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As applications, we get some inverse forms of Pachpatte’s inequalities which were established in 1998.. In particular, Pachpatte11 proved some inequalities similar to Hilbert’s integral

Volume 2010, Article ID 820857, 7 pages

doi:10.1155/2010/820857

Research Article

On Hilbert-Pachpatte Multiple

Integral Inequalities

1 Department of Mathematics, College of Science, China Jiliang University,

Hangzhou 310018, China

2 Department of Mathematics, The University of Hong Kong, Pokfulam Road,

Hong Kong, China

Received 11 March 2010; Revised 16 July 2010; Accepted 28 July 2010

Academic Editor: N Govil

Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We establish some multiple integral Hilbert-Pachpatte-type inequalities As applications, we get some inverse forms of Pachpatte’s inequalities which were established in 1998

1 Introduction

In 1934, Hilbert1 established the following well-known integral inequality

If f ∈ L p 0, ∞, g ∈ L p 0, ∞, f, g ≥ 0, p > 1 and 1/p  1/q  1, then

∞

0

∞

0

f xgx

π

sin

∞

0

f p xdx

1/p∞

0

g q xdx

1/q

, 1.1

where π/ sinπ/p is the best value.

In recent years, considerable attention has been given to various extensions and improvements of the Hilbert inequality form different viewpoints 2 10 In particular, Pachpatte11 proved some inequalities similar to Hilbert’s integral inequalities in 1998 In this paper, we establish some new multiple integral Hilbert-Pachpatte-type inequalities

2 Main Results

Theorem 2.1 Let h i ≥ 1, let fiσi ∈ C1xi , 0 , 0, ∞, i  1, , n, where xi are positive real numbers, and define F isi 0

s i f iσidσi , for s i ∈ xi , 0  Then for 1/αi  1/βi  1, 0 < βi < 1 and

n

i11/αi  1/α,

0

x1

· · ·

0

x n

n

i1F h i

i si



i11/αi−si1/α ds1· · · dsn≥ n

i1

−xi 1/α i h i

0

x i

si −xiF h i−1

i sifisi β i

ds i

1/β i

.

2.1

easy to observe that

n

i1

F h i

i si  n

i1

h i

0

s i

F h i−1

i σifiσidσi

≥ n

i1

h i−si 1/α i

0

s i

F h i−1

i σifiσi β i dσ i

1/β i

, s i ∈ xi , 0 , i  1, , n.

2.2

Let us note the following means inequality:

n

i1

m 1/α i

α

n



i1

1

α i m i

1/α

We obtain that

n

i1F h i

i si



i11/αi−si1/α ≥ n

i1

h i

0

s i

F h i−1

i σifiσi β i

dσ i

1/β i

Integrating both sides of2.4 over si from xi i  1, 2, , n to 0 and using the special case

of inverse H ¨older integral inequality, we observe that

0

x1

· · ·

0

x n

n

i1F h i

i si



i11/αi−si1/α ds1· · · dsn

≥ n

i1

h i

0

x i

0

s i

F h i−1

i σifiσi β i

dσ i

1/β i

ds i

≥ n

i1

h i−xi 1/α i

0

x i

0

s i

F h i−1

i σifiσi β i dσ i



ds i

1/β i

 n

i1

−xi 1/α i h i

0

x i

si − xiF h i−1

i sifisi β i

ds i

1/β i

.

2.5

The proof is complete

Remark 2.2 Taking n  2, βi  1/2 to 2.1, 2.1 changes to

0

x1

0

x2

F h1

1 s1F h2

2 s2

s1 s2−2 ds1ds2

≥ 4h1h2x1x2−10

x1

s1− x1F h1 −1

1 s1f1s1 1/2 ds1

2

×

0

x2

s2− x2F h2 −1

2 s2f2s2 1/2 ds2

2

.

2.6

This is just an inverse inequality similar to the following inequality which was proved by Pachpatte11:

x

0

y

0

F h sG l t

1

2hl



xy1/2x

0

x − sF h−1sfs 2ds

1/2

×

y

0

y − tG l−1tgt 2dt

1/2

.

2.7

Theorem 2.3 Let f iσi, Fisi, αi , and β i be as in Theorem 2.1 Let p iσi be n positive functions

defined for σ i ∈ xi , 0  i  1, 2, , n, and define Pisi  0

s i p iσidσi , where x i are positive real numbers Let φ i i  1, 2, , n be n real-valued nonnegative, concave, and super-multiplicative

functions defined on R Then

0

x1

· · ·

0

x n

n

i1φ iFisi



i11/αi−si1/α ds1· · · dsn

≥ Lx1, , x n n

i1

0

x i

si − xi



p isiφi

f isi

p isi

βi

ds i

1/β i

,

2.8

where

L x1, , x n  n

i1

0

x

φ iPisi

P isi

αi

ds i

1/α i

Proof By using Jensen integral inequalitysee 11 and inverse H¨older integral inequality

see 12 and noticing that φi i  1, 2, , n are n real-valued super-multiplicative

functions, it is easy to observe that

φ iFisi  φi

⎛

⎝P isi

0

s i p iσif iσi/piσidσ i

0

s i p iσidσi

⎞

⎠

≥ φiPisiφi

⎛

⎝

0

s i p iσif iσi/piσidσ i

0

s i p iσidσi

⎞

⎠

≥ φ iPisi

P isi

0

s i

p iσiφi



f iσi

p iσi



dσ i

≥



φ iPisi

P isi



−si 1/α i

0

s i



p iσiφi



f iσi

p iσi

βi

dσ i

1/β i

.

2.10

In view of the means inequality and integrating two sides of 2.10 over si from x i i 

1, 2, , n to 0 and noticing H¨older integral inequality, we observe that

0

x1

· · ·

0

x n

n

i1φ iFisi



i11/αi−si1/α ds1· · · dsn

≥ n

i1

0

x i



φ iPisi

P isi

0

s i



p iσiφi



f iσi

p iσi

βi

dσ i

1/β i

ds i

≥ n

i1

0

x i

φ iPisi

P isi

αi

ds i

1/α i0

x i

0

s i



p iσiφi

f iσi

p iσi

βi

dσ i ds i

1/β i

 Lx1, , x n n

i1

0

x i

si − xi



p isiφi

f isi

p isi

βi

ds i

1/β i

.

2.11

This completes the proof ofTheorem 2.3

Remark 2.4 Taking n  2, βi  1/2 to 2.8, 2.8 changes to

0

x1

0

x2

φ1F1s1φ2F2s2

s1 s2−2 ds1ds2

≥ Lx1, x2

⎛

⎝0

x1

s1− x1



p1s1φ1

f

1s1

p1s1

1/2

ds1

⎞

⎠

2

×

0

x

s2− x2



p2s2φ2

f

2s2

p2s2

1/2

ds2

2

,

2.12

L x1, x2  4

0

x1

φ

1P1s1

P1s1

−1

ds1

−10

x2

φ

2P2s2

P2s2

−1

ds2

−1

This is just an inverse inequality similar to the following inequality which was proved by Pachpatte11:

x

0

y

0

φ FsψGt



x, yx

0

x − s



p sφ

f s

p s

2

ds

1/2

×

y

0



y − tq tψ



g t

q t

2

dt

1/2

,

2.14

where

L

x, y

 1 2

x

0

φ Ps

P s

2

ds

1/2y

0

ψ Qt

Q t

2

dt

1/2

Theorem 2.5 Let f iσi, piσi, Piσi, αi , and β i be as Theorem 2.3 , and define F isi 

1/Pisi0

s i p iσifiσidσi for σ i , s i ∈ xi , 0 , where xi are positive real numbers Let φ i i 

1, 2, , n be n real-valued, nonnegative, and concave functions on R Then

0

x1

· · ·

0

x n

n

i1P isiφiFisi



αn

i11/αi−si1/α ds1· · · dsn

≥ n

i1

x 1/α i

i

0

x i

si − xip isiφifisiβ i

ds i

1/β i

.

2.16

Proof From the hypotheses and by using Jensen integral inequality and the inverse H ¨older

integral inequality, we have

φ iFisi  φi

1

P isi

0

s i

p iσifiσidσi



≥ 1

P isi

0

s i

p iσiφif iσidσ i

≥ 1

P isi −si 1/α i

0

s



p iσiφif iσiβ i dσ i

1/β i

.

2.17

0

x1

· · ·

0

x n

n

i1P isiφiFisi



i11/αi−si1/α ds1· · · dsn

≥ n

i1

0

x i

0

s i



p iσiφifiσiβ i dσ i

1/β i

ds i

≥ n

i1

x 1/α i

i

0

x i

0

s i



p iσiφifiσiβ i

dσ i ds i

1/β i

 n

i1

−xi 1/α i

0

x i

si − xip isiφifisiβi ds i

1/β i

.

2.18

Remark 2.6 Taking n  2, βi  1/2 to 2.16, 2.16 changes to

0

x1

0

x2

P1s1P2s2φ1F1s1φ2F2s2

s1 s2−2 ds1ds2

≥ 4x1x2−1

0

x1

s1− x1p1s1φ1



f1s11/2

ds1

2

×

0

x2

s2− x2p2s2φ2



f2s21/2

ds2

2

.

2.19

This is just an inverse inequality similar to the following inequality which was proved by Pachpatte11:

x

0

y

0

P sQtφFsψGt

≤ 1

2



xy1/2x

0

x − sp sφf s2ds

1/2y

0



y − tq tψg t2dt

1/2

.

2.20

Remark 2.7 In 2.20, if p1s1  p2s2  1, then P1s1  s1, P2s2  s2 Therefore2.20 changes to

0

x1

0

x2

φ1F1s1φ2F2s2

s1 s2−2 ds1ds2

≥ 4x1x2−1

0

x1

s1− x1φ1



f1s11/2

ds1

20

x2

s2− x2φ2



f2s21/2

ds2

2

.

2.21

This is just an inverse inequality similar to the following Inequality which was proved by Pachpatte11:

x

0

y

0

φ FsψGt

st−1s  t ds dt

≤ 1

2



xy1/2x

0

x − sφ

f s2

ds

1/2y

0



y − tψ

g t2

dt

1/2

.

2.22

Acknowledgments

This paper is supported by the National Natural Sciences Foundation of China10971205 This paper is partially supported by the Research Grants Council of the Hong Kong SAR, ChinaProject no HKU7016/07P and an HKU Seed Grant for Basic Research

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Proof By using Jensen integral inequalitysee 11 and inverse Hăolder integral inequality

see...

2.5

The proof is complete

Remark 2.2 Taking n  2, βi  1/2 to 2.1, 2.1...

2

,

2.12

L x1, x2