# Some New Hilbert’s Type Inequalities

## Full text

(1)

Journal of Inequalities and Applications Volume 2009, Article ID 851360,10pages doi:10.1155/2009/851360

## Some New Hilbert’s Type Inequalities

1

### and Wing-Sum Cheung

2

1Department of Information and Mathematics Sciences, College of Science, China Jiliang University, Hangzhou 310018, China

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

Correspondence should be addressed to Chang-Jian Zhao,chjzhao@163.com Received 25 December 2008; Accepted 24 April 2009

Recommended by Peter Pang

Some new inequalities similar to Hilbert’s type inequality involving series of nonnegative terms are established.

Copyrightq2009 C.-J. Zhao and W.-S. Cheung. This 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.

### 1. Introduction

In recent years, several authors 1–10 have given considerable attention to Hilbert’s type inequalities and their various generalizations. In particular, in1, Pachpatte proved some new inequalities similar to Hilbert’s inequality11, page 226involving series of nonnegative terms. The main purpose of this paper is to establish their general forms.

### 2. Main Results

In1, Pachpatte established the following inequality involving series of nonnegative terms.

Theorem A. Letp≥1, q≥1,and let{am}and{bn}be two nonnegative sequences of real numbers defined form 1, . . . , k,andn 1, . . . , r, wherek, r are natural numbers. LetAm m

s1asand Bnn

t1bt. Then

k m1

r n1

ApmBqn mnC

p, q, k, r k

m1

k−m1

amAp−1m 2 1/2

× r

n1

r−n1

bnBq−1n 2 1/2

, 2.1

(2)

where

C

p, q, k, r 1

2pqkr1/2. 2.2

We first establish the following general form of inequality2.1.

Theorem 2.1. Letp ≥ 1, q ≥ 1, t > 0,and 1/α1/β 1, α > 1. Let {am1,...,mn},and{bn1,...,nn} be positive sequences of real numbers defined for mi 1,2, . . . , ki, and ni 1,2, . . . , ri, where ki, ri i 1, . . . , n are natural numbers. Let Am1,...,mn m1

s11· · ·mn

sn1as1,...,sn, and Bn1,...,nn n1

t11· · ·nn

tn1bt1,...,tn.Then

k1

m11

· · · kn

mn1 r1

n11

· · ·rn

nn1

αβt1/βApm1,...,mnBnq1,...,nn m1· · ·mnβn1· · ·nnαt

L

k1, . . . , kn, r1, . . . , rn, p, q, α, β

× k1

m11· · · kn

mn1

n j1

kjmj1

am1,...,mnAp−1m1,...,mnβ 1/β

×

r1

n11

· · ·rn

nn1

n j1

rjnj1

bn1,...,nnBnq−11,...,nnα

1/α

,

2.3

where

L

k1, . . . , kn, r1, . . . , rn, p, q, α, β

pqk1· · ·kn1/αr1· · ·rn1/β. 2.4

Proof. By using the following inequalitysee12:

n 1

m11

· · · nn

mn1

zm1,...,mn

p

pn1

m11

, . . . ,nn

mn1

zm1,...,mn

m 1

k11

· · ·mn

kn1

zk1,...,kn

p−1

, 2.5

wherep≥1 is a constant, andzm1,...,mn ≥0, mi1,2, . . . , ki,i1,2, . . . , n, we obtain

Apm1,...,mnpm1

s11

· · ·mn

sn1

as1,...,snAp−1s1,...,sn. 2.6

Similarly, we have

Bnq1,...,nnqn1

t11

· · ·nn

tn1

bt1,...,tnBq−1t1,...,tn. 2.7

(3)

From2.6and2.7, using H ¨older’s inequality13and the elementary inequality:

a1/αb1/βa

αt1/β t1/αb

β , 2.8

where 1/α1/β1, α >1, b >0, a >0, andt >0,we have

Apmi,...,mnBqni,...,nnpq m

1

s11

· · ·mn

sn1

as1,...,snAp−1s1,...,sn n1

t11

· · ·nn

tn1

bt1,...,tnBtq−11,...,tn

pqm1, . . . , mn1/α m

1

s11

· · ·mn

sn1

as1,...,snAp−1s1,...,snβ 1/β

×n1, . . . , nn1/β n

1

t11

· · ·nn

tn1

bt1,...,tnBq−1t1,...,tnα 1/α

pq

m1· · ·mn

αt1/β n1· · ·nnt1/α β

m1

s11

· · ·mn

sn1

as1,...,snAp−1s1,...,sn

β 1/β

× n

1

t11

. . .nn

tn1

bt1,...,tnBq−1t1,...,tnα 1/α .

2.9

Dividing both sides of2.9bym1· · ·mnβn1· · ·nnαt/αβt1/β,summing up overnifrom 1 tori i1,2, . . . , nfirst, then summing up overmifrom 1 toki i1,2, . . . , n, using again H ¨older’s inequality, then interchanging the order of summation, we obtain

k1

m11

· · · kn

mn1 r1

n11

· · ·rn

nn1

αβt1/βApm1,...,mnBnq1,...,nn m1· · ·mnβn1· · ·nnαt

pq

⎧⎨

k1

m11

· · · kn

mn1

m 1

s11

· · ·mn

sn1

as1,...,snAp−1s1,...,sn

β 1/β

×

⎧⎨

r1

n11

· · ·rn

nn1

n 1

t11

· · ·nn

tn1

bt1,...,tnBq−1t1,...,tnα 1/α

pqk1· · ·kn1/α k

1

m11

· · · kn

mn1

m 1

s11

· · ·mn

sn1

as1,...,snAp−1s1,...,snβ 1/β

×r1· · ·rn1/β k

1

n11

· · ·rn

nn1

n 1

t11

· · ·nn

tn1

bt1,...,tnBtq−11,...,tnα 1/α

(4)

L

k1, . . . , kn, r1, . . . , rn, p, q, α, β

× k

1

s11

· · ·kn

sn1

as1,...,snAp−1s1,...,snβ k 1

m1s1

· · · kn

mnsn

1

1/β

× r

1

t11

· · ·rn

tn1

bt1,...,tnBq−1t1,...,tnα r 1

n1t1

· · · rn

nntn

1

1/α

L

k1, . . . , kn, r1, . . . , rn, p, q, α, β

×

⎧⎨

k1

s11

· · ·kn

sn1

n j1

kjsj1

as1,...,snAp−1s1,...,sn

β

1/β

×

⎧⎨

r1

t11

· · ·rn

tn1

n j1

rjtj1

bt1,...,tnBtq−11,...,tnα

1/α

L

k1, . . . , kn, r1, . . . , rn, p, q, α, β

×

⎧⎨

k1

m11

· · · kn

mn1

n j1

kjmj1

am1,...,mnAp−1m1,...,mnβ

1/β

×

⎧⎨

r1

n11

· · ·rn

nn1

n j1

rjnj1

bn1,...,nnBnq−11,...,nnα

1/α

.

2.10

This completes the proof.

Remark 2.2. Takingαβnj2,2.3becomes

k1

m1 k2

m21

r 1

n11 r2

n21

Apm1,m2Bqn1,n2 m1m2t−1/2n1n2t1/2

≤ 1 2pq

k1k2r1r2

k 1

m1 k2

m21

k1m11k2m21

am1,m2Ap−1m1,m22 1/2

× r

1

n1 r2

n21

r1n11r2n21

bn1,n2Bnp−11,n22 1/2 .

2.11

Takingt1,and changing{am1,m2},{bn1,n2},{Am1,m2},and{Bn1,n2}into{am},{bn},{Am},and {Bn},respectively, and with suitable changes,2.11reduces to Pachpatte1, inequality1.

In1, Pachpatte also established the following inequality involving series of nonneg- ative terms.

(5)

Theorem B. Let{am},{bn}, Am, Bn be as defined in Theorem A. Let {pm}and {qn} be positive sequences form1, . . . , k,andn1, . . . , r,wherek, r are natural numbers. DefinePm m

s1ps, and Qn n

t1qt. Let φ and ψ be real-valued, nonnegative, convex, submultiplicative functions defined onR 0,∞.Then

k m1

r n1

φAmψBn

mnMk, r

k

m1

k−m1

pmφ am

pm

2 1/2

× r

n1

r−n1

qnφ bn

qn

2 1/2

,

2.12

where

Mk, r 1 2

k

m1

φPm Pm

2 1/2 r

n1

φQn Qn

2 1/2

. 2.13

Inequality2.12can also be generalized to the following general form.

Theorem 2.3. Let{am1,...,mn},{bn1,...,nn},α, β, t, Am1,...,mn,andBn1,...,nn be as defined inTheorem 2.1.

Let{pm1,...,mn}and {qn1,...,nn}be positive sequences formi 1,2, . . . , ki,and ni 1,2, . . . , ri i 1,2, . . . , n. DefinePm1,...,mn m1

s11· · ·mn

sn1ps1,...,sn,andQn1,...,nn n1

t11· · ·nn

tn1qt1,...,tn.Letφ and ψ be real-valued, nonnegative, convex, submultiplicative functions defined onR 0,∞.

Then

k1

m11

· · · kn

mn1 r1

n11

· · ·rn

nn1

αβt1/βφAm1,...,mnψBn1,...,nn m1· · ·mnβn1· · ·nnαt

M

k1, . . . , kn, r1, . . . , rn, α, β

×

⎧⎨

k1

m11

· · · kn

mn1

n j1

kjmj1

pm1,...,mnφ

am1,...,mn pm1,...,mn

β

1/β

×

⎧⎨

r1

n11

· · ·rn

nn1

n j1

rjnj1

qn1,...,nnψa

bn1,...,nn qn1,...,mn

α

1/α

,

2.14

where

M

k1, . . . , kn, r1, . . . , rn, α, β

k 1

m11

· · · kn

mn1

φPm1,...,mn Pm1,...,mn

α 1/α r 1

n11

· · ·rn

nn1

ψQn1,...,nn Qn1,...,nn

β 1/β

. 2.15

(6)

Proof. By the hypotheses, Jensen’s inequality, and H ¨older’s inequality, we obtain

φAm1,...,mn φ

Pm1,...,mn

m1

s11. . .mn

sn1ps1,...,sn

as1,...,sn/ps1,...,sn

m1

s11· · ·mn

sn1ps1,...,sn

φPm1,...,mnφ m1

s11· · ·mn

sn1ps1,···,sn

as1,...,sn/ps1,...,sn m1

s11· · ·mn

sn1ps1,...,sn

φPm1,...,mn Pm1,...,mn

m1

s11

· · ·mn

sn1

ps1,...,snφ

as1,...,sn

ps1,...,sn

φPm1,...,mn

Pm1,...,mn m1· · ·mn1/α m

1

s11

· · ·mn

sn1

ps1,...,snφ

as1,...,sn

ps1,...,sn

β 1/β .

2.16

Similarly,

φBn1,...,nnφQn1,...,nn

Qn1,...,nn n1· · ·nn1/β n

1

n11

· · ·nn

nn1

qt1,...,tnψ

bt1,...,tn qt1,...,tn

α 1/α

. 2.17

By2.16and2.17, and using the elementary inequality:

a1/αb1/βa

αt1/β t1/αb

β , 2.18

where 1/α1/β1, α >1, b >0, a >0,andt >0,we have

φAm1,...,mnφBn1,...,nn

m1· · ·mn

αt1/β n1· · ·nnt1/α β

× φPm1,...,mn Pm1,...,mn

m 1

s11

· · ·mn

sn1

ps1,...,snφ

as1,...,sn

ps1,...,sn

β 1/β

× φQn1,...,nn Qn1,...,nn

n 1

t11

. . .nn

tn1

qt1,...,tnψ

bt1,...,tn

qt1,...,tn

α 1/α .

2.19

(7)

Dividing both sides of2.19bym1· · ·mnβn1· · ·nnαt/αβt1/β,and summing up overni

from 1 tori i1,2, . . . , nfirst, then summing up overmifrom 1 toki i1,2, . . . , n, using again inverse H ¨older’s inequality, and then interchanging the order of summation, we obtain

k1

m11

· · · kn

mn1 r1

n11

· · ·rn

nn1

αβt1/βφAm1,...,mnψBn1,...,nn m1. . . mnβn1. . . nnαt

k1

m11

. . . kn

mn1

φPm1,...,mn Pm1,...,mn

m 1

s11

· · ·mn

sn1

ps1,...,snφ

as1,...,sn ps1,...,sn

β 1/β

×r1

n11

· · ·rn

nn1

φQn1,...,nn Qn1,...,nn

n 1

n11

· · ·nn

nn1

qt1,...,tnψ

bt1,...,tn qt1,...,tn

α 1/α

k

1

m11

· · · kn

mn1

φPm1,...,mn Pm1,...,mn

α 1/α

× k

1

m11

· · · kn

mn1

m 1

s11

· · ·mn

sn1

ps1,...,snφ

as1,...,sn ps1,...,sn

β 1/β

× r

1

n11

· · ·rn

nn1

ψQn1,...,nn Qn1,...,nn

β 1/β

× r

1

n11

· · ·rn

nn1

n 1

t11

· · ·nn

tn1

qt1,...,tnψ

bt1,...,tn qt1,...,tn

α 1/α

M

k1, . . . , kn, r1, . . . , rn, α, β

×

⎧⎨

k1

m11

· · · kn

mn1

n j1

kjmj1

pm1,...,mnφ

am1,...,mn

pm1,...,mn

β

1/β

×

⎧⎨

r1

n11

· · ·rn

nn1

n j1

rjnj1

qn1,...,nnψ

bn1,...,nn

qn1,...,mn α

1/α

.

2.20 The proof is complete.

(8)

Remark 2.4. Takingαβnj2,2.14becomes

k1

m1 k2

m21

r 1

n11 r2

n21

φAm1,m2ψBn1,n2 m1m2t−1/2n1n2t1/2

Mk1, k2, r1, r2

× k

1

m1 k2

m21

k1m11k2m21

pm1,m2φ

am1,...,mn pm1,...,mn

2 1/2

× r

1

n1 r2

n21

r1n11r2n21

qn1,n2ψ

bn1,...,nn qn1,...,nn

2 1/2

,

2.21

where

Mk1, k2, r1, r2

k 1

m11 k2

m21

φPm1,m2 Pm1,m2

2 1/2 r 1

n11 r2

n21

ψQn1,n2 Qn1,n2

2 1/2

. 2.22

Takingt1,and changing{am1,m2},{bn1,n2},{Am1,m2},and{Bn1,n2}into{am},{bn},{Am},and {Bn},respectively, and with suitable changes,2.21reduces to Pachpatte1, Inequality7.

Theorem 2.5. Let{am1,...,mn},{bn1,...,nn},{pm1,...,mn},{qn1,...,nn},Pm1,...,mn,andQn1,...,nn,α, β, t,be as defined inTheorem 2.3. Define

Am1,...,mn 1 Pm1,...,mn

m1

s11

· · ·mn

sn1

ps1,...,snas1,...,sn, Bn1,...,n2 1

Qn1,...,nn

n1

t11

· · ·nn

tn1

qt1,...,tnbt1,...,tn,

2.23

formi 1,2, . . . , ki,and ni 1,2, . . . , ri i 1,2, . . . , n,where ki, ri i 1, . . . , n are natural numbers. Letφandψbe real-valued, nonnegative, convex functions defined onR 0,∞.Then

k1

m11

· · · kn

mn1 r1

n11

· · ·rn

nn1

αβt1/βPm1,...,mnQn1,...,nnφAm1,...,mnψBn1,...,nn m1· · ·mnβn1· · ·nnαt

k1· · ·kn1/αr1· · ·rn1/β

×

⎧⎨

k1

m11

· · · kn

mn1

n j1

kjmj1

pm1,...,mnφam1,...,mnβ

1/β

×

⎧⎨

r1

n11

· · ·rn

nn1

n j1

rjnj1

qn1,...,nnψbn1,...,nnα

1/α

.

2.24

(9)

Proof. By the hypotheses, Jensen’s inequality, and H ¨older’s inequality, it is easy to observe that

φAm1,...,mn φ 1

Pm1,...,mn

m1

s11

· · ·mn

sn1

ps1,...,snas1,...,sn

≤ 1

Pm1,...,mn

m1

s11

· · ·mn

sn1

ps1,...,snφas1,...,sn

≤ 1

Pm1,...,mnm1· · ·mn1/α m

1

s11

· · ·mn

sn1

ps1,...,snφas1,...,snβ 1/β ,

2.25

ψBn1,...,nn ψ 1

Qn1,...,nn

n1

t11

· · ·nn

tn1

qt1,...,tnbt1,...,tn

≤ 1

Qn1,...,nn

n1

t11

· · ·nn

tn1

qt1,...,tnψbt1,...,tn

≤ 1

Qn1,...,nnn1· · ·nn1/β n

1

t11

· · ·nn

tn1

qt1,...,tnψbt1,...,tnα 1/α .

2.26

Proceeding now much as in the proof of Theorems2.1and2.3, and with suitable modifica- tions, it is not hard to arrive at the desired inequality. The details are omitted here.

Remark 2.6. In the special case wherej1, t1, αβ2,andn1,Theorem 2.5reduces to the following result.

Theorem C. Let {am},{bn},{pm},{qn}, Pm, Qn be as defined in Theorem B. Define Am 1/Pmm

s1psas,andBn 1/Qnn

t1qtbt form 1, . . . , k,andn 1, . . . , r, wherek, r are natural numbers. Letφandψbe real-valued, nonnegative, convex functions defined onR 0,∞.

Then

k m1

r n1

PmQnφAmψBn

mn ≤ 1

2kr1/2 k

m1

k−m1

pmφam2 1/2

× r

n1

r−n1qnψbn2

1/2

.

2.27

This is the new inequality of Pachpatte in [1, Theorem 4].

Remark 2.7. Takingj 1, t1, pm1, qn1, αβ2,andn1 inTheorem 2.5, and in view ofPmm, Qnn, we obtain the following theorem.

Updating...

## References

Related subjects :