Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 30, 2006
AN EXTENDED HARDY-HILBERT INEQUALITY AND ITS APPLICATIONS
JIA WEIJIAN AND GAO MINGZHE
D EPARTMENT OF M ATHEMATICS AND C OMPUTER S CIENCE
N ORMAL C OLLEGE , J ISHOU U NIVERSITY
J ISHOU H UNAN 416000,
P EOPLE ’ S R EPUBLIC OF C HINA .
mingzhegao@163.com
Received 17 February, 2004; accepted 10 November, 2005
Communicated by L. Pick
A BSTRACT. 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 form ax1+x (a >
π
0, x ∈ [0, +∞)), and the coefficient (a)1/q (b)1/p
is shown to be the best possible constant
sin π/p
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.
Key words and phrases: Power-exponent function, Weight function, Hardy-Hilbert’s integral inequality, Hardy-Littlewood’s
integral inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
The famous Hardy-Hilbert’s integral inequality is
1q
Z ∞
p1 Z ∞
Z ∞Z ∞
f (x) g (y)
π
p
q
g (y) dy
,
(1.1)
dxdy ≤
f (x) dx
x+y
sin πp
0
0
0
0
where p > 1, q = p/(p − 1) and the constant sinπ π is best possible (see [1]). In particular, when
p
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
(1.2)
dxdy ≤ π
f (x) dx
g (y) dy ,
x+y
0
0
0
0
where the coefficient π is best possible.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
The author is grateful to the referees for valuable suggestions and helpful comments in this subject.
032-04
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J IA W EIJIAN
AND
G AO M INGZHE
Recently, the following result was given by introducing the power function in [2]:
Z bZ b
f (x) g (y)
dxdy
(1.3)
xt + y t
a
a
p1 1q
Z b
Z b
≤ ω (t, p, q)
x1−t f p (x) dx
ω (t, q, p)
x1−t g q (x) dx ,
a
a
where t is a parameter which is independent of x and y, ω (t, p, q) =
function ϕ is defined by
Z a/b t−2+1/r
u
du, r = p, q.
ϕ (r) =
1 + ut
0
π
t sin
π
pt
− ϕ (q) and here the
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
π
ax1+x +by 1+y , then we can obtain a new inequality and show that the coefficient (a)1/q (b)1/p
sin π/p
is the best 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 Hardy-Littlewood 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 ϕ (x) 0 ≤ ϕ (x) <
ln x
1
such that h (x) = 2 − ϕ (x).
Proof. Consider the function defined by
s (x) =
1+x
+ ln x,
x
x ∈ (0, +∞).
It is easy to see that the minimum of s (x) is 2. Hence s (x) ≥ 2, and h (x) = s−1 (x) ≤ 12 .
1
> 0. We can define a nonnegative function ϕ by
Obviously h (x) = s(x)
ϕ (x) =
(1.4)
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.
2. M AIN R ESULTS
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
J. Inequal. Pure and Appl. Math., 7(1) Art. 30, 2006
Z
0<
∞
ω (q, x) g q (x) dx < +∞,
0
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1
2
,
A N E XTENDED H ARDY-H ILBERT I NEQUALITY AND I TS A PPLICATIONS
3
the weight function ω (r, x) is defined by (2.1), p1 + 1q = 1, and p ≥ q > 1. Then
Z ∞Z ∞
f (x) g (y)
(2.2)
dxdy
ax1+x + by 1+y
0
0
Z ∞
p1 Z ∞
1q
µπ
q
p
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 two functions
by
1
1
0 F (x) (by 1+y ) p ax1+x pq
(2.3)
and
α=
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
Let us apply Hölder’s inequality to estimate the right hand side of (2.2) as follows:
Z ∞Z ∞
Z ∞Z ∞
f (x) g (y)
dxdy =
(2.4)
αβdxdy
ax1+x + by 1+y
0
0
0
0
Z ∞ Z ∞
p1 Z ∞ Z ∞
1q
p
q
≤
α dxdy
.
β dxdy
0
0
0
0
It is easy to deduce that
1+x 1q
Z ∞Z ∞
Z ∞Z ∞
0
(by 1+y )
ax
p
α dxdy =
F p (x) dxdy
1+x
1+y
1+y
ax
+ by
by
0
0
Z0 ∞ 0
=
ωq F p (x) dx.
0
We compute the weight function ωq as follows:
1+x 1q
Z ∞
0
(by 1+y )
ax
ωq =
dy
ax1+x + by 1+y by 1+y
0
1+x 1q
Z ∞
1
ax
1+y
=
d
by
.
ax1+x + by 1+y by 1+y
0
Let t = by 1+y /ax1+x . Then we have
1q
Z ∞
1
1
π
π
ωq =
dt =
.
π =
1+t t
sin q
sin πp
0
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
(2.5)
α dxdy =
ax
f (x) dx,
sin πp 0
0
0
and ,similarly,
Z
∞
Z
(2.6)
0
0
∞
π
β dxdy =
sin πp
q
J. Inequal. Pure and Appl. Math., 7(1) Art. 30, 2006
Z
∞
by 1+y
0 1−q
g q (y) dy.
0
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AND
Substituting (2.5) and (2.6) into (2.3), we obtain
Z
∞
∞
Z
(2.7)
0
0
f (x) g (y)
π
dxdy
≤
ax1+x + by 1+y
sin πp
∞
Z
ax
1−p
1+x 0
p1
p
f (x) dx
0
Z
×
∞
by
1−q
1+y 0
q
g (y) dy
1q
.
0
We need to show that the constant factor sinπ π contained in (2.7) is best possible.
p
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 +∞ q
2p
ax
y ∈ (0, 1)
0,
g̃ (y) =
−(1+ε)/q
(by 1+y )
0
(by 1+y ) , y ∈ [1, +∞)
.
(p ≥ q > 1). Then
1−p
1+x 0
f˜p (x) dx =
0
Z
+∞
ax1+x
−1−ε
1
1
d ax1+x = .
ε
Similarly, we have
Z
∞
0
If
π
sin π
p
by 1+y
0 1−q
1
g̃ q (y) dy = .
ε
is not best possible, then there exists k > 0, k <
Z
∞
Z
(2.8)
0
0
∞
π
sin π
p
such that
Z ∞ p1
1−p p
f˜ (x) g̃ (y)
1+x 0
dxdy < k
ax
f˜ (x) dx
ax1+x + by 1+y
0
Z ∞ 1q
1−q q
k
1+y 0
×
by
g̃ (y) dy
= .
ε
0
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+x − p
1+y
1+x 0
=
d
by
ax
ax
dx
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
J. Inequal. Pure and Appl. Math., 7(1) Art. 30, 2006
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A N E XTENDED H ARDY-H ILBERT I NEQUALITY AND I TS A PPLICATIONS
5
If the lower limit b/ax1+x of this integral is replaced by zero, then the resulting error is smaller
α
(b/ax1+x )
than
, where α is positive and independent of ε. In fact, we have
α
1+ε
Z b/ax1+x
Z b/ax1+x
β
1
1 q
(b/ax1+x )
−(1+ε)/q
dt <
t
dt =
1+t t
β
0
0
where β = 1 − (1 + ε)/q. If 0 < ε <
q
,
2p
then we may take α such that
α=1−
Consequently, we get
Z ∞Z
(2.9)
0
∞
0
1 + q/2p
1
= .
q
2p
f˜ (x) g̃ (y)
1
dxdy >
1+x
1+y
ax
+ by
ε
(
π
+ o (1)
sin πp
)
(ε → 0).
Clearly, when ε is small enough, the inequality (2.7) is in contradiction with (2.9). Therefore,
π
is the best possible value for which the inequality (2.7) is valid.
sin π
p
Let u = ax1+x and v = by 1+y . Then
1+x
0
1+x
+ ln x = ax1+x h−1 (x) .
u = ax
x
Similarly, we have v 0 = by 1+y h−1 (y). Substituting them into (2.7) and then using Lemma 1.1,
µπ
the inequality (2.2) yields after simplifications. The constant factor sin
π is best possible, where
p
1/q
µ = (1/a)
1/p
(1/b)
. Thus the proof of the theorem is completed.
It is known from (2.1) that
r−1 r−1
1
1
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.
(1+x)(1−r)
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 < +∞,
0
then
Z
∞
Z
(2.10)
0
0
∞
f (x) g (y)
dxdy
ax1+x + by 1+y
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 ,
0
where µ = (1/a)1/q (1/b)1/p and the constant factor
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.
J. Inequal. Pure and Appl. Math., 7(1) Art. 30, 2006
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J IA W EIJIAN
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Corollary 2.3. If
∞
1
0<
x
− ϕ (x) f 2 (x) dx < +∞
2
0
Z ∞
1
−(1+y)
0<
y
− ϕ (y) g 2 (y) dy < +∞,
2
0
Z
−(1+x)
and
where ϕ (x) is a function defined by (1.4), then
Z ∞
12
Z ∞Z ∞
π
1
f (x) g (y)
− ϕ (x) f 2 (x) dx
(2.11)
dxdy ≤ √
x−(1+x)
2
ax1+x + by 1+y
ab
0
0
0
Z ∞
12
1
−(1−y)
2
×
y
− ϕ (y) g (y) dy ,
2
0
where the constant factor √πab is best possible.
Corollary 2.4. Let ϕ (x) be a function defined by (1.4). If
Z ∞
1
−(1+x)
0<
− ϕ (x) f 2 (x) dx < +∞,
x
2
0
then
Z ∞
Z ∞Z ∞
f (x) f (y)
π
1
−(1+x)
(2.12)
dxdy ≤ √
x
− ϕ (x) f 2 (x) dx,
1+x + by 1+y
ax
2
ab
0
0
0
where the constant factor √πab 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
Z
∞
∞
Z
(2.13)
0
0
f (x) g (y)
π
dxdy ≤ √
1+x
1+y
ax
+ by
2 ab
12
∞
Z
x
−(1+x)
2
(1 − 2ϕ (x)) f (x) dx
0
Z
∞
×
y
−(1+y)
2
(1 − 2ϕ (y)) g (y) dy
12
,
0
where the constant factor
π
√
2 ab
is best possible.
Similarly, an equivalent proposition to Corollary 2.4 is:
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 + ∞,
0
then
Z
∞
Z
∞
f (x) f (y)
π
(2.14)
dxdy ≤ √
1+x
1+y
ax
+ by
2 ab
0
0
π
where the constant factor 2√ab is best possible.
J. Inequal. Pure and Appl. Math., 7(1) Art. 30, 2006
Z
∞
x−(1+x) (1 − 2ϕ (x)) f 2 (x) dx,
0
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A N E XTENDED H ARDY-H ILBERT I NEQUALITY AND I TS A PPLICATIONS
7
3. A PPLICATION
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
Z 1
∞
X
2
(3.1)
an < π
f 2 (x) dx,
0
n=0
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
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
0
We will further extend the inequality (3.3), some new results can be obtained by further
extending inequality (3.3).
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) = x . Also, let ϕ (x) be a weight function defined by (1.4), (r = p, q), p1 + 1q = 1
and p ≥ q > 1. If
r−1
Z ∞
1
(1+x)(1−r)
0<
x
− ϕ (x)
f r (x) dx < +∞,
2
0
1+x
then
Z
(3.4)
∞
f 2 (x) dx
2
0
where the constant factor
! p1
p−1
1
x(1+x)(1−p)
− ϕ (x)
f p (x) dx
2
0
Z ∞
1q Z 1
1
(1+y)(1−q)
q
×
y
− ϕ (y) f (y) dy
th2 (t) dt,
2
0
0
µπ
<
sin πp
µπ
sin π
p
Z
∞
in (3.4) is best possible, and µ = (1/a)1/q (1/b)1/p .
Proof. Let us write f 2 (x) in the form:
2
Z
f (x) =
1
f (x) tu(x) |h (t)| dt.
0
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J IA W EIJIAN
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We apply, in turn, Schwarz’s inequality and Theorem 2.1 to obtain
Z ∞
2 Z ∞ Z 1
2
2
u(x)
f (x) dx =
f (x) t
|h (t)| dt dx
0
0
Z
0
1
Z
∞
u(x)−1/2
=
f (x) t
Z
0
0
1 Z ∞
1/2
dx t
2
Z
1
dx dt
th2 (t) dt
0
0
0
Z ∞
Z 1
Z 1 Z ∞
u(x)−1/2
u(y)−1/2
=
f (x) t
dx
f (y) t
dy dt
th2 (t) dt
0
0
0
0
Z 1
Z 1 Z ∞ Z ∞
=
f (x) f (y) tu(x)+u(y)−1 dxdy dt
th2 (t) dt
0
0
0
0
Z ∞ Z ∞
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
≤
(3.5)
u(x)−1/2
2
|h (t)| dt
f (x) t
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
Then
Z
∞
2
f (x) dx
(3.6)
0
2
µπ
<
2 sin πp
Z
×
p1
∞
Z
x
(1+x)(1−p)
p−1
(1 − 2ϕ (x))
p
f (x) dx
0
1q Z
∞
y
(1+y)(1−q)
q−1
(1 − 2ϕ (y))
q
and the constant factor
th2 (t) dt,
0
0
µπ
sin π
p
1
f (y) dy
1/q
in (3.6) is best possible, where µ = (1/a)
1/p
(1/b)
.
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
where ϕ (x) is a function defined by (1.4). Then
Z ∞
2
Z ∞
Z 1
π
1
2
−(1+x)
2
(3.7)
f (x) dx < √
x
− ϕ (x) f (x) dx
th2 (t) dt,
2
ab
0
0
0
J. Inequal. Pure and Appl. Math., 7(1) Art. 30, 2006
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A N E XTENDED H ARDY-H ILBERT I NEQUALITY AND I TS A PPLICATIONS
and the constant factor
√π
ab
9
in (3.7) is best possible.
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
where ϕ (x) is a function defined by (1.4). Then
Z ∞
2
Z ∞
Z 1
π
2
−(1+x)
2
x
(1 − 2ϕ (x)) f (x) dx
th2 (t) dt,
(3.8)
f (x) dx < √
2 ab
0
0
0
π
and the constant factor 2√ab 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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