Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 4, Article 96, 2004
ON THE EXTENDED HILBERT’S INTEGRAL INEQUALITY
BICHENG YANG
D EPARTMENT OF M ATHEMATICS ,
G UANGDONG I NSTITUTE OF E DUCATION ,
G UANGZHOU , G UANGDONG 510303
P EOPLE ’ S R EPUBLIC OF C HINA .
bcyang@pub.guangzhou.gd.cn
Received 21 May, 2003; accepted 21 October, 2004
Communicated by S.S. Dragomir
A BSTRACT. This paper gives two distinct generalizations of the extended Hilbert’s integral inequality with the same best constant factor involving the β function. As applications, we consider
some equivalent inequalities.
Key words and phrases: Hilbert’s inequality, weight function, β function, Hölder’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
R∞
R∞
If f, g ≥ 0, such that 0 < 0 f 2 (x)dx < ∞ and 0 < 0 g 2 (x)dx < ∞, then the famous
Hilbert’s integral inequality is given by
Z ∞
12
Z ∞Z ∞
Z ∞
f (x)g(y)
2
2
(1.1)
dxdy < π
f (x)dx
g (x)dx ,
x+y
0
0
0
0
where the constant factor π is the best possible (see [2]). Inequality (1.1) had been generalized
by Hardy-Riesz [1] as:
R∞
R∞
If p > 1, p1 + 1q = 1, 0 < 0 f p (x)dx < ∞ and 0 < 0 g q (x)dx < ∞, then
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
π
p
q
f (x)dx
g (x)dx ,
(1.2)
dxdy <
x+y
0
0
0
0
sin πp
π
where the constant factor sin(π/p)
is the best possible. When p = q = 2, inequality (1.2) reduces
to (1.1). We call (1.2) Hardy-Hilbert’s integral inequality, which is important in analysis and its
applications (see [4]).
In recent years, by introducing a parameter λ and the β function, Yang [7, 8] gave an extension of (1.2) as:
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
Research support by the Science Foundation of Professor and Doctor of Guangdong Institute of Education.
068-03
2
B ICHENG YANG
R∞
R∞
If λ > 2 − min{p, q}, 0 < 0 x1−λ f p (x)dx < ∞ and 0 < 0 x1−λ g q (x)dx < ∞, then
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
1−λ p
1−λ q
(1.3)
dxdy < kλ (p)
x f (x)dx
x g (x)dx ,
(x + y)λ
0
0
0
0
p+λ−2
where the constant factor kλ (p) = B p+λ−2
,
is the best possible (B(u, v) is the β
p
p
function). Its equivalent inequality is (see [9, (2.12)]):
p
Z ∞
f (x)
p
(1.4)
y
dx dy < [kλ (p)]
x1−λ f p (x)dx,
λ
(x + y)
0
0
0
ip
h where the constant factor [kλ (p)]p = B p+λ−2
is the best possible.
, p+λ−2
p
p
When λ = 1, inequality (1.3) reduces to (1.2), and (1.4) reduces to the equivalent form of
(1.2) as:

p
p
Z ∞ Z ∞
Z ∞
π 
f (x)

dx dy <
(1.5)
f p (x)dx.
π
x+y
0
0
0
sin p
Z
∞
(λ−1)(p−1)
Z
∞
For p = q = 2, by (1.3), we have λ > 0, and
Z ∞
12
Z ∞
Z ∞Z ∞
λ λ
f (x)g(y)
1−λ 2
1−λ 2
(1.6)
dxdy < B
,
x f (x)dx
x g (x)dx .
(x + y)λ
2 2
0
0
0
0
We define (1.6) as the extended Hilbert’s integral inequality. Recently, Yang et al. [10] provided an extensive account of the above results and Yang [6] gave a reverse of (1.4) with the
same best constant factor. The main objective of this paper is to build two distinct generalizations of (1.6), with the same best constant factor but different from (1.3). As applications, we
consider some equivalent inequalities.
For this, we need some lemmas.
2. S OME L EMMAS
We have the formula of the β function as (see [5]):
Z ∞
tu−1
(2.1)
B(u, v) =
dt = B(v, u)
(1 + t)u+v
0
(u, v > 0).
Lemma 2.1 (see [3]). If p > 1, p1 + 1q = 1, ω(σ) > 0, f, g ≥ 0, f ∈ Lpω (E) and g ∈ Lqω (E) ,
then the weighted Hölder’s inequality is as follows:
Z
p1 Z
1q
Z
p
q
(2.2)
ω(σ)f (σ)g(σ)dσ ≤
ω(σ)f (σ)dσ
ω(σ)g (σ)dσ ,
E
E
E
where the equality holds if and only if there exists non-negative real numbers A and B, such
that they are not all zero and Af p (σ) = Bg q (σ), a.e. in E.
Lemma 2.2. If r > 1, and λ > 0, define the weight function ωλ (r, x) as
Z ∞
1
λ(1− r1 )
y (λ−r)/r dy.
(2.3)
ωλ (r, x) := x
λ
(x + y)
0
Then we have
λ
1
(2.4)
ωλ (r, x) = B
,λ 1 −
.
r
r
J. Inequal. Pure and Appl. Math., 5(4) Art. 96, 2004
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O N THE E XTENDED H ILBERT ’ S I NTEGRAL I NEQUALITY
3
Proof. Setting y = xu in the integral of (2.3), we find
Z ∞
(xu)(λ−r)/r
λ(1− r1 )
ωλ (r, x) = x
xdu
xλ (1 + u)λ
0
Z ∞
λ
1
u r −1 du.
=
λ
(1 + u)
0
By (2.1), we have (2.4) and the lemma is proved.
Note. It is obvious that for p > 1, p1 +
1
q
= 1 and λ > 0, one has
λ λ
ωλ (p, x) = B
,
= ωλ (q, x).
p q
(2.5)
Lemma 2.3. If p > 1, p1 +
(2.6)
1
q
= 1 and 0 < ε < λ, one has
Z ∞
Z ∞
λ−q−ε
λ−p−ε
1
p
I1 :=
y q
x
dxdy
(x + y)λ
1
1
2
1
λ−ε λ ε
p
> B
, +
−
.
ε
p
q p
λ−ε
Proof. Setting x = yu in I1 , in view of (2.1), one has
Z ∞
Z ∞
λ−p−ε
1
−1−ε
u p du dy
I1 =
y
λ
1
1/y (1 + u)
Z ∞
Z ∞
λ−ε
1
−1
−1−ε
p
=
y
u
du dy
(1 + u)λ
1
0
"Z 1
#
Z ∞
y
λ−ε
1
−
y −1−ε
u p −1 du dy
λ
(1
+
u)
1
0
Z ∞
Z 1
y
λ−ε
1
λ−ε λ ε
−1
> B
, +
−
y
u p −1 dudy.
ε
p
q p
1
0
By calculating the above integral, one has (2.6). The lemma is proved.
Lemma 2.4. If p > 1, p1 +
(2.7)
1
q
= 1 and 0 < ε < λ(p − 1), one has
Z ∞
Z ∞
λ+ε
1
−1
λ− λ+ε
q
xλ− p −1 dxdy
I2 :=
y
λ
(x + y)
1
1
−2
1
λ ε λ ε
λ ε
> B
− , +
−
−
.
ε
q
p p p
q
p
Proof. Setting x = yu in I2 , in view of (2.1), one has
Z ∞
Z ∞
1
λ− λ+ε
−1
−1−ε
p
I2 =
y
u
du dy
λ
1
1/y (1 + u)
Z ∞
Z ∞
1
λ− λ+ε
−1
−1−ε
p
=
y
u
du dy
(1 + u)λ
1
0
#
"Z 1
Z ∞
y
λ+ε
1
uλ− p −1 du dy
−
y −1−ε
λ
(1
+
u)
0
1
Z ∞
Z 1
y
λ+ε
1
λ ε λ ε
−1
> B
− , +
−
y
uλ− p −1 dudy.
ε
q
p p p
1
0
J. Inequal. Pure and Appl. Math., 5(4) Art. 96, 2004
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4
B ICHENG YANG
By calculating the above integral, one has (2.7). The lemma is proved.
3. M AIN R ESULTS AND A PPLICATIONS
R∞
Theorem 3.1. If f, g ≥ 0, p > 1, p1 + 1q = 1, λ > 0, such that 0 < 0 xp−1−λ f p (x)dx < ∞
R∞
and 0 < 0 xq−1−λ g q (x)dx < ∞, then
Z ∞Z ∞
f (x)g(y)
(3.1)
dxdy
(x + y)λ
0
0
Z ∞
p1 Z ∞
1q
λ λ
q−1−λ q
p−1−λ p
x
g (x)dx ;
<B
,
x
f (x)dx
p q
0
0
p
p Z ∞
f (x)
λ λ
(3.2)
y
dx dy < B
,
xp−1−λ f p (x)dx,
λ
(x + y)
p q
0
0
0
h ip
are all the best possible. Inequality (3.2)
where the constant factors B λp , λq and B λp , λq
is equivalent to (3.1). In particular, for λ = 1, one has the following two equivalent inequalities:
Z
Z
∞
∞
λ(p−1)−1
∞
Z
(3.3)
0
0
y
(3.4)
p−2
Z
0
0
Proof. By (2.2), one has
Z ∞Z
J1 :=
(3.5)
∞
f (x)g(y)
π
dxdy <
x+y
sin πp
∞
Z
Z
∞
f (x)
dx
x+y
p
p1 Z
∞
Z
x
p−2 p
f (x)dx
x
0
q−2 q
g (x)dx
1q
;
0
p

dy < 
∞
sin
π 
π
p
Z
∞
xp−2 f p (x)dx.
0
∞
f (x)g(y)
dxdy
(x + y)λ
0
0
"
# "
#
1
1
Z ∞Z ∞
y q−λ pq
1
xp−λ pq
=
f (x)
g(y) dxdy
(x + y)λ
y q−λ
xp−λ
0
0
) p1
(Z "Z
p−λ 1q #
∞
∞
1
x
≤
dy f p (x)dx
λ
q−λ
(x
+
y)
y
0
0
(Z "Z
) 1q
q−λ p1 #
∞
∞
1
y
.
×
dx g q (y)dy
λ
p−λ
(x
+
y)
x
0
0
If (3.5) takes the form of an equality, then by Lemma 2.1, there exist real numbers A and B,
such that they are not all zero, and
q−λ p1
p−λ 1q
1
y
1
x
p
g q (y), a.e. in (0, ∞) × (0, ∞).
f (x) = B
A
λ
λ
q−λ
(x + y)
xp−λ
(x + y)
y
Hence we find
Axp−λ f p (x) = By q−λ g q (y), a.e. in (0, ∞) × (0, ∞).
J. Inequal. Pure and Appl. Math., 5(4) Art. 96, 2004
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O N THE E XTENDED H ILBERT ’ S I NTEGRAL I NEQUALITY
5
It follows that there exists a constant C, such that
Axp−λ f p (x) = C,
a.e. in (0, ∞);
By q−λ g q (y) = C,
a.e. in (0, ∞).
Without loss of generality, suppose that A 6= 0. One has
xp−λ−1 f p (x) =
C
,
Ax
a.e. in (0, ∞),
R∞
which contradicts the fact that 0 < 0 xp−1−λ f p (x)dx < ∞. Hence, (3.5) takes the form of
strict inequality, and by (2.3), we may rewrite (3.5) as
1q
Z ∞
p1 Z ∞
p−1−λ p
q−1−λ q
(3.6)
J1 <
ωλ (q, x)x
f (x)dx
ωλ (p, y)y
g (y)dy
.
0
0
Hence by (2.5), one has (3.1).
∼
∼
For 0 < ε < λ, setting f (x) and g(y) as:
∼
∼
f (x) = g(y) = 0,
∼
f (x) = x
λ−p−ε
p
x, y ∈ (0, 1);
∼
, g(y) = y
λ−q−ε
q
x, y ∈ [1, ∞),
,
then we find
Z
∞
x
(3.7)
p−1−λ
p1 Z
∼p
∞
f (x)dx
0
x
q−1−λ ∼q
1q
g (x)dx
0
1
= .
ε
If there exists λ > 0, such that the constant
in (3.1) is not the best possible, then there
factor
λ λ
exists a positive number K ( with K < B p , q ), such that (3.1) is still valid if one replaces
λ λ
B p , q by K. In particular, one has
Z
∞
Z
∞
∼
∼
f (x)g(y)
εI1 = ε
dxdy
(x + y)λ
0
0
Z ∞
p1 Z
∼p
p−1−λ
< εK
x
f (x)dx
∞
x
1q
q−1−λ ∼q
g (x)dx
.
0
0
Hence by (2.6) and (3.7), one has
2
λ−ε λ ε
p
B
, +
−ε
< K,
p
q p
λ−ε
and then B λp , λq ≤ K (ε → 0+ ). This contradicts the fact that K < B λp , λq . It follows
that the constant
in (3.1) is the best possible.
R ∞ factor
p−1−λ p
Since 0 < 0 x
f (x)dx < ∞, there exists T0 > 0, such that for any T > T0 , one has
R T p−1−λ p
0< 0 x
f (x)dx < ∞. We set
g(y, T ) := y
λ(p−1)−1
Z
0
J. Inequal. Pure and Appl. Math., 5(4) Art. 96, 2004
T
f (x)
dx
(x + y)λ
p−1
,
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6
B ICHENG YANG
and use (3.1) to obtain
Z T
0<
y q−1−λ g q (y, T )dy
0
Z T
p
Z T
f (x)
λ(p−1)−1
y
dx dy
=
λ
0 (x + y)
0
Z TZ T
f (x)g(y, T )
=
dxdy
(x + y)λ
0
0
1q
Z T
p1 Z T
λ λ
p−1−λ p
q−1−λ q
(3.8)
y
g (y, T )dy .
,
x
f (x)dx
<B
p q
0
0
Hence we find
Z T
1− 1q
q−1−λ q
0<
y
g (y, T )dy
0
Z
=
T
y λ(p−1)−1
0
<B
(3.9)
R∞
λ λ
,
p q
Z
p p1
f (x)
dx dy
λ
0 (x + y)
p1
p−1−λ p
x
f (x)dx .
Z
T
T
0
q−1−λ q
It follows that 0 < 0 y
g (y, ∞)dy < ∞. Hence (3.8) and (3.9) are strict inequalities as
T → ∞. Thus inequality (3.2) holds.
On the other hand, if (3.2) is valid, by Hölder’s inequality (2.2), one has
Z ∞Z ∞
f (x)g(y)
dxdy
(x + y)λ
0
0
h
Z ∞
Z ∞
i
λ+1−q
f
(x)
− λ+1−q
q
=
y q
dx
y
g(y)
dy
(x + y)λ
0
0
Z ∞
Z ∞
p p1 Z ∞
1q
f (x)
λ(p−1)−1
q−1−λ q
≤
y
dx dy
(3.10)
y
g (y)dy .
(x + y)λ
0
0
0
Hence by (3.2), one has (3.1). It follows that (3.2) is equivalent to (3.1).
If the constant factor in (3.2) is not the best possible, one can get a contradiction that the
constant factor in (3.1) is not the best possible by using (3.10). The theorem is thus proved. R∞
Theorem 3.2. If f, g ≥ 0, p > 1, p1 + 1q = 1, λ > 0, such that 0 < 0 x(p−1)(1−λ) f p (x)dx < ∞
R∞
and 0 < 0 x(q−1)(1−λ) g q (x)dx < ∞, then
Z ∞Z ∞
f (x)g(y)
(3.11)
dxdy
(x + y)λ
0
0
Z ∞
p1 Z ∞
1q
λ λ
(p−1)(1−λ) p
(q−1)(1−λ) q
<B
,
x
f (x)dx
x
g (x)dx ;
p q
0
0
p
p Z ∞
f (x)
λ λ
(3.12)
y
dx dy < B
,
x(p−1)(1−λ) f p (x)dx,
λ
(x + y)
p q
0
0
0
h ip
where the constant factors B λp , λq and B λp , λq
are the best possible. Inequality (3.12)
is equivalent to (3.11). In particular, for λ = p > 1, one has the following two equivalent
Z
∞
λ−1
Z
∞
J. Inequal. Pure and Appl. Math., 5(4) Art. 96, 2004
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O N THE E XTENDED H ILBERT ’ S I NTEGRAL I NEQUALITY
7
inequalities:
∞
Z
∞
Z
(3.13)
0
0
f (x)g(y)
1
dxdy <
p
(x + y)
p−1
∞
Z
f p (x)
dx
x(p−1)2
0
p1 Z
∞
0
g q (x)
dx
x
1q
and
Z
∞
y
(3.14)
p−1
0
∞
Z
0
f (x)
dx
(x + y)p
p
dy <
1
p−1
p Z
0
∞
f p (x)
dx.
x(p−1)2
Proof. By (2.2), one has
Z ∞Z ∞
f (x)g(y)
J1 =
dxdy
(x + y)λ
0
0
"
!
#"
!
#
Z ∞Z ∞
2
2
x(q−λ)/q
1
y (p−λ)/p
=
f (x)
g(y) dxdy
(x + y)λ
y (p−λ)/p2
x(q−λ)/q2
0
0
(Z "Z
! #
) p1
2
∞
∞
x(q−λ)p/q
1
dy f p (x)dx
≤
λ
(p−λ)/p
(x
+
y)
y
0
0
(Z "Z
! #
) 1q
2
∞
∞
1
y (p−λ)q/p
×
dx g q (y)dy
(3.15)
.
λ
(q−λ)/q
(x
+
y)
x
0
0
Following the same manner as (3.6), one has
Z ∞
p1 Z
(p−1)(1−λ) p
(3.16) J1 <
ωλ (p, x)x
f (x)dx
0
∞
ωλ (q, x)x
(q−1)(1−λ) q
g (x)dx
1q
.
0
Hence by (2.5), one has (3.11).
∼
∼
For 0 < ε < λ(p − 1), setting f (x) and g(y) as:
∼
∼
x, y ∈ (0, 1);
f (x) = g(y) = 0,
∼
f (x) = xλ−1−
λ+ε
p
∼
, g(y) = y λ−1−
λ+ε
q
,
x, y ∈ [1, ∞),
by Lemma 2.4 and the same way of Theorem 3.1, we can show that the constant factor in (3.11)
is the best possible.
In a similar fashion to Theorem 3.1, we can show that (3.12) is valid, which is equivalent to
(3.11). By the equivalence of (3.11) and (3.12), we may conclude that the constant factor in
(3.12) is the best possible. The theorem is proved.
Remark 3.3.
(i) For p = q = 2, both inequalities (3.1) and (3.11) reduce to (1.6). Inequalities (3.1) and (3.11) are distinct generalizations of (1.6) with the same best constant
factor B λp , λq , but different from (1.3).
(ii) Since inequalities (3.3) and (1.2) are different, we may conclude that inequality (3.1) is
not a generalization of (1.3).
(iii) Since all the given inequalities are with the best constant factors, we have obtained some
new results.
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8
B ICHENG YANG
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