Journal of Inequalities and Applications Volume 2009, Article ID 851360,10pages doi:10.1155/2009/851360
Research Article
Some New Hilbert’s Type Inequalities
Chang-Jian Zhao
1and Wing-Sum Cheung
21Department 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 mn ≤C
p, q, k, r k
m1
k−m1
amAp−1m 2 1/2
× r
n1
r−n1
bnBq−1n 2 1/2
, 2.1
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
kj−mj1
am1,...,mnAp−1m1,...,mnβ 1/β
×
⎛
⎝r1
n11
· · ·rn
nn1
n j1
rj−nj1
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,...,mn ≤pm1
s11
· · ·mn
sn1
as1,...,snAp−1s1,...,sn. 2.6
Similarly, we have
Bnq1,...,nn ≤qn1
t11
· · ·nn
tn1
bt1,...,tnBq−1t1,...,tn. 2.7
From2.6and2.7, using H ¨older’s inequality13and the elementary inequality:
a1/αb1/β≤ a
αt1/β t1/αb
β , 2.8
where 1/α1/β1, α >1, b >0, a >0, andt >0,we have
Apmi,...,mnBqni,...,nn ≤pq 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
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
kj−sj1
as1,...,snAp−1s1,...,sn
β⎫
⎬
⎭
1/β
×
⎧⎨
⎩
r1
t11
· · ·rn
tn1
n j1
rj−tj1
bt1,...,tnBtq−11,...,tnα⎫
⎬
⎭
1/α
L
k1, . . . , kn, r1, . . . , rn, p, q, α, β
×
⎧⎨
⎩
k1
m11
· · · kn
mn1
n j1
kj−mj1
am1,...,mnAp−1m1,...,mnβ⎫
⎬
⎭
1/β
×
⎧⎨
⎩
r1
n11
· · ·rn
nn1
n j1
rj−nj1
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
k1−m11k2−m21
am1,m2Ap−1m1,m22 1/2
× r
1
n1 r2
n21
r1−n11r2−n21
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.
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
mn ≤Mk, 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
kj−mj1
pm1,...,mnφ
am1,...,mn pm1,...,mn
β⎫
⎬
⎭
1/β
×
⎧⎨
⎩
r1
n11
· · ·rn
nn1
n j1
rj−nj1
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
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,...,nn≤ φQn1,...,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/β≤ a
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
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
kj−mj1
pm1,...,mnφ
am1,...,mn
pm1,...,mn
β⎫
⎬
⎭
1/β
×
⎧⎨
⎩
r1
n11
· · ·rn
nn1
n j1
rj−nj1
qn1,...,nnψ
bn1,...,nn
qn1,...,mn α⎫
⎬
⎭
1/α
.
2.20 The proof is complete.
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
k1−m11k2−m21
pm1,m2φ
am1,...,mn pm1,...,mn
2 1/2
× r
1
n1 r2
n21
r1−n11r2−n21
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
kj−mj1
pm1,...,mnφam1,...,mnβ⎫
⎬
⎭
1/β
×
⎧⎨
⎩
r1
n11
· · ·rn
nn1
n j1
rj−nj1
qn1,...,nnψbn1,...,nnα⎫
⎬
⎭
1/α
.
2.24
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.