Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 256796, 8 pages
doi:10.1155/2010/256796
Research Article
On a New Hilbert-Type Intergral Inequality with
the Intergral in Whole Plane
Zheng Zeng1 and Zitian Xie2
1
2
Department of Mathematics, Shaoguan University, Shaoguan, Guangdong 512005, China
Department of Mathematics, Zhaoqing University, Zhaoqing, Guangdong 526061, China
Correspondence should be addressed to Zitian Xie, gdzqxzt@163.com
Received 5 May 2010; Accepted 14 July 2010
Academic Editor: Andrea Laforgia
Copyright q 2010 Z. Zeng and Z. 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.
By introducing some parameters and estimating the weight functions, we build a new Hilbert’s
inequality with the homogeneous kernel of 0 order and the integral in whole plane. The equivalent
inequality and the reverse forms are considered. The best constant factor is calculated using
Complex Analysis.
1. Introduction
If fx, gx ≥ 0 and satisfy that 0 <
1
∞
0
∞
0
f 2 xdx < ∞ and 0 <
fxgx
dx dy < π
xy
∞
0
f 2 xdx
∞
∞
0
g 2 xdx < ∞, then we have
1/2
g 2 xdx
,
1.1
0
where the constant factor π is the best possible. Inequality 1.1 is well known as Hilbert’s
integral inequality, which has been extended by Hardy-Riesz as 2.
∞ p
∞ q If p > 1, 1/p 1/q 1, fx, gx ≥ 0, such that 0 < 0 f xdx < ∞ and 0 <
g xdx < ∞, then we have the following Hardy-Hilbert’s integral inequality:
0
∞
0
∞
1/p ∞
1/q
fxg y
π
p
dx dy <
f xdx
g q xdx
,
xy
sin π/p
0
0
where the constant factor π/ sinπ/p also is the best possible.
1.2
2
Journal of Inequalities and Applications
Both of them are important in Mathematical Analysis and its applications 3. It
attracts some attention in recent years. Actually, inequalities 1.1 and 1.2 have many
generalizations and variations. Equation 1.1 has been strengthened by Yang and others
including double series inequalities 4–21.
In 2008, Xie and Zeng gave a new Hilbert-type Inequality 4 as follows.
If a > 0, b > 0, c > 0, p > 1, 1/p 1/q 1, fx, gx ≥ 0 such that 0 <
∞ −1−p/2
∞
x
f p xdx < ∞ and 0 < 0 x−1−q/2 g q xdx < ∞, then
0
∞
0
fxg y
dx dy
x a2 y x b2 y x c2 y
∞
1/p ∞
1/q
−1−p/2 p
−1−q/2 q
<K
x
f xdx
x
g xdx
,
0
1.3
0
where the constant factor K π/a ba cb c is the best possible.
The main purpose of this paper is to build a new Hilbert-type inequality with
homogeneous kernel of degree 0, by estimating the weight function. The equivalent
inequality is considered.
In the following, we always suppose that: 1/p 1/q 1, p > 1, r ∈ −1, 0, 0 < α <
β < π.
2. Some Lemmas
We start by introducing some lemmas.
∞
∞
Lemma 2.1. If k1 : 0 u−1r ln1 2u cos α u2 /1 2u cos β u2 du, k2 : 0 u−1r ln1 −
2u cos β u2 /1 − 2u cos α u2 du, then
4π sin r β − α /2 sin r α β /2
,
k1 r sin rπ
4π sin r β − α /2 sin rπ − r α β /2
,
k2 r sin rπ
∞
1 2u cos α u2 −1r k :
|u|
ln
du
2
1
2u
cos
β
u
−∞
4π sin r β − α /2 cos r/2 π − α − β
k1 k2 .
r cosrπ/2
2.1
Proof. We have
∞
A :
x
0
−
2
r
r−1
∞
0
2
: − B.
r
∞
1 r 2
ln x 2x cos α 1 dx x ln x 2x cos α 1 r
0
2
xr x cos α
dx
x2 2x cos α 1
2.2
Journal of Inequalities and Applications
3
Setting fz zr z cos α/z2 2z cos α 1, z1 −eiα , z2 −e−iα , then
2πi Res f, z1 Res f, z2
2πri
1−e
r
z1 z1 cos α zr2 z2 cos α
π cos rα
2πi
−
z1 − z2
z2 − z1
sin rπ
1 − e2πri
B
2.3
we find that A −2B/r 2π cos rα/r sin rπ, then
4π sin r β − α /2 sin r α β /2
1 2u cos α u2
,
k1 :
u
ln
du r sin rπ
1 2u cos β u2
0
∞
∞
1 2u cos π − β u2
1 − 2u cos β u2
−1r
−1r
k2 :
u
ln
du u
ln
du
1 − 2u cos α u2
1 2u cosπ − α u2
0
0
4π sin r β − α /2 sin r/2 2π − α − β
,
r sin rπ
∞
1 2u cos α u2 −1r k
|u|
ln
du
1 2u cos β u2 −∞
∞
0
1 2u cos β u2
1 2u cos α u2
−1r
−1r
u
ln
du
ln
du
−u
1 2u cos β u2
1 2u cos α u2
0
−∞
4π sin r β − α /2 cos r/2 π − α − β
.
k1 k2 r cosrπ/2
∞
−1r
2.4
The lemma is proved.
Lemma 2.2. Define the weight functions as follow:
|x|−r x2 2xy cos α y2 wx :
dy,
1−r ln 2
x 2xy cos β y2 −∞ y ∞ r y x2 2xy cos α y2 w
y :
ln 2
dx,
1r x 2xy cos β y2 −∞ |x|
∞
2.5
then wx wy
k 4π sinrβ − α/2 cosr/2π − α − β/r cosrπ/2.
Proof. We only prove that wx k for x ∈ −∞, 0.
Using Lemma 2.1, setting y ux andy −ux,
0
∞
x2 2xy cos αy2
x2 2xy cos βy2
−x−r
−x−r
ln
dy
ln 2
dy
1−r
2
2
1−r
x 2xy cos βy
x 2xy cos αy2
y
−∞ −y
0
∞
∞
1−2u cos βu2
1 2u cos α u2
−1r
u−1r ln
du
u
ln
du k1 k2 k.
1 2u cos β u2
1 − 2u cos αu2
0
0
wx and the lemma is proved.
2.6
4
Journal of Inequalities and Applications
g as follows:
Lemma 2.3. For ε > 0, and r − max{2ε/p, 2ε/q} ∈ −1, 0, define both functions f,
⎧
⎪
x−r−1−2ε/p ,
⎪
⎪
⎨
fx
0,
⎪
⎪
⎪
⎩
−x−r−1−2ε/p ,
if x ∈ 1, ∞,
if x ∈ −1, 1,
if x ∈ −∞, −1,
⎧
⎪
if x ∈ 1, ∞,
xr−1−2ε/q ,
⎪
⎪
⎨
g x 0,
if x ∈ −1, 1,
⎪
⎪
⎪
⎩
−xr−1−2ε/q , if x ∈ −∞, −1,
2.7
then
∞
Iε : ε
: ε
Iε
−∞
|x|
pr1−1 p
f xdx
1/p ∞
−∞
|x|
qr−1−1 q
g xdx
x2 2xy cos α y2 g y ln
fx
dx dy −→ k
x2 2xy cos β y2 −∞
∞
1/q
1,
2.8
ε −→ 0 .
Proof. Easily, we get the following:
1/p ∞
1/q
∞
−1 −2ε
−1 −2ε
x x dx
x x dx
1.
2
Iε ε 2
1
2.9
1
Let y −Y , using f−x
fx,
g −x g x and
f−x
∞
x2 2xY cos α Y 2 x2 − 2xy cos α y2 g y ln 2
g Y ln 2
dy fx
dY,
x − 2xy cos β y2 x 2xY cos β Y 2 −∞
−∞
∞
2.10
we have that fx
on x, then
∞
−∞
g y| lnx2 2xy cos αy2 /x2 2xy cos βy2 |dy is an even function
x2 2xy cos α y2 g y ln 2
dy dx
x 2xy cos β y2 0
−∞
−1
∞
r−1−2ε/q x2 2xy cos β y2
−r−1−2ε/p
2ε
x
ln 2
dy dx
−y
x 2xy cos α y2
1
−∞
∞
∞
x2 2xy cos α y2
−r−1−2ε/p
r−1−2ε/q
x
y
ln 2
dy dx
x 2xy cos β y2
1
1
2ε
Iε
∞
fx
: I1 I2 .
∞
2.11
Journal of Inequalities and Applications
5
Setting y tx then
x2 − 2xy cos β y2
x
y
ln 2
dy dx
I1 2ε
x − 2xy cos α y2
1
1
∞
∞
1 − 2t cos β t2
−1−2ε
r−1−2ε/q
x
t
ln
dt dx
2ε
1 − 2t cos α t2
1
1/x
∞
∞
1 − 2t cos β t2
−1−2ε
r−1−2ε/q
x
t
ln
dt dx
2ε
1 − 2t cos α t2
1
1
1
∞
1 − 2t cos β t2
−1−2ε
r−1−2ε/q
x
t
ln
dt dx
1 − 2t cos α t2
1
1/x
∞
∞
−r−1−2ε/p
tr−1−2ε/q ln
1
2ε
1
t
∞
tr−1−2ε/q ln
1
∞
r−1−2ε/q
1 − 2t cos β t2
dt
1 − 2t cos α t2
r−1−2ε/q
0
∞
1 − 2t cos β t2
ln
1 − 2t cos α t2
2.12
∞
1 − 2t cos β t2
dt 1 − 2t cos α t2
x
−1−2ε
dx dt
1/t
1
tr−12ε/p ln
0
1 − 2t cos β t2
dt
1 − 2t cos α t2
1
1 − 2t cos β t
1 − 2t cos β t2
2ε/p
−2ε/q r−1
t
t
dt
−
t
ln
dt
1 − 2t cos α t2
1 − 2t cos α t2
0
0
4π sin r − 2ε/q β − α /2 sin r − 2ε/q 2π − α − β /2
ηε,
r − 2ε/q sin r − 2ε/q π
2
tr−1−2ε/q ln
where limε → 0 ηε 0, and we have I1 → k2 ε → 0 .
Similarly, I2 → k1 ε → 0 . The lemma is proved.
Lemma 2.4. If fx is a nonnegative measurable function and 0 <
∞
J :
−∞
pr−1
y
∞
−∞
|x|p1r−1 f p xdx < ∞, then
p
∞
x2 2xy cos α y2 p
fxln 2
|x|p1r−1 f p xdx.
dx dy ≤ k
x 2xy cos β y2 −∞
−∞
∞
2.13
Proof. By Lemma 2.2, we find that
p
∞ x2 2xy cos α y2 ln 2
dx
2
x 2xy cos β y −∞
1−r/p p
∞ x2 2xy cos α y2 |x|1r/q
y
dx
fx
ln 2
x 2xy cos β y2 y1−r/p
|x|1r/q
−∞ 6
Journal of Inequalities and Applications
∞ 1rp−1
2
2
x 2xy cos α y |x|
≤
f p xdx
ln 2
x 2xy cos β y2 y1−r
−∞ p−1
∞ x2 2xy cos α y2 y1−rq−1
dx
×
ln 2
x 2xy cos β y2 |x|1r
−∞ ∞ 1rp−1
2
2
x
2xy
cos
α
y
−rp1
|x|
kp−1 y
f p xdx,
ln 2
x 2xy cos β y2 y1−r
−∞ ∞ ∞ 1rp−1
2
2
x 2xy cos α y |x|
p−1
p
f xdx dy
J≤k
ln 2
x 2xy cos β y2 y1−r
−∞
−∞ ∞ ∞ 1rp−1
2
2
x 2xy cos α y |x|
p−1
dy f p xdx
k
ln 2
x 2xy cos β y2 y1−r
−∞
−∞ k
p
∞
−∞
|x|p1r−1 f p xdx.
2.14
3. Main Results
Theorem 3.1. If both functions, fx and gx,
∞ are nonnegative measurable functions and
∞
satisfy 0 < −∞ |x|p1r−1 f p xdx < ∞ and 0 < −∞ |x|q1−r−1 g q xdx < ∞, then
x2 2xy cos α y2 fxg y ln 2
dx dy
x 2xy cos β y2 −∞
∞
∗
I :
∞
<k
J
−∞
∞
−∞
< kp
|x|
p1r−1 p
pr−1
y
∞
−∞
f xdx
1/p ∞
−∞
|x|
q1−r−1 q
3.1
1/q
g xdx
,
p
x2 2xy cos α y2 fxln 2
dx dy
2
x
2xy
cos
β
y
−∞
∞
3.2
|x|p1r−1 f p xdx.
Inequalities 3.1 and 3.2 are equivalent, and where the constant factors k and kp are the best
possibles.
Proof. If 2.13 takes the form of equality for some y ∈ −∞, 0 ∪ 0, ∞, then there exists
constants M and N, such that they are not all zero, and
1−rq−1
y
|x|1rp−1 p
M 1−r f x N
y
|x|1r
a.e. in −∞, ∞ × −∞, ∞.
3.3
Journal of Inequalities and Applications
7
Hence, there exists a constant C, such that
1−rq
M|x|1rp f p x N y
C
3.4
a.e. in −∞, ∞ × −∞, ∞.
We claim that M 0. In fact, if M
0, then |x|p1r−1 f p x C/M|x|−1 a.e. in −∞, ∞
∞/
which contradicts the fact that 0 < −∞ |x|p1r−1 f p xdx < ∞. In the same way, we claim that
N 0. This is too a contradiction and hence by 2.13, we have 3.2.
By Hölder’s inequality with weight 22 and 3.2, we have the following:
∞ x2 2xy cos α y2 −1r1/q ∞
y1−r−1/q g y dy
dx
y
fxln 2
I x 2xy cos β y2 −∞
−∞
∗
≤ J1/p
∞
−∞
q1−r−1 q
y
g ydy
1/q
3.5
.
Using 3.2, we have 3.1. ∞
Setting
|y|rp−1 −∞ fx| lnx2 2xy cos α y2 /x2 2xy cos β y2 |dxp−1 ,
∞ gy
q1−r−1 q
g ydy by 2.13, we have J < ∞. If J 0 then 3.2 is proved. If
then J −∞ |y|
0 < J < ∞, by 3.1, we obtain
∞
q1−r−1 q y
g y dy J I ∗
−∞
∞
1/p ∞
1/q
<k
,
|x|p1r−1 f p xdx
|x|q1−r−1 g q xdx
0 <
∞
−∞
−∞
3.6
−∞
|x|q1−r−1 g q xdx
1/p
J 1/p < k
∞
−∞
|x|p1r−1 f p xdx
1/p
.
Inequalities 3.1 and 3.2 are equivalent.
If the constant factor k in 3.1 is not the best possible, then there exists a positive h
with h < k, such that
x2 2xy cos α y2 fxg y ln 2
dx dy
x 2xy cos β y2 −∞
∞
∞
1/p ∞
1/q
p1r−1 p
q1−r−1 q
<h
f xdx
g xdx
.
|x|
|x|
−∞
3.7
−∞
For ε > 0, by 3.7, using Lemma 2.3, we have
1/p ∞
1/q
∞
−1 p
−1 q
xdx
g
xdx
k.
f
k o1 < εh
|x|
|x|
−∞
3.8
−∞
Hence, we find k o1 < h. For ε → 0 , it follows that k ≤ h, which contradicts the fact that
h < k. Hence the constant k in 3.1 is the best possible.
Thus we complete the proof of the theorem.
8
Journal of Inequalities and Applications
Remark 3.2. For α π/4, β π/3 in 3.1, we have the following particular result:
√
x2 2xy y2 fxg y ln
dx dy
x2 xy y2 −∞
∞
<
4π sinπr/24 sin5πr/24
r sinπr/2
∞
−∞
|x|p1r−1 f p xdx
1/p ∞
−∞
|x|q1−r−1 g q xdx
1/q
.
3.9
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