m + n)1

C(p,q,k,r; )

km=1

(k m + 1)(Ap1m am)21

2

rn=1

(r n + 1)(Bq1n bn)21

2

,

unless{

am}

or{

bn}

is null, where C(p,q,k,r; ) = 12 1 pqkr.

An integral analogue of (1.3) is given in the following result.Letp 1, q 1, > 0 and f() 0, g() 0 for (0, x), (0, y), where

x, y are positive real numbers and define F(s) =s0

f()d and G(t) =t0

g()d,for s (0, x), t (0, y). Then

(1.4)x0

y0

Fp(s)Gq(t)dsdt

(s + t)1

D(p,q,x,y; )

x

0

(x s)(Fp1(s)f(s))2ds

12

y

0

(y t)(Gq1(t)g(t))2dt1

2

,

unless f() 0 or g() 0, where D(p,q,x,y; ) =12

1 pq

xy.


5/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 5 of 28

Go Back

Full Screen

Close

The purpose of the present paper is to derive some new generalized inequalities(1.1) and (1.2) that are similar to (1.3) and (1.4). By applying an elementary inequal-ity, we also obtain some new inequalities similar to some results in [1, 9].


6/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 6 of 28

Go Back

Full Screen

Close

2. Main Results

Now we give our results as follows in this paper.

Theorem 2.1. Let p 1, q 1, > 1, > 1 and{am} and{bn} be two non-negative sequences of real numbers defined form = 1, 2, . . . , k andn = 1, 2, . . . , r,where

kand

rare natural numbers and define

Am =m

s=1 asand

Bn =n

t=1 bt.

Then

(2.1)k

m=1

rn=1

ApmBqn

m(1)(+)

+ n(1)(+)

C(p,q,k,r; , )

k

m=1(k m + 1)(Ap1m am)

1

r

n=1(r n + 1)(Bq1n bn)

1

,

unless {am} or{bn} is null, where C(p,q,k,r; , ) = pq+k1

r1

.

Proof. The idea for the proof Theorem 2.1 comes from Theorem 1 of [1] and The-orem 2.1 of [9]. From the hypotheses of Theorem 2.1 and using the following in-equality (see [10, 11]),

(2.2)

n

m=1zm

n

m=1zm

m

k=1zk

1,

where 1 is a constant and zm 0, (m = 1, 2, . . . , n), it is easy to observe thatApm p

ms=1

Ap1s as, m = 1, 2, . . . , k ,(2.3)

Bqn qnt=1

Bq1t bt, n = 1, 2, . . . , r .


7/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 7 of 28

Go Back

Full Screen

Close

From (2.3) and Hlders inequality, we have

(2.4)ms=1

Ap1s as m1

ms=1

(Ap1s as)

1

, m = 1, 2, . . . , k ,

and

(2.5)nt=1

Bq1t bt n1

nt=1

(Bq1t bt)

1

, n = 1, 2, . . . , r .

Using the inequality of means [12]

(2.6) n

i=1 s

i

i

1n

1

n

n

i=1 is

r

i

1r

for r > 0, i > 0,n

i=1 i = n, we observe that

(2.7) (s11 s22 )

r/(1+2) 11 + 2

(1sr1 + 2s

r2) .

Let s1 = m1, s2 = n1, 1 = 1 , 2 =1

and r = 1 + 2, from (2.3) (2.5)

and (2.7), we have

ApmBqn pqm

1 n

1

ms=1

(Ap1s as)

1

nt=1

(Bq1t bt)

1

(2.8)

pq +

m

(1)(+)

+

n(1)(+)


8/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 8 of 28

Go Back

Full Screen

Close

ms=1

(Ap1s as)

1

nt=1

(Bq1t bt)

1

,

for m = 1, 2, . . . , k , n = 1, 2, . . . , r . From (2.8), we observe that

(2.9) Ap

mB

q

n

m(1)(+)

+ n(1)(+)

pq +

ms=1

(Ap1s as)

1

nt=1

(Bq1t bt)

1

,

for m = 1, 2, . . . , k , n = 1, 2, . . . , r . Taking the sum on both sides of (2.9) firstover n from 1 to r and then over m from 1 to k of the resulting inequality and using

Hlders inequality with indices , /( 1) and , /( 1) and interchangingthe order of summations, we observe that

km=1

rn=1

ApmBqn

m(1)(+)

+ n(1)(+)

pq

+

k

m=1

m

s=1

(Ap1s as)

1

r

n=1

n

t=1

(Bq1t bt)

1

pq

+ k1

k

m=1

ms=1

(Ap1s as)

1

r1

r

n=1

nt=1

(Bq1t bt)

1


9/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 9 of 28

Go Back

Full Screen

Close

=pq

+ k1 r

1

k

m=1

(k m + 1)(Ap1m am) 1

r

n=1(r n + 1)(Bq1n bn)

1

.

Remark 1. In Theorem 2.1, setting = = 2, we have (1.1). In Theorem 2.1,setting 1

+ 1

= 1, we have

k

m=1r

n=1ApmB

qn

m1 + n1 C(p,q,k,r; , )

km=1

(k m + 1)(Ap1m am) 1

rn=1

(r n + 1)(Bq1n bn) 1

,

unless {am} or {bn} is null, where C(p,q,k,r; , ) = pq+k1 r

1 .

Remark2. In Theorem 2.1, setting p = q = 1, we have

(2.10)

km=1

rn=1

AmBnm

(1)(+) + n

(1)(+)

C(1, 1, k , r; , )

km=1

(k m + 1)am 1

rn=1

(r n + 1)bn 1

,

unless {am} or {bn} is null, where C(1, 1, k , r; , ) = 1+k1 r

1 .
http://jipam.vu.edu.au/http://close/http://fullscreen/http://goback/http://prevpage/http://lastpage/mailto:[email protected]://jipam.vu.edu.au/http://jipam.vu.edu.au/


10/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 10 of 28

Go Back

Full Screen

Close

In the following theorem we give a further generalization of the inequality (2.10)obtained in Remark 2. Before we give our result, we point out that {pm} and {qn}should be two positive sequences for m = 1, 2, . . . , k and n = 1, 2, . . . , r in Theorem2.3 of [9].

Theorem 2.2. Let > 1, > 1 and{am} and{bn} be two nonnegative sequences ofreal numbers and{pm} and{qn} be positive sequences defined form = 1, 2, . . . , kandn = 1, 2, . . . , r, where k andr are natural numbers and define Am =

ms=1 as,

Bn =n

t=1 bt, Pm =m

s=1 ps andQn =n

t=1 qt. Let and be two real-valued,nonnegative, convex, and submultiplicative functions defined on R+ = [0,). Then

(2.11)k

m=1

rn=1

(Am)(Bn)

m(1)(+)

+ n(1)(+)

M(k, r; , )

km=1

(k m + 1)pm

ampm

1

rn=1

(r n + 1)

qn

bnqn

1,

where

M(k, r; , ) =1

+

k

m=1

(Pm)

Pm

1

1

r

n=1

(Qn)

Qn

1

1

.

Proof. From the hypotheses of and and by using Jensens inequality and Hlders


11/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 11 of 28

Go Back

Full Screen

Close

inequality, it is easy to observe that

(Am) =

Pm

ms=1 psas/psms=1 ps

(2.12)

(Pm)

ms=1 psas/psm

s=1 ps

(Pm)

Pm

m

s=1

psas

ps

(Pm)Pm

m1

ms=1

ps

asps

1,

and similarly,

(2.13) (Bn) (Qn)

Qn n

1 n

t=1

qt

btqt

1

.

Let s1 = m1, s2 = n1, 1 = 1 , 2 =1

and r = 1 + 2, from (2.7), (2.12)and (2.13), we have

(Am)(Bn) m1 n1

(Pm)

Pm

m

s=1ps

asps

1

(2.14)

(Qn)

Qn

nt=1

qt

btqt

1

+

m

(1)(+)

+

n(1)(+)


12/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 12 of 28

Go Back

Full Screen

Close

(Pm)

Pm

ms=1

ps

asps

1(Qn)

Qn

nt=1

qt

btqt

1for m = 1, 2, . . . , k , n = 1, 2, . . . , r . From (2.14), we observe that

(2.15)

(Am)(Bn)

m(1)(+) + n

(1)(+)

1 +

(Pm)

Pm

ms=1

ps

asps

1

(Qn)

Qn

n

t=1

qtbtqt

1

for m = 1, 2, . . . , k, n = 1, 2, . . . , r . Taking the sum on both sides of (2.15) firstover n from 1 to r and then over m from 1 to k of the resulting inequality and usingHlders inequality with indices , /( 1) and , /( 1) and interchangingthe order of summations, we observe thatk

m=1

r

n=1

(Am)(Bn)

m

(1)(+)

+ n

(1)(+)

1 +

km=1

(Pm)

Pm

ms=1

ps

asps

1 rn=1

(Qn)

Qn

nt=1

qt

btqt

1

1 +

k

m=1

(Pm)

Pm

1

1

k

m=1

ms=1

ps

asps

1


13/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 13 of 28

Go Back

Full Screen

Close

rn=1

(Qn)

Qn

1

1

r

n=1

nt=1

qt

btqt

1

=1

+

k

m=1(Pm)

Pm

1

1

r

n=1 (Qn)

Qn

1

1

km=1

(k m + 1)ps

asps

1 rn=1

(r n + 1)

qt

btqt

1.

Remark3. From the inequality (2.7), we obtain

(2.16) s11

s22

1

1 + 2

1s1+21

+ 2s1+22

for 1 > 0, 2 > 0. If we apply the elementary inequality (2.16) on the right-handsides of (2.1) in Theorem 2.1 and (2.11) in Theorem 2.2, then we get the followinginequalities

k

m=1r

n=1ApmB

qn

m(1)(+)

+ n(1)(+)

C(p,q,k,r; , ) +

1

k

m=1

(k m + 1)(Ap1m am)

+

+1

r

n=1

(r n + 1)(Bq1n bn)+

,


14/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page14of 28

Go Back

Full Screen

Close

where C(p,q,k,r; , ) = pq+k1 r

1 . Also,

km=1

rn=1

(Am)(Bn)

m(1)(+)

+ n(1)(+)

M(k, r; , ) +

1

k

m=1

(k m + 1)pm

ampm

+

+1

r

n=1

(r n + 1)

qn

bnqn

+ ,where

M(k, r; , ) =1

+

k

m=1

(Pm)

Pm

1

1

r

n=1

(Qn)

Qn

1

1

.

The following theorems deal with slight variants of the inequality (2.11) given inTheorem 2.2.

Theorem 2.3. Let > 1, > 1 and

{am

}and

{bn

}be two nonnegative sequences

of real numbers defined for m = 1, 2, . . . , k andn = 1, 2, . . . , r, where k andr arenatural numbers and define Am =

1m

ms=1 as andBn =

1n

nt=1 bt. Let and be

two real-valued, nonnegative, convex functions defined on R+ = [0,). Thenk

m=1

rn=1

mn(Am)(Bn)

m(1)(+)

+ n(1)(+)

C(1, 1, k , r; , )


15/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page15of 28

Go Back

Full Screen

Close

km=1

(k m + 1)(am) 1

rn=1

(r n + 1)(bn) 1

,

where C(1, 1, k , r; , ) = 1+k1 r

1 .

Proof. From the hypotheses and by using Jensens inequality and Hlders inequal-ity, it is easy to observe that

(Am) =

1

m

ms=1

as

1

m

ms=1

(as) 1m

m1

ms=1

(as)

1

,

(Bn) =

1n

nt=1

bt 1n

nt=1

(bt) 1nn1

n

t=1(bt)

1.

The rest of the proof can be completed by following the same steps as in theproofs of Theorems 2.1 and 2.2 with suitable changes and hence we omit the details.

Theorem 2.4. Let > 1, > 1 and{am} and{bn} be two nonnegative sequences ofreal numbers and

{pm}

and

{qn}

be positive sequences defined form = 1, 2, . . . , kandn = 1, 2, . . . , r, where k andr are natural numbers and define Pm =

ms=1 ps,

Qn =n

t=1 qt, Am =1Pm

ms=1 pmas andBn =

1Qn

nt=1 qnbt. Let and be two

real-valued, nonnegative, convex functions defined on R+ = [0,). Thenk

m=1

rn=1

PmQn(Am)(Bn)

m(1)(+)

+ n(1)(+)


16/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 16 of 28

Go Back

Full Screen

Close

C(1, 1, k , r; , )

km=1

(k m + 1) [pm (am)] 1

r

n=1(r n + 1) [qn (bn)]

1

,

where C(1, 1, k , r; , ) = 1+

k1 r

1 .

Proof. From the hypotheses and by using Jensens inequality and Hlders inequal-ity, it is easy to observe that

(Am) =

1

Pm

m

s=1psas

1Pm

ms=1

ps(as) 1Pm

m1

ms=1

[ps(as)]

1

,

(Bn) =

1

Qn

nt=1

qtbt

1Qn

nt=1

qt(bt) 1Qn

n1

nt=1

[qt(bt)]

1

.

The rest of the proof can be completed by following the same steps as in theproofs of Theorems 2.1 and 2.2 with suitable changes and hence we omit the details.


17/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 17 of 28

Go Back

Full Screen

Close

3. Integral Analogues

Now we give the integral analogues of the inequalities in Theorems 2.1 2.4.An integral analogue of Theorem 2.1 is given in the following theorem.

Theorem 3.1. Let p 0, q 0, > 1, > 1 and f() 0, g() 0 for (0, x),

(0, y) , where x, y are positive real numbers, define F(s) = s

0f()d,

G(t) =t0 g()d fors (0, x), t (0, y). Then

(3.1)x0

y0

Fp(s)Gq(t)

s(1)(+)

+ t(1)(+)

dsdt

D(p,q,x,y; , )x

0

(x s)(Fp1(s)f(s))ds 1

y

0

(y t)(Gq1(t)g(t))dt 1

,

unless f() 0 org() 0, where D(p,q,x,y; , ) = pq+

x1 y

1 .

Proof. From the hypotheses ofF(s) and G(t), it is easy to observe that

Fp(s) = ps

0

Fp1()f()d, s

(0, x),(3.2)

Gq(t) = q

t0

Gq1()g()d, t (0, y).

From (3.2) and Hlders inequality, we have

(3.3)x0

Fp1()f()d s1s

0

(Fp1()f())d

1

, s (0, s),


18/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 18 of 28

Go Back

Full Screen

Close

and

(3.4)y0

Gq1(t)g(t)dt t1t

0

(Gq1()g())d

1

, t (0, t).

Let s1 = s1, s2 = t1, 1 = 1 , 1 =1, r = 1 + 2, from (3.2) (3.4) and

(2.7), we observe thatFp(s)Gq(t)(3.5)

pqs1 t1s

0

(Fp1()f())d

1t

0

(Gq1()g())d

1

pq +

m(1)(+)

+

n(1)(+)

s0

(Fp1()f())d

1t

0

(Gq1()g())d

1

for s (0, x), t (0, y). From (3.5), we observe that

(3.6)Fp(s)Gq(t)

s

(1)(+)

+ t

(1)(+)

pq +

s0

(Fp1()f())d

1t

0

(Gq1()g())d

1

for s (0, x), t (0, y). Taking the integral on both sides of (3.6) first over t from0 to y and then over s from 0 to x of the resulting inequality and using Hldersinequality with indices , /( 1) and , /( 1) and interchanging the order


19/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 19 of 28

Go Back

Full Screen

Close

of integrals, we observe thatx0

y0

Fp(s)Gq(t)

s(1)(+)

+ t(1)(+)

dsdt

pq

+

x

0

s

0

(Fp1()f())d1

ds y

0

t

0

(Gq1()g())d1

dt pq

+ x1

x0

s0

(Fp1()f())dds

1

y1

y0

t0

(Gq1()g())ddt

1

=pq

+ x1 y

1

x0

(x s)(Fp1(s)f(s))ds 1

t

0


.

Remark 4. In Theorem 3.1, setting = = 2, we have (1.2). In Theorem 3.1,setting 1 +

1 = 1, we havex

0

y0

Fp(s)Gq(t)

s1 + t1dsdt

D(p,q,x,y; , )x

0

(x

s)(Fp1(s)f(s))ds1

y

0


,

unless f() 0 or g() 0, where D(p,q,x,y; , ) = pq+

x1 y

1 .


20/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page20of 28

Go Back

Full Screen

Close

Remark5. In Theorem 3.1, setting p = q = 1, we have

(3.7)x0

y0

F(s)G(t)

s(1)(+)

+ t(1)(+)

dsdt

D(1, 1, x , y; , )

x

0

(x

s)f(s)ds

1

y

0

(y

t)g(t)dt

1

,

unless f() 0 or g() 0, where D(1, 1, x , y; , ) = 1+

x1 y

1 .

In the following theorem we give a further generalization of the inequality (3.7)obtained in Remark 5.

Theorem 3.2. Let > 1, > 1 and f() 0, g() 0, p() > 0 and q() > 0for (0, x), (0, y) , where x, y are positive real numbers. Define F(s) =s0 f()d and G(t) = t0 g()d, P(s) = s0 p()d and Q(t) = t0 q()d fors (0, x), t (0, y). Let and be two real-valued, nonnegative, convex, andsubmultiplicative functions defined on R+ = [0,). Then

(3.8)x0

y0

(F(s))(G(t))

s(1)(+)

+ t(1)(+)

dsdt

L(x, y; , )x

0

(x s) p(s)f(s)

p(s)

ds1

y

0

(y t)

q(t)

g(t)

q(t)

dt

1

,

where

L(x, y; , ) =1

+

x0

(P(s))

P(s)

1

ds

1

y0

(Q(t))

Q(t)

1

dt

1

.


21/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 21 of 28

Go Back

Full Screen

Close

Proof. From the hypotheses of and and by using Jensens inequality and Hldersinequality, it is easy to see that

(F(s))) =

P(s)

s0

p()f()p()

d

s

0p()d

(3.9)

(P(s))

s0

p()f()p()

ds

0p()d

(P(s))P(s)

s0

p()

f()

p()

d

(P(s))

P(s)s1

s

0p()f()

p()

d1

,

and similarly,

(3.10) (G(t)) (Q(t))Q(t)

t1

t0

q()

g()

q()

d

1

.

Let s1 = s1, s2 = t1, 1 = 1 , 1 =1

, r = 1 + 2, from (3.9), (3.10) and

(2.7), we observe that

(F(s))(G(t)) s1 t1

(P(s))

P(s)

s0

p()

f()

p()

d

1

(3.11)

(Q(t))

Q(t)

t0

q()

g()

q()

d

1


22/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 22 of 28

Go Back

Full Screen

Close

+

m

(1)(+)

+

n(1)(+)

(P(s))

P(s)

s0

p()

f()

p()

d

1

(Q(t))

Q(t)

t

0

q()

g()

q()

d

1

for s (0, x), t (0, y). From (3.11), we observe that

(3.12)(F(s))(G(t))

s(1)(+)

+ t(1)(+)

1 +

(P(s))

P(s)

s

0

p()

f()

p()

d

1

(Q(t))

Q(t)

t0

q()

g()

q()

d

1

for s (0, x), t (0, y). Taking the integral on both sides of (3.12) first over t from0 to y and then over s from 0 to x of the resulting inequality and using Hldersinequality with indices , /( 1) and , /( 1) and interchanging the orderof integrals, we observe thatx0

y0

(F(s))(G(t))

s(1)(+)

+ t(1)(+)

dsdt


23/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 23 of 28

Go Back

Full Screen

Close

1 +

x0

(P(s))

P(s)

s0

p()

f()

p()

d

1

ds

y

0

(Q(t))

Q(t)

t0

q()

g()

q()

d

1

dt

1 +

x0

(P(s))

P(s)

1

ds

1 x

0

s0

p()

f()

p()

dds

1

y

0

(Q(t))

Q(t)

1

dt

1 y

0

t0

q()

g()

q()

ddt

1

=

1

+ x

0(P(s))

P(s)

1

ds

1 y

0(Q(t))

Q(t)

1

dt

1

x

0

(x s)p(s)

f(s)

p(s)

ds

1y

0

(y t)

q(t)

g(t)

q(t)

dt

1

.

Remark6. From the inequality (2.7), we obtain

(3.13) s11 s22 11 + 2

1s

1+21 + 2s

1+22

for 1 > 0, 2 > 0. If we apply the elementary inequality (3.13) on the right-handsides of (3.1) in Theorem 3.1 and (3.8) in Theorem 3.2, then we get the following


24/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 24 of 28

Go Back

Full Screen

Close

inequalities

(3.14)x0

y0

Fp(s)Gq(t)

s(1)(+)

+ t(1)(+)

dsdt

D(p,q,x,y; , )

+ 1

x

0

(x

s)(Fp1(s)f(s))ds

+

+1

y0

(y t)(Gq1(t)g(t))dt+

,

where D(p,q,x,y; , ) = pq+

x1 y

1 . Also,

x0

y0

(F(s))(G(t))

s (

1)(+) + t (

1)(+)

dsdt

L(x, y; , ) +

1

x0

(x s)p(s)

f(s)

p(s)

ds

+

+1

y0

(y t)

q(t)

g(t)

q(t)

dt

+

,

where

L(x, y; , ) =1

+

x0

(P(s))

P(s)

1

ds

1 y

0

(Q(t))

Q(t)

1

dt

1

.

The following theorems deal with slight variants of (3.8) given in Theorem 3.2.Before we state our next theorem, we point out that F(s) =

s

0f()d and G(t) =


25/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page25of 28

Go Back

Full Screen

Close

t0 g()d are replaced by F(s) =

1s

s0 f()d and G(t) =

1t

t0 g()d in The-

orem 3.4 in [9].

Theorem 3.3. Let > 1, > 1 and f() 0, g() 0 for (0, x), (0, y), where x, y are positive real numbers. Define F(s) = 1s

s0

f()d, G(t) =1t

t

0 g()d fors (0, x), t (0, y). Let and be two real-valued, nonnegative,convex functions defined on

R+ = [0,). Thenx

0

y0

st(F(s))(G(t))

s(1)(+)

+ t(1)(+)

dsdt

D(1, 1, x , y; , )x

0

(x s)(f(s))ds 1

y

0

(y t)(g(t))dt 1

,

where D(1, 1, x , y; , ) = 1+x1 y

1 .

Theorem 3.4. Let > 1, > 1 and f() 0, g() 0, p() > 0 andq() > 0 for (0, x), (0, y) , where x, y are positive real numbers. De-

fine P(s) =s0 p()d, Q(t) =

t0 q()d, F(s) =

1P(s)

s0 p()f()d andG(t) =

1Q(t)

t0

q()g()d for s (0, x), t (0, y). Let and be two real-valued,nonnegative, convex functions defined on R+ = [0,). Then

x0

y0

P(s)Q(t)(F(s))(G(t))

s (1)(+) + t (

1)(+) dsdt

D(1, 1, x , y; , )x

0

(x s) [p(s) (f(s))] ds 1

y

0

(y t) [q(t) (g(t))]dt 1

,


26/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 26 of 28

Go Back

Full Screen

Close

where D(1, 1, k , r; , ) = 1+x1 y

1 .

The proofs of Theorems 3.3 and 3.4 are similar to the proof of Theorem 3.2 andsimilar to the proofs of Theorems 2.3 and 2.4. Hence, we leave out the details.


27/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 27 of 28

Go Back

Full Screen

Close

References

[1] B.G. PACHPATTE, On some new inequalities similar to Hilberts inequality, J.Math. Anal. and Appl., 226 (1998), 116179.

[2] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, London:Cambridge University Press, 1952.

[3] G.D. HANDLEY, J.J. KOLIHA AN D J. PECARIC, Hilbert-Pachpatte type mul-tidimensional integral inequalities, J. Ineq. Pure and Appl. Math., 5(2) (2004),Art. 34. [ONLINE http://jipam.vu.edu.au/article.php?sid=389].

[4] C.J. ZHAO AND W.S. CHEUNG, Inverses of new Hilbert-Pachatte type in-equalities, J. Inequ. and Appl., 2006 (2006), 111.

[5] B.G. PACHPATTE, Inequalities similar to certain extensions of Hilberts in-equality, J. Math. Anal. and Appl., 243 (2000), 217227.

[6] Z.X. L, Some new inequalities similar to Hilbert-Pachpatte type inequali-ties, J. Ineq. Pure and Appl. Math., 4(2) (2003), Art. 33. [ONLINE http://jipam.vu.edu.au/article.php?sid=271].

[7] J. PECARIC, I. PERIC AND P. VUKVIC, Hilbert-Pachpatte type inequalitiesfrom Bonsalls form of Hilberts inequality, J. Ineq. Pure and Appl. Math., 9(1)(2008), Art. 9. [ONLINE http://jipam.vu.edu.au/article.php?sid=949].

[8] S.S. DRAGOMIR AN D YOUNG-HO KIM, Hilbert-Pachpatte type integral in-equalities and their improvement, J. Ineq. Pure and Appl. Math., 4(1) (2003),Art. 16. [ONLINE http://jipam.vu.edu.au/article.php?sid=252].
http://jipam.vu.edu.au/article.php?sid=252http://jipam.vu.edu.au/article.php?sid=252http://jipam.vu.edu.au/article.php?sid=949http://jipam.vu.edu.au/article.php?sid=949http://jipam.vu.edu.au/article.php?sid=271http://jipam.vu.edu.au/article.php?sid=271http://jipam.vu.edu.au/article.php?sid=389http://jipam.vu.edu.au/article.php?sid=389http://jipam.vu.edu.au/http://close/http://fullscreen/http://goback/http://prevpage/http://lastpage/mailto:[email protected]://jipam.vu.edu.au/http://jipam.vu.edu.au/


28/28


Wengui Yang

vol. 10, iss. 1, art. 26, 2009

Title Page

Contents

Page 28 of 28

Go Back

Full Screen

Close

[9] YOUNG-HO KIM, An improvement of some inequalities similar to Hilbertsinequality, Int. J. Math. and Math. Sci., 28(4) (2001), 211221.

[10] G.S. DAVIS AN D G.M. PETERSON, On an inequality of Hardys (II), Quart.J. Math. Oxford Ser., 15 (1964), 3540.

[11] J. NMETH, Generalizations of the Hardy-Littlewood inequality, Acta Sci.Math. (Szeged), 32 (1971), 295299.

[12] D.S. MITRINOVIC, J.E. PECARIC AN D A.M. FINK, Classical and New In-equalities in Analysis, Kluwer Academic, Dordrecht, 1993.

[13] YOUNG-HO KIM AND BYUNG-IL KIM, An analogue of Hilberts inequalityand its extensions, Bull. Korean Math. Soc., 39(3) (2002), 377388.

Hilbert Pachpatte Inequalities

Documents

Transcript of Hilbert Pachpatte Inequalities