Journal of Inequalities in Pure and
Applied Mathematics
SOME EXTENSIONS OF HILBERT’S TYPE INEQUALITY
AND ITS APPLICATIONS
volume 7, issue 2, article 61,
2006.
YONGJIN LI AND BING HE
Department of Mathematics
Sun Yat-sen University
Guangzhou, 510275 P. R. China.
Received 28 November, 2005;
accepted 19 January, 2006.
Communicated by: B. Yang
EMail: stslyj@mail.sysu.edu.cn
EMail: hzs314@163.com
Abstract
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Victoria University
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Abstract
By introducing parameters λ and µ, we give a generalization of the Hilbert’s
type integral inequality. As applications, we give its equivalent form.
2000 Mathematics Subject Classification: 26D15.
Key words: Hilbert’s integral inequality, Weight function.
Some Extensions of Hilbert’s
Type Inequality and its
Applications
The authors are grateful to the referee for the careful reading of the manuscript and
for several useful remarks. The work was partially supported by the Foundation of
Sun Yat-sen University Advanced Research Centre.
Yongjin Li and Bing He
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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1.
Introduction
1
p
If f, g ≥ 0, p > 1,
∞, then
∞
Z
∞
Z
1
q
= 1, 0 <
0
π
<
sin(π/p)
∞
Z
Z
∞
(1.2)
0
R∞
f p (x)dx < ∞ and 0 <
0
R∞
0
g q (x)dx <
f (x)g(y)
dxdy
x+y
(1.1)
0
+
0
f (x)
dx
x+y
p
p1 Z
∞
Z
p
1q
∞
q
f (x)dx
g (x)dx
0
,
Some Extensions of Hilbert’s
Type Inequality and its
Applications
0
π
dy <
sin(π/p)
p Z
∞
Yongjin Li and Bing He
f p (x)dx,
0
π
where the constant factor sin(π/p)
is the best possible. Inequality (1.1) is known
as Hardy-Hilbert’s integral inequality (see [1]); it is important in analysis and
its applications (see [4]). Under the same condition of (1.1), we have the HardyHilbert type inequality similar to (1.1):
Z
∞
∞
Z
(1.3)
0
0
f (x)g(y)
dxdy < pq
max{x, y}
Z
∞
f p (x)dx
p1 Z
0
∞
g q (x)dx
0
1q
,
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Z
∞
Z
(1.4)
0
0
∞
f (x)
dx
max{x, y}
p
dy < (pq)p
Z
∞
f p (x)dx,
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J. Ineq. Pure and Appl. Math. 7(2) Art. 61, 2006
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where the constant factor pq is the best possible. The corresponding inequality
for series is:
! p1 ∞ ! 1q
∞ X
∞
∞
X
X
X
am b n
(1.5)
< pq
apn
bqn
,
max{m,
n}
n=1 m=1
n=1
n=1
where the constant factor pq is also the best possible. In particular, when p =
q = 2, we have the well-known Hilbert type inequality:
12
Z ∞
12 Z ∞
Z ∞Z ∞
f (x)g(y)
(1.6)
g 2 (x)dx .
dxdy < 4
f 2 (x)dx
max{x, y}
0
0
0
0
In recent years, Kuang (see [3]) gave a strengthened form as:
(1.7)
∞ X
∞
X
am b n
max{m, n}
n=1 m=1
(∞
) p1 ( ∞
) 1q
X
X
[pq − G(q, n)]bqn
,
<
[pq − G(p, n)]apn
n=1
n=1
where G(r, n) = r+1/3r−4/3
> 0 (r = p, q).
(2n+1)1/r
Yang (see [5, 8]) gave: for λ > 2 − min{p, q}
Z ∞Z ∞
f (x)g(y)
pqλ
dxdy <
(1.8)
λ
λ
max{x , y }
(p + λ − 2)(q + λ − 2)
0
0
Z ∞
p1 Z ∞
1q
(p−1)(2−λ)−1 p
(q−1)(2−λ)−1 q
×
x
f (x)dx
x
g (x)dx
0
0
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
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and
∞
Z
Z
(1.9)
0
0
∞
f (x)g(y)
pqλ
dxdy
<
max{xλ , y λ }
(p + λ − 2)(q + λ − 2)
Z ∞
p1 Z ∞
1q
1−λ p
1−λ q
x f (x)dx
×
x g (x)dx .
0
0
At the same time, Yang (see [6, 7]) considered the refinement of other types of
Hilbert’s inequalities.
In this paper, we give a generalization of Hilbert’s type inequality and an
improvement as:
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
Z
0
∞
Z
0
∞
f (xλ )g(y µ )
dxdy
max{x, y}
1q
Z ∞
p1 Z ∞
1
1
pq
−1 q
−1 p
µ
λ
< 1 1
x
f (x)dx
x
g (x)dx ,
0
0
λp µq
where λ > 0 and µ > 0.
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2.
Main Results
Lemma 2.1. Suppose r > 1, 1s +
x
(2.1)
= 1, λ, µ, ε > 0. Then
1
∞
Z
1
r
µ
−ε( 1s + λr
)−1
x− λ
Z
1
−1−µε
1
t r dtdx = O(1)
max{1, t}
0
Proof. There exist n ∈ N which is large enough, such that 1 +
1
ε ∈ (0, µn
] and x ≥ 1, we have
1
Z
0
x− λ
(ε → o+ ).
−1−µε
r
1
−1−µε
1
t r dt =
max{1, t}
Z
x− λ
t
1+
−1−µε
r
x
1
−λ
dt =
0
Since for a ≥ 1 the function g(y) =
1
−1−µε
r
1+ −1−µε
r
1
yay
≤
1
1+
−1−µε
r
x
1
−λ
1
1+
−1−1/n
r
x
1
−λ
1+ −1−1/n
r
1
∞
1
µ
x−ε( s + λr )−1
0<
1
Z
<
0
∞
x
1
−1
1
1
=
λ
x− λ
Z
1+
1
1+
−1−1/n
r
!2
−1−1/n
r
1+ −1−µε
r
.
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
(y ∈ (0, ∞)) is decreasing, we find
so
Z
> 0 for
−1−µ
1
t r dtdx
max{1, t}
1 1+ −1−1/n
r
x− λ
.
Hence relation (2.1) is valid. The lemma is proved.
,
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Now we study the following inequality:
Theorem 2.2. Suppose f (x), g(x) ≥ 0, p > 1, p1 + 1q = 1, λ > 0, µ > 0 and
Z ∞
Z ∞
1
1
−1
p
0<
x λ f (x)dx < ∞, 0 <
x µ −1 g q (x)dx < ∞.
0
0
∞
f (xλ )g(y µ )
dxdy
max{x, y}
1q
p1 Z ∞
Z ∞
1
1
pq
−1
−1
p
q
µ
< 1 1
x λ f (x)dx
x
g (x)dx ,
0
0
λp µq
Then
Z
∞
Z
(2.2)
0
0
where the constant factor
pq
1
1
λp µq
is the best possible for λ = µ.
Proof. By Hölder’s inequality, we have
Z ∞Z ∞
f (xλ )g(y µ )
dxdy
max{x, y}
0
0
ih
i
h
Z ∞ Z ∞ x λ1 −1 f (x) y µ1 −1 g(y)
11
n 1 1o
=
dxdy
λµ 0
0
max x λ , y µ

! p1
Z ∞Z ∞
1
1
−1
(1− λ
)p/q
λ
11
x
f
(x)
x

=

1
n 1 1 o p1
λµ 0
0
y 1− µ
max x λ , y µ
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
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x
y
1
λ
1
µ
! pq1

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

J. Ineq. Pure and Appl. Math. 7(2) Art. 61, 2006
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

× 
y
1
−1
µ
g(y)
n 1 1 o 1q
max x λ , y µ
y
x
! p1
1
(1− µ
)
1
(1− λ
)p/q
y
x

Z Z
1
1 1  ∞ ∞ p( 1 −1)
f p (x)
x(1− λ )p/q
λ
n 1 1o
x
≤
1
λµ 0
0
y 1− µ
max x λ , y µ

Z

×
∞
0
Z
∞
1
y q( µ −1)
0
1
µ
! pq1

 dxdy
1
λ
x
y
1
λ
g (y)
y
n 1 1o
1
x1− λ
max x λ , y µ
 p1
! 1q
dxdy 
1
µ
1
(1− µ
)q/p
q

y
1
µ
1
xλ
! p1
 1q
dxdy  .
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
Define the weight function ϕ(x), ψ(y) as
Z
∞
n 1 1o ·
max x λ , y µ
ϕ(x) :=
0
Z
ψ(y) :=
0
1
∞
1
n 1 1o ·
max x λ , y µ
1
x(1− λ )p/q
y
1
1− µ
1
y (1− µ )q/p
1
x1− λ
1
xλ
y
! 1q
dy,
1
µ
1
yµ
1
xλ
x ∈ (0, ∞),
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! p1
dx,
y ∈ (0, ∞),
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then above inequality yields
Z ∞Z ∞
f (xλ )g(y µ )
dxdy
max{x, y}
0
0
Z ∞
p1 Z ∞
1q
1
1
11
−1) q
q( µ
p( λ
−1) p
≤
ϕ(x)x
f (x)dx
g (y)dy .
ψ(y)y
λµ 0
0
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1
1
For fixed x, let y µ = x λ t, we have
ϕ(x) := µx
1
(p−1)(1− λ
)
∞
Z
0
1
1
t p −1 dt
max{1, t}
1
(p−1)(1− λ
)
= µpqx
1
= µpqx(p−1)(1− λ ) .
1
By the same token, ψ(y) = λpqy (q−1)(1− µ ) , thus
Z ∞Z ∞
f (xλ )g(y µ )
(2.3)
dxdy
max{x, y}
0
0
1q
Z ∞
p1 Z ∞
1
1
pq
−1
−1
p
q
≤ 1 1
x µ g (x)dx .
x λ f (x)dx
0
0
λp µq
If (2.3) takes the form of the equality, then there exist constants c and d, such
that Kuang (see [2])
! 1q
! p1
1
1
1
1
1
1
xp( λ −1) f p (x) x(1− λ )p/q x λ
y q( µ −1) g q (y) y (1− µ )q/p y µ
n 1 1o ·
c
·
=d
1
1
1
1
1
1
xλ
x1− λ
max{x λ , y µ }
y 1− µ
yµ
max x λ , y µ
a.e. on (0, ∞) × (0, ∞).
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
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Then we have
1
1
cx λ f p (x) = dy µ g q (y)
a.e. on (0, ∞) × (0, ∞).
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Hence we have
1
1
cx λ f p (x) = dy µ g q (y) = constant
a.e. on (0, ∞) × (0, ∞),
which contradicts the facts that
Z ∞
1
0<
x λ f p (x)dx < ∞
∞
Z
0
1
y µ g q (x)dx < ∞.
and 0 <
0
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Hence (2.3) takes the form of strict inequality. So we have (2.2).
For 0 < ε < 12 , setting fε (x) = x
−ε−1/λ
p
, for x ∈ [1, ∞); fε (x) = 0, for
−ε−1/µ
q
x ∈ (0, 1), and gε (y) = y
, for y ∈ [1, ∞); gε (y) = 0, for y ∈ (0, 1).
Assume that the constant factor 1pq 1 in (2.2) is not the best possible, then there
λp µq
exists a positive number K with K <
pq
, such that (2.2) is valid by changing
to K. We have
1 1
λp µq
Z
pq
1 1
λp µq
∞
0
Z
0
∞
f (xλ )g(y µ )
dxdy
max{x, y}
p1 Z
Z ∞
1
−1
p
<K
x λ f (x)dx
0
Since
Z
0
∞
Yongjin Li and Bing He
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∞
1
x µ −1 g q (x)dx
0
1q
=
K
.
ε
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−1−εµ
1
t q dt = pq + o(1)
max{1, t}
+
(ε → 0 ),
J. Ineq. Pure and Appl. Math. 7(2) Art. 61, 2006
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1
1
setting y µ = x λ t, by (2.1), we find
Z ∞Z ∞
f (xλ )g(y µ )
dxdy
max{x, y}
0
0
h
ih
i
Z ∞ Z ∞ x λ1 −1 f (x) y µ1 −1 g(y)
11
n 1 1o
dxdy
=
λµ 0
µ
0
λ
max x , y
=
11
λµ
Z
∞
1
Z
Z
∞
1
∞
Z
∞
x
1
(λ
−1)+
−ε− 1
λ
p
1
(µ
−1)+
y
n 1 1o
max x λ , y µ
1
(λ
−1)+
−ε− 1
λ
p
µ
µ
λ
1
−ε− µ
Some Extensions of Hilbert’s
Type Inequality and its
Applications
q
dxdy
1
(µ
−1)+
Yongjin Li and Bing He
1
−ε− µ
11
x
(t x )
n 1
o
x µtµ−1 dxdt
1
1
−
λµ 1
x λ
max x λ , tx λ
Z
Z ∞
µ
−1−εµ
1
1 ∞ −ε( p1 + λq
)−1
q
x
t
dtdx
=
1
λ 1
x− λ max{1, t}
Z ∞
Z
µ
−1−εµ
1 ∞ −ε( p1 + λq
1
)−1
=
x
t q dtdx
λ 1
max{1, t}
0

1
Z x− λ
−1−εµ
1
t q dtdx
−
max{1, t}
0
"
#
1
pq
=
µ + o(1) .
λε p1 + λq
=
q
µ
λ
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Since for ε > 0 small enough, we have
"
#
pq
1
µ + o(1) < K.
λ p1 + λq
1
1
It is obvious that λ p µ q ≤
λ
p
+
µ
q
(i.e.
1
λ
1
µ
1
+ λq
p
≤
1
1
1
λp µq
) by Young’s inequality.
Consider the case of taking the form of the equality for Young’s inequality, we
p−1
1
get µ q = λ p , i.e. λ = µ, Then
"
#
pq
pq
1
1
1 + o(1) < K.
µ + o(1) =
1
λ p + λq
λp µq
Thus we get
factor
pq
1
1
λp µq
pq
1 1
λp µq
≤ K, which contradicts the hypothesis. Hence the constant
in (2.2) is best possible for λ = µ.
Remark 1. For λ = µ, inequality (2.2) becomes
Z
∞
Z
(2.4)
0
0
∞
f (xλ )g(y λ )
dxdy
max{x, y}
1q
Z ∞
p1 Z ∞
1
1
pq
−1
q
−1
p
<
x λ f (x)dx
x λ g (x)dx ,
λ
0
0
where the constant factor
pq
λ
is the best possible.
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
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R∞ 1
Theorem 2.3. Suppose f ≥ 0, p > 1, λ > 0 and 0 < 0 x λ −1 f p (x)dx < ∞.
Then
p
Z ∞ Z ∞
Z ∞
1
f (xλ )
1
p
(2.5)
dx dy < (pq)
x λ −1 f (x)p dx,
max{x, y}
λ
0
0
0
where the constant factor λ1 (pq)p is the best possible. Inequality (2.5) is equivalent to (2.4).
Proof. Setting g(y) as

p−1
Z ∞
1
−1
x λ f (x)

n 1 1 o dx
> 0,
0
max x λ , y λ
Some Extensions of Hilbert’s
Type Inequality and its
Applications
y ∈ (0, ∞).
then by (2.4), we find
p
Z ∞
Z ∞ Z ∞
1
f (xλ )
−2
−1
q
p−1
dx dy
λ
y λ g (y)dy = λ
max{x, y}
0
0
0
Z ∞Z ∞
f (xλ )g(y λ )
=
dxdy
max{x, y}
0
0
Z ∞
p1 Z ∞
1q
1
1
pq
≤
x λ −1 f p (x)dx
y λ −1 g q (y)dy .
(2.6)
λ
0
0
Hence we obtain
Z ∞
Z
1
−1 q
p
p
λ
g (y)dy ≤ λ (pq)
(2.7)
0<
y
0
0
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∞
1
x λ −1 f p (x)dx < ∞.
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By (2.4), both (2.6) and (2.7) take the form of strict inequality, so we have (2.5).
On the other hand, suppose that (2.5) is valid. By Hölder’s inequality, we
find
Z ∞Z ∞
f (xλ )g(y λ )
dxdy
max{x, y}
0
0
Z ∞ Z ∞
f (xλ )
=
dx g(y λ )dy
max{x,
y}
0
0
1q
p p1 Z ∞
Z ∞ Z ∞
f (xλ )
q λ
(2.8)
g (y )dy .
dx dy
≤
max{x, y}
0
0
0
Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
Then by (2.5), we have (2.4). Thus (2.4) and (2.5) are equivalent.
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1
(pq)p
λ
If the constant
in (2.5) is not the best possible, by (2.8), we may get
a contradiction that the constant factor in (2.4) is not the best possible. Thus we
complete the proof of the theorem.
Remark 2.
(i) For λ = µ = 1, (2.2) and (2.5) reduce respectively to (1.3) and (1.4). It follows that (2.2) is a new extension of (1.6) and (1.3) with some parameters
and the equivalent form (2.4) is a new extension of (1.4).
(ii) It is amazing that (2.4) and (1.9) are different, although both of them are
the extensions of (1.6) with (p, q)−parameter and the best constant factor.
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References
[1] G.H. HARDY, J.E. LITTLEWOOD
bridge Univ Press Cambridge, 1952.
AND
G. POLYA, Inequalities, Cam-
[2] J. KUANG, Applied Inequalities (Chinese), Hunan Jiaoyu Chubanshe,
Changsha, 1993.
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Some Extensions of Hilbert’s
Type Inequality and its
Applications
Yongjin Li and Bing He
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[7] B. YANG, A new Hardy-Hilbert type inequality and its applications (Chinese), Acta Math. Sinica, 46(6) (2003), 1079–1086.
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[8] B. YANG, Best generalization of a Hilbert-type inequality (Chinese), J. Jilin
Univ. Sci., 42(1) (2004), 30–34.
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