Volume 2009, Article ID 820176, 8 pagesdoi:10.1155/2009/820176 Research Article A Hilbert’s Inequality with a Best Constant Factor Zheng Zeng1 and Zi-tian Xie2 1 Department of Mathematic
Trang 1Volume 2009, Article ID 820176, 8 pages
doi:10.1155/2009/820176
Research Article
A Hilbert’s Inequality with a Best Constant Factor
Zheng Zeng1 and Zi-tian Xie2
1 Department of Mathematics, Shaoguan University, Shaoguan, Guangdong 512005, China
2 Department of Mathematics, Zhaoqing University, Zhaoqing, Guangdong 526061, China
Correspondence should be addressed to Zi-tian Xie,gdzqxzt@163.com
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
Copyrightq 2009 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
If p > 1, 1/p 1/q 1, a n, bn > 0 such that ∞ >∞
n1 a p n > 0 and ∞ >∞
n1 b q n > 0, then the
well-known Hardy-Hilbert’s inequality and its equivalent form are given by
∞
n1
∞
m1
ambn
m n <
π
sin
π/p
∞
n1
a p n
1/p∞
n1
b n q
1/q
∞
n1
∞
m1
am
m n
p
< π
sin
π/p
p∞
n1
a p n
where the constant factors are all the best possible 1 It attracted some attention in the recent years Actually, inequalities1.1 and 1.2 have many generalizations and variants Equation1.1 has been strengthened by Yang and others including integral inequalities
2 11
Trang 2In 2006, Yang gave an extension of2 as follows.
If p > 1, 1/p 1/q 1, r > 1, 1/r 1/s 1, t ∈ 0, 1, 2 − min{r, s}t min{r, s} ≥ λ >
2 − min{r, s}t, such that ∞ > ∞n1 n p1−t2t−λ/r−1 a p n > 0, ∞ > ∞
n1 n q1−t2t−λ/s−1 b q n > 0,
then
∞
n1
∞
m1
ambn
m n λ
< B
r − 2t λ
r ,
s − 2t λ
s
∞
n1
n p1−t2t−λ/r−1 a p n
1/p∞
n1
n q1−t2t−λ/s−1 b q n
1/q
.
1.3
Bu, v is the Beta function.
In 2007 Xie gave a new Hilbert-type Inequality3 as follows
If p > 1, 1/p1/q 1, a, b, c > 0, 2/3 ≥ μ > 0, and the right of the following inequalities
converges to some positive numbers, then
∞
m1
∞
n1
ambn
n μ a2m μ n μ b2m μ n μ a2m μ
< π
μ a bb cc a
∞
n1
n 1−3μ/2p−1 a p n
1/p∞
n1
n 1−3μ/2q−1 b q n
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
1 1/p 1/q 1, p > 1, a ≥ 0, −1 < α < 1,
2 both functions ux and vx are differentiable and strict increasing in n0− 1, ∞
andm0− 1, ∞, respectively,
3 ux/u α x, vx/v α x are strictly increasing in n0 − 1, ∞ and m0 − 1, ∞,
respectively.{unvm/u2n 2au nvm v2
m u α
n v m α } is strict decreasing on n and m,
4 un u n, un0 u0, un0− 1 vm0− 1 0, u∞ ∞, v∞ ∞, un
un, vm vm, vm0 v0, vm v
m
Trang 32 Some Lemmas
Lemma 2.1 Define the weight coefficients as follows:
W
p, m : ∞
nn0
1
u2
n 2au nvm v2
m
·v
αp−1
m
u α · un
v
ω
p, m :
∞
n o−1
1
u2x 2auxv m v2
m
·v
αp−1
m
u α x ·
ux
v
mp−1 dx, 2.2
W
q, n : ∞
mm0
1
u2
n 2au nvm v2
m
·u
αq−1
n
v α m
· v m
u
ω
q, n :
∞
m0 −1
1
u2
n 2au nv
y
v2
y ·u
αq−1
n
v α
y · v
y
u
nq−1 dy, 2.4
then
W
p, m
< ω
p, m
Kv
pα−2α−1 m
v
mp−1 , Wq, n< ωq, n Ku qα−2α−1 n
u
nq−1 , 2.5
where
K
∞
0
dσ
1 2aσ σ2σ α
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
π
2√
a2− 1 sin απ a
√
a2− 1α− 1
a √a2− 1α
, if α / 0, a > 1,
πcscθcsc απ sinαθ, if α / 0, a cos θ, 0 < θ < π,
1
√
a2− 1ln
a √
a2− 1, if α 0, a > 1, θcscθ, if α 0, a cos θ, 0 < θ < π
2 ,
2.6
Trang 4Proof Let fz 1/1 2az z2z α 1/z − z1z − z2z α then K 2πi/1 −
e −2απi Resf, z1 Resf, z2 if a > 1 then z1 −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
⎤
⎥
2√
a2− 1 sin απ
⎡
⎢
⎣a
a2− 1α− 1
a √
a2− 1α
⎤
⎥
⎦,
2.7
if a cos θ 0 < θ < π/2, then z1 −e iθ , z2 −e −iθ
K 2πi
1− e −2απi
1
−2i sin θ−e iθα 1
2i sin θ−e −iθα
πcscθcscαπ sinαθ. 2.8
On the other hand, Wp, m < ωp, m Setting ux v mσ, then ωp, m Kv m pα−2α−1 /
v
mp−1 Similarly, Wq, n < ωq, n Ku qα−2α−1 n /unq−1
Lemma 2.2 For 0 < ε < min{p, p1 − α} one has
∞ 0
dσ
1 2aσ σ2σ αε/p K o1 ε −→ 0. 2.9
Proof.
∞
0
1
1 2aσ σ2σ αε/p dσ − K
≤
1 0
σ −α
1− σ −ε/p
1 2aσ σ2 dσ
∞ 1
σ −α
1− σ −ε/p
1 2aσ σ2 dσ
≤
1 0
σ −α
1− σ −ε/p
dσ
∞ 1
σ −2−α
1− σ −ε/p
dσ
1− α1 − 1
1− α − ε/p
1 α1 − 1
1 α ε/p
−→ 0 for ε −→ 0.
2.10
The lemma is proved
Trang 5Lemma 2.3 Setting w n u n (or vm and w0 n0(or m0, resp.), then k > 0 {τ w /τ k
w } is strictly
decreasing, then
N
ww0
τ w
τ k w
N
w0
τx
τ k x dx A. 2.11
There A ∈ 0, τ w0/τ k
w0, for any N).
Proof We have
N
w0
τx
τ k x dx <
N
ww0
τ w
τ w k
τ w0
τ w k0
N
ww0 1
τ w
τ w k
< τ
w0
τ w k0
N
w0
τx
τ k x dx. 2.12 Easily, A had up bounded when N → ∞.
3 Main Results
Theorem 3.1 If a n > 0, bn > 0, 0 < ∞
n1 v pα−2α−1 m /v m p−1
a p n < ∞, 0 < ∞
nn0u qα−2α−1 n /
u
nq−1
b q n < ∞, then
∞
nn0
∞
mm0
ambn
u2
n 2au nvm v2
m
< K
∞
mm0
v pα−2α−1 m
v
mp−1 a p m
1/p∞
nn0
u qα−2α−1 n
u
nq−1 b q n
1/q
, 3.1
∞
nn0
u pαp−2α−1 n un
∞
mm0
am
u2
n 2au nvm v2
m
p
< K p
∞
mm0
v pα−2α−1 m
v
mp−1 a p m. 3.2
K is defined by Lemma 2.1
Proof By H ¨older’s inequality12 and 2.5,
J :
∞
nn0
∞
mm0
ambn
u2n 2au nvm v2
m
∞
nn0
∞
mm0
1
u2
n 2au nvm v2
m
·v
α/q m
u α/p n
· un1/p
v
m1/q am·u
α/p n
v α/q m
·v m 1/q
u
n1/p bn
≤
∞
mm0
Wp, ma p m
1/p∞
nn0
Wq, nb q n
1/q
< K
∞
mm
v m pα−2α−1
v
mp−1 a p m
1/p∞
nn
u qα−2α−1 n
u
nq−1 b n q
1/q
,
3.3
Trang 6setting b n u pα−2αp−1 n un∞mm0am/u2
n 2au nvm v2
mp−1 > 0 By3.1 we have
∞
nn0
u qα−2α−1 n
u
nq−1 b q n ∞
nn0
u pα−2αp−1 n un
∞
mm0
am
u2
n 2au nvm v2
m
p
J ≤ K
∞
mm0
v pα−2α−1 m
v
mp−1 a p m
1/p∞
nn0
u qα−2α−1 n
u
nq−1 b q n
1/q
.
3.4
By 0 <∞
nn0u qα−2α−1
n /unq−1 b q
n < ∞ and 3.4 taking the form of strict inequality, we have
3.1 By H¨older’s inequality12, we have
J
∞
nn0
u −α2α/q1/q n u
n−11/q∞
mm0
am
u2
n 2au nvm v2
m
u α−2α/q−1/q n bn
u
n1−1/q
≤
∞
nn0
u pα−2αp−1 n un
∞
mm0
am
u2
n 2au nvm v2
m
p1/p∞
nn0
u qα−2α−1 n
u
nq−1 b q n
1/q
.
3.5
as 0 < {∞
nn0u qα−2α−1
n /unq−1 b q
n}1/q < ∞ By 3.2, 3.5 taking the form of strict inequality,
we have3.1
Theorem 3.2 If α 0, then both constant factors, K and K p of3.1 and 3.2, are the best possible
Proof We only prove that K is the best possible If the constant factor K in 3.1 is not the best
possible, then there exists a positive H with H < K, such that
J < H
∞
mm0
v m−1
v
mp−1 a p m
1/p∞
nn0
u−1n
u
nq−1 b q n
1/q
For 0 < ε < min{p, q}, setting am v m −ε/p v m , bn u −ε/q n un, then
∞
mm
v m−1
v
mp−1 a p m
1/p∞
nn
u−1n
u
nq−1 b q n
1/q
∞
mm
vm
v1εm
1/p∞
nn
un
u1εn
1/q
. 3.7
Trang 7On the other handux σvy and vy τ,
∞
mm0
∞
nn0
u −ε/p n un v m −ε/q v m
u2
n 2au nvm v2
m
>
∞
m0
∞
n0
u −ε/p xuxdx
u2x 2auxvy
v2
y
vy −ε/q v
y
dy
∞
m0
∞
u0/vy
σ −ε/p dσ
σ2 2aσ 1
vy −1−ε v
y
dy
∞
v0
∞ 0
σ −ε/p dσ
σ2 2aσ 1
τ −1−ε dτ
−
∞
v0
u0/τ 0
σ −ε/p dσ
σ2 2aσ 1
τ −1−ε dτ
≥ K o1
∞
v0
τ −1−ε dτ −
∞
v0
τ−1
u0/τ 0
σ −ε/p dσ
dτ
K o1
∞
v0
τ −1−ε dτ − u
1−ε/p
0 v0−1ε/p
1 − ε/p2
K o1
∞
v0
τ −1−ε dτ − O 1.
3.8
By3.6, 3.7, 3.8, andLemma 2.3, we have
K o1 − ∞O1
v0τ −1−ε dτ < H
∞
mm0
vm /v m1ε
∞
v0τ −1−ε dτ
1/p∞
nn0
un /u1εn
∞
v0τ −1−ε dτ
1/q
K o1 − ∞O1
v0τ −1−ε dτ < H
1 ∞O1
v0τ −1−ε dτ
1/p
1 ∞O1
v0τ −1−ε dτ
1/q
We have K ≤ H, ε → 0 This contracts the fact that H < K.
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...u1εn
1/q
. 3.7
Trang 7On the other handux...
3.3
Trang 6setting b n u pα−2αp−1 n un∞mm0am/u2
n... 0.
2.10
The lemma is proved
Trang 5Lemma 2.3 Setting