Journal of Inequalities in Pure and
Applied Mathematics
A RELATION TO HARDY-HILBERT’S INTEGRAL INEQUALITY
AND MULHOLLAND’S INEQUALITY
volume 6, issue 4, article 112,
2005.
BICHENG YANG
Department of Mathematics
Guangdong Institute of Education
Guangzhou, Guangdong 510303
P. R. China.
Received 14 November, 2004;
accepted 25 August, 2005.
Communicated by: L. Pick
EMail: bcyang@pub.guangzhou.gd.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
This paper deals with a relation between Hardy-Hilbert’s integral inequality and
Mulholland’s integral inequality with a best constant factor, by using the Beta
function and introducing a parameter λ. As applications, the reverse, the equivalent form and some particular results are considered.
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
2000 Mathematics Subject Classification: 26D15.
Key words: Hardy-Hilbert’s integral inequality; Mulholland’s integral inequality; β
function; Hölder’s inequality.
Bicheng Yang
Contents
1
2
3
4
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Some Particular Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
The first reversible form . . . . . . . . . . . . . . . . . . . . . . . .
4.2
The second reversible form . . . . . . . . . . . . . . . . . . . . . .
4.3
The form which does not have a reverse . . . . . . . . . . .
References
3
6
9
16
16
19
23
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1.
Introduction
R∞
If p > 1, p1 + 1q = 1, f, g ≥ 0 satisfy 0 < 0 f p (x)dx < ∞ and 0 <
R∞ q
g (x)dx < ∞, then one has two equivalent inequalities as (see [1]):
0
Z ∞Z ∞
f (x)g(y)
(1.1)
dxdy
x+y
0
0
Z ∞
p1 Z ∞
1q
π
p
q
<
f (x)dx
g (x)dx ;
0
0
sin πp
Z
∞
Z
(1.2)
0
0
∞
f (x)
dx
x+y
p

p
dy < 
h
sin
π 
π
p
ip
Z
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
∞
p
f (x)dx,
0
π
π
where the constant factors sin(π/p)
and sin(π/p)
are all the best possible. Inequality (1.1) is called Hardy- Hilbert’s integral inequality, which is important
in analysisRand its applications (cf. Mitrinovic
R ∞ 1 etq al. [2]).
∞ 1 p
If 0 < 1 x F (x)dx < ∞ and 0 < 1 y G (y)dy < ∞, then the Mulholland’s integral inequality is as follows (see [1, 3]):
Z ∞Z ∞
F (x)F (y)
(1.3)
dxdy
xy ln xy
1
1
Z ∞ p
p1 Z ∞ q
1q
π
F (x)
G (y)
<
dx
dy ,
x
y
1
1
sin πp
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π
where the constant factor sin(π/p)
is the best possible. Setting f (x) = F (x)/x,
and g(y) = G(y)/y in (1.3), by simplification, one has (see [12])
Z ∞Z ∞
f (x)g(y)
(1.4)
dxdy
ln xy
1
1
Z ∞
p1 Z ∞
1q
π
<
xp−1 f p (x)dx
xq−1 g q (x)dx .
π
1
1
sin p
We still call (1.4) Mulholland’s integral inequality.
In 1998, Yang [11] first introduced an independent parameter λ and the β
function for given an extension of (1.1) (for p = q = 2). Recently, by introducing a parameter λ, Yang [8] and Yang et al. [10] gave some
R ∞ extensions of (1.1)
and (1.2)Ras: If λ > 2 − min{p, q}, f, g ≥ 0 satisfy 0 < 0 x1−λ f p (x)dx < ∞
∞
and 0 < 0 x1−λ g q (x)dx < ∞, then one has two equivalent inequalities as:
Z ∞Z ∞
f (x)g(y)
(1.5)
dxdy
(x + y)λ
0
0
p1 Z ∞
1q
Z ∞
1−λ p
1−λ q
< kλ (p)
x f (x)dx
x g (x)dx
0
0
∞
y
(1.6)
0
(p−1)(λ−1)
Z
0
∞
f (x)
dx
(x + y)λ
p
Bicheng Yang
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and
Z
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
p
Z
dy < [kλ (p)]
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∞
x
1−λ p
f (x)dx,
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0
p
p+λ−2 q+λ−2
, q
p
where the constant factors kλ (p) and [kλ (p)] (kλ (p) = B
,
B(u, v) is the β function) are all the best possible. By introducing a parameter
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α, Kuang [5] gave an extension of (1.1), and
[9] gave an improvement
R ∞Yang
(p−1)(1−α) p
of [5]
If α > 0, f, g ≥ 0 satisfy 0 < 0 x
f (x)dx < ∞ and
R ∞as: (q−1)(1−α)
q
0< 0 x
g (x)dx < ∞, then
Z
∞
Z
∞
f (x)g(y)
dxdy
xα + y α
0
0
Z ∞
p1 Z ∞
1q
π
<
x(p−1)(1−α) f p (x)dx
x(q−1)(1−α) g q (x)dx ,
π
0
0
α sin p
(1.7)
π
where the constant α sin(π/p)
is the best possible. Recently, Sulaiman [6] gave
some new forms of (1.1) and Hong [14] gave an extension of Hardy-Hilbert’s
inequality by introducing two parameters λ and α. Yang et al. [13] provided an
extensive account of the above results.
The main objective of this paper is to build a relation to (1.1) and (1.4) with
a best constant factor, Rby Rintroducing the β function and a parameter λ, related
b b
f (x)g(y)
to the double integral a a (u(x)+u(y))
λ dxdy (λ > 0). As applications, the reversion, the equivalent form and some particular results are considered.
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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2.
Some Lemmas
First, we need the formula of the β function as (cf. Wang et al. [7]):
Z ∞
1
(2.1)
B(u, v) :=
tu−1 dt = B(v, u)
(u, v > 0).
u+v
(1
+
t)
0
Lemma 2.1 (cf. [4]). If p > 1, p1 + 1q = 1, ω(σ) > 0, f, g ≥ 0, f ∈ Lpω (E) and
g ∈ Lqω (E), then one has the Hölder’s inequality with weight as:
Z
Z
ω(σ)f (σ)g(σ)dσ ≤
(2.2)
E
ω(σ)f p (σ)dσ
p1 Z
E
ω(σ)g q (σ)dσ
1q
;
E
if p < 1 (p 6= 0), with the above assumption, one has the reverse of (2.2),
where the equality (in the above two cases) holds if and only if there exists nonnegative real numbers c1 and c2 , such that they are not all zero and c1 f p (σ) =
c2 g q (σ), a. e. in E.
Lemma 2.2. If p 6= 0, 1, p1 + 1q = 1, φr = φr (λ) > 0 (r = p, q), φp +φq = λ, and
u(t) is a differentiable strict increasing function in (a, b) (−∞ ≤ a < b ≤ ∞)
such that u(a+) = 0 and u(b−) = ∞, for r = p, q, define ωr (x) as
Z b
0
(u(y))φr −1 u (y)
λ−φr
(2.3)
ωr (x) := (u(x))
dy
(x ∈ (a, b)).
λ
a (u(x) + u(y))
Then for x ∈ (a, b), each ωr (x) is constant, that is
(2.4)
ωr (x) = B(φp , φq )
(r = p, q).
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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Proof. For fixed x, setting v =
u(y)
u(x)
in (2.3), one has
0
b
(u(y))φr −1 u (y)
ωr (x) = (u(x))
dy
λ
λ
a (u(x)) (1 + u(y)/u(x))
Z ∞
(vu(x))φr −1
λ−φr
= (u(x))
u(x)dv
(u(x))λ (1 + v)λ
0
Z ∞ φr −1
v
=
dv
(r = p, q).
(1 + v)λ
0
λ−φr
Z
By (2.1), one has (2.4). The lemma is proved.
Lemma 2.3. If p > 1, p1 + 1q = 1, φr > 0 (r = p, q), satisfy φp + φq = λ, and
u(t) is a differentiable strict increasing function in (a, b) (−∞ ≤ a < b ≤ ∞)
satisfying u(a+) = 0 and u(b−) = ∞, then for c = u−1 (1) and 0 < ε < qφp ,
(2.5)
ε
0
(u(x))φq − p −1 u (x)
φp − qε −1 0
(u(y))
u (y)dxdy
I :=
(u(x) + u(y))λ
c
c
1
ε
ε
> B φp − , φq +
− O(1) ;
ε
q
q
Z bZ
b
if 0 < p < 1 (or p < 0), with the above assumption and 0 < ε < −qφq (or
0 < ε < qφp ), then
ε
ε
1
(2.6)
I < B φp − , φq +
.
ε
q
q
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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Proof. For fixed x, setting v =
b
Z
I :=
φq − pε −1
(u(x))
u(y)
u(x)
in I, one has
"Z
0
u (x)
c
c
Z
b
c
Z
∞
#
ε
(u(y))φp − q −1 0
u (y)dy dx
(u(x) + u(y))λ
ε
1
v φp − q −1 dvdx
λ
(1 + v)
1
u(x)
φp − qε −1
Z ∞
0
u (x)dx
v
dv
=
1+ε
(1 + v)λ
0
c (u(x))
Z 1 φp − qε −1
Z b
0
u(x) v
u (x)
dvdx
−
1+ε
(1 + v)λ
0
c (u(x))
"Z 1
#
ε
Z
Z b 0
u(x)
ε
1 ∞ v φp − q −1
u (x)
v φp − q −1 dv dx
>
dv −
λ
ε 0 (1 + v)
0
c (u(x))
−2
Z ∞ φp − qε −1
v
ε
1
=
dv − φp −
.
λ
ε 0 (1 + v)
q
Z
(2.7)
0
(u(x))−1−ε u (x)
=
b
b
By (2.1), inequality (2.5) is valid. If 0 < p < 1 (or p < 0), by (2.7), one has
Z b
Z ∞
0
ε
u (x)
1
I<
dx
v φp − q −1 dv,
1+ε
λ
(1 + v)
c (u(x))
0
and then by (2.1), inequality (2.6) follows. The lemma is proved.
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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3.
Main Results
Theorem 3.1. If p > 1, p1 + 1q = 1, φr > 0 (r = p, q), φp + φq = λ, u(t) is a
differentiable strict increasing function in (a, b) (−∞ ≤ a < b ≤ ∞), such that
Rb
p(1−φq )−1
u(a+) = 0 and u(b−) = ∞, and f, g ≥ 0 satisfy 0 < a (u(x))
f p (x)dx <
0
(u (x))p−1
Rb
q(1−φp )−1
∞ and 0 < a (u(x))
g q (x)dx < ∞, then
0
(u (x))q−1
Z bZ
(3.1)
a
a
b
f (x)g(y)
dxdy
(u(x) + u(y))λ
Z b
p1
(u(x))p(1−φq )−1 p
< B(φp , φq )
f (x)dx
(u0 (x))p−1
a
Z b
1q
(u(x))q(1−φp )−1 q
×
g (x)dx ,
(u0 (x))q−1
a
where the constant factor B(φp , φq ) is the best possible. If p < 1 (p 6= 0),
{λ; φr > 0 (r = p, q), φp + φq = λ} 6= φ, with the above assumption, one has
the reverse of (3.1), and the constant is still the best possible.
Proof. By (2.2), one has
Z bZ b
f (x)g(y)
dxdy
J :=
λ
a
a (u(x) + u(y))
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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0
(u(x))(1−φq )/q (u (y))1/p
1
=
f (x)
λ (u(y))(1−φp )/p (u0 (x))1/q
a
a (u(x) + u(y))
0
(u(y))(1−φp )/p (u (x))1/q
×
g(y) dxdy
(u(x))(1−φq )/q (u0 (y))1/p
Z b Z b
p1
0
(u(y))φp −1 u (y)
(u(x))(p−1)(1−φq ) p
≤
dy
f (x)dx
λ
(u0 (x))p−1
a
a (u(x) + u(y))
1q
Z b Z b
0
(u(y))(q−1)(1−φp ) q
(u(x))φq −1 u (x)
×
dx
g (y)dy
.
λ
(u0 (y))q−1
a
a (u(x) + u(y))
Z bZ
(3.2)
b
If (3.2) takes the form of equality, then by (2.2), there exist non-negative numbers c1 and c2 , such that they are not all zero and
0
0
u (x)(u(y))(q−1)(1−φp ) q
u (y)(u(x))(p−1)(1−φq ) p
f
(x)
=
c
g (y),
c1
2
(u(y))1−φp (u0 (x))p−1
(u(x))1−φq (u0 (y))q−1
a.e. in (a, b) × (a, b).
It follows that
c1
(u(y))q(1−φp ) q
(u(x))p(1−φq ) p
f
(x)
=
c
g (y) = c3 , a.e. in (a, b) × (a, b),
2
0
(u (x))p
(u0 (y))q
where c3 is a constant. Without loss of generality, suppose c1 6= 0. One has
0
(u(x))p(1−φq )−1 p
c3 u (x)
f
(x)
=
, a.e. in (a, b),
0
(u (x))p−1
c1 u(x)
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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which contradicts 0 <
b
Z
(3.3) J <
a
Rb
a
(u(x))p(1−φq )−1 p
f (x)dx
(u0 (x))p−1
< ∞. Then by (2.3), one has
p1
(u(x))p(1−φq )−1 p
ωp (x)
f (x)dx
(u0 (x))p−1
1q
Z ∞
(u(x))q(1−φp )−1 q
×
ωq (x)
g (x)dx ,
(u0 (x))q−1
0
and in view of (2.4), it follows that (3.1) is valid.
For 0 < ε < qφp , setting fε (x) = gε (x) = 0, x ∈ (a, c) (c = u−1 (1));
ε
0
fε (x) = (u(x))φq − p −1 u (x),
ε
0
gε (x) = (u(x))φp − q −1 u (x),
x ∈ [c, b), we find
Z
(3.4)
a
b
p1
(u(x))p(1−φq )−1 p
fε (x)dx
(u0 (x))p−1
Z b
1q
(u(x))q(1−φp )−1 q
1
×
gε (x)dx
= .
0
q−1
(u (x))
ε
a
If the constant factor B(φp , φq ) in (3.1) is not the best possible, then, there exists
a positive constant k < B(φp , φq ), such that (3.1) is still valid if one replaces
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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B(φp , φq ) by k. In particular, by (2.6) and (3.4), one has
ε
ε
B φp − , φq +
− εO(1)
q
q
Z bZ b
fε (x)gε (y)
dxdy
<ε
λ
a
a (u(x) + u(y))
Z b
p1 Z b
1q
(u(x))p(1−φq )−1 p
(u(x))q(1−φp )−1 q
< εk
fε (x)dx
gε (x)dx
= k,
(u0 (x))p−1
(u0 (x))q−1
a
a
and then B(φp , φq ) ≤ k (ε → 0+ ). This contradicts the fact that k < B(φp , φq ).
Hence the constant factor B(φp , φq ) in (3.1) is the best possible.
For 0 < p < 1 (or p < 0), by the reverse of (2.2) and using the same
procedures, one can obtain the reverse of (3.1). For 0 < ε < −qφq (or 0 < ε <
qφp ), setting fε (x) and gε (x) as the above, we still have (3.4). If the constant
factor B(φp , φq ) in the reverse of (3.1) is not the best possible, then, there exists
a positive constant K > B(φp , φq ), such that the reverse of (3.1) is still valid if
one replaces B(φp , φq ) by K. In particular, by (2.7) and (3.4), one has
ε
ε
B φp − , φq +
q
q
Z bZ b
fε (x)gε (y)
>ε
dxdy
λ
a
a (u(x) + u(y))
Z b
p1 Z b
1q
(u(x))p(1−φq )−1 p
(u(x))q(1−φp )−1 q
> εK
fε (x)dx
gε (x)dx
= K,
(u0 (x))p−1
(u0 (x))q−1
a
a
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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and then B(φp , φq ) ≥ K (ε → 0+ ). This contradiction concludes that the
constant in the reverse of (3.1) is the best possible. The theorem is proved.
Theorem 3.2. Let the assumptions of Theorem 3.1 hold.
(i) If p > 1, p1 +
follows
b
Z
(3.5)
a
1
q
= 1, one obtains the equivalent inequality of (3.1) as
0
u (y)
(u(y))1−pφp
p
f (x)
dx dy
λ
a (u(x) + u(y))
Z b
(u(x))p(1−φq )−1 p
p
< [B(φp , φq )]
f (x)dx;
(u0 (x))p−1
a
Z
b
(ii) If 0 < p < 1, one obtains the reverse of (3.5) equivalent to the reverse of
(3.1);
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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(iii) If p < 0, one obtains inequality (3.5) equivalent to the reverse of (3.1),
where the constants in the above inequalities are all the best possible.
Proof. Set
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0
u (y)
g(y) :=
(u(y))1−pφp
Z
a
b
f (x)
dx
(u(x) + u(y))λ
p−1
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and use (3.1) to obtain
Z b
(u(y))q(1−φp )−1 q
0<
g (y)dy
(u0 (y))q−1
a
p
Z b
Z b
0
u (y)
f (x)
=
dx dy
1−pφp
λ
a (u(y))
a (u(x) + u(y))
Z bZ b
f (x)g(y)
=
dxdy ≤ B(φp , φq )
λ
a
a (u(x) + u(y))
p1
Z b
(u(x))p(1−φq )−1 p
×
f (x)dx
(u0 (x))p−1
a
Z b
1q
(u(y))q(1−φp )−1 q
(3.6)
×
g (y)dy ;
(u0 (y))q−1
a
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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Z
b
q(1−φp )−1
1− 1q
(u(y))
g q (y)dy
0
q−1
(u
(y))
a
(Z
Z b
p ) p1
0
b
u (y)
f (x)
=
dx dy
1−pφp
λ
a (u(y))
a (u(x) + u(y))
Z b
p1
(u(x))p(1−φq )−1 p
f (x)dx
≤ B(φp , φq )
< ∞.
(u0 (x))p−1
a
0<
(3.7)
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It follows that (3.6) takes the form of strict inequality by using (3.1); so does
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(3.7). Hence one can get (3.5). On the other hand, if (3.5) is valid, by (2.2),
Z bZ b
f (x)g(y)
dxdy
λ
a
a (u(x) + u(y))
#"
#
1
1
Z b
Z b"
0
(u (y)) p
(u(y)) p −φp
f (x)
=
dx
g(y) dy
1
1
λ
a
(u(y)) p −φp a (u(x) + u(y))
(u0 (y)) p
(Z
Z b
p ) p1
0
b
u (y)
f (x)
≤
dx dy
1−pφ
λ
p
a (u(x) + u(y))
a (u(y))
Z b
1q
(u(y))q(1−φp )−1 q
(3.8)
×
g (y)dy .
(u0 (y))q−1
a
Hence by (3.5), (3.1) yields. It follows that (3.1) and (3.5) are equivalent.
If the constant factor in (3.5) is not the best possible, one can get a contradiction that the constant factor in (3.1) is not the best possible by using (3.8).
Hence the constant factor in (3.5) is still the best possible.
If 0 < p < 1 (or p < 0), one can get the reverses of (3.6), (3.7) and (3.8),
and thus concludes the equivalence. By (3.6), for 0 < p < 1, one can obtain the
reverse of (3.5); for p < 0, one can get (3.5). If the constant factor in the reverse
of (3.5) (or simply (3.5)) is not the best possible, then one can get a contradiction
that the constant factor in the reverse of (3.1) is not the best possible by using
the reverse of (3.8). Thus the theorem is proved.
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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4.
Some Particular Results
We point out that the constant factors in the following particular results of Theorems 3.1 – 3.2 are all the best possible.
4.1.
The first reversible form
Corollary 4.1. Let the assumptions of Theorems 3.1 – 3.2 hold. For
1
φr = 1 −
(λ − 2) + 1
(r = p, q),
r
b
Z
0<
a
and
(u(x))1−λ p
f (x)dx < ∞
(u0 (x))p−1
b
(u(x))1−λ q
g (x)dx < ∞,
0
q−1
a (u (x))
q+λ−2
setting kλ (p) = B p+λ−2
,
,
p
q
Z
0<
(i) If p > 1, p1 + 1q = 1, λ > 2 − min{p, q} , then we have the following two
equivalent inequalities:
Z bZ
(4.1)
a
a
b
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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f (x)g(y)
dxdy
(u(x) + u(y))λ
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Z
< kλ (p)
a
b
(u(x))1−λ p
f (x)dx
(u0 (x))p−1
p1 Z
a
b
(u(x))1−λ q
g (x)dx
(u0 (x))q−1
1q
and
Z
(4.2)
a
b
0
u (y)
(u(y))(p−1)(1−λ)
Z
a
b
p
f (x)
dx dy
(u(x) + u(y))λ
Z b
(u(x))1−λ p
p
f (x)dx.
< [kλ (p)]
0
p−1
a (u (x))
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
(ii) If 0 < p < 1 and 2 − p < λ < 2 − q, one obtains two equivalent reverses
of (4.1) and (4.2),
Bicheng Yang
(iii) If p < 0 and 2 − q < λ < 2 − p, we have the reverse of (4.1) and the
inequality (4.2), which are equivalent. In particular, by (4.1),
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α
(a) setting u(x) = x (α > 0, x ∈ (0, ∞)), one has
Z
(4.3)
0
∞
Z
∞
f (x)g(y)
dxdy
(xα + y α )λ
0
Z ∞
p1
1
p−1+α(2−λ−p) p
< kλ (p)
x
f (x)dx
α
0
Z ∞
1q
q−1+α(2−λ−q) q
×
x
g (x)dx ;
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(b) setting u(x) = ln x, x ∈ (1, ∞), one has
Z
∞
∞
Z
f (x)g(y)
dxdy
(ln xy)λ
1
Z ∞
p1
p−1
1−λ p
< kλ (p)
x (ln x) f (x)dx
(4.4)
1
1
Z
×
∞
x
q−1
1−λ q
(ln x)
1q
g (x)dx
;
1
(c) setting u(x) = ex , x ∈ (−∞, ∞), one has
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
Z
∞
Z
∞
f (x)g(y)
dxdy
x
y λ
−∞ (e + e )
Z ∞
p1
(2−p−λ)x p
< kλ (p)
e
f (x)dx
(4.5)
−∞
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−∞
Z
∞
×
e(2−q−λ)x g q (x)dx
1q
;
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(d) setting u(x) = tan x, x ∈ (0, π2 ), one has
Z
π
2
Z
(4.6)
0
0
π
2
f (x)g(y)
dxdy
(tan x + tan y)λ
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(Z
π
2
< kλ (p)
0
tan1−λ x p
f (x)dx
sec2(p−1) x
) p1 (Z
π
2
0
tan1−λ x q
g (x)dx
sec2(q−1) x
) 1q
;
(e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has
Z
(4.7)
0
4.2.
π
2
Z
π
2
f (x)g(y)
dxdy
λ
0 (sec x + sec y − 2)
) p1
(Z π
2 (sec x − 1)1−λ
f p (x)dx
< kλ (p)
p−1
0 (sec x tan x)
) 1q
(Z π
2 (sec x − 1)1−λ
.
g q (x)dx
×
q−1
0 (sec x tan x)
Corollary 4.2. Let the assumptions of Theorems 3.1 – 3.2 hold. For
λ−1 1
+
2
r
b
Z
0<
a
Bicheng Yang
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The second reversible form
φr =
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
(r = p, q),
1−λ
(u(x))p 2 p
f (x)dx < ∞
(u0 (x))p−1
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and
Z
0<
a
b
q 1−λ
2
(u(x))
g q (x)dx < ∞,
0
q−1
(u (x))
J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005
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setting keλ (p) = B
pλ−p+2 qλ−q+2
, 2q
2p
,
(i) If p > 1, p1 + 1q = 1, λ > 1 − 2 min{ p1 , 1q } , then one can get two equivalent
inequalities as follows:
Z bZ
(4.8)
a
a
b
f (x)g(y)
dxdy
(u(x) + u(y))λ
) p1
(Z
b
p 1−λ
2
(u(x))
< keλ (p)
f p (x)dx
0
p−1
(u
(x))
a
) 1q
(Z
1−λ
b
(u(x))q 2 q
;
g (x)dx
×
0
q−1
a (u (x))
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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Z
(4.9)
a
b
0
u (y)
p 1−λ
2
(u(y))
Z
a
b
f (x)
dx
(u(x) + u(y))λ
Contents
p
dy
ip Z
e
< kλ (p)
h
a
b
1−λ
(u(x))p 2 p
f (x)dx,
(u0 (x))p−1
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(ii) If 0 < p < 1, 1 − p2 < λ < 1 − 2q , one can get two equivalent reversions of
(4.8) and (4.9),
(iii) If p < 0, 1 − 2q < λ < 1 − p2 , one can get the reversion of (4.8) and
inequality (4.9), which are equivalent. In particular, by (4.8),
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(a) setting u(x) = xα (α > 0, x ∈ (0, ∞)), one has
Z
∞
(4.10)
0
∞
Z
f (x)g(y)
dxdy
(xα + y α )λ
0
Z ∞
p1
1e
p−1+α(1−p 1+λ
)
p
2
< kλ (p)
x
f (x)dx
α
0
1q
Z ∞
q−1+α(1−q 1+λ
)
q
2
×
x
g (x)dx ;
0
(b) setting u(x) = ln x, x ∈ (1, ∞), one has
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
Z
∞
(4.11)
1
∞
Z
f (x)g(y)
dxdy
(ln xy)λ
1
p1
Z ∞
p
p−1
p 1−λ
e
x (ln x) 2 f (x)dx
< kλ (p)
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1
Z
×
∞
q 1−λ
2
xq−1 (ln x)
g q (x)dx
1q
;
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(c) setting u(x) = ex , x ∈ (−∞, ∞), one has
Z
∞
Z
∞
(4.12)
−∞
−∞
f (x)g(y)
dxdy
(ex + ey )λ
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Z
∞
< keλ (p)
(1−p 1+λ
)x p
2
e
p1 Z
∞
f (x)dx
(1−q 1+λ
)x q
2
e
−∞
1q
g (x)dx
;
−∞
(d) setting u(x) = tan x, x ∈ (0, π2 ), one has
Z
π
2
Z
(4.13)
0
0
π
2
f (x)g(y)
dxdy
(tan x + tan y)λ
) p1
(Z π
1−λ
2 tanp 2 x
f p (x)dx
< keλ (p)
2(p−1)
x
0 sec
) 1q
(Z π
1−λ
2 tanq 2 x
;
g q (x)dx
×
2(q−1)
x
0 sec
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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(e) setting u(x) = sec x − 1, x ∈ (0,
Z
(4.14)
0
π
2
Z
π
2
π
),
2
one has
f (x)g(y)
dxdy
λ
0 (sec x + sec y − 2)
(Z π
) p1
1−λ
2 (sec x − 1)p 2
< keλ (p)
f p (x)dx
p−1
(sec
x
tan
x)
0
) 1q
(Z π
1−λ
2 (sec x − 1)q 2
g q (x)dx
×
.
q−1
(sec
x
tan
x)
0
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4.3.
The form which does not have a reverse
Corollary 4.3. Let the assumptions of Theorems 3.1 – 3.2 hold. For
φr =
λ
1 1
(r = p, q),
if p > 1, + = 1, λ > 0,
r
p q
Z b
(u(x))(p−1)(1−λ) p
0<
f (x)dx < ∞
(u0 (x))p−1
a
and
b
(u(x))(q−1)(1−λ) q
0<
g (x)dx < ∞,
(u0 (x))q−1
a
then one can get two equivalent inequalities as:
Z bZ b
f (x)g(y)
(4.15)
dxdy
λ
a
a (u(x) + u(y))
Z b
p1
λ λ
(u(x))(p−1)(1−λ) p
<B
,
f (x)dx
p q
(u0 (x))p−1
a
Z b
1q
(u(x))(q−1)(1−λ) q
×
g (x)dx ;
(u0 (x))q−1
a
Z
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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Z
(4.16)
a
b
0
u (y)
(u(y))1−λ
Z
a
b
p
f (x)
dx dy
(u(x) + u(y))λ
p Z b
λ λ
(u(x))(p−1)(1−λ) p
< B
,
f (x)dx.
p q
(u0 (x))p−1
a
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In particular, by (4.15),
(a) setting u(x) = xα (α > 0; x ∈ (0, ∞)), one has
Z
∞
(4.17)
0
∞
Z
f (x)g(y)
dxdy
(xα + y α )λ
0
Z ∞
p1
λ λ
1
(p−1)(1−αλ) p
< B
,
x
f (x)dx
α
p q
0
Z ∞
1q
(q−1)(1−αλ) q
×
x
g (x)dx ;
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
0
Bicheng Yang
(b) setting u(x) = ln x, x ∈ (1, ∞), one has
Z
(4.18)
∞
∞
Z
f (x)g(y)
dxdy
(ln xy)λ
1
1
Z ∞
p1
λ λ
p−1
(p−1)(1−λ) p
<B
,
x (ln x)
f (x)dx
p q
1
Z ∞
1q
q−1
(q−1)(1−λ) q
×
x (ln x)
g (x)dx ;
1
(c) setting u(x) = ex , x ∈ (−∞, ∞), one has
Z
∞
Z
∞
(4.19)
−∞
−∞
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f (x)g(y)
dxdy
(ex + ey )λ
J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005
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<B
λ λ
,
p q
Z
∞
(1−p)λx p
e
p1 Z
∞
f (x)dx
−∞
(1−q)λx q
e
g (x)dx
1q
;
−∞
(d) setting u(x) = tan x, x ∈ (0, π2 ), one has
Z
π
2
(4.20)
0
Z
π
2
f (x)g(y)
dxdy
λ
0 (tan x + tan y)
) p1
(Z π
2 tan(p−1)(1−λ) x
λ λ
f p (x)dx
<B
,
2(p−1) x
sec
p q
0
) 1q
(Z π
2 tan(q−1)(1−λ) x
;
g q (x)dx
×
2(q−1) x
sec
0
(e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has
Z
(4.21)
π
2
Z
π
2
f (x)g(y)
dxdy
λ
0
0 (sec x + sec y − 2)
) p1
(Z π
2 (sec x − 1)(p−1)(1−λ)
λ λ
<B
,
f p (x)dx
p−1
p q
(sec
x
tan
x)
0
(Z π
) 1q
2 (sec x − 1)(q−1)(1−λ)
×
g q (x)dx
.
q−1
(sec
x
tan
x)
0
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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Remark 1. For α = 1, (4.3) reduces to (1.5). For λ = 1, inequalities (4.3),
(4.10) and (4.17) reduce to (1.7), and inequalities (4.4), (4.11) and (4.18) reduce
to (1.4). It follows that inequality (3.5) is a relation between (1.4) and (1.7)(ro
(1.1)) with a parameter λ. Still for λ = 1, (4.5), (4.12) and (4.19) reduce to
Z ∞Z ∞
f (x)g(y)
(4.22)
dxdy
x
y
−∞ −∞ e + e
Z ∞
p1 Z ∞
1q
π
(1−q)x q
(1−p)x p
e
g (x)dx ,
<
e
f (x)dx
−∞
−∞
sin πp
(4.6), (4.13) and (4.20) reduce to
Z πZ π
2
2
f (x)g(y)
dxdy
(4.23)
0
0 tan x + tan y
(Z π
) p1 (Z π
) 1q
2
2
π
<
cos2(p−1) xf p (x)dx
cos2(q−1) xg q (x)dx
,
π
0
0
sin p
and (4.7), (4.14) and (4.21) reduce to
Z πZ π
2
2
f (x)g(y)
dxdy
(4.24)
0
0 sec x + sec y − 2
(Z π ) p1 (Z π ) 1q
p−1
q−1
2
2
2
cos2 x
π
cos
x
<
f p (x)dx
g q (x)dx
.
π
sin
x
sin
x
0
0
sin p
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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References
[1] G. H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, Cambridge, 1952.
[2] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Inequalities Involving Functions and Their Integrals and Derivatives, Kluwer Academic Publishers, Boston, 1991.
[3] H.P. MULHOLLAND, Some theorems on Dirichlet series with positive
coefficients and related integrals, Proc. London Math. Soc., 29(2) (1929),
281–292.
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
[4] JICHANG KUANG, Applied Inequalities, Shangdong Science and Technology Press, Jinan, 2004.
Bicheng Yang
[5] JICHANG KUANG, On a new extension of Hilbert’s integral inequality,
J. Math. Anal. Appl., 235 (1999), 608–614.
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[6] W.T. SULAIMAN, On Hardy-Hilbert’s integral inequality, J. Inequal. in
Pure and Appl. Math., 5(2) (2004), Art. 25. [ONLINE: http://jipam.
vu.edu.au/article.php?sid=385]
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[7] ZHUXI WANG AND DUNRIN GUO, An Introduction to Special Functions, Science Press, Beijing, 1979.
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[8] BICHENG YANG, On a general Hardy-Hilbert’s inequality with a best
value, Chinese Annals of Math., 21A(4) (2000), 401–408.
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[9] BICHENG YANG, On an extension of Hardy-Hilbert’s inequality, Chinese Annals of Math., 23A(2) (2002), 247–254.
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[10] BICHENG YANG, On Hardy-Hilbert’s integral inequality, J. Math. Anal.
Appl., 261 (2001), 295–306.
[11] BICHENG YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl.,
220 (1998), 778–785.
[12] BICHENG YANG, On new inequality similar to Hardy-Hilbert’s inequality, Math. Ineq. and Appl., 6(1) (2003), 37–44.
[13] BICHENG YANG AND Th.M. RASSIAS, On the way of weight coefficient and research for the Hilbert-type inequalities, Math. Ineq. Appl., 6(4)
(2003), 625–658.
[14] HONG YONG, A extension and improvement of Hardy-Hilbert’s double
series inequality, Math. Practice and Theory, 32(5) (2002), 850–854.
A Relation to Hardy-Hilbert’s
Integral Inequality and
Mulholland’s Inequality
Bicheng Yang
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