Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2011, Article ID 401428, 11 pages
doi:10.1155/2011/401428
Research Article
A Hilbert-Type Integral Inequality in the Whole
Plane with the Homogeneous Kernel of Degree −2
Dongmei Xin and Bicheng Yang
Department of Mathematics, Guangdong Education Institute, Guangzhou, Guangdong 510303, China
Correspondence should be addressed to Dongmei Xin, [email protected]
Received 20 December 2010; Accepted 29 January 2011
Academic Editor: S. Al-Homidan
Copyright q 2011 D. Xin and B. Yang. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any m edium, provided the original work is properly cited.
By applying the way of real and complex analysis and estimating the weight functions, we build
a new Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree
−2 involving some parameters and the best constant factor. We also consider its reverse. The
equivalent forms and some particular cases are obtained.
1. Introduction
If fx,gx ≥ 0, satisfying 0 <
∞
0
f
2
xdx < ∞ and 0 <
∞
0
g
2
x
dx
1/2
,
1.1
where the constant factor π is the best possible. Inequality 1.1 is well known as Hilbert’s
integral inequality, which is important in analysis and in its applications 1, 2. In recent
years, by using the way of weight functions, a number of extensions of 1.1 were given
by Yang 3. Noticing that inequality 1.1 is a Homogenous kernel of degree −1, in 2009,
a survey of the study of Hilbert-type inequalities with the homogeneous kernels of degree
negative numbers and some parameters is given by 4. Recently, some inequalities with the
homogenous kernels of degree 0 and nonhomogenous kernels have been studied see 5–9.
2 Journal of Inequalities and Applications
All of the above inequalities are built in the quarter plane. Yang 10 built a new Hilbert-type
integral inequality in the whole plane as follows:
∞
−∞
f
x
g
y
in the whole plane.
By applying the method of 10, 11 and using the way of real and complex analysis,
the main objective of this paper is to give a new Hilbert-type integral inequality in the whole
plane with the homogeneous kernel of degree −2 involving some parameters and a best
constant factor. The reverse form i s considered. As applications, we also obtain the equivalent
forms and some particular cases.
2. Some Lemmas
Lemma 2.1. If |λ| < 1, 0 <α
1
<α
2
<π, define the weight functions ωx and yx, y ∈
−∞, ∞ as follow:
ω
x
:
∞
−∞
min
i∈
{
1,2
}
1
x
2
1
x
2
2xy cos α
i
y
2
y
1−λ
|
x
|
−λ
dx.
2.1
Then we have ωxykλx, y
/
0,where
k
λ
:
π
sin λπ
λ
α
1
sin α
1
π − α
2
sin α
2
.
2.2
Proof. For x ∈ −∞, 0, setting u y/x, u −y/x, respectively, in the following first and
second integrals, we have
ω
x
0
−∞
1
x
2
y
λ
dy
∞
0
u
−λ
u
2
2u cos α
1
1
du
∞
0
u
−λ
u
2
− 2u cos α
2
1
du.
2.3
Journal of Inequalities and Applications 3
Setting a complex function as fz1/z
2
1−λ−1
du
u
2
2u cos α
1
1
2πi
1 − e
2π1−λi
Re s
z
−λ
f
z
,z
1
Re s
z
−λ
f
z
−λ
sin π
1 −λ
·
−1
1−λ
cos
−λ
α
1
i sin
−λ
α
1
−2i sin α
1
cos λα
1
i sin λα
1
0
u
−λ
u
2
− 2u cos α
2
1
du
∞
0
u
−λ
u
2
2u cos
π −α
2
1
du
π ·sin λ
π −α
2
sin λπ · sin α
1
x
2
2xy cos α
2
y
2
·
x
1λ
−y
λ
dy
∞
0
1
x
2
2xy cos α
1
y
2
·
x
1λ
y
Bythesameway,westillcanfindthatyωxkλy,x
/
0; |λ| < 1.The
lemma is proved.
Note 1. 1 It is obvious that ω000; 2 If α
1
α
2
α ∈ 0,π, then it follows
that
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
y
2
1
x
2
2xy cos α y
2
2
<π,andfx is a nonnegative measurable
function in −∞, ∞,thenwehave
J :
∞
−∞
y
p1−λ−1
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
y
2
¨
older’s inequality 13,wehave
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
y
2
f
x
p
∞
y
λ/p
|
x
|
−λ/q
dx
p
≤
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
1
x
2
2xy cos α
i
y
2
y
q−1λ
|
x
|
−λ
dx
p−1
k
p−1
λ
y
p
x
dx.
2.11
Then by Fubini theorem, it follows that
J ≤ k
p−1
λ
∞
−∞
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
x
|
x
|
−pλ−1
f
p
x
dx
k
p
λ
∞
−∞
|
x
|
−pλ−1
f
p
x
dx.
{
1,2
}
1
x
2
2xy cos α
i
y
2
f
x
g
y
dx dy
<k
λ
∞
−∞
|
3.1
6 Journal of Inequalities and Applications
J
∞
−∞
y
p1−λ−1
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
y
2
Inequality 3.1 and 3.2 are equivalent.
Proof. If 2.11 takes the form of equality for a y ∈ −∞, 0∪0, ∞, then there exist constants A
and B, such that they are not all zero, and A|x|
1−pλ
/|y|
λ
f
p
xB|y|
q−1λ
/|x|
−λ
g
q
y a.e.
in −∞, 0 ∪ 0, ∞.Hence,thereexistsaconstantC,suchthatA·|x|
−pλ
f
p
xB·|y|
qλ
g
q
y
C a.e. in 0, ∞. We suppose A
/
0 otherwise B A 0.Then|x|
−pλ−1
f
p
1,2
}
1
x
2
2xy cos α
i
y
2
f
x
dx
y
λ−1/q
g
y
dy
y
p1−λ−1
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
y
2
f
x
dx
p−1
,
∞
−∞
|
x
|
−pλ−1
f
p
x
dx
1/p
∞
−∞
y
qλ−1
g
q
∞
−∞
|
x
|
−pλ−1
f
p
x
dx
1/p
.
3.5
Hencewehave3.2, which is equivalent to 3.1.
Journal of Inequalities and Applications 7
For ε>0, define functions
fx, gx as follows:
f
x
:
−∞, −1
,
g
x
:
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
x
−λ−2ε/q
,x∈
1, ∞
,
0,x∈
−∞
|y|
qλ−1
g
q
ydy}
1/q
1/ε and
I :
∞
−∞
min
i∈
{
1,2
}
1
x
2
2xy cos α
i
y
2
f
−∞
−y
−λ−2ε/q
x
2
2xy cos α
1
y
2
dy
dx,
I
2
:
−1
−∞
−x
λ−2ε/p
∞
1
y
−λ−2ε/q
2
dy
dx,
I
4
:
∞
1
x
λ−2ε/p
∞
1
y
−λ−2ε/q
x
2
2xy cos α
1
y
2
dy
dx.
3.8
By Fubini theorem 14,weobtain
I
−1−2ε
1
1/x
u
−λ−2ε/q
du
u
2
2u cos α
1
1
∞
1
u
−λ−2ε/q
du
u
2
2u cos α
1
1
dx
1
1
8 Journal of Inequalities and Applications
1
2ε
1
0
u
−λ2ε/p
u
2
2u cos α
1
1
du
∞
1
u
−λ−2ε/q
u
2
2u cos α
1
1
du
,
.
3.9
In view of the above results, if the constant factor kλ in 3.1 is not the best possible, then
exists a positive number K with K<kλ,suchthat
1
0
u
−λ2ε/p
u
2
2u cos α
1
1
du
∞
1
u
−λ−2ε/q
u
2
2u cos α
1
1
du
1
λ
∞
0
u
−λ
u
2
2u cos α
1
1
du
∞
0
u
−λ
u
2
− 2u cos α
2
1
du
1
0
lim
u
−λ2ε/p
u
2
− 2u cos α
2
1
du
∞
1
lim
ε →0
u
−λ−2ε/q
u
2
− 2u cos α
2
1
du
≤ lim
ε →0
1
0
du
∞
1
u
−λ−2ε/q
u
2
− 2u cos α
2
1
du
≤ K,
3.11
which contradicts the fact that K<kλ. Hence the constant factor kλ in 3.1 is the best
possible.
If the constant factor in 3.2 is not the best possible, then by 3.3,wemaygeta
contradiction that the constant factor in 3.1 is not the best possible. Thus the theorem is
proved.
Journal of Inequalities and Applications 9
In view of Note 2 and Theorem 3.1, we still have the following theorem.
Theorem 3.2. If p>1, 1/p 1/q 1, |λ| < 1, 0 <α<π,andf, g ≥ 0, satisfying
0 <
∞
−∞
|x|
−pλ−1
f
α −π/2
cos
λπ/2
sin α
∞
−∞
|
x
|
−pλ−1
f
p
x
dx
1/p
∞
−∞
y
dx
p
dy
<
π cos λ
α −π/2
cos
λπ/2
sin α
p
∞
−∞
|
x
|
−pλ−1
f
p
x
3cos
λπ/2
∞
−∞
|
x
|
−pλ−1
f
p
x
dx
1/p
∞
−∞
y
qλ−1
dx
p
dy
<
2π cos
λπ/6
√
3cos
λπ/2
p
∞
−∞
|
x
|
−pλ−1
f
p
x
2
− 2u cos α
2
1
u
−λ2ε/p
du
∞
1
1
u
2
2u cos α
1
1
1
u
2
− 2u cos α
2
1
u
−λ−2ε/q
du > K.
1
u
2
2u cos α
1
1
1
u
2
− 2u cos α
2
1
u
−λ
du.
3.15
For 0 <ε<ε
0
, q<0, such that |λ 2ε
0
/q| < 1, since
u
−λ−2ε/q
≤ u
−λ−2ε
0
/q
,u∈
0
q
< ∞,
3.16
then by Lebesgue control convergence theorem 14,forε → 0
,wehave
∞
1
1
u
2
2u cos α
1
1
1
u
2
− 2u cos α
2
1
u
−λ−2ε/q
du
−→
Theorem 3.4. By the assumptions of Theorem 3.2,replacingp>1 by 0 <p<1,wehavethe
equivalent reverses of 3.12 with the best constant factors.
Journal of Inequalities and Applications 11
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olya, Inequalities, The University Press, Cambridge, UK, 2nd
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c, J. E. Pe
ˇ
cari
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