Journal of Inequalities in Pure and
Applied Mathematics
AN EXTENDED HARDY-HILBERT INEQUALITY
AND ITS APPLICATIONS
volume 7, issue 1, article 30,
2006.
JIA WEIJIAN AND GAO MINGZHE
Department of Mathematics and Computer Science
Normal College, Jishou University
Jishou Hunan 416000,
People’s Republic of China.
EMail: mingzhegao@163.com
Received 17 February, 2004;
accepted 10 November, 2005.
Communicated by: L. Pick
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
032-04
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Abstract
In this paper, it is shown that an extended Hardy-Hilbert’s integral inequality
with weights can be established by introducing a power-exponent function of the
π
form ax1+x (a > 0, x ∈ [0, +∞)), and the coefficient 1/q 1/p
is shown to
(a)
(b)
sin π/p
be the best possible constant in the inequality. In particular, for the case p = 2,
some extensions on the classical Hilbert’s integral inequality are obtained. As
applications, generalizations of Hardy-Littlewood’s integral inequality are given.
2000 Mathematics Subject Classification: 26D15.
Key words: Power-exponent function, Weight function, Hardy-Hilbert’s integral inequality, Hardy-Littlewood’s integral inequality.
The author is grateful to the referees for valuable suggestions and helpful comments
in this subject.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
References
An Extended Hardy-Hilbert
Inequality and Its Applications
Jia Weijian and Gao Mingzhe
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1.
Introduction
The famous Hardy-Hilbert’s integral inequality is
Z ∞Z ∞
f (x) g (y)
(1.1)
dxdy
x+y
0
0
Z ∞
p1 Z ∞
1q
π
p
q
f (x) dx
g (y) dy ,
≤
sin πp
0
0
where p > 1, q = p/(p − 1) and the constant sinπ π is best possible (see [1]).
p
In particular, when p = q = 2, the inequality (1.1) is reduced to the classical
Hilbert integral inequality:
Z ∞
12 Z ∞
12
Z ∞Z ∞
f (x) g (y)
2
2
g (y) dy ,
(1.2)
dxdy ≤ π
f (x) dx
x+y
0
0
0
0
where the coefficient π is best possible.
Recently, the following result was given by introducing the power function
in [2]:
Z bZ b
f (x) g (y)
(1.3)
dxdy
xt + y t
a
a
p1
Z b
≤ ω (t, p, q)
x1−t f p (x) dx
a
Z
× ω (t, q, p)
a
b
x1−t g q (x) dx
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1q
,
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where t is a parameter which is independent of x and y, ω (t, p, q) =
ϕ (q) and here the function ϕ is defined by
Z a/b t−2+1/r
u
du,
ϕ (r) =
1 + ut
0
π
t sin
π
pt
−
r = p, q.
However, in [2] the best constant for (1.3) was not determined.
Afterwards, various extensions on the inequalities (1.1) and (1.2) have appeared in some papers (such as [3, 4] etc.). The purpose of the present paper is
to show that if the denominator x + y of the function on the left-hand side of
(1.1) is replaced by the power-exponent function ax1+x + by 1+y , then we can
π
obtain a new inequality and show that the coefficient (a)1/q (b)1/p
is the best
sin π/p
constant in the new inequality. In particular if p = 2 then several extensions
of (1.2) follow. As its applications, it is shown that extensions on the HardyLittlewood integral inequality can be established.
Throughout this paper we stipulate that a > 0 and b > 0.
For convenience, we give the following lemma which will be used later.
x
Lemma 1.1. Let h (x) = 1+x+x
, x ∈ (0, +∞), then there exists a function
ln x
1
ϕ (x) 0 ≤ ϕ (x) < 2 , such that h (x) = 12 − ϕ (x).
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Proof. Consider the function defined by
1+x
+ ln x,
s (x) =
x
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x ∈ (0, +∞).
It is easy to see that the minimum of s (x) is 2. Hence s (x) ≥ 2, and h (x) =
1
s−1 (x) ≤ 12 . Obviously h (x) = s(x)
> 0. We can define a nonnegative function
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ϕ by
(1.4)
ϕ (x) =
Hence we have h (x) =
1
2
1 − x + x ln x
2 (1 + x + x ln x)
x ∈ (0, +∞) .
− ϕ (x). The lemma follows.
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Jia Weijian and Gao Mingzhe
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2.
Main Results
Define a function by
ω (r, x) = x
(2.1)
(1+x)(1−r)
r−1
1
− ϕ (x)
2
x ∈ (0, +∞),
where r > 1 and ϕ (x) is defined by (1.4).
Theorem 2.1. Let
Z ∞
0<
ω (p, x) f p (x) dx < +∞,
0
Z
0<
An Extended Hardy-Hilbert
Inequality and Its Applications
∞
ω (q, x) g q (x) dx < +∞,
the weight function ω (r, x) is defined by (2.1), p1 + 1q = 1, and p ≥ q > 1. Then
Z
(2.2)
0
∞
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0
Z
∞
f (x) g (y)
dxdy
ax1+x + by 1+y
0
Z ∞
p1 Z ∞
1q
µπ
p
q
≤
ω (p, x) f (x) dx
ω (q, x) g (x) dx ,
sin πp
0
0
where µ = (1/a)1/q (1/b)1/p and the constant factor
µπ
sin π
p
is best possible.
1
1
0
0
Proof. Let f (x) = F (x) (ax1+x ) q and g (y) = G (y) (by 1+y ) p . Define
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two functions by
1
1
0 F (x) (by 1+y ) p ax1+x pq
α=
and
1
1+y
(ax1+x + by 1+y ) p by
1
1
0 G (y) (ax1+x ) q by 1+y pq
β=
.
1
1+x
(ax1+x + by 1+y ) q ax
(2.3)
Let us apply Hölder’s inequality to estimate the right hand side of (2.2) as follows:
Z ∞Z ∞
f (x) g (y)
(2.4)
dxdy
ax1+x + by 1+y
0
0
Z ∞Z ∞
=
αβdxdy
0
0
Z
∞Z
∞
αp dxdy
≤
0
p1 Z
0
∞Z
0
∞
β q dxdy
1q
.
0
It is easy to deduce that
Z
∞
Z
∞
∞
Z
α dxdy =
0
0
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Z
p
An Extended Hardy-Hilbert
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0
Z
=
0
∞
0
(by 1+y )
ax1+x + by 1+y
∞
ωq F p (x) dx.
ax1+x
by 1+y
1q
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p
F (x) dxdy
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We compute the weight function ωq as follows:
Z
∞
ωq =
0
Z
∞
=
0
0
(by 1+y )
ax1+x + by 1+y
1
1+x
ax
+ by 1+y
ax1+x
by 1+y
1q
ax1+x
by 1+y
1q
dy
d by 1+y .
Let t = by 1+y /ax1+x . Then we have
Z
ωq =
0
∞
1q
π
1
π
1
.
dt =
π =
sin πp
sin q
1+t t
0 −1/q
Notice that F (x) = (ax1+x )
f (x). Hence we have
Z ∞
Z ∞Z ∞
1−p p
π
p
1+x 0
ax
f (x) dx,
(2.5)
α dxdy =
sin πp 0
0
0
and ,similarly,
Z ∞Z
(2.6)
0
0
∞
Z
(2.7)
0
0
∞
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∞
β q dxdy =
π
sin πp
Z
∞
by
1−q
1+y 0
0
Substituting (2.5) and (2.6) into (2.3), we obtain
Z
An Extended Hardy-Hilbert
Inequality and Its Applications
f (x) g (y)
dxdy
ax1+x + by 1+y
g q (y) dy.
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Z
π
≤
sin πp
∞
ax
1−p
1+x 0
p1
p
f (x) dx
0
Z
∞
×
by
1−q
1+y 0
q
1q
g (y) dy
.
0
π
sin π
p
We need to show that the constant factor
contained in (2.7) is best possible.
Define two functions by
(
0,
x ∈ (0, 1)
f˜ (x) =
−(1+ε)/p
0
(ax1+x )
(ax1+x ) , x ∈ [1, +∞)
and
(
Assume that 0 < ε <
Z
+∞
ax
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y ∈ (0, 1)
0,
g̃ (y) =
An Extended Hardy-Hilbert
Inequality and Its Applications
−(1+ε)/q
(by 1+y )
q
2p
0
(by 1+y ) , y ∈ [1, +∞)
.
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(p ≥ q > 1). Then
1−p
1+x 0
f˜p (x) dx =
0
Z
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+∞
ax1+x
−1−ε
1
1
d ax1+x = .
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Similarly, we have
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Z
0
∞
by 1+y
1−q
0
1
g̃ q (y) dy = .
ε
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If
π
sin π
p
is not best possible, then there exists k > 0, k <
Z
∞
Z
(2.8)
0
0
∞
π
sin π
p
such that
f˜ (x) g̃ (y)
dxdy
ax1+x + by 1+y
Z ∞ p1
1−p p
1+x 0
˜
<k
ax
f (x) dx
0
Z
×
0
∞
by
1−q
1+y 0
q
g̃ (y) dy
1q
=
k
.
ε
On the other hand, we have
Z ∞Z ∞ ˜
f (x) g̃ (y)
dxdy
1+x + by 1+y
0
0 ax
n
on
o
1+ε
1+x 0
1+y − q
1+y 0
Z ∞ Z ∞ (ax1+x )− 1+ε
p
(ax )
(by )
(by )
=
dxdy
ax1+x + by 1+y
1
1
)
Z ∞ (Z ∞
− 1+ε
n
1+ε
o
(by 1+y ) q
1+y
1+x − p
1+x 0
=
d
by
ax
dx
ax
ax1+x + by 1+y
1
1
)
− 1+ε
Z ∞ (Z ∞
q
−1−ε
1
1
=
dt ax1+x
d ax1+x
t
1
b/ax1+x 1 + t
− 1+ε
Z
q
1 ∞
1
1
=
dt.
ε b/ax1+x 1 + t t
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If the lower limit b/ax1+x of this integral is replaced by zero, then the resulting
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α
(b/ax1+x )
error is smaller than
, where α is positive and independent of ε. In
α
fact, we have
1+ε
Z b/ax1+x
Z b/ax1+x
β
1
1 q
(b/ax1+x )
dt <
t−(1+ε)/q dt =
1+t t
β
0
0
where β = 1 − (1 + ε)/q. If 0 < ε <
α=1−
q
,
2p
then we may take α such that
1 + q/2p
1
= .
q
2p
Consequently, we get
(
)
Z ∞Z ∞ ˜
1
π
f (x) g̃ (y)
+ o (1)
(2.9)
dxdy >
1+x
1+y
ax
+ by
ε sin πp
0
0
An Extended Hardy-Hilbert
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Jia Weijian and Gao Mingzhe
(ε → 0).
Clearly, when ε is small enough, the inequality (2.7) is in contradiction with
(2.9). Therefore, sinπ π is the best possible value for which the inequality (2.7) is
p
valid.
Let u = ax1+x and v = by 1+y . Then
1+x
0
1+x
u = ax
+ ln x = ax1+x h−1 (x) .
x
0
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1+y −1
Similarly, we have v = by h (y). Substituting them into (2.7) and then
using Lemma 1.1, the inequality (2.2) yields after simplifications. The constant
1/q
µπ
factor sin
(1/b)1/p . Thus the proof of the
π is best possible, where µ = (1/a)
p
theorem is completed.
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It is known from (2.1) that
r−1 r−1
1
1
(1+x)(1−r)
ω (r, x) = x
− ϕ (x)
=
x(1+x)(1−r) (1 − 2ϕ (x))r−1 .
2
2
The following result is equivalent to Theorem 2.1.
Theorem 2.2. Let ϕ (x) be a function defined by (1.4), p1 + 1q = 1 and p ≥ q > 1.
If
Z ∞
0<
x(1+x)(1−p) (1 − 2ϕ (x))p−1 f p (x) dx < +∞ and
Z0 ∞
0<
y (1+y)(1−q) (1 − 2ϕ (y))q−1 g q (y) dy < +∞,
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then
Z
(2.10)
∞
Z
∞
f (x) g (y)
dxdy
ax1+x + by 1+y
0
0
Z ∞
p1
µπ
p−1 p
(1+x)(1−p)
≤
x
(1 − 2ϕ (x)) f (x) dx
2 sin πp
0
Z ∞
1q
q−1 q
(1+y)(1−q)
×
y
(1 − 2ϕ (y)) g (y) dy ,
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where µ = (1/a)1/q (1/b)1/p and the constant factor
Contents
µπ
2 sin π
p
is best possible.
In particular, for case p = 2, some extensions on (1.2) are obtained. According to Theorem 2.1, we get the following results.
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Corollary 2.3. If
Z ∞
1
−(1+x)
0<
− ϕ (x) f 2 (x) dx < +∞
x
2
Z0 ∞
1
−(1+y)
− ϕ (y) g 2 (y) dy < +∞,
0<
y
2
0
and
where ϕ (x) is a function defined by (1.4), then
Z ∞Z ∞
f (x) g (y)
(2.11)
dxdy
ax1+x + by 1+y
0
0
Z ∞
12
1
π
2
−(1+x)
− ϕ (x) f (x) dx
≤√
x
2
ab
0
Z ∞
12
1
×
y −(1−y)
− ϕ (y) g 2 (y) dy ,
2
0
where the constant factor
√π
ab
is best possible.
Corollary 2.4. Let ϕ (x) be a function defined by (1.4). If
Z ∞
1
−(1+x)
0<
x
− ϕ (x) f 2 (x) dx < +∞,
2
0
then
Z
∞
Z
(2.12)
0
0
∞
f (x) f (y)
dxdy
ax1+x + by 1+y
Z ∞
π
1
−(1+x)
≤√
x
− ϕ (x) f 2 (x) dx,
2
ab 0
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where the constant factor
√π
ab
is best possible.
A equivalent proposition of Corollary 2.3 is:
Corollary 2.5. Let ϕ (x) be a function defined by (1.4),
Z ∞
0<
x−(1+x) (1 − 2ϕ (x)) f 2 (x) dx < +∞
Z0 ∞
0<
y −(1+y) (1 − 2ϕ (y)) g 2 (y) dy < +∞,
and
0
then
An Extended Hardy-Hilbert
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Jia Weijian and Gao Mingzhe
Z
(2.13)
0
∞
Z
∞
f (x) g (y)
dxdy
ax1+x + by 1+y
0
Z ∞
12
π
−(1+x)
2
≤ √
x
(1 − 2ϕ (x)) f (x) dx
2 ab
0
Z ∞
21
−(1+y)
2
×
y
(1 − 2ϕ (y)) g (y) dy ,
0
where the constant factor
π
√
2 ab
is best possible.
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Similarly, an equivalent proposition to Corollary 2.4 is:
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Corollary 2.6. Let ϕ (x) be a function defined by (1.4). If
Z ∞
0<
x−(1+x) (1 − 2ϕ (x)) f 2 (x) dx + ∞,
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then
Z
∞
Z
(2.14)
0
0
∞
f (x) f (y)
dxdy
ax1+x + by 1+y
Z ∞
π
≤ √
x−(1+x) (1 − 2ϕ (x)) f 2 (x) dx,
2 ab 0
where the constant factor
π
√
2 ab
is best possible.
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3.
Application
In this section, we will give various extensions of Hardy-Littlewood’s integral
inequality.
Let f (x) ∈ L2 (0, 1) and f (x) 6= 0. If
Z 1
an =
xn f (x) dx, n = 0, 1, 2, . . .
0
then we have the Hardy-Littlewood’s inequality (see [1]) of the form
∞
X
(3.1)
n=0
a2n
Z
<π
1
f 2 (x) dx,
0
R1
where f (x) = 0 tx h (x) dx, x ∈ [0, +∞).
Afterwards the inequality (3.2) was refined into the form in the paper [6]:
Z ∞
Z 1
2
(3.3)
f (x) dx ≤ π
th2 (t) dt.
0
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where π is the best constant that keeps (3.1) valid. In our previous paper [5],
the inequality (3.1) was extended and the following inequality established:
Z ∞
Z 1
2
(3.2)
f (x) dx < π
h2 (x) dx,
0
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We will further extend the inequality (3.3), some new results can be obtained
by further extending inequality (3.3).
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Theorem 3.1. Let h (t) ∈ L2 (0, 1), h (t) 6= 0. Define a function by
Z 1
f (x) =
tu(x) |h (t)| dt, x ∈ [0, +∞),
0
where u (x) = x1+x . Also, let ϕ (x) be a weight function defined by (1.4),
(r = p, q), p1 + 1q = 1 and p ≥ q > 1. If
Z
0<
∞
x
(1+x)(1−r)
0
r−1
1
f r (x) dx < +∞,
− ϕ (x)
2
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then
Z
2
∞
2
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f (x) dx
(3.4)
0
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! p1
p−1
µπ
1
<
− ϕ (x)
f p (x) dx
x(1+x)(1−p)
π
sin p
2
0
Z ∞
1q Z 1
1
×
y (1+y)(1−q)
− ϕ (y) f q (y) dy
th2 (t) dt,
2
0
0
Z
∞
where the constant factor
µπ
sin π
p
1/q
in (3.4) is best possible, and µ = (1/a)
Proof. Let us write f 2 (x) in the form:
Z 1
2
f (x) =
f (x) tu(x) |h (t)| dt.
0
1/p
(1/b)
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We apply, in turn, Schwarz’s inequality and Theorem 2.1 to obtain
2
Z ∞
2
(3.5)
f (x) dx
0
Z
∞
Z
1
=
u(x)
f (x) t
0
Z
1
Z
2
|h (t)| dt dx
0
∞
=
u(x)−1/2
f (x) t
1/2
2
|h (t)| dt
dx t
2 Z 1
Z
u(x)−1/2
≤
f (x) t
dx dt
th2 (t) dt
0
0
0
Z 1 Z ∞
=
f (x) tu(x)−1/2 dx
0
Z0 ∞
Z 1
u(y)−1/2
×
f (y) t
dy dt
th2 (t) dt
0
0
Z 1
Z 1 Z ∞ Z ∞
u(x)+u(y)−1
=
f (x) f (y) t
dxdy dt
th2 (t) dt
0
0Z ∞ Z0 ∞ 0
Z 1
f (x) f (y)
dxdy
th2 (t) dt
=
u
(x)
+
u
(y)
0
0
0
(Z
) p1
p−1
∞
1
µπ
x(1+x)(1−p)
− ϕ (x)
f p (x) dx
≤
sin πp
2
0
(Z
) 1q Z
q−1
∞
1
1
×
y (1+y)(1−q)
− ϕ (y)
f q (y) dy
th2 (t) dt.
2
0
0
0
0
1 Z ∞
An Extended Hardy-Hilbert
Inequality and Its Applications
Jia Weijian and Gao Mingzhe
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Since h (t) 6= 0, f 2 (x) 6= 0. It is impossible to take equality in (3.5). We
therefore complete the proof of the theorem.
An equivalent proposition to Theorem 3.1 is:
Theorem 3.2. Let the functions h (t) , f (x) and u (x) satisfy the assumptions
of Theorem 3.1, and assume that
Z ∞
0<
x(1+x)(1−r) (1 − 2ϕ (x))r−1 f r (x) dx < +∞ (r = p, q).
0
An Extended Hardy-Hilbert
Inequality and Its Applications
Then
Z
Jia Weijian and Gao Mingzhe
2
∞
2
f (x) dx
(3.6)
0
Z ∞
p1
µπ
p−1 p
(1+x)(1−p)
x
(1 − 2ϕ (x)) f (x) dx
<
2 sin πp
0
Z ∞
1q Z 1
q−1 q
(1+y)(1−q)
×
y
(1 − 2ϕ (y)) f (y) dy
th2 (t) dt,
0
0
1/q
µπ
and the constant factor sin
(1/b)1/p .
π in (3.6) is best possible, where µ = (1/a)
p
In particular, when p = q = 2, we have the following result.
Corollary 3.3. Let the functions h (t) , f (x) and u (x) satisfy the assumptions
of Theorem 3.1, and assume that
Z ∞
1
−(1+x)
0<
x
− ϕ (x) f 2 (x) dx < +∞,
2
0
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where ϕ (x) is a function defined by (1.4). Then
Z
2
∞
2
f (x) dx
Z ∞
Z 1
π
1
−(1+x)
2
<√
x
− ϕ (x) f (x) dx
th2 (t) dt,
2
ab
0
0
(3.7)
0
√π
ab
and the constant factor
in (3.7) is best possible.
An Extended Hardy-Hilbert
Inequality and Its Applications
A result equivalent to Corollary 3.3 is:
Corollary 3.4. Let the functions h (t) , f (x) and u (x) satisfy the assumptions
of Theorem 3.1, and assume that
Z ∞
0<
x−(1+x) (1 − 2ϕ (x)) f 2 (x) dx < +∞,
0
∞
2
2
0
π
< √
2 ab
and the constant factor
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f (x) dx
(3.8)
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where ϕ (x) is a function defined by (1.4). Then
Z
Jia Weijian and Gao Mingzhe
Z
∞
x
−(1+x)
Z
(1 − 2ϕ (x)) f (x) dx
0
π
√
2 ab
2
Close
1
2
th (t) dt,
0
in (3.8) is best possible.
The inequalities (3.4), (3.6), (3.7) and (3.8) are extensions of (3.3).
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References
[1] G.H. HARDY, J.E. LITTLEWOOD
bridge Univ. Press, Cambridge 1952.
AND
G. POLYA, Inequalities, Cam-
[2] JICHANG KUANG, On new extensions of Hilbert’s integral inequality, J.
Math. Anal. Appl., 235(2) (1999), 608–614.
[3] BICHENG YANG AND L. DEBNATH, On the extended Hardy-Hilbert’s
inequality, J. Math. Anal. Appl., 272(1) (2002), 187–199.
[4] BICHENG YANG, On a general Hardy-Hilbert’s inequality with a best
value, Chinese Ann. Math. (Ser. A ), 21(4) (2000), 401–408.
[5] MINGZHE GAO, On Hilbert’s inequality and its applications, J. Math.
Anal. Appl., 212(1) (1997), 316–323.
[6] MINGZHE GAO, LI TAN AND L. DEBNATH, Some improvements on
Hilbert’s integral inequality, J. Math. Anal. Appl., 229(2) (1999), 682–689.
An Extended Hardy-Hilbert
Inequality and Its Applications
Jia Weijian and Gao Mingzhe
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