Journal of Inequalities and Applications Volume 2009, Article ID 820176,8pages doi:10.1155/2009/820176
Research Article
A Hilbert’s Inequality with a Best Constant Factor
Zheng Zeng
1and Zi-tian Xie
21Department of Mathematics, Shaoguan University, Shaoguan, Guangdong 512005, China
2Department of Mathematics, Zhaoqing University, Zhaoqing, Guangdong 526061, China
Correspondence should be addressed to Zi-tian Xie,[email protected] Received 6 February 2009; Revised 3 May 2009; Accepted 23 July 2009 Recommended by Yong Zhou
We give a new Hilbert’s inequality with a best constant factor and some parameters.
Copyrightq2009 Z. Zeng and Z.-t. Xie. 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
Ifp >1, 1/p1/q1,an, bn >0 such that∞ >∞
n1apn >0 and∞>∞
n1bqn > 0, then the well-known Hardy-Hilbert’s inequality and its equivalent form are given by
∞ n1
∞ m1
ambn
mn < π sin
π/p ∞
n1
apn
1/p∞
n1
bnq
1/q
, 1.1
∞ n1
∞
m1
am
mn
p
<
π sin
π/p p∞
n1
apn
, 1.2
where the constant factors are all the best possible 1. It attracted some attention in the recent years. Actually, inequalities1.1and 1.2 have many generalizations and variants.
Equation1.1has been strengthened by Yang and othersincluding integral inequalities 2–11.
In 2006, Yang gave an extension of2as follows.
Ifp >1, 1/p1/q1,r >1,1/r1/s1, t∈0,1,2−min{r, s}tmin{r, s} ≥λ >
2−min{r, s}t,such that∞ > ∞
n1np1−t2t−λ/r−1apn > 0,∞ > ∞
n1nq1−t2t−λ/s−1bqn >0, then
∞ n1
∞ m1
ambn
mnλ
< B
r−2tλ
r ,s−2tλ s
∞
n1
np1−t2t−λ/r−1apn
1/p∞
n1
nq1−t2t−λ/s−1bqn
1/q . 1.3
Bu, vis the Beta function.
In 2007 Xie gave a new Hilbert-type Inequality3as follows.
Ifp >1,1/p1/q1, a, b, c >0,2/3≥μ >0, and the right of the following inequalities converges to some positive numbers, then
∞ m1
∞ n1
ambn
nμa2mμnμb2mμnμa2mμ
< π
μabbcca ∞
n1
n1−3μ/2p−1apn
1/p∞
n1
n1−3μ/2q−1bqn
1/q .
1.4
The main objective of this paper is to build a new Hilbert’s inequality with a best constant factor and some parameters.
In the following, we always suppose that
11/p1/q1, p >1,a≥0,−1< α <1,
2both functionsuxandvxare differentiable and strict increasing inn0−1,∞ andm0−1,∞,respectively,
3ux/uαx, vx/vαx are strictly increasing in n0 − 1,∞ and m0 − 1,∞, respectively.{unvm/u2n2aunvmvm2uαnvmα}is strict decreasing onnandm,
4un un, un0 u0, un0−1 vm0−1 0, u∞ ∞, v∞ ∞, un un, vm vm, vm0 v0, vm vm.
2. Some Lemmas
Lemma 2.1. Define the weight coefficients as follows:
W p, m
: ∞
nn0
1
u2n2aunvmvm2
·vαp−1m
uαn
· un
vm p−1, 2.1
ω p, m
: ∞
no−1
1
u2x 2auxvmv2m
·vmαp−1
uαx · ux
vmp−1dx, 2.2
W q, n
: ∞
mm0
1
u2n2aunvmvm2
·uαq−1n
vαm
· vm
unq−1, 2.3
ω
q, n :
∞
m0−1
1 u2n2aunv
y v2
y·uαq−1n
vα
y · v y
unq−1dy, 2.4
then
W p, m
< ω p, m
Kvpα−2α−1m
vmp−1 , W q, n
<ω q, n
Kuqα−2α−1n
unq−1 , 2.5
where
K ∞
0
dσ 12aσσ2σα
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
π 2√
a2−1 sinαπ
a√
a2−1α− 1 a√
a2−1α
, ifα /0, a >1,
|απ|/sin|απ|, ifα /0, a1,
πcscθcscαπsinαθ, ifα /0, acosθ,0< θ < π,
√ 1
a2−1ln a√
a2−1
, ifα0, a >1,
θcscθ, ifα0, acosθ,0< θ < π
2 ,
1, ifα0, a1,
2.6
Proof. Let fz 1/1 2az z2zα 1/z − z1z − z2zα then K 2πi/1 − e−2απiResf, z1 Resf, z2ifa >1 thenz1−a−√
a2−1, z2 −a√ a2−1
K 2πi
1−e−2απi
⎡
⎢⎣
−a−√
a2−1−α
−2√
a2−1
−a√
a2−1−α 2√
a2−1
⎤
⎥⎦
π 2√
a2−1 sinαπ
⎡
⎢⎣
a
a2−1α
− 1
a√
a2−1α
⎤
⎥⎦,
2.7
ifacosθ0< θ < π/2, thenz1−eiθ, z2−e−iθ
K 2πi
1−e−2απi
1
−2isinθ−eiθα 1 2isinθ−e−iθα
πcscθcscαπsinαθ. 2.8
On the other hand, Wp, m < ωp, m. Setting ux vmσ, then ωp, m Kvmpα−2α−1/ vmp−1.Similarly,Wq, n <ωq, n Kuqα−2α−1n /unq−1.
Lemma 2.2. For 0< ε <min{p, p1−α}one has ∞
0
dσ
12aσσ2σαε/p Ko1 ε−→0. 2.9
Proof.
∞
0
1
12aσσ2σαε/pdσ−K
≤
1
0
σ−α
1−σ−ε/p 12aσσ2 dσ
∞
1
σ−α
1−σ−ε/p 12aσσ2 dσ
≤
1
0
σ−α
1−σ−ε/p dσ
∞
1
σ−2−α
1−σ−ε/p dσ
1
1−α− 1 1−α−ε/p
1
1α− 1
1αε/p
−→0 for ε−→0.
2.10
The lemma is proved.
Lemma 2.3. Settingwn un(orvmandw0 n0(orm0, resp.), thenk > 0.{τw /τwk}is strictly decreasing, then
N ww0
τw τwk
N
w0
τx
τkxdxA. 2.11
ThereA∈0, τw0/τwk0,for anyN).
Proof. We have N
w0
τx τkxdx <
N ww0
τw τwk
τw0 τwk0
N
ww01
τw τwk
< τw0 τwk0
N
w0
τx
τkxdx. 2.12
Easily,Ahad up bounded whenN → ∞.
3. Main Results
Theorem 3.1. If an > 0, bn > 0, 0 < ∞
n1vpα−2α−1m /vm p−1apn < ∞, 0 < ∞
nn0uqα−2α−1n / unq−1bqn<∞, then
∞ nn0
∞ mm0
ambn
u2n2aunvmv2m
< K ∞
mm0
vpα−2α−1m
vm p−1apm
1/p∞
nn0
uqα−2α−1n
unq−1bqn
1/q
, 3.1
∞ nn0
upαp−2α−1n un ∞
mm0
am
u2n2aunvmv2m
p
< Kp ∞ mm0
vpα−2α−1m
vmp−1apm. 3.2
Kis defined byLemma 2.1.
Proof. By H ¨older’s inequality12and2.5,
J: ∞
nn0
∞ mm0
ambn
u2n2aunvmvm2
∞
nn0
∞ mm0
1
u2n2aunvmvm2
·vα/qm
uα/pn
· un1/p
vm 1/qam·uα/pn
vα/qm
·vm 1/q un1/pbn
≤ ∞
mm0
Wp, mapm
1/p∞
nn0
Wq, nbqn
1/q
< K ∞
mm0
vmpα−2α−1
vmp−1apm
1/p∞
nn0
uqα−2α−1n
unq−1bnq
1/q
,
3.3
settingbnupα−2αp−1n un∞
mm0am/u2n2aunvmv2mp−1>0.By3.1we have
∞ nn0
uqα−2α−1n
unq−1bqn ∞
nn0
upα−2αp−1n un ∞
mm0
am
u2n2aunvmvm2 p
J≤K ∞
mm0
vpα−2α−1m
vmp−1apm
1/p∞
nn0
uqα−2α−1n
unq−1bqn
1/q
.
3.4
By 0<∞
nn0uqα−2α−1n /unq−1bqn<∞and3.4taking the form of strict inequality, we have 3.1. By H ¨older’s inequality12, we have
J ∞
nn0
u−α2α/q1/qn un−11/q∞
mm0
am
u2n2aunvmv2m
uα−2α/q−1/qn bn
un1−1/q
≤ ∞
nn0
upα−2αp−1n un ∞
mm0
am
u2n2aunvmv2m
p1/p∞
nn0
uqα−2α−1n
unq−1 bqn
1/q
.
3.5
as 0<{∞
nn0uqα−2α−1n /unq−1bnq}1/q<∞. By3.2,3.5taking the form of strict inequality, we have3.1.
Theorem 3.2. Ifα0, then both constant factors,KandKpof3.1and3.2, are the best possible.
Proof. We only prove thatKis the best possible. If the constant factorKin3.1is not the best possible, then there exists a positiveHwithH < K, such that
J < H ∞
mm0
vm−1 vmp−1apm
1/p∞
nn0
u−1n unq−1bqn
1/q
. 3.6
For 0< ε < min{p, q}, settingamvm−ε/pvm ,bnu−ε/qn un, then
∞
mm0
vm−1 vmp−1apm
1/p∞
nn0
u−1n unq−1bnq
1/q
∞
mm0
vm v1εm
1/p∞
nn0
un u1εn
1/q
. 3.7
On the other handux σvyandvy τ, ∞
mm0
∞ nn0
u−ε/pn unvm−ε/qvm u2n2aunvmv2m
>
∞
m0
∞
n0
u−ε/pxuxdx u2x 2auxv
y v2
y vy−ε/qv y
dy
∞
m0
∞
u0/vy
σ−ε/pdσ
σ22aσ1 vy−1−εv y
dy
∞
v0
∞
0
σ−ε/pdσ
σ22aσ1 τ−1−εdτ
− ∞
v0
u0/τ
0
σ−ε/pdσ
σ22aσ1 τ−1−εdτ
≥Ko1 ∞
v0
τ−1−εdτ− ∞
v0
τ−1 u0/τ
0
σ−ε/pdσ dτ
Ko1 ∞
v0
τ−1−εdτ−u1−ε/p0 v0−1ε/p 1−ε/p2 Ko1
∞
v0
τ−1−εdτ−O1.
3.8
By3.6,3.7,3.8, andLemma 2.3, we have
Ko1−∞O1
v0τ−1−εdτ < H ∞
mm0
vm/vm1ε ∞
v0τ−1−εdτ
1/p∞
nn0
un/u1εn ∞
v0τ−1−εdτ 1/q
, 3.9
Ko1−∞O1
v0τ−1−εdτ < H
1∞O1
v0τ−1−εdτ 1/p
1∞O1
v0τ−1−εdτ 1/q
. 3.10
We haveK≤H,ε → 0. This contracts the fact thatH < K.
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