Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2008, Article ID 381650, 8 pages
doi:10.1155/2008/381650
Research Article
On a Generalization of Hilbert-Type
Integral Inequality
Sun Baoju
Basic Courses Department, Zhejiang Water Conservancy & Hydropower College,
Hangzhou 310018, China
Correspondence should be addressed to Sun Baoju, sunbj@mail.zjwchc.com
Received 1 March 2008; Accepted 9 May 2008
Recommended by Feng Qi
By introducing some parameters, we establish generalizations of the Hilbert-type inequality. As
applications, the reverse and its equivalent form are considered.
Copyright q 2008 Sun Baoju. 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
Considerable attention has been given to Hilbert inequalities and Hilbert-type inequalities by
several authors including Gao and Yang 1, Yang 2–4, Jichang and Debnath 5, Pachpatte
6, Zhao 7, Brnetić and Pečarić 8. In 2007, Li et al. 9 gave a new inequality similar to
Hilbert inequality for integrals:
∞
∞
If fx, gx ≥ 0, 0 < 0 f 2 xdx < ∞, 0 < 0 g 2 xdx < ∞, then one has
∞
0
| ln x − ln y|
fxgydx dy < 4
x y |x − y|
∞
0
f 2 xdx
∞
g 2 xdx
1/2
,
1.1
0
where constant factor 4 is the best possible.
An equivalent inequality is
∞ ∞
0
0
∞
2
| ln x − ln y|
fxdx dy < 16 f 2 xdx.
x y |x − y|
0
1.2
In this paper, by introducing some parameters we generalize 1.1, 1.2, and we obtain
the reverse form for each of them. The equivalent forms are also considered.
2
International Journal of Mathematics and Mathematical Sciences
2. Main results
Lemma 2.1. Suppose that λ > 0, p > 1 1/p 1/q 1, define weight functions wλ x, q, wλ y, p,
respectively as
∞
λ/q
| ln x − ln y|
x
1
dy,
1−λ
λ y λ x λ − y λ y
y
x
0
∞
λ/p
y
| ln x − ln y|
1
wλ y, p dx.
1−λ
λ
λ
λ
λ
x
x
0 x y x −y
wλ x, q 2.1
One has wλ x, q wλ y, p 1/2λ2 p2 q2 .
Proof. Letting t y λ /xλ , we have
1
wλ x, q 2
λ
∞
0
| ln t|
1
· t−1/q dt 2
1 t |1 − t|
λ
1
0
− ln t −1/q
dt ·t
2
∞
1
ln t −1/q
1 dt 2 p2 q2 .
·t
2t
2λ
2.2
By symmetry we have
wλ y, p 1 2
p q2 .
2
2λ
2.3
The lemma is proved.
Lemma 2.2. Let p > 1 or 0 < p < 1, 1/p 1/q 1, λ > 0, and 0 < ε < qλ/2p, setting
∞
| ln x − ln y|
x−1εp−11−λ/p y −1εq−11−λ/q dx dy.
Jε λ y λ xλ − y λ x
1
2.4
Then for ε→0 , one gets
p2 q2 p2 q2 1 o1 − O1 < Jε <
1 o1 .
2
2
2λ ε
2λ ε
2.5
Proof. Letting t y λ /xλ , we have
∞
1 ∞ 1
| ln t|
1/p−1−ε/λq
Jε 2
dt
dx
·
t
λ 1 x1ε 1/xλ 1 t |1 − t|
∞
| ln t|
1 ∞ 1
1/p−1−ε/λq
·
t
dt
dx
2
λ 1 x1ε 0 1 t |1 − t|
1/xλ
1 ∞ 1
| ln t|
1/p−1−ε/λq
− 2
dt
dx
·
t
1 t |1 − t|
λ 1 x1ε 0
1/xλ
| ln t|
| ln t|
1 ∞
1 ∞ 1
1/p−1−ε/λq
1/p−1−ε/λq
2
dt − 2
dt dx
·t
·t
1 t |1 − t|
λ ε 1 1 t |1 − t|
λ 1 x1ε 0
: I1 − I2 .
2.6
Sun Baoju
3
Now, observe that
∞
0
p2 q2 | ln t|
· t1/p−1−ε/λq dt 1 o1 , ε −→ 0 , 0 <
1 t |1 − t|
2
1/xλ
− ln t 1/p−1−ε/λq
·t
dt
2
0
1/xλ
− ln t 1/2p−1
·t
<
dt
2
0
1/xλ
0
| ln t|
· t1/p−1−ε/λq dt
1 t |1 − t|
pλx−λ/2p ln x 2p2 x−λ/2p ,
2.7
then
p2 q2 1 o1 ,
2λ2 ε
∞
∞
8p3
1
−1−λ/2p
2
−1−λ/2p
ln x dx 2p
x
dx 3 .
0 < I2 < 2 pλ x
λ
λ
1
1
I1 2.8
We get
Jε >
p2 q2 1 o1 − O1.
2λ2 ε
2.9
On the other hand,
Jε ∞
1
| ln x − ln y|
x−1εp−11−λ/p y −1εq−11−λ/q dx dy
xλ y λ xλ − y λ ∞ ∞
<
1
0
| ln x − ln y|
x−1εp−11−λ/p y −1εq−11−λ/q dy dx
xλ y λ xλ − y λ 2.10
p2 q2 1 o1 .
2λ2 ε
Hence, 2.5 is valid. The lemma is proved.
Theorem
2.3. Let p > 1, 1/p 1/q 1, λ > 0, fx ≥ 0, gy ≥ 0. If 0 <
∞
0 < 0 y q−11−λ g q ydy < ∞, then one has
∞
0
∞
0
xp−11−λ f p xdx < ∞,
| ln x − ln y|
fxgydx dy
x λ y λ x λ − y λ 1/p ∞
1/q
∞
p2 q2
p−11−λ p
q−11−λ q
<
x
f
xdx
y
g
ydy
,
2λ2
0
0
2.11
4
International Journal of Mathematics and Mathematical Sciences
where constant factor p2 q2 /2λ2 is the best possible. In particular, for λ 1, inequality 2.11
reduces to
∞
0
| ln x − ln y|
p2 q2
fxgydx dy <
x y |x − y|
2
∞
f p xdx
1/p ∞
0
g q ydy
1/q
.
2.12
0
Proof. Applying Hölder’s inequality and Lemma 2.1, we have
∞
0
xλ
| ln x − ln y|
fxgydx dy
y λ xλ − y λ ∞ 0
| ln x − ln y|
λ
x y λ x λ − y λ ×
≤
∞
0
×
1/p λ/pq 1−λ/q
x
x
fx
y
y 1−λ/p
| ln x − ln y|
xλ y λ xλ − y λ 1/q λ/pq 1−λ/p
y
y
fy
dx dy
x
x1−λ/q
λ/q p/q1−λ
1/p
| ln x − ln y|
x
x
p
f xdx dy
y 1−λ
xλ y λ xλ − y λ y
∞
0
p2 q2
≤
2λ2
2.13
λ/p q/p1−λ
1/q
y
| ln x − ln y|
y
q
g
ydx
dy
x1−λ
x λ y λ x λ − y λ x
∞
x
p−11−λ p
f xdx
1/p ∞
y
0
q−11−λ q
g ydy
1/q
.
0
If 2.13 takes the form of equality, then there exist constants A and B, such that they are
not all zero, and
λ/q p/q1−λ
λ/p q/p1−λ
y
y
| ln x − ln y|
| ln x − ln y|
x
x
p
A λ
f x B λ
g q y,
y 1−λ
x1−λ
x y λ xλ − y λ y
x y λ xλ − y λ x
2.14
a.e. x, y in 0, ∞ × 0, ∞. It follows that there exists a constant C, such that
Ax · xp−11−λ f p x By · y q−11−λ g q y C,
a.e. x, y in 0, ∞ × 0, ∞.
2.15
Without lose of generality, suppose A /
0, then we have
xp−11−λ f p x 1
,
Ax
2.16
∞
which contradicts the fact that 0 < 0 xp−11−λ f p xdx < ∞, hence 2.13 takes the form of
strict inequality, so we obtain 2.11.
Sun Baoju
5
Assume that the constant factor p2 q2 /2λ2 in 2.11 is not the best possible, then there
exists a positive number k with k < p2 q2 /2λ2 such that 2.11 is still valid if one replaces
p2 q2 /2λ2 by k. In particular, for 0 < ε < qλ/2p, setting f and g as fx
g x 0 for
−1εp−11−λ/p
−1εq−11−λ/q
x ∈ 0, 1, fx x
, g x x
for x ∈ 1, ∞, then we have
∞
k
xp−11−λ fp xdx
1/p ∞
0
y q−11−λ g q ydy
1/q
0
k
.
ε
2.17
By using Lemma 2.2, we find
∞
0
| ln x − ln y|
g ydx dy
fx
xλ y λ xλ − y λ ∞
1
>
| ln x − ln y|
x−1εp−11−λ/p y −1εq−11−λ/q dx dy
x λ y λ x λ − y λ 2.18
p2 q2 1 o1 − O1.
2λ2 ε
Therefore, we get
p2 q2 k
1 o1 − O1 <
2
ε
2λ ε
2.19
p2 q2 1 o1 − εO1 < k.
2
2λ
2.20
or
For ε→0 , it follows that p2 q2 /2λ2 ≤ k. This contradicts the fact that k < p2 q2 /2λ2 .
Hence, the constant factor in 2.11 is the best possible. Theorem 2.3 is proved.
∞
Theorem
Let 0 < p < 1, 1/p1/q 1, λ > 0, fx ≥ 0, gy ≥ 0. If 0 < 0 xp−11−λ f p xdx <
∞ 2.4.
∞, 0 < 0 y q−11−λ g q ydy < ∞, then one has
∞
0
| ln x − ln y|
fxgydx dy
x λ y λ x λ − y λ p2 q2
>
2λ2
∞
xp−11−λ f p xdx
1/p ∞
0
y q−11−λ g q ydy
2.21
1/q
,
0
where the constant factor p2 q2 /2λ2 is the best possible. In particular, for λ 1, the inequality
reduces to
∞
0
| ln x − ln y|
p2 q2
fxgydx dy >
x y |x − y|
2
∞
0
f p xdx
1/p ∞
0
g q ydy
1/q
.
2.22
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International Journal of Mathematics and Mathematical Sciences
Proof. Applying reverse Hölder’s inequality and the same arguments as before, we have 2.21.
If the constant factor p2 q2 /2λ2 in 2.21 is not the best possible, then there exists
a positive number h with h > p2 q2 /2λ2 , such that 2.21 is still valid if one replaces
p2 q2 /2λ2 by h. In particular, for 0 < ε < qλ/2p, setting f and g as in Theorem 2.3, we have
1/p ∞
1/q
∞
h
p−11−λ p
f xdx
x
y q−11−λ g q ydy
.
h
ε
0
0
2.23
By using Lemma 2.2, we find
∞
| ln x − ln y|
g ydx dy
fx
λ y λ xλ − y λ x
0
∞
1
<
xλ
| ln x − ln y|
x−1εp−11−λ/p y −1εq−11−λ/q dx dy
y λ x λ − y λ 2.24
p2 q2 I o1 .
2
2λ ε
Therefore, we get
h
p2 q2 1 o1 >
2
ε
2λ ε
or
p2 q2 1 o1 > h
2
2λ
2.25
for ε→0 , and it follows that p2 q2 /2λ2 ≥ h. This contradicts the fact that h > p2 q2 /2λ2 .
Hence, the constant factor in 2.21 is the best possible. Theorem 2.4 is proved.
∞
Theorem 2.5. If p > 1, 1/p 1/q 1, λ > 0, fx ≥ 0, 0 < 0 xp−11−λ f p xdx < ∞, then one has
∞
y
λ−1
∞
0
0
p
2
| ln x − ln y|
p q2 p ∞ p−11−λ p
fxdx dy <
x
f xdx,
2λ2
xλ y λ xλ − y λ 0
2.26
where the constant factorp2 q2 /2λ2 p is the best possible. Inequality 2.26 is equivalent to 2.11.
Proof. Setting
gy y λ−1
∞
0
| ln x − ln y|
fxdx
x λ y λ x λ − y λ p−1
,
2.27
then by 2.11, we find
∞
y q−11−λ g q ydy
0
∞
∞
p
| ln x − ln y|
| ln x − ln y|
fxdx dy fxgydx dy
λ
λ
λ
λ
λ
λ
λ
λ
0
0 x y x −y
0 x y x −y
∞
∞
1/p ∞
p 1/q
p2 q2
| ln x − ln y|
p−11−λ p
λ−1
≤
x
f
xdx
y
dy
.
fxdx
λ
λ
λ
λ
2λ2
0
0
0 x y |x − y |
2.28
∞
y λ−1
Sun Baoju
7
Hence, we obtain
∞
y q−11−λ g q ydy ≤
0
p2 q2
2λ2
p ∞
xp−11−λ f p xdx,
2.29
0
Thus, by 2.11, both 2.28 and 2.29 keep the form of strict inequalities, then we have 2.26.
Applying Hölder’s inequality, we have
∞
0
| ln x − ln y|
fxgydx dy
x λ y λ x λ − y λ ∞
y λ−1/p
∞
0
≤
0
∞
y
λ−1
0
∞
0
| ln x − ln y|
fxdx y 1−λ/p gy dy
λ
λ
λ
λ
x y x − y p 1/p ∞
1/q
| ln x − ln y|
q−11−λ
q
fxdx dy
×
y
g ydy
.
xλ y λ xλ − y λ 0
2.30
Therefore, by 2.26 we have 2.11. It follows that inequality 2.26 is equivalent to 2.11, and
the constant factors in 2.26 are the best possible. The theorem is proved.
∞
Theorem 2.6. If 0 < p < 1, 1/p 1/q 1, λ > 0, fx ≥ 0, 0 < 0 xp−11−λ f p xdx < ∞, then one
has
∞
2
∞
p
p q2 p ∞ p−11−λ p
| ln x − ln y|
fxdx dy >
y λ−1
x
f xdx,
2.31
λ
λ
λ
λ
2λ2
0
0 x y x −y
0
where the constant factor p2 q2 /2λ2 p is the best possible. Inequality 2.31 is equivalent to 2.21.
The proof of Theorem 2.6 is similar to that of Theorem 2.5, so we omit it.
Acknowledgment
The author would like to thank the anonymous referees for their suggestions and corrections.
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8
International Journal of Mathematics and Mathematical Sciences
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