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, xdm77108@gdei.edu.cn
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 medium, 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 <
1
∞
0
∞
0
f 2 xdx < ∞ and 0 <
∞
0
g 2 xdx < ∞, then we have see
∞
∞
1/2
fxg y
2
2
dx dy < π
f xdx
g xdx
,
xy
0
0
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:
∞
∞
1/2
fxg y
−x 2
−x 2
dx
dy
<
π
e
f
e
g
,
xdx
x
xy
−∞ 1 e
−∞
−∞
∞
1.2
where the constant factor π is the best possible. Zeng and Xie 11 also give a new inequality
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 is considered. As applications, we also obtain the equivalent
forms and some particular cases.
2. Some Lemmas
Lemma 2.1. If |λ| < 1, 0 < α1 < α2 < π, define the weight functions ωx and y x, y ∈
−∞, ∞ as follow:
∞
ωx :
y :
min
−∞ i∈{1,2}
∞
1
2
x 2xy cos αi y2
min
−∞ i∈{1,2}
x2
1
2xy cos αi y2
|x|1λ
λ dy,
y
1−λ
y
|x|−λ
2.1
dx.
Then we have ωx y kλ x, y /
0, where
sin λα1 sin λπ − α2 π
sin λπ sin α1
sin α2
kλ :
0 < |λ| < 1;
α1
π − α2
.
k0 : lim kλ λ→0
sin α1
sin α2
2.2
Proof. For x ∈ −∞, 0, setting u y/x, u −y/x, respectively, in the following first and
second integrals, we have
ωx 0
1
−x1λ
·
dy
x2 2xy cos α1 y2 −yλ
−∞
∞
0
∞
0
1
−x1λ
·
dy
x2 2xy cos α2 y2
yλ
u−λ
du u2 2u cos α1 1
∞
0
u−λ
du.
u2 − 2u cos α2 1
2.3
Journal of Inequalities and Applications
3
Setting a complex function as fz 1/z2 2z cos α1 1, where z1 −eiα1 and z2 −e−iα1
are the first-order poles of fz, and z ∞ is the first-order zero point of fz, in view of the
theorem of obtaining real integral by residue 12, it follows for 0 < |λ| < 1 that
∞
0
u−λ du
u2 2u cos α1 1
∞
0
u1−λ−1 du
u2 2u cos α1 1
2πi
−λ
−λ
Re
s
z
Re
s
z
fz,
z
fz,
z
1
2
1 − e2π1−λi
−λ
z1
z−λ
2πi
2
1 − e2π1−λi z1 − z2 z2 − z1
−π · −1−λ
sin π1 − λ · −11−λ
cos−λα1 i sin−λα1 cos λα1 i sin λα1
−2i sin α1
2i sin α1
π sin λα1
.
sin πλ sin α1
2.4
For λ 0, we can find by the integral formula that
∞
0
u2
1
α1
.
du sin
α1
2u cos α1 1
2.5
Obviously, we find that for 0 < |λ| < 1,
∞
0
u−λ
du u2 − 2u cos α2 1
∞
0
u2
∞
0
u−λ
du
u2 2u cosπ − α2 1
π · sin λπ − α2 ;
sin λπ · sin α2
π − α2
1
du .
sin α2
− 2u cos α2 1
Hence we find ωx kλ x ∈ −∞, 0.
for λ 0,
2.6
4
Journal of Inequalities and Applications
For x ∈ 0, ∞, setting u −y/x, u y/x, respectively, in the following first and
second integrals, we have
ωx 0
x2
−∞
∞
x2
0
1
x1λ
· λ dy
2
2xy cos α2 y
−y
∞
0
1
x1λ
· λ dy
2
2xy cos α1 y
y
u−λ
du u2 − 2u cos α2 1
∞
0
2.7
u−λ
du kλ.
u2 2u cos α1 1
By the same way, we still can find that y ωx kλ y, x /
0; |λ| < 1. The
lemma is proved.
Note 1. 1 It is obvious that ω0 0 0; 2 If α1 α2 α ∈ 0, π, then it follows
that
min
i∈{1,2}
1
x2 2xy cos αi y2
1
,
x2 2xy cos α y2
2.8
and by Lemma 2.1, we can obtain
π cos λα − π/2
y ωx cosλπ/2 sin α
y, x /
0 .
2.9
Lemma 2.2. If p > 1, 1/p 1/q 1, |λ| < 1, 0 < α1 < α2 < π, and fx is a nonnegative measurable
function in −∞, ∞, then we have
∞
J :
−∞
p1−λ−1
y
≤ kp λ
∞
−∞
∞
min
−∞ i∈{1,2}
p
1
fxdx dy
x2 2xy cos αi y2
2.10
|x|−pλ−1 f p xdx.
Journal of Inequalities and Applications
5
Proof. By Lemma 2.1 and Hölder’s inequality 13, we have
∞
min
−∞ i∈{1,2}
≤
p
1
fx
x2 2xy cos αi y2
∞
min
−∞ i∈{1,2}
∞
min
−∞ i∈{1,2}
1
2
x 2xy cos αi y2
1
2
x 2xy cos αi y2
∞
min
×
pλ−11
kp−1 λ
y
∞
min
−∞ i∈{1,2}
λ/p p
y
dx
|x|−λ/q
|x|1−pλ p
λ f xdx
y
1
x2 2xy cos αi y2
−∞ i∈{1,2}
|x|−λ/q
λ/p fx
y
q−1λ
y
|x|−λ
2.11
p−1
dx
1
2
x 2xy cos αi y2
|x|1−pλ p
λ f xdx.
y
Then by Fubini theorem, it follows that
J≤k
p−1
λ
kp−1 λ
kp λ
∞ ∞
−∞
∞
−∞
∞
−∞
min
−∞ i∈{1,2}
1
2
x 2xy cos αi y2
|x|1−pλ p
λ f xdx dy
y 2.12
ωx|x|−pλ−1 f p xdx
|x|−pλ−1 f p xdx.
The lemma is proved.
3. Main Results and Applications
Theorem
3.1. If p > 1, 1/p 1/q
|λ| < 1, 0 < α1 < α2 < π, f, g ≥ 0, satisfying 0 <
∞ 1,
∞
−pλ−1 p
qλ−1 q
|x|
f
xdx
<
∞
and
0
<
|y|
g ydy < ∞, then we have
−∞
−∞
∞
I :
min
−∞ i∈{1,2}
1
fxg y dx dy
x2 2xy cos αi y2
1/q
∞
1/p ∞
qλ−1 q −pλ−1 p
y
f xdx
g y dy
,
< kλ
|x|
−∞
−∞
3.1
6
Journal of Inequalities and Applications
J
∞
−∞
p1−λ−1
y < kp λ
∞
−∞
∞
min
−∞ i∈{1,2}
p
1
dy
fxdx
x2 2xy cos αi y2
3.2
|x|−pλ−1 f p xdx,
where the constant factor kλ and kp λ are the best possible and kλ is defined by Lemma 2.1.
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 x B|y|q−1λ /|x|−λ g q y a.e.
in −∞, 0 ∪ 0, ∞. Hence, there exists a constant C, such that A·|x|−pλ f p x B·|y|qλ g q y C a.e. in 0, ∞. We suppose A /
0 otherwiseB A 0. Then |x|−pλ−1 f p x C/A|x| a. e. in
∞
−∞, ∞, which contradicts the fact that 0 < −∞ |x|−pλ−1 f p xdx < ∞. Hence 2.11 takes the
form of strict inequality, so does 2.10, and we have 3.2.
By the Hlder’s inequality 13, we have
I
∞ −∞
≤ J 1/p
1/q−λ
y ∞
−∞
∞
min
−∞ i∈{1,2}
λ−1/q 1
y fxdx
g y dy
2
2
x 2xy cos αi y
qλ−1 q y g y dy
3.3
1/q
.
By 3.2, we have 3.1. On the other hand, suppose that 3.1 is valid. Setting
p1−λ−1
g y y ∞
min
−∞ i∈{1,2}
p−1
1
fxdx
,
x2 2xy cos αi y2
3.4
∞
then it follows J −∞ |y|qλ−1 g q ydy. By 2.10, we have J < ∞. If J 0, then 3.2 is obvious
value; if 0 < J < ∞, then by 3.1, we obtain
∞
0<
−∞
qλ−1 q y
g y dy J I
∞
1/q
1/p ∞
qλ−1 q y < kλ
g y dy
,
|x|−pλ−1 f p xdx
−∞
J 1/p ∞
−∞
qλ−1 q y g y dy
−∞
1/p
∞
1/p
< kλ
.
|x|−pλ−1 f p xdx
Hence we have 3.2, which is equivalent to 3.1.
−∞
3.5
Journal of Inequalities and Applications
7
For ε > 0, define functions fx,
g x as follows:
⎧
⎪
xλ−2ε/p ,
⎪
⎪
⎪
⎪
⎨
fx
: 0,
⎪
⎪
⎪
⎪
⎪
⎩
−xλ−2ε/p ,
x ∈ 1, ∞,
x ∈ −1, 1,
x ∈ −∞, −1,
3.6
⎧
⎪
x−λ−2ε/q ,
x ∈ 1, ∞,
⎪
⎪
⎪
⎪
⎨
g x : 0,
x ∈ −1, 1,
⎪
⎪
⎪
⎪
⎪
⎩
−x−λ−2ε/q , x ∈ −∞, −1.
: {
Then L
∞
−∞
I :
|x|−pλ−1 fp xdx}1/p {
∞
min
−∞ i∈{1,2}
∞
−∞
|y|qλ−1 g q ydy}1/q 1/ε and
1
g y dx dy I1 I2 I3 I4 ,
fx
2
2
x 2xy cos αi y
3.7
where
−1
I1 :
−∞
−1
I2 :
−∞
∞
I3 :
−x
λ−2ε/p
−x
λ−2ε/p
−∞
∞
1
xλ−2ε/p
−1
−∞
1
∞
I4 :
−1
x
λ−2ε/p
∞
1
1
−λ−2ε/q
−y
dy dx,
x2 2xy cos α1 y2
y−λ−2ε/q
dy dx,
x2 2xy cos α2 y2
3.8
−λ−2ε/q
−y
dy dx,
x2 2xy cos α2 y2
y−λ−2ε/q
dy dx.
x2 2xy cos α1 y2
By Fubini theorem 14, we obtain
I1 I4 ∞
x
−1−2ε
1
1/x
∞
x
−1−2ε
1
u−λ−2ε/q
du
u2 2u cos α1 1
1
1/x
1 ∞
0
∞
1/u
u−λ−2ε/q du
u2 2u cos α1 1
x−1−2ε dy
u
∞
1
y
x
u−λ−2ε/q du
u2 2u cos α1 1
1
u−λ−2ε/q du
2
u 2u cos α1 1 2ε
∞
1
u2
dx
u−λ−2ε/q du
2u cos α1 1
8
Journal of Inequalities and Applications
1
∞
u−λ2ε/p
u−λ−2ε/q
1
du du ,
2
2
2ε
0 u 2u cos α1 1
1 u 2u cos α1 1
1
I2 I 3 2ε
1
0
u−λ2ε/p
du 2
u − 2u cos α2 1
∞
1
u−λ−2ε/q
du .
u2 − 2u cos α2 1
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, such that
1
u−λ2ε/p
du 2
u 2u cos α1 1
0
1
0
u2
∞
u2
1
u−λ2ε/p du
− 2u cos α2 1
u−λ−2ε/q
du
2u cos α1 1
3.10
∞
u2
1
u−λ−2ε/q du
K.
εI < εK L
− 2u cos α2 1
By Fatou lemma 14 and 3.10, we have
kλ ∞
0
1
0
u−λ
du u2 2u cos α1 1
∞
0
u−λ2ε/p
du lim 2
ε → 0 u 2u cos α1 1
1
0
≤ lim
ε → 0
u−λ
du
u2 − 2u cos α2 1
∞
lim
1 ε→0
u−λ2ε/p
du lim 2
ε → 0 u − 2u cos α2 1
1
0
u−λ2ε/p
du u2 2u cos α1 1
1
0
u−λ−2ε/q
du
u2 2u cos α1 1
∞
lim
1 ε→0
∞
1
u−λ−2ε/q
du
u2 − 2u cos α2 1
3.11
u−λ−2ε/q
du
u2 2u cos α1 1
u−λ2ε/p
du 2
u − 2u cos α2 1
∞
1
u−λ−2ε/q
du ≤ K,
u2 − 2u cos α2 1
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, we may get a
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, and f, g ≥ 0, satisfying
∞
∞
0 < −∞ |x|−pλ−1 f p xdx < ∞ and 0 < −∞ |y|qλ−1 g q ydy < ∞, then we have
∞
−∞
x2
1
fxg y dx dy
2
2xy cos α y
π cos λα − π/2
<
cosλπ/2 sin α
∞
−∞
p1−λ−1
y ∞
−∞
∞
−∞
|x|
−pλ−1 p
f xdx
1/p ∞
−∞
1
fxdx
x2 2xy cos α y2
π cos λα − π/2
<
cosλπ/2 sin α
p ∞
−∞
qλ−1 q
y g ydy
1/q
,
3.12
p
dy
|x|−pλ−1 f p xdx,
where the constant factors π cos λα − π/2/ cosλπ/2 sin α and π cos λα − π/2/
cosλπ/2 sin αp are the best possible. Inequality 3.12 is equivalent.
In particular, for α π/3, we have the following equivalent inequalities:
∞
−∞
x2
1
fxg y dx dy
2
xy y
2π cosλπ/6
< √
3 cosλπ/2
∞
−∞
p1−λ−1
y
∞
∞
−∞
2π cosλπ/6
< √
3 cosλπ/2
−∞
|x|−pλ−1 f p xdx
1
fxdx
x2 xy y2
p ∞
−∞
1/p ∞
−∞
qλ−1 q y g y dy
1/q
,
3.13
p
dy
|x|−pλ−1 f p xdx.
Theorem 3.3. As the assumptions of Theorem 3.1, replacing p > 1 by 0 < p < 1, we have the
equivalent reverses of 3.1 and 3.2 with the best constant factors.
Proof. By the reverse Hölder’s inequality 13, we have the reverse of 2.10 and 3.3. It is
easy to obtain the reverse of 3.2. In view of the reverses of 3.2 and 3.3, we obtain the
reverse of 3.1. On the other hand, suppose that the reverse of 3.1 is valid. Setting the same
gy as Theorem 3.1, by the reverse of 2.10, we have J > 0. If J ∞, then the reverse of 3.2
is obvious value; if J < ∞, then by the reverse of 3.1, we obtain the reverses of 3.5. Hence
we have the reverse of 3.2, which is equivalent to the reverse of 3.1.
10
Journal of Inequalities and Applications
If the constant factor kλ in the reverse of 3.1 is not the best possible, then there
exists a positive constant K with K > kλ, such that the reverse of 3.1 is still valid as we
replace kλ by K. By the reverse of 3.10, we have
1
0
1
1
u−λ2ε/p du
u2 2u cos α1 1 u2 − 2u cos α2 1
∞
1
1
1
u−λ−2ε/q du > K.
u2 2u cos α1 1 u2 − 2u cos α2 1
3.14
For ε → 0 , by the Levi’s theorem 14, we find
1
0
−→
1
1
u−λ2ε/p du
u2 2u cos α1 1 u2 − 2u cos α2 1
1
0
1
1
u2 2u cos α1 1 u2 − 2u cos α2 1
u−λ du.
3.15
For 0 < ε < ε0 , q < 0, such that |λ 2ε0 /q| < 1, since
u−λ−2ε/q ≤ u−λ−2ε0 /q ,
∞
1
u ∈ 1, ∞,
1
1
2ε0
−λ−2ε0 /q
u
< ∞,
du
≤
k
λ
q
u2 2u cos α1 1 u2 − 2u cos α2 1
3.16
then by Lebesgue control convergence theorem 14, for ε → 0 , we have
∞
1
−→
1
1
u−λ−2ε/q du
u2 2u cos α1 1 u2 − 2u cos α2 1
∞
1
1
1
u−λ du.
u2 2u cos α1 1 u2 − 2u cos α2 1
3.17
By 3.14, 3.15, and 3.17, for ε → 0 , we have kλ ≥ K, which contradicts the fact that
kλ < K. Hence the constant factor kλ in the reverse of 3.1 is the best possible.
If the constant factor in reverse of 3.2 is not the best possible, then by the reverse of
3.3, we may get a contradiction that the constant factor in the reverse of 3.1 is not the best
possible. Thus the theorem is proved.
By the same way of Theorem 3.3, we still have the following theorem.
Theorem 3.4. By the assumptions of Theorem 3.2, replacing p > 1 by 0 < p < 1, we have the
equivalent reverses of 3.12 with the best constant factors.
Journal of Inequalities and Applications
11
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