Journal of Inequalities in Pure and
Applied Mathematics
ON HARDY-HILBERT INTEGRAL INEQUALITIES WITH
SOME PARAMETERS
volume 6, issue 4, article 92,
2005.
YONG HONG
Department of Mathematics
Guangdong Business College
Guangzhou 510320
People’s Republic of China
Received 07 April, 2005;
accepted 06 June, 2005.
Communicated by: B. Yang
EMail: hongyong59@sohu.com
Abstract
Contents
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
In this paper, we give a new Hardy-Hilbert’s integral inequality with some parameters and a best constant factor. It includes an overwhelming majority of
results of many papers.
2000 Mathematics Subject Classification: 26D15.
Key words: Hardy-Hilbert’s integral inequality, Weight, Parameter, Best constant factor, β-function, Γ-function.
Contents
1
Introduction and Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Weight Function and Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3
Proofs of the Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4
Some Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
References
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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1.
Let
R∞
0
Introduction and Main Result
R∞
+ 1q = 1(p > 1), f ≥ 0, g ≥ 0, 0 < 0 f p (x)dx < +∞, 0 <
g (x)dx < +∞, then we have the well known Hardy-Hilbert inequality
1
p
q
∞
Z
∞
Z
f (x)g(x)
dxdy
x+y
(1.1)
0
0
<
sin
π
π
p
p1 Z
∞
Z
p
1q
∞
q
g (x)dx
f (x)dx
0
;
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
0
and an equivalent form as:
∞
Z
Z
(1.2)
0
0
∞
f (x)
dx
x+y
Yong Hong
p
p

dy < 
sin
π 
π
p
∞
Z
f p (x)dx.
Title Page
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Contents
In recent years, many results have been obtained in the research of these two
inequalities (see [1] – [13]). Yang [1] and [2] gave:
Z
∞
Z
(1.3)
0
0
∞
f (x)g(y)
dxdy
(x + y)λ
p+λ−2 p+λ−2
,
<B
p
q
Z ∞
p1 Z
1−λ p
×
x f (x)dx
0
0
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∞
x
1−λ q
g (x)dx
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1q
,
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where B(r, s) is the β-function; and Kuang [3] gave:
Z
∞
Z
∞
f (x)g(x)
dxdy
xλ + y λ
0
0
Z ∞
p1 Z ∞
1q
π
1−λ p
1−λ q
1 <
x f (x)dx
x g (x)dx .
1
π
π
0
0
λ sin p pλ
sin q qλ
(1.4)
Recently, Hong [4] gave:
Z
∞
(1.5)
0
Z
∞
f (x)g(x)
p
dxdy
x2 + y 2
0
Z ∞
p1 Z ∞
1q
1
1
1
p
q
≤ √ Γ
Γ
f (x)dx
g (x)dx .
2p
2q
2 π
0
0
∞
Z
Yong Hong
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And Yang [5] gave:
Z
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
∞
f (x)g(x)
dxdy
xλ + y λ
0
0
Z ∞
p1 Z ∞
1q
π
(p−1)(1−λ) p
(q−1)(1−λ) q
<
x
f (x)dx
x
g (x)dx ;
0
0
λ sin πp
(1.6)
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Z
(1.7)
0
∞
y λ−1
Z
0
∞
f (x)
dx
xλ + y λ
p
dy
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p

<
π

λ sin πp
Z
∞
x(p−1)(1−λ) f p (x)dx.
0
These results generalize and improve (1.1) and (1.2) in a certain degree.
In this paper, by introducing a few parameters, we obtain a new HardyHilbert integral inequality with a best constant factor, which is a more extended
inequality, and includes all the results above and the overwhelming majority of
results of many recent papers.
Our main result is as follows:
Theorem 1.1. If p1 + 1q = 1 (p > 1), α > 0, λ > 0, m, n ∈ R, such that
0 < 1 − mp < αλ, 0 < 1 − nq < αλ, and f ≥ 0, g ≥ 0, satisfy
Z ∞
(1.8)
0<
x(1−αλ)+p(n−m) f p (x)dx < ∞,
0
Z
0<
(1.9)
y
(1−αλ)+q(m−n) q
g (y)dy < ∞,
(1.10)
0
Z
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then
∞
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∞
0
Z
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
∞
f (x)g(x)
dxdy
(xα + y α )λ
0
Z
< Hλ,α (m, n, p, q)
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∞
x(1−αλ)+p(n−m) f p (x)dx
Close
p1
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0
Z
×
∞
y
0
(1−αλ)+q(m−n) q
g (y)dy
Page 5 of 22
1q
;
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and
Z
∞
y
(1.11)
Z
(1−αλ)+q(m−n)
1−q
0
∞
0
f (x)
dx
(xα + y α )λ
p
dy
Z
∞
x(1−αλ)+p(n−m) f p (x)dx,
e λ,α (m, n, p, q)
<H
0
where
1 1
Hλ,α (m, n, p, q) = B p
α
1 − mp
1 − mp
,λ −
α
α
B
1
q
1 − nq
1 − nq
,λ −
α
α
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
and
e λ,α (m, n, p, q) = 1 B
H
αp
1 − mp
1 − mp
,λ −
α
α
B
p−1
1 − nq
1 − nq
,λ −
α
α
.
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Theorem 1.2. If p > 1, p1 + 1q = 1, α > 0, λ > 0, m, n ∈ R, such that
0 < 1 − mp < αλ, mp + nq = 2 − αλ, and f (x) ≥ 0, g(y) ≥ 0, satisfy
Z ∞
(1.12)
0<
xn(p+q)−1 f p (x)dx < ∞,
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Z
(1.13)
0<
0
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∞
y m(p+q)−1 g q (y)dy < ∞,
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then
Z
∞
∞
Z
(1.14)
0
0
f (x)g(x)
1
1 − mp
1 − mp
dxdy < B
,λ −
(xα + y α )λ
α
α
α
Z ∞
p1 Z ∞
1q
n(p+q)−1 p
m(p+q)−1 q
×
x
f (x)dx
y
g (y)dy ;
0
0
p
f (x)
(1.15)
dx dy
y
(xα + y α )λ
0
0
Z ∞
1 p 1 − mp
1 − mp
,λ −
xn(p+q)−1 f p (x)dx,
< pB
α
α
α
0
1−mp
where the constant factors α1 B 1−mp
in (1.14) and
,
λ
−
α
α
1−mp
1
p 1−mp
B
,λ − α
in (1.15) are the best possible.
αp
α
Z
∞
m(p+q)−1
1−q
Z
∞
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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2.
Weight Function and Lemmas
The weight function is defined as follows
Z ∞
1
yn
ωλ,α (m, n, y) =
·
dx,
(xα + y α )λ xm
0
y ∈ (0, +∞).
Lemma 2.1. If α > 0, λ > 0, m ∈ R, 0 < 1 − m < αλ, then
1−m
1−m
1 (1−αλ)+(n−m)
(2.1)
ωλ,α (m, n, y) = y
B
,λ −
.
α
α
α
Proof. Setting t =
xα
,
yα
then
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
Z
1−m
1 ∞
1
ωλ,α (m, n, y) =
y (1−αλ)+(n−m) t α −1 dt
λ
α 0 (1 + t)
Z
1−m
1
1 (1−αλ)+(n−m) ∞
= y
t α −1 dt
λ
α
(1 + t)
0
1 (1−αλ)+(n−m)
1−m
1−m
B
,λ −
.
= y
α
α
α
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Hence (2.1) is valid. The lemma is proved.
Close
Lemma 2.2. If α > 0, λ > 0, β < 1, a ∈ R, then
Z
(2.2)
1
∞
1
x1+ε
Z
0
1
xα
1−β
1
−1−aε
α
t
dtdx = O(1),
λ
(1 + t)
Quit
(ε → 0+ ).
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Proof. Since (1 − β)/α > 0, for ε small enough, such that
Z
1
∞
1
x1+ε
Z
0
1
xα
− aε > 0, then
Z 1
1 xα ( 1−β
t α −aε)−1 dtdx
x 0
1
Z ∞
1
=
xβ+aεα−2 dx
1 − β − aεα 1
1
=
.
(1 − β − aεα)2
1−β
1
−1−aε
α
t
dtdx <
λ
(1 + t)
Hence (2.2) is valid.
1−β
α
Z
∞
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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3.
Proofs of the Theorems
Proof of Theorem 1.1. By Hölder’s inequality and Lemma 2.1, we have
Z ∞Z ∞
f (x)g(y)
G=
dxdy
(xα + y α )λ
0
0
Z ∞Z ∞
f (x)
xn
g(y)
ym
=
dxdy
(xα + y α )λ/p y m (xα + y α )λ/q xn
0
0
Z ∞ Z ∞
1 Z ∞ Z ∞
1q
f p (x) xnp p
g q (x) y mq
(3.1) ≤
dxdy
,
(xα + y α )λ y mp
(xα + y α )λ xnq
0
0
0
0
according to the condition of taking equality in Hölder’s inequality, if (3.1) takes
equality, then there exists a constant C, such that
f p (x) xnp
g q (y) y mq
≡ C, a.e. (x, y) ∈ (0, +∞)×(0, +∞)
(xα + y α )λ y mp
(xα + y α )λ xnq
it follows that
p
f (x)x
hence
Z
0
n(p+q)
q
≡ Cg (y)y
m(p+q)
≡ C1 (constant), a.e. (x, y) ∈ (0, +∞)×(0, +∞)
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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∞
x(1−αλ)+p(n−m) f p (x)dx
Z ∞
=
x(1−αλ)+n(p+q)−nq−mp f p (x)dx
0
Z ∞
= C1
x(1−αλ)−np−mq dx
0
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Z
1
= C1
x
(1−αλ)−np−mq
∞
Z
x(1−αλ)−np−mq dx = +∞,
dx + C1
0
1
which contradicts (1.8) and (1.9). Hence, by (3.1), we have
p1
1
xnp
p
dy f (x)dx
G<
(xα + y α )λ y mp
0
0
Z ∞ Z ∞
1q
1
y mq
q
×
dy g (y)dy
(xα + y α )λ xnq
0
0
p1
Z ∞
1 − mp
1 − mp
1
(1−αλ)+p(n−m)
p
=
x
B
,λ −
f (x)dx
α 0
α
α
Z ∞
1q
1
1
−
nq
1
−
nq
×
y (1−αλ)+q(m−n) B
,λ −
g q (y)dy
α 0
α
α
Z ∞
p1
(1−αλ)+p(n−m) p
= Hλ,α (m, n, p, q)
x
f (x)dx
Z
∞
Z
∞
0
Z
×
∞
y (1−αλ)+q(m−n) g q (y)dy
1q
.
0
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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Hence (1.10) is vaild.
Let β = [(1 − αλ) + q(m − n)]/(1 − q), and
ge(y) = y β
Z
0
∞
f (x)
dx
+ y α )λ
(xα
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pq
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.
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By (1.10), we have
Z ∞
y (1−αλ)+q(m−n) geq (y)dy
0
Z ∞
=
y β(1−q) geq (y)dy
p
Z0 ∞ Z ∞
f (x)
β
=
y
dx dy
(xα + y α )λ
0
0
pq Z ∞
Z ∞ Z ∞
f (x)
f (x)
β
dx dy
=
y
dx
(xα + y α )λ
(xα + y α )λ
0
0
0
Z ∞Z ∞
f (x)e
g (y)
dxdy
=
α
(x + y α )λ
0
0
Z ∞
p1
(1−αλ)+p(n−m) p
< Hλ,α (m, n, p, q)
x
f (x)dx
∞
×
0
It follows that
Z
Z ∞
(1−αλ)+q(m−n)
1−q
y
0
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Z
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
1q
(1−αλ)+q(m−n) q
y
ge (y)dy| ,
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∞
0
f (x)
dx
+ y α )λ
p
Close
dy
(xα
Z
e λ,α (m, n, p, q)
<H
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∞
x
(1−αλ)+p(n−m) p
f (x)dx.
Page 12 of 22
0
Hence, (1.11) is valid.
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Proof of Theorem 1.2. Since 0 < 1 − mp < αλ, mp + nq = 2 − αλ, then
0 < 1 − nq < αλ,
(1 − αλ) + p(n − m) = n(p + q) − 1,
(1 − αλ) + q(m − n) = m(p + q) − 1,
1 − mp
1 − nq
=λ−
.
α
α
By Theorem 1.1, (1.14) and (1.15) are valid.
For ε > 0, setting
( [−n(p+q)−ε]/p
x
, x ≥ 1;
f0 (x) =
0,
0 ≤ x < 1,
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
Title Page
and
(
g0 (y) =
Contents
y [−m(p+q)−ε]/q , y ≥ 1;
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J
0 ≤ y < 1.
0,
We have
∞
Z
(3.2)
xn(p+q)−1 f0p (x)dx
0<
∞
Z
x−1−ε dx =
=
0
1
1
< ∞,
ε
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Z
(3.3)
0<
0
∞
y m(p+q)−1 g0q (y)dy =
Z
1
∞
y −1−ε dy =
1
< ∞.
ε
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Z
0
∞
∞
Z
f0 (x)g0 (y)
dxdy
(xα + y α )λ
0
Z ∞Z ∞
n(p+q)+ε
m(p+q)+ε
1
−
−
p
q
=
x
y
dxdy
(xα + y α )λ
1
1
Z ∞
Z ∞
n(p+q)+ε
m(p+q)+ε
1
−
−
p
q
=
y
x
dydx
(xα + y α )λ
1
1
Z
Z ∞
1−mp
ε
1
1 ∞ 1
−1− qα
α
t
dtdx
=
α 1 x1+ε x1α (1 + t)λ
Z ∞
Z ∞
1−mt
ε
1
1
1
−1− qα
α
=
t
dtdx
α 1 x1+ε 0 (1 + t)λ
#
Z 1α
Z ∞
x
1−mt
ε
1
1
t α −1− qα dtdx .
−
λ
1+ε
(1 + t)
x
0
1
By Lemma 2.2, when ε → 0, we have
Z
1
∞
1
x1+ε
Z
0
1
xα
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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1−mp
ε
1
−1− qα
α
t
dtdx = O(1).
(1 + t)λ
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Since
Z
0
∞
1−mp
ε
1
t α −1− qα dt = B
λ
(1 + t)
1 − mp
1 − mp
,λ −
α
α
Close
+ o(1),
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we have
Z ∞Z ∞
f0 (x)g0 (y)
1 − mp
1 1
1 − mp
dxdy =
B
,λ −
+ o(1) − O(1)
(xα + y α )λ
α ε
α
α
0
0
1 1
1 − mp
1 − mp
=
B
,λ −
− o(1)
ε α
α
α
1 − mp
1 − mp
1
(3.4)
=
B
,λ −
(1 − o(1)).
εα
α
α
1−mp
If the constant α1 B 1−mp
,
λ
−
in (1.14) is not the best possible, then
α
α
1−mp
1−mp
1
there exists a K < α B α , λ− α , such that (1.14) still is valid when
we replace α1 B 1−mp
by K. By (3.2), (3.3) and (3.4), we find
, λ − 1−mp
α
α
1
B
εα
1 − mp
1 − mp
,λ −
(1 − o(1))
α
α
Z ∞
p1 Z
n(p+q)−1 p
<K
x
f0 (x)dx
0
0
1−mp
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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∞
y m(p+q)−1 g0q (y)dy
1q
1
=K .
ε
For ε → 0+ , we have α1 B 1−mp
,λ − α
≤ K, which contradicts
α
the fact
1−mp
1−mp
1−mp
1−mp
1
1
that K < α B α , λ − α . It follows that α B α , λ − α in (1.14)
is the best possible.
Since (1.14) is equivalent to (1.15), then the constant α1p B p 1−mp
, λ − 1−mp
α
α
in (1.15) is the best possible. The theorem is proved.
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4.
Some Corollaries
When we take the appropriate parameters, many new inequalities can be obtained as follows:
Corollary 4.1. If p1 + 1q = 1 (p > 1), α > 0, λ > 0, f ≥ 0, g ≥ 0, and
x(1−αλ)(p−1)/p f (x) ∈ Lp (0, +∞), x(1−αλ)(q−1)/q g(x) ∈ Lq (0, +∞), then
Z
∞
Z
(4.1)
0
0
∞
f (x)g(y)
dxdy
(xα + y α )λ
p1
Γ λp Γ λq Z ∞
(1−αλ)(p−1) p
<
x
f (x)dx
αΓ(λ)
0
Z ∞
1q
(1−αλ)(q−1) q
×
x
g (x)dx ;
0
Z
∞
y
(4.2)
0
α−1
Z
0
∞
p
f (x)
dx dy
(xα + y α )λ
 p
Z ∞
Γ λp Γ λq


<
x(1−αλ)(p−1) f p (x)dx,
αΓ(λ)
0
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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.
h .
ip
where the constants Γ λp Γ λq
(αΓ(λ)) in (4.1) and Γ λp Γ λq
(αΓ(λ))
in (4.2) are the best possible.
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Proof. If we take m =
be obtained.
1
p
−
αλ
,
p2
n=
1
q
−
αλ
q2
in Theorem 1.2, (4.1) and (4.2) can
Corollary 4.2. If p1 + 1q = 1 (p > 1), λ > 0, f ≥ 0, g ≥ 0 and x(1−λ)(p−1)/p f (x) ∈
Lp (0, +∞), x(1−λ)(q−1)/q g(x) ∈ Lq (0, +∞), then
Z
∞
Z
∞
f (x)g(y)
dxdy
(x + y)λ
0 0 p1 Z ∞
1q
Γ λp Γ λq Z ∞
(1−λ)(p−1) p
(1−λ)(q−1) q
<
x
f (x)dx
x
g (x)dx ,
Γ(λ)
0
0
(4.3)
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
Z
∞
∞
Z
(4.4)
0
0
p
f (x)
dx dy
(x + y)λ
 p
Z ∞
Γ λp Γ λq


<
x(1−λ)(p−1) f p (x)dx,
Γ(λ)
0
.
ip
h .
where Γ λp Γ λq
Γ(λ) in (4.3) and Γ λp Γ λq
Γ(λ) in (4.4) are
the best possible.
Proof. If we take α = 1 in Corollary 4.1, (4.3) and (4.4) can be obtained.
Corollary 4.3. If
1
p
+
1
q
= 1 (p > 1), λ > 0, p + λ − 2 > 0, q + λ − 2 > 0,
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f ≥ 0, g ≥ 0, and x(1−λ)/p f (x) ∈ Lp (0, +∞), x(1−λ)/q g(x) ∈ Lq (0, +∞), then
Z ∞Z ∞
f (x)g(y)
(4.5)
dxdy
(x + y)λ
0
0
Z ∞
p1 Z ∞
1q
p+λ−2 q+λ−2
1−λ p
1−λ q
<B
,
x f (x)
x g (x)dx ,
p
q
0
0
p
f (x)
(4.6)
y
dx dy
(x + y)λ
0
0
Z ∞
p+λ−2 q+λ−2
p
<B
,
x1−λ f p (x)dx,
p
q
0
where B p+λ−2
in (4.5) and B p p+λ−2
in (4.6) are the best
, q+λ−2
, q+λ−2
p
q
p
q
possible.
Z
∞
1−λ
1−q
Z
∞
Proof. If we take α = 1, m = n =
obtained.
2−λ
pq
in Theorem 1.2, (4.5) and (4.6) can be
Corollary 4.4. If p1 + 1q = 1 (p > 1), α > 0, f ≥ 0, g ≥ 0, and x(1−α)/p f (x) ∈
Lp (0, +∞), x(1−α)/q g(x) ∈ Lq (0, +∞), then
Z ∞Z ∞
f (x)g(x)
(4.7)
dxdy
xα + y α
0
0
Z ∞
p Z ∞
q
π
1−α p
1−α q
1 <
x f (x)dx
x g (x)dx ,
1
π
π
0
0
α sin p pα
sin q qα
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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Z
∞
y
(4.8)
1−α
1−q
Z
0
∞
0
f (x)
dx
α
x + yα

<
p
α sin
Proof. If we take λ = 1, m = n =
obtained.
dy
p
1
p
π
π
pα
1
,
pq
Z
sin
1
q

π
qα
∞
x1−α f p (x)dx.
0
in Theorem 1.1, (4.7) and (4.8) can be
Corollary 4.5. If p1 + 1q = 1 (p > 1), α > 0, f ≥ 0, g ≥ 0, and f (x) ∈
Lp (0, +∞), g(x) ∈ Lq (0, +∞), then
Z ∞Z ∞
f (x)g(x)
(4.9)
1 dxdy
(xα + y α ) α
0
0
1
1
Z ∞
p1 Z ∞
1q
Γ pα
Γ qα
<
f p (x)dx
g q (x)dx ,
αΓ α1
0
0
Z
∞
Z
(4.10)
0
0
∞
!p
f (x)
(xα
1
+ yα) α
dx
dy
 p
1
1
Z ∞
Γ pα
Γ qα


<
f p (x)dx,
αΓ α1
0
.
h .
ip
1
1
1
1
where Γ pα
Γ qα
αΓ α1 in (4.9) and Γ pα
Γ qα
αΓ α1
in (4.10) are the best possible.
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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Proof. If we take λ = α1 , m = n =
obtained.
1
pq
in Theorem 1.2, (4.9) and (4.10) can be
Remark 1. (4.1) and (4.2) are respectively generalizations of (1.6) and (1.7).
For α = 1 in (4.1) and (4.2), (1.6) and (1.7) can be obtained.
Remark 2. (4.3) and (4.4) are respectively generalizations of (1.1) and (1.2) .
Remark 3. (4.5) is the result of [1] and [2]. (4.6) is a new inequality.
Remark 4. (4.7) is the result of [3]. (1.7) is a new inequality.
Remark 5. (4.9) is a generalization of (1.5). (4.10) is a new inequality.
For other appropriate values of parameters taken in Theorems 1.1 and 1.2,
many new inequalities and the inequalities of [6] – [13] can yet be obtained.
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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References
[1] BICHENG YANG, A generalized Hardy-Hilbert’s inequality with a best
constant factor, Chin. Ann. of Math. (China), 21A(2000), 401–408.
[2] BICHENG YANG, On Hardy-Hilbert’s intrgral inequality, J. Math. Anal.
Appl., 261 (2001), 295–306.
[3] JICHANG KUANG, On new extensions of Hilbert’s integtal inequality,
Math. Anal. Appl., 235 (1999), 608–614.
[4] YONG HONG, All-sided Generalization about Hardy-Hilbert’s integral
inequalities, Acta Math. Sinica (China), 44 (2001), 619–626.
[5] BICHENG YANG, On a generalization of Hardy-Hilbert’s inequality,
Ann. of Math. (China), 23 (2002), 247–254.
[6] BICHENG YANG, On a generalization of Hardy-Hilbert’s integral ineguality, Acta Math. Sinica (China), 41(4) (1998), 839–844.
[7] KE HU, On Hilbert’s inequality, Ann. of Math. (China), 13B (1992), 35–
39
[8] KE HU, On Hilbert’s inequality and it’s application, Adv. in Math.
(China), 22 (1993), 160–163.
[9] BICHENG YANG AND MINGZHE GAO, On a best value of HardyHilbert’s inequality, Adv. in Math. (China), 26 (1999), 159–164.
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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[10] MINGZHE GAO, An improvement of Hardy-Riesz’s extension of the
Hilbert inequality, J. Mathematical Research and Exposition, (China) 14
(1994), 255–359.
[11] MINGZHE GAO AND BICHENG YANG, On the extended Hilbert’s inequality, Proc. Amer. Math. Soc., 126 (1998), 751–759.
[12] BICHENG YANG AND L. DEBNATH, On a new strengtheaed HardyHilbert’s intequality, Internat. J. Math. & Math. Sci., 21 (1998): 403–
408.
[13] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J. Math. Anal. & Appl., 226 (1998), 166–179.
On Hardy-Hilbert Integral
Inequalities with Some
Parameters
Yong Hong
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