Theoretical Mathematics & Applications, vol.1, no.1, 2011, 103-113
ISSN: 1792- 9687 (print), 1792-9709 (online)
International Scientific Press, 2011
A Hilbert-type Inequality
with a non-homogeneous
and the Integral in Whole Plane
Zitian Xie1 and Zheng Zeng2
Abstract
By the way of weight function, this paper gives a new Hilberttype inequality with the integral in whole plane and with some parameters, which is an extension of Hilbert-type inequality with a nonhomogeneous, and gives its equivalent form. Also in the paper, by the
way of Complex Analysis, the best constant factor is calculated.
Mathematics Subject Classification: 26D15
Keywords: Hilbert-type integral inequality, weight function, Hölder’s inequality.
1
Introduction
R∞
R∞
If f (x), g(x) ≥ 0, such that 0 < 0 f 2 (x)dx < ∞ and 0 < 0 g 2 (x)dx < ∞,
then [1]
Z ∞
1/2
Z ∞
Z ∞Z ∞
f (x)g(x)
2
2
dxdy < π
g (x)dx
f (x)dx
,
(1.1)
x+y
0
0
0
0
1
2
e-mail: gdzqxzt@163.com
e-mail: zz@sgu.edu.cn
Article Info: Revised : October 24, 2011.
Published online : October 31, 2011
104
A Hilbert-type inequality with a non-homogeneous ...
where the constant factor π is the best possible. In 1925, Hardy give an
extension of (1.1) by introducing one pair of conjugates (p,q),(i.e. p1 + 1q = 1) :
If p > 1, p1 + 1q = 1, f (x), g(x) ≥ 0, such that
Z ∞
Z ∞
p
0<
f (x)dx < ∞, and 0 <
g q (x)dx < ∞,
0
0
then we have the following Hardy-Hilbert’s integral inequality:
Z ∞
1/p Z ∞
1/q
Z ∞Z ∞
π
f (x)g(y)
p
q
dxdy <
f (x)dx
g (x)dx
, (1.2)
x+y
sin(π/p)
0
0
0
0
π
also is the best possible.
where the constant factor
sin(π/p)
In recent years,by introducing some parameters and estimating the way of
weight function,inequalities (1.1)and(1.2) have many generalizations and variants. (1.1) has been strengthened by Yang and others. ( including double
series inequalities ) [2-12].
In 2008 Xie gave a new Hilbert-type Inequality [2] as follows :
If p > 1, p1 + 1q = 1, f (x), g(x) ≥ 0 such that
Z ∞
Z ∞
−1−p/2 p
0<
x
f (x)dx < ∞, and 0 <
x−1−q/2 g q (x)dx < ∞,
0
then
0
Z
∞Z ∞
f (x)g(y)
dxdy
+ b2 y)(x + c2 y)
0 (x +
0
Z ∞
1/p Z ∞
1/q
−1−p/2 p
−1−q/2 q
<K
x
f (x)dx
x
g (x)dx
,
a2 y)(x
0
where the constant factor K =
(1.3)
0
π
(a+b)(a+c)(b+c)
is the best possible.
In this paper, by using the way of weight function and the technic of real
analysis and by the way of complex analysis, a new Hilbert-type inequality
with the integral in whole plane is given.
In the following, we always suppose that
π
a = cos θ, θ ∈ (0, ); p > 1, 1/p + 1/q = 1, r ∈ (−3, 1).
2
105
Zitian Xie and Zheng Zeng
2
Some lemmas
Lemma 2.1. If
k1 :=
and
k :=
then
k1 = k(θ, r) =





k2 = k(π−θ, r) =
and
Z
Z
∞
0
∞
−∞





f (z) =
k :=
Z
∞
0
(t2
k2 :=
(
Z
∞
0
(t2
dt
,
− 2at + 1)2 tr
dt
,
+ 2at + 1)2 |t|r
if r 6= −2, −1, 0,
r = −2, and α = 0,
if r = −1.
π[sin r(π−θ)−r sin θ cos(1+r)(π−θ)]
,
2 sin3 θ sin rπ
2(π−θ)+sin2θ
,
4 sin3 θ
sin θ+(π−θ) cos θ
,
2 sin3 θ
Proof. Let
∗
(t2
π[sin rθ−r sin θ cos(1+r)θ]
,
2 sin3 θ sin rπ
2θ−sin2θ
,
if
4 sin3 θ
sin θ−θ cos θ
,
2 sin3 θ
k = k1 + k2 =
then
dt
,
2
(t + 2at + 1)2 tr
(2.1)
if r 6= −2, −1, 0,
if r = −2, and r = 0, (2.2)
if r = −1.
π[cos r( π2 −θ)+r sin θ cos( π2 −θ)(1+r)]
, if r 6= −1,
2 sin3 θ cos rπ
2
2 sin θ+(π−2θ) cos θ
,
if r = −1,
2 sin3 θ
1
1
=
,
2
2
r
2
(1 + 2bz + z ) z
(z − z1 ) (z − z2 )2 z r
2πi
dt
=
[Res(f, z1 ) + Res(f, z2 )];
2
r
+ 2bt + 1) t
1 − e−2rπi
if b = cos ϑ (0 < ϑ < π), and z1 = −eiϑ , z2 = −e−iϑ , then
Res(f, z1 ) = (
k∗
1
1
−2
r
)0 |z=z1 =
[
− r+1 ]
2
r
2
r
(z − z2 ) z
(z1 − z2 ) (z1 − z2 )z
z1
2πi
1
−2
−2
[
+
−2rπi
2
iϑ
r
1−e
(−2i sin ϑ) (−2i sin ϑ)(−e )
(2i sin ϑ)(−e−iϑ )r
r
r
−
−
]
iϑ
r+1
−iϑ
(−e )
(−e )r+1
 π[sin rϑ−r sin ϑ cos(1+r)ϑ

,
if r 6= −2, −1, 0,

2 sin3 ϑ sin rπ
2ϑ−sin2ϑ
=
,
if r = −2, and r = 0,
4 sin3 ϑ

 sin ϑ−ϑ cos ϑ
,
if r = −1.
2 sin3 ϑ
=
106
A Hilbert-type inequality with a non-homogeneous ...
Setting ϑ = θ and ϑ = π − θ, we have (2.1)and (2.2).
On the other hand,
Z 0
dt
dt
+
k =
2
2
r
2
2
r
(t + 2at + 1) t
−∞ (t + 2at + 1) (−t)
0
= k1 + k2 = k(θ, r) + k(π − θ, r)
Z
=
=











(
∞
π[sin
rπ
2
π(1+r)
2
cos r( π2 −θ)−r sin θ cos
sin3 θ sin rπ
π
,
2 sin3 θ
2 sin θ+(π−2θ) cos θ
,
2 sin3 θ
cos( π2 −θ)(1+r)]
, if r 6= −2, −1, 0,
if r = −2, and r = 0,
if r = −1,
π[cos r( π2 −θ)+r sin θ cos( π2 −θ)(1+r)]
, if r 6= −1,
2 sin3 θ cos rπ
2
2 sin θ+(π−2θ) cos θ
,
if r = −1,
2 sin3 θ
The lemma is proved.
Lemma 2.2. Define the weight functions as follow:
w(x) :=
w(y)
e
:=
then w(x) = w(y)
e
= k.
Z
Z
∞
−∞
∞
−∞
1
|x|−r+1
dy,
r
2
2
|y| (x y + 2xy cos θ + 1)2
1
|y|−r+1
dx.
r
2
2
|x| (x y + 2xy cos θ + 1)2
Proof. We only prove that w(x) = k, for x ∈ (−∞, 0).
Z
0
1
|x|−r+1
dy
r
2
2
|y| (x y + 2xy cos θ + 1)2
−∞
Z ∞ −r+1
|x|
1
+
dy
r
2
2
|y| (x y + 2xy cos θ + 1)2
0
Z 0
1
(−x)−r+1
dy
=
(−y)r (x2 y 2 + 2xy cos θ + 1)2
−∞
Z ∞
(−x)−r+1
1
+
dy := w1 + w2 ,
r
2
2
y
(x y + 2xy cos θ + 1)2
0
w(x) =
(2.3)
107
Zitian Xie and Zheng Zeng
Setting u = xy, then
Z ∞
Z 0
1
1
(−x)−r+1
−r
u
dy
=
du = k1 .
w1 =
(−y)r (x2 y 2 + 2xy cos θ + 1)2
(u2 + 2u cos θ + 1)2
0
−∞
Similarly, setting y = −tx,
Z ∞
u−r
w2 =
0
(u2
1
du = k2 ,
− 2u cos θ + 1)2
and
w(x) = k1 + k2 = k.
We have (2.3).
, 2ε }) ∈ (−3, 1),
Lemma 2.3. For 0 < ε, and (r + max{ 2ε
p q
functions, fe, ge as follow:

−r−2ε/p

,
if x ∈ (1, ∞),
 x
e
f (x) =
0,
if x ∈ [−1, 1],


−r−2ε/p
(−x)
,
if x ∈ (−∞, −1);


if x ∈ (1, ∞),
 0,
−r+2ε/q
ge(x) =
x
,
if x ∈ [−1, 1],


0,
if x ∈ (−∞, −1),
define both
then
I(ε) := 2ε
Z
∞
−∞
e :=
I(ε)
|x|
Z
pr−1 ep
f (x)dx
∞Z ∞
1/p Z
fe(x)e
g (y)
−∞ −∞
∞
−∞
|x|
qr−1 q
ge (x)dx
1/q
= 1;
1
dxdy
(x2 y 2 + 2xy cos θ + 1)2
= k + o(1).(f or ε → 0+ )
(2.5)
Proof. Easily,
I(ε) = 2ε
Z
∞
1
−1 −2ε
x x
(2.4)
dx
1/p Z
1
−1 −2ε
x x
0
dx
1/q
= 1;
Let y = −Y ,using fe(−x) = fe(x), ge(−x) = ge(x), and
Z ∞
1
e
f (−x)
ge(y) 2 2
dy
(x y − 2xy cos θ + 1)2
−∞
Z ∞
1
e
= f (x)
ge(Y ) 2 2
dY
(x Y + 2xY cos θ + 1)2
−∞
108
A Hilbert-type inequality with a non-homogeneous ...
we have that fe(x)
R∞
ge(y) (x2 y2 +2xy1 cos θ+1)2 dy is an even function,then
−∞
Z
1
ge(y) 2 2
dy dx
(x y + 2xy cos θ + 1)2
0
−∞
Z ∞
Z 0
2ε
1
−r− 2ε
−r+
p
q
dy dx
= 2ε
x
(−y)
(x2 y 2 + 2xy cos θ + 1)2
1
−1
Z 1
Z ∞
1
−r− 2ε
−r+ 2ε
p
q
+
x
y
dy dx
(x2 y 2 + 2xy cos θ + 1)2
1
0
:= I1 + I2 .
e
I(ε)
= 2ε
Z
∞
fe(x)
∞
Setting y = u/x, then
Z 1
Z ∞
1
−r+ 2ε
−r− 2ε
p
q
y
dy dx
I1 = 2ε
x
(x2 y 2 − 2xy cos θ + 1)2
0
1
Z x
Z ∞
1
−r+ 2ε
−1−2ε
q
= 2ε
x
u
du dx
(u2 − 2u cos θ + 1)2
1
0
Z ∞
Z 1
1
−r+ 2ε
−1−2ε
q
du dx +
= 2ε
x
u
(u2 − 2u cos θ + 1)2
1
0
Z x
Z ∞
1
−r+ 2ε
−1−2ε
q
x
+
u
du dx
(u2 − 2u cos θ + 1)2
1
1
Z 1
2ε
1
u−r+ q 2
=
du
(u − 2u cos θ + 1)2
0
Z ∞
Z ∞
1
−r+ 2ε
−1−2ε
q
du
x
dx du
+ 2ε
u
(u2 − 2u cos θ + 1)2
u
1
Z 1
Z ∞
1
1
−r+ 2ε
−r− 2ε
q
p
=
u
du
+
u
du
(u2 − 2u cos θ + 1)2
(u2 − 2u cos θ + 1)2
0
1
Z ∞
Z 1
2ε
2ε
1
1
−r− 2ε
p
=
u
du + (u q − u− p )u−r 2
2
2
(u − 2u cos θ + 1)
(u − 2u cos θ + 1)2
0
Z0 ∞
2ε
1
=
u−r− p 2
du
(u + 2u cos(π − θ) + 1)2
0
Z 1
2ε
2ε
1
+ (u q − u− p )u−r 2
(u − 2u cos θ + 1)2
0
2ε
= k(π − θ, r + ) + η(ε)
p
there limε→0+ η(ε) = 0, and we have I1 → k2 (for ε → 0+ )
Similarly I2 → k1 (for ε → 0+ ). The lemma is proved.
109
Zitian Xie and Zheng Zeng
R∞
Lemma 2.4. If 0 < −∞ |x|pr−1 f p (x)dx < ∞,we have
p
Z ∞
Z ∞
1
−pr+p−1
dx dy
J :=
|y|
f (x) 2 2
(x y + 2xy cos θ + 1)2
−∞
−∞
Z ∞
p
|x|pr−1 f p (x)dx.
≤k
(2.6)
−∞
Proof. By Lemma 2.2, we find
Z ∞
p
1
f (x) 2 2
dx
(x y + 2xy cos θ + 1)2
−∞
Z ∞
r/q
r/p p
1
|x|
|y|
=
dx
f (x)
2
2
2
r/p
|y|
|x|r/q
−∞(x y + 2xy cos θ + 1)
Z ∞
1
|x|r(p−1) p
≤
f (x)dx
2 2
2
|y|r
−∞(x y + 2xy cos θ + 1)
p−1
Z ∞
|y|r(q−1)
1
dx
×
2 2
2
|x|r
−∞(x y + 2xy cos θ + 1)
Z ∞
|x|r(p−1) p
1
p−1
pr−p+1
f (x)dx,
= k |y|
2 2
2
|y|r
−∞(x y + 2xy cos θ + 1)
Z ∞Z ∞
1
|x|r(p−1) p
p−1
J ≤k
f (x)dx dy
2 2
2
|y|r
−∞ −∞(x y + 2xy cos θ + 1)
Z ∞Z ∞
|x|r(p−1)
1
p−1
=k
dy f p (x)dx
2 y 2 + 2xy cos θ + 1)2
r
(x
|y|
Z ∞−∞ −∞
= kp
|x|pr−1 f p (x)dx.
(2.7)
−∞
3
Main results
Theorem 3.1. If both functions, f (x) and g(x) , are nonnegative measurable functions, and satisfy
Z ∞
Z ∞
pr−1 p
|x|qr−1 g q (x)dx < ∞,
|x|
f (x)dx < ∞ and 0 <
0<
−∞
−∞
then,
∗
I :=
Z
∞Z ∞
f (x)g(y)
−∞ −∞
(x2 y 2
1
dxdy
+ 2xy cos θ + 1)2
110
A Hilbert-type inequality with a non-homogeneous ...
<k
and
J=
Z
Z
∞
−∞
∞
−∞
|y|
|x|
pr−1 p
f (x)dx
−pr+p−1
(
Z
<k
1/p Z
∞
f (x)
−∞
p
Z
∞
−∞
(x2 y 2
∞
−∞
|x|
qr−1 q
g (x)dx
1/q
,
(3.1)
1
dx)p dy
+ 2xy cos θ + 1)2
|x|pr−1 f p (x)dx.
(3.2)
Inequalities (3.1) and (3.2) are equivalent, and the constant factors in the two
forms are all the best possible.
Proof. If (2.7) takes the form of equality for some y ∈ (−∞, 0) ∪ (0, ∞) ,
then there exist constants M and N , such that they are not all zero, and
M
|x|r(p−1) p
f (x), ×(−∞, ∞).
|y|r
Hence, there exists a constant C, such that
M |x|rp f p (x) = N |y|rq = C,
a.e. in (−∞, ∞).
We claim that M = 0. In fact, if M 6= 0, then |x|pr−1 f p (x) = MC|x| a.e. in
R∞
(−∞, ∞) which contradicts the fact that 0 < −∞ |x|pr−1 f p (x)dx < ∞. In
the same way, we claim that N = 0. This is too a contradiction and hence by
(2.7), we have (3.2).
By Hölder’s inequality with weight and (3.2), we have,
h
Z ∞
Z ∞
i
1
−r+ 1q
r− 1q
∗
g(y)
dy
f (x) 2 2
|y|
I =
dx
|y|
(x y + 2xy cos θ + 1)2
−∞
−∞
Z ∞
1/q
1
qr−1 q
p
|y|
g (y)dy
≤ (J)
.
(3.3)
−∞
Using (3.2), we have (3.1).
Setting
g(y) = |y|
−pr+p−1
(
Z
∞
−∞
then
J=
by (2.7) we have J < ∞.
f (x)
Z
∞
−∞
(x2 y 2
1
dx)p−1 ,
2
+ 2xy cos θ + 1)
|y|qr−1 g q (y)dy
111
Zitian Xie and Zheng Zeng
If J = 0 then (3.2) is proved; if 0 < J < ∞, by (3.1), we obtain
Z ∞
|y|qr−1 g q (y)dy = J = I ∗
0 <
−∞
< k
Z
Z
∞
−∞
∞
−∞
|x|
|x|
pr−1 p
f (x)dx
qr−1 q
g (x)dx
1/p Z
1/p
=J
∞
−∞
1/p
|x|
qr−1 q
<k
g (x)dx
Z
∞
−∞
|x|
1/q
,
pr−1 p
f (x)dx
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
Z ∞Z ∞
1
dxdy
f (x)g(y) 2 2
(x y + 2xy cos θ + 1)2
−∞ −∞
<h
Z
∞
−∞
|x|
pr−1 p
f (x)dx
1/p Z
∞
−∞
|x|
qr−1 q
g (x)dx
1/q
.
(3.4)
= h.
(3.5)
For ε > 0, by (3.4), using Lemma 2.3, we have
< εh
Z
∞
−∞
|x|
pr−1 ep
e = k + o(1)
I(ε)
f (x)dx
1/p Z
∞
−∞
|x|
qr−1 q
ge (x)dx
1/q
Hence we find, k + o(1) < 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. As (3.1) and (3.2) are equivalent, if the constant factor in (3.2) is
not the best possible, then by using (3.2), we can get a contradiction that the
constant factor in (3.1) is not the best possible. Thus we complete the prove
of the theorem.
Remark 3.2. For θ = π4 , in (3.1), we have the following particular result:
Z
∞Z ∞
−∞
1
√
dxdy <
f (x)g(y)
2
2
(x y + 2xy + 1)2
−∞
Z
∞
−∞
|x|
pr−1 p
f (x)dx
1/p
×
Z
∞
−∞
|x|
√
qr−1 q
√
2π[sin rπ
+ 22r cos (1+r)π
]
4
4
πr
cos 4
g (x)dx
1/q
.
(3.6)
112
A Hilbert-type inequality with a non-homogeneous ...
References
[1] G.H. Hardy, J.E. Littlewood and G. Pólya, Inequalities, Cambridge University Press, Cambridge, 1952.
[2] Zitian Xie and Zeng Zheng, A Hilbert-type integral inequality whose kernel is a homogeneous form of degree −3, J. Math. Appl., [J], 339, (2008),
324-331.
[3] Bicheng Yang, On the the norm of an intergal operator and applications,
J. Math. Anal. Appl., 321, (2006), 182-192.
[4] Zitian Xie and Zheng Zeng,A Hilbert-type Integral Inequality with a nonhomogeneous form and a best constant factor, Advances and Applications
in Mathematical Sciens, 3(1), (2010), 61-71.
[5] Zheng Zeng and Zi-tian Xie, A Hilbert’s inequality with a best constant factor, Journal of Inequalities and Applications, Article ID 820176,
(2009), 8 pages, doi:10.1155/2009/820176.
[6] Xie Zitian,Zeng Zheng, A Hilbert-type integral inequality with the homogeneous kernel of real number-degree and its equivalent, Journal of Zhejiang
University (Science Edition).
[7] Zitian Xie, Zheng Zeng, The Hilbert-type integral inequality with the
system kernel of -λ degree homogeneous form, Kyungpook Mathematical
Journal, 50, (2010), 297-306.
[8] Zitian Xie,Bicheng Yang, Zheng Zeng,A New Hilbert-type integral inequality with the homogeneous kernel of real number-degree, Journal of
Jilin University (Science Edition), 48(6), (2010), 941-945.
[9] Xie Zitian,Zeng Zheng, A new Hilberttype integral inequality with the
homogeneous kernel of degree 0, Journal of Southwest University, (Natturnal Science Edition), 238, (2011), 137-141.
[10] Zitian Xie and Benlu Fu, A new Hilbert-type integral inequality with the
best constant factor, J. Wuhan Univ. (Nat.Sci.Ed), 55(6), (2009), 637640.
Zitian Xie and Zheng Zeng
113
[11] Xie Zitian,Zeng Zheng,A Hilbert-type inequality with some parameters
and the integral in whole plane, Advances in Pure Mathematics, 1(3),
(2011), 84-89.
[12] Bicheng Yang, On the norm of an integral operatorand applications, J.
Math. Anal. Appl., 321, (2006), 182-192.