Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2008, Article ID 918534, 12 pages
doi:10.1155/2008/918534
Research Article
The Generalizations of Hilbert’s Inequality
Liubin Hua,1 Huazhong Ke,2 and Yongjin Li2
1
2
Guangzhou Sontan Polytechnic College, Guangzhou 511370, China
Department of Mathematics, Sun Yat-Sen University, Guangzhou 510275, China
Correspondence should be addressed to Yongjin Li, stslyj@mail.sysu.edu.cn
Received 24 June 2008; Accepted 2 October 2008
Recommended by Feng Qi
By introducing some parameters and estimating the weight functions, we establish a new Hilberttype inequalities with best constant factors. The equivalent inequalities are also considered.
Copyright q 2008 Liubin Hua et al. 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 f, g are real functions such that 0 <
see 1
∞ ∞
0
0
∞
0
f 2 xdx < ∞ and 0 <
fxgy
dx dy < π
xy
∞
f 2 xdx
∞
0
∞
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 the well-known Hilbert’s
inequality. Inequality 1.1 had been generalized by Hardy-Riesz see 2 in 1925 as if f, g
∞
∞
are real functions such that 0 < 0 f p xdx < ∞ and 0 < 0 g q xdx < ∞, then
∞ ∞
0
0
fxgy
π
dx dy <
xy
sinπ/p
∞
0
f p xdx
1/p ∞
1/q
g q xdx
,
1.2
0
where the constant factor c π/ sinπ/p is the best possible. When p q 2, 1.2
reduces to 1.1. Inequality 1.2 is named after Hardy-Hilbert’s integral inequality, which
is important in the analysis and its applications see 3, it has been studied and generalized
in many directions by a number of mathematicians see 4–8.
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International Journal of Mathematics and Mathematical Sciences
Under the same condition of 1.2, we have Hardy-Hilbert’s type inequality see 1,
Theorems 341 and 342:
∞ ∞
0
∞ ∞
0
0
0
∞
∞
1/2
fxgy
2
dx dy < 4
f xdx g 2 xdx
,
max{x, y}
0
0
ln x − ln y
fxgydx dy < π 2
x−y
∞
f 2 xdx
0
∞
1.3
1/2
g 2 xdx
,
0
where the constant factors 4 and π 2 are both the best possible.
Recently, Li et al. 9, by introducing the function | ln x − ln y|/x y |x − y|, establish
new inequalities similar to Hilbert-type inequality for integrals.
∞
∞
Theorem 1.1. If fx, gx ≥ 0, 0 < 0 f 2 xdx < ∞, 0 < 0 g 2 xdx < ∞. Then, one has
∞ ∞
0
0
∞
1/2 ∞
1/2
| ln x − ln y|
2
2
fxgydx dy < 4
f xdx
g xdx
,
x y |x − y|
0
0
1.4
where the constant factor 4 is the best possible.
In this paper, we give further analogs of Hilbert-type inequality and its applications.
The main result unifies and generalizes the classical results as follows.
Assume that r > 0, s > − min{1, r}, t 0, p, q > 1, and 1/p 1/q 1. If f, g 0, such
∞
∞
that 0 < 0 f p xdx < ∞, 0 < 0 g q xdx < ∞, then
∞ ∞
0
0
| ln x − ln y|t
fxgydx dy < K1
x ry s|x − y|
∞
f xdx
p
1/p ∞
0
1/q
g q xdx
,
1.5
0
where the constant factor
K1 ∞
0
| ln u|t
u−1/q du
1 ru s|1 − u|
1.6
is the best possible.
2. Some lemmas
Our results will be based on the following results. In the following lemmas, assume that
r > 0, s > − min{1, r}, t 0, p, q > 1, and 1/p 1/q 1.
Liubin Hua et al.
3
Lemma 2.1. Define the following weight functions:
ω1 x ∞
0
ω2 y ∞
0
1/q
| ln x − ln y|t
x
dy,
x ry s|x − y| y
x > 0,
1/p
y
| ln x − ln y|
dx,
x ry s|x − y| x
y > 0.
2.1
t
Then, ω1 ω2 K1 , where K1 is defined by 1.1.
Proof. For ω1 x, let u y/x, and we have
ω1 ∞
0
| ln u|t
u−1/q du K1 .
1 ru s|1 − u|
2.2
For ω2 y, first let v x/y, and then let u 1/v. Thus, we have
ω2 ∞
0
| ln v|t
v−1/p dv 1 rv s|v − 1|
∞
0
| ln u|t
u−1/q du K1 .
1 ru s|1 − u|
2.3
Hence, the lemma is proved.
Lemma 2.2. Assume that ε > 0, then
∞
0
| ln u|t
u−1ε/q du K1 o1
1 ru s|1 − u|
ε → 0 .
2.4
Proof. Since
lim u1/2p | ln u|t 0,
u→0
lim
u1/2p
u → ∞ | ln u|t
∞,
2.5
it follows that there exists δ1 ∈ 0, 1, such that if u ∈ 0, δ1 , then u−1/2p > | ln u|t . Moreover,
there exists δ2 ∈ 1, ∞, such that if u ∈ δ2 , ∞, then u1/2q > | ln u|t . Using the expression
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International Journal of Mathematics and Mathematical Sciences
of K1 , if s 0, then
∞
| ln u|t
−1ε/q
u
du
−
K
1
1 ru s|1 − u|
0
∞
−1ε/q
| ln u|t
−1/q
−u
du
u
1 ru s|1 − u|
0
δ1
δ2
0
δ1
<
0
∞
δ2
δ1
−1ε/q
| ln u|t
u
− u−1/q du
1 ru s|1 − u|
| ln u|t −1ε/q
u
− u−1/q du 1s
δ2
−1ε/q
| ln u|t
u
− u−1/q du
δ1 1 ru s|1 − u|
∞
| ln u|t −1/q
− u−1ε/q du
u
δ2 r su
δ1
<
0
u−1/2p −1ε/q
− u−1/q du u
1s
δ2
−1ε/q
| ln u|t
u
− u−1/q du
δ1 1 ru s|1 − u|
∞
u1/2q −1/q
− u−1ε/q du
u
δ2 r su
1
1
1
1−1/2p−1ε/q
1−1/2p−1/q
δ
δ
−
1 s 1 − 1/2p − 1 ε/q 1
1 − 1/2p − 1/q 1
δ2
−1ε/q
2q −12ε/2q
1
| ln u|t
−1/2q
−1/q du u
δ
−u
−
2qδ2
rs
1 2ε 2
δ1 1 ru s|1 − u|
o1 ε → 0 .
2.6
Therefore, Lemma 2.2 is proved for s 0. If s > 0, then we can replace the right-hand side of
the first strict inequality above with
δ1
0
| ln u|t −1ε/q
− u−1/q du u
1
δ2
−1ε/q
| ln u|t
u
− u−1/q du
δ1 1 ru s|1 − u|
∞
| ln u|t −1/q
− u−1ε/q du.
u
ru
δ2
2.7
By the same way, we can show that the lemma is valid for s > 0. Hence, the lemma is proved.
Lemma 2.3. Assume that ε > 0, then
∞
x
1
−1−ε
x−1
dx
0
| ln u|t
u−1ε/q du O1
1 ru s|1 − u|
ε → 0 .
2.8
Liubin Hua et al.
5
Proof. Using δ1 introduced in the proof of Lemma 2.2, if s 0, then
∞
0<
x−1−ε dx
x−1
1
0
∞
x−1
x−1−ε dx
1
0
∞
x−1
<
x−1−ε dx
1
| ln u|t
u−1ε/q du
1 ru s1 − u
| ln u|t −1ε/q
u
du
1s
0
1/δ1
∞ x
1/δ1
x−1−ε dx
δ1
x−1−ε dx
δ1
0
∞
x−1
1/δ1
<
x−1−ε dx
0
x−1−ε dx
1
δ1
0
∞
x−1−ε dx
1/δ1
dx
δ1
| ln u|t −1ε/q
u
du 1s
x−1
1/δ1
∞
x−1−ε dx
1/δ1
x−1−ε dx
x−1
1
δ1
x−1
0
| ln u|t −1ε/q
u
du
1s
| ln u|t −1ε/q
u
du
1s
| ln u|t −1ε/q
u
du
1s
u−1/2p −1ε/q
u
du 1s
0
x−1 | ln u|t −1ε/q
u
du
1s
| ln u|t −1ε/q
u
du 1s
1
1/δ1
0
1/δ1
x−1
0
1
−1−ε
1/δ1
1
| ln u|t
u−1ε/q du
1 ru s|1 − u|
1/δ1
x−1−ε dx
x−1
1
δ1
| ln u|t −1ε/q
u
du
1s
u−1/2p −1ε/q
u
du
1s
1−1/2p−1ε/q
1 − δ1 δ 1
ε1 s 1 − 1/2p − 1 ε/q
1/δ1
x−1−ε dx
1
x−1
δ1
| ln u|t −1ε/q
u
du
1s
12ε/2p
2pδ1
1 2ε1 s 1 − 1/2p − 1 ε/q
O1 ε → 0 .
2.9
Therefore, Lemma 2.3 is proved for s 0. If s > 0, then we can replace the right-hand side of
the second strict inequality mentioned above with
∞
x
1
−1−ε
x−1
dx
0
| ln u|t −1ε/q
u
du.
1
2.10
By the same way, we can show that the lemma is valid for s > 0. Therefore, the lemma is
proved.
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International Journal of Mathematics and Mathematical Sciences
3. Integral case
In this section, we will state our main results.
Proposition 3.1. Assume
r}, t 0, p, q > 1, and 1/p 1/q 1. If
∞ that r > 0, s > − min{1,
∞
f, g 0, such that 0 < 0 f p xdx < ∞, 0 < 0 g q xdx < ∞, then
∞ ∞
0
0
| ln x − ln y|t
fxgydx dy < K1
x ry s|x − y|
∞
f p xdx
1/p ∞
0
1/q
g q xdx
,
3.1
0
where the constant factor
K1 ∞
0
| ln u|t
u−1/q du
1 ru s|1 − u|
3.2
is the best possible.
Proof. Using Hölder’s inequality, we have
∞ ∞
0
0
| ln x − ln y|t
fxgydx dy
x ry s|x − y|
∞ ∞
0
0
| ln x − ln y|t
x ry s|x − y|
∞ ∞
0
×
0
1/p
1/q
| ln x − ln y|t
x
p
f xdx dy
x ry s|x − y| y
∞ ∞
0
∞
1/pq
1/pq
y
x
fx
gy dx dy
y
x
0
1/p
1/q
y
| ln x − ln y|
q
g ydx dy
x ry s|x − y| x
ω1 xf p xdx
1/p ∞
0
K1
3.3
t
1/q
ω2 yg q ydy
0
∞
0
f xdx
p
1/p ∞
1/q
g xdx
q
.
0
If 3.3 takes the form of equality, then there are constants a and b, such that they are
not all zero, and
a
1/q
1/p
| ln x − ln y|t
| ln x − ln y|t
y
x
f p x b
g q y,
x ry s|x − y| y
x ry s|x − y| x
a.e. in 0, ∞ × 0, ∞.
3.4
Liubin Hua et al.
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Hence,
axf p x byg q y,
a.e. in 0, ∞ × 0, ∞.
3.5
Therefore, there is a constant c, such that
axf p x byg q y c,
a.e. in 0, ∞ × 0, ∞.
3.6
We claim that a 0. In fact, if a /
0, then f p x c/ax, a.e., in 0, ∞, which contradicts the
∞ p
fact that 0 < 0 f xdx < ∞. In the same way, we claim that b 0. This is a contradiction.
Hence, by 3.3, we have 3.1.
If the constant factor K1 in 3.1 is not the best possible factor, then there exists a
positive constant H with H < K1 , such that 3.1 is still valid if we replace K1 with H.
For ε > 0 sufficiently small, construct the following functions:
0,
fε x x−1ε/p ,
0,
gε x x−1ε/q ,
x ∈ 0, 1,
x ∈ 0, ∞,
3.7
x ∈ 0, 1,
x ∈ 0, ∞,
Thus, we obtain
∞
H
p
fε xdx
1/p ∞
0
∞ ∞
I :
0
0
q
1/q
gε xdx
0
H
,
ε
t
| ln x − ln y|
fε xgε ydx dy
x ry s|x − y|
∞ ∞
| ln x − ln y|t
x−1ε/p y−1ε/q dx dy
x
ry
s|x
−
y|
1 1
∞
∞
| ln x − ln y|t
−1ε/p
y−1ε/q dy.
x
dx
1
1 x ry s|x − y|
3.8
Setting u y/x, we have
I
∞
x
−1−ε
∞
dx
1
∞
1
x−1−ε dx
| ln u|t
u−1ε/q du
x−1 1 ru s|1 − u|
∞
0
| ln u|t
u−1ε/q du −
1 ru s|1 − u|
1
K1 o1 O1
ε
∞
1
x−1−ε dx
x−1
0
| ln u|t
u−1ε/q du
1 ru s|1 − u|
ε → 0 .
3.9
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International Journal of Mathematics and Mathematical Sciences
Since
I<
H
,
ε
3.10
it follows that
K1 o1 < H
ε → 0 .
3.11
Hence,
K1 H.
3.12
This contradicts the fact that H < K1 . So the constant factor K1 in 3.1 is the best possible.
Then Proposition 3.1 is proved.
Proposition 3.2. Under the same assumptions of Proposition 3.1, one has
∞ ∞
0
0
∞
p
| ln x − ln y|t
p
fxdx dy < K1
f p xdx,
x ry s|x − y|
0
3.13
p
where the constant factor K1 is the best possible. Inequalities 3.1 and 3.13 are equivalent.
Proof. Setting
gy ∞
0
| ln x − ln y|t
fxdx
x ry s|x − y|
p−1
,
y ∈ 0, ∞,
3.14
by 3.1, we have
∞
g q ydy 0
∞ ∞
0
0
p
| ln x − ln y|t
fxdx dy
x ry s|x − y|
∞ ∞
| ln x − ln y|t
fxgydx dy
0 0 x ry s|x − y|
1/p ∞
1/q
∞
< K1
f p xdx
g q xdx
.
0
3.15
0
As a result,
∞
0
∞
1/p
g xdx
q
< K1
0
1/p
f xdx
p
.
3.16
Liubin Hua et al.
9
Hence,
∞
g xdx <
q
0
p
K1
∞
f p xdx.
3.17
0
Then, we have 3.13.
Conversely, by Hölder’s inequality, we have
∞ ∞
0
| ln x − ln y|t
fxgydx dy
x ry s|x − y|
0
∞ ∞
0
0
| ln x − ln y|t
fxdx gydy
x ry s|x − y|
∞ ∞
0
0
∞
< K1
0
p 1/p ∞
1/q
| ln x − ln y|t
fxdx dy
g q xdx
x ry s|x − y|
0
f xdx
p
1/p ∞
3.18
1/q
g xdx
q
.
0
Then, by 3.13, we have 3.1. Hence inequalities 3.1 and 3.13 are equivalent.
p
If the constant factor K1 in 3.13 is not the best possible, then by 3.18 we can get a
contradiction that the constant factor K1 in 3.1 is not the best possible. Hence Proposition 3.2
is proved.
Remark 3.3. In 3.1, let r 1, s 0, t 0, p q 2, and we have Hilbert’s integral inequality
∞ ∞
0
0
fxgy
dx dy < π
xy
∞
f xdx
2
0
∞
1/2
g xdx
2
3.19
.
0
Let r 1, s 1, t 0, p q 2; we have Hardy-Hilbert’s classical inequality
∞ ∞
0
0
∞
∞
1/2
fxgy
dx dy < 2
f 2 xdx g 2 xdx
.
2 max{x, y}
0
0
3.20
Let r 1, s 1/3, t 0, p q 2, and we can combine the above two inequalities as follows:
∞ ∞
0
0
fxgy
dx dy < K1
2/3 x y max{x, y}
∞
0
f 2 xdx
∞
0
1/2
g 2 xdx
.
3.21
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International Journal of Mathematics and Mathematical Sciences
Let r 1, s −1/3, t 0, and we can get the following inequality:
∞ ∞
0
0
fxgy
dx dy < K1
2/3 x y min{x, y}
∞
f xdx
2
∞
0
1/2
g xdx
2
.
3.22
0
4. Discrete case
We also give results for the discrete case.
Proposition 4.1. Assume that r > 0, s ∈ − min{1, r}, min{1, r}, p, q > 1 and 1/p 1/q 1. If
am , bn 0, such that
0<
∞
p
am < ∞,
0<
m1
∞
q
bn < ∞,
4.1
n1
then one has
∞
∞ am bn
< K1
m rn s|m − n|
n1 m1
∞ ∞
n1
∞
p
am
1/p ∞
m1
am
m
rn
s|m − n|
m1
1/q
q
bn
,
4.2
n1
p
∞
p
< K1
p
am ,
4.3
m1
where the constant factor
K1 ∞
0
1
u−1/q du
1 ru s|1 − u|
4.4
is the best possible, and inequalities 4.2 and 4.3 are equivalent.
Proof. Define the following weight functions:
1/q
1
m
,
m
rn
s|m
−
n|
n
n1
m 1, 2, . . . ,
1/p
1
n
2 n ,
m rn s|m − n| m
m1
n 1, 2, . . . ,
1 m ∞
∞
4.5
then,
1 m < ω1 m K1 ,
2 n < ω2 n K1 ,
4.6
Liubin Hua et al.
11
where ω1 , ω2 are defined in Lemma 2.1. By Hölder’s inequality, we have
∞
∞ am bn
m
rn
s|m − n|
n1 m1
≤
∞
∞ 1
m
rn
s|m − n|
n1 m1
m
n
1/pq
am
n
m
1/pq
bn
∞
∞ 1/q 1/p 1/p 1/q
∞
∞ 1
1
m
n
p
q
am
bn
m rn s|m − n| n
m rn s|m − n| m
n1 m1
n1 m1
∞
p
1 mam
1/p ∞
m1
< K1
1/q
q
2 nbn
n1
1/p 1/q
∞
∞
p
q
am
bn
.
m1
n1
4.7
The last strict inequality holds because both series {am } and {bn } have positive terms. Thus,
we have 4.2.
If the constant factor K1 in 4.2 is not the best possible, then there exists a positive
constant H with H < K1 , such that 4.2 is still valid if we replace K1 by H. For ε > 0 small
enough, construct series
bn n−1ε/q .
am m−1ε/p ,
4.8
Then, we have
am bn
>
m rn s|m − n|
n1 m1
∞
∞ ∞
m1
p
am
1/p ∞
n1
q
bn
∞ ∞
0
0
fε xgε ydx dy 1 K1 o1 O1,
x ry s|x − y|
ε
1/q
∞
−1−ε
n
<1
n1
∞
1
1
x−1−ε dx 1 ,
ε
4.9
4.10
where fε x and gε x are defined in the proof of Proposition 3.1. Since
1
1
K1 o1 O1 < H 1 ,
ε
ε
4.11
K1 o1 < H.
4.12
it follows that
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International Journal of Mathematics and Mathematical Sciences
Hence,
K1 H.
4.13
This contradicts with the fact that H < K1 . So the constant factor K1 in 4.2 is the best
possible one.
Setting
∞
am
bn m
rn
s|m − n|
m1
p−1
,
n 1, 2, . . . ,
4.14
we find
∞ ∞
n1
am
m
rn
s|m − n|
m1
p
∞
n1
q
bn ∞
∞ am bn
.
m
rn
s|m − n|
n1 m1
4.15
By the same argument used in the proof of Proposition 3.2, we can show that 4.3 is valid,
p
the constant factor K1 in 4.3 is the best possible one and inequalities 4.2 and 4.3 are
equivalent.
Remark 4.2. Proposition 4.1 is the corresponding series form of Propositions 3.1 and 3.2 for
s ∈ − min{1, r}, min{1, r}, t 0, and it is also a generalization of Hilbert’s inequality. Here,
we restrict the constants s, t so that we can use the monotony of functions to obtain 4.6 and
4.9.
Acknowledgment
This work was partially supported by the National Natural Science Foundation of China
Grant no. 10871213.
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