Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 4, Article 112, 2005
A RELATION TO HARDY-HILBERT’S INTEGRAL INEQUALITY AND
MULHOLLAND’S INEQUALITY
BICHENG YANG
D EPARTMENT OF M ATHEMATICS
G UANGDONG I NSTITUTE OF E DUCATION
G UANGZHOU , G UANGDONG 510303
P. R. C HINA
bcyang@pub.guangzhou.gd.cn
Received 14 November, 2004; accepted 25 August, 2005
Communicated by L. Pick
A BSTRACT. 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.
Key words and phrases: Hardy-Hilbert’s integral inequality; Mulholland’s integral inequality; β function; Hölder’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
R∞
R∞
If p > 1, p1 + 1q = 1, f, g ≥ 0 satisfy 0 < 0 f p (x)dx < ∞ and 0 < 0 g q (x)dx < ∞, then
one has two equivalent inequalities as (see [1]):
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
π
q
p
(1.1)
dxdy <
f (x)dx
g (x)dx ;
π
x+y
0
0
0
0
sin p
Z
∞
Z
(1.2)
0
0
∞
f (x)
dx
x+y
p
p

dy < 
sin
π 
π
p
Z
∞
f p (x)dx,
0
h
ip
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 analysis and its applications
(cf. Mitrinovic et al. [2]).
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
223-04
2
B ICHENG YANG
R∞
R∞
If 0 < 1 x1 F p (x)dx < ∞ and 0 < 1 y1 Gq (y)dy < ∞, then the Mulholland’s integral
inequality is as follows (see [1, 3]):
p1 Z ∞ q
1q
Z ∞ p
Z ∞Z ∞
π
F (x)
G (y)
F (x)F (y)
dx
dy ,
(1.3)
dxdy <
x
y
xy ln xy
1
1
1
1
sin π
p
π
sin(π/p)
where the constant factor
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 ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
π
q−1 q
p−1 p
(1.4)
dxdy <
x g (x)dx .
x f (x)dx
π
ln xy
1
1
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. R[10] gave some extensions of (1.1)
R ∞ and (1.2) as: 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 ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
1−λ p
1−λ q
(1.5)
dxdy < kλ (p)
x f (x)dx
x g (x)dx
(x + y)λ
0
0
0
0
and
Z
∞
y
(1.6)
0
(p−1)(λ−1)
Z
0
∞
f (x)
dx
(x + y)λ
p
p
Z
dy < [kλ (p)]
∞
x1−λ f p (x)dx,
0
q+λ−2
where the constant factors kλ (p) and [kλ (p)]p (kλ (p) = B p+λ−2
,
, B(u, v) is the β
p
q
function) are all the best possible. By introducing a parameter α, Kuang [5] gave an extension
of [5] as: If α > 0, f, g ≥ 0 satisfy
R ∞ of (1.1), and Yang [9] gave an improvement
R∞
0 < 0 x(p−1)(1−α) f p (x)dx < ∞ and 0 < 0 x(q−1)(1−α) g q (x)dx < ∞, then
Z ∞Z ∞
f (x)g(y)
(1.7)
dxdy
xα + y α
0
0
Z ∞
p1 Z ∞
1q
π
(p−1)(1−α) p
(q−1)(1−α) q
x
f (x)dx
x
g (x)dx ,
<
0
0
α sin πp
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
by introducing the β function and a parameter λ, related to the double integral
R b R b factor,
f (x)g(y)
dxdy (λ > 0). As applications, the reversion, the equivalent form and some
a a (u(x)+u(y))λ
particular results are considered.
2. S OME L EMMAS
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
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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A R ELATION TO H ARDY-H ILBERT ’ S I NTEGRAL I NEQUALITY
AND
M ULHOLLAND ’ S I NEQUALITY
3
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)
p1 Z
p
ω(σ)f (σ)dσ
E
E
1q
q
ω(σ)g (σ)dσ
;
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 non-negative 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
(2.3)
λ−φr
Z
b
ωr (x) := (u(x))
a
0
(u(y))φr −1 u (y)
dy
(u(x) + u(y))λ
(x ∈ (a, b)).
Then for x ∈ (a, b), each ωr (x) is constant, that is
ωr (x) = B(φp , φq )
(2.4)
Proof. For fixed x, setting v =
u(y)
u(x)
(r = p, q).
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
ε
(u(x))φq − p −1 u (x)
0
I :=
(u(y))φp − q −1 u (y)dxdy
λ
(u(x) + u(y))
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
(2.6)
1
ε
ε
I < B φp − , φq +
.
ε
q
q
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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4
B ICHENG YANG
u(y)
u(x)
Proof. For fixed x, setting v =
b
Z
I :=
φq − pε −1
(u(x))
in I, one has
"Z
0
u (x)
c
c
Z
b
Z
b
#
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
(1 + v)
c
u(x)
Z ∞ φp − qε −1
Z 1 φp − qε −1
Z b 0
Z b
0
u(x) v
v
u (x)
u (x)dx
dvdx
dv
−
=
1+ε
1+ε
(1 + v)λ
(1 + v)λ
0
c (u(x))
0
c (u(x))
"Z 1
#
ε
Z
Z b 0
u(x)
ε
u (x)
1 ∞ v φp − q −1
v φp − q −1 dv dx
>
dv −
ε 0 (1 + v)λ
(u(x))
c
0
−2
Z ∞ φp − qε −1
1
ε
v
dv − φp −
.
=
λ
ε 0 (1 + v)
q
=
(2.7)
0
(u(x))−1−ε u (x)
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.
3. M AIN R ESULTS
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 u(a+) = 0 and u(b−) = ∞, and
R b (u(x))q(1−φp )−1 q
Rb
p(1−φq )−1
p
f
(x)dx
<
∞
and
0
<
g (x)dx < ∞,
f, g ≥ 0 satisfy 0 < a (u(x))
0
a
(u (x))p−1
(u0 (x))q−1
then
Z bZ
(3.1)
a
b
f (x)g(y)
dxdy
λ
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
< B(φp , φq )
f (x)dx
g (x)dx ,
(u0 (x))p−1
(u0 (x))q−1
a
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)
J :=
dxdy
λ
a
a (u(x) + u(y))
Z bZ b
0
1
(u(x))(1−φq )/q (u (y))1/p
=
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
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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A R ELATION TO H ARDY-H ILBERT ’ S I NTEGRAL I NEQUALITY
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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 (u(x) + u(y))
Z b Z b
1q
0
(u(x))φq −1 u (x)
(u(y))(q−1)(1−φp ) q
×
dx
g (y)dy
.
λ
(u0 (y))q−1
a
a (u(x) + u(y))
Z b Z
≤
a
(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 (y)(u(x))(p−1)(1−φq ) p
u (x)(u(y))(q−1)(1−φp ) q
c1
f
(x)
=
c
g (y),
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
(u(y))q(1−φp ) q
(u(x))p(1−φq ) p
c1
f
(x)
=
c
g (y) = c3 , a.e. in (a, b) × (a, b),
2
(u0 (x))p
(u0 (y))q
where c3 is a constant. Without loss of generality, suppose c1 6= 0. One has
0
c3 u (x)
(u(x))p(1−φq )−1 p
f (x) =
, a.e. in (a, b),
0
p−1
(u (x))
c1 u(x)
Rb
p(1−φq )−1
which contradicts 0 < a (u(x))
f p (x)dx < ∞. Then by (2.3), one has
0
(u (x))p−1
Z
(3.3) J <
a
b
(u(x))p(1−φq )−1 p
ωp (x)
f (x)dx
(u0 (x))p−1
p1 Z
0
∞
(u(x))q(1−φp )−1 q
g (x)dx
ωq (x)
(u0 (x))q−1
1q
,
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 b
p1 Z b
1q
(u(x))p(1−φq )−1 p
(u(x))q(1−φp )−1 q
1
(3.4)
fε (x)dx
gε (x)dx
= .
0
0
p−1
q−1
(u (x))
(u (x))
ε
a
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 B(φp , φq ) by k. In particular,
by (2.6) and (3.4), one has
ε
ε
− εO(1)
B φp − , φq +
q
q
Z bZ b
fε (x)gε (y)
<ε
dxdy
λ
a
a (u(x) + u(y))
Z b
1q
p1 Z b
(u(x))p(1−φq )−1 p
(u(x))q(1−φp )−1 q
gε (x)dx
= k,
< εk
fε (x)dx
(u0 (x))q−1
(u0 (x))p−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
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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6
B ICHENG YANG
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
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 + 1q = 1, one obtains the equivalent inequality of (3.1) as follows
Z
(3.5)
a
b
0
u (y)
(u(y))1−pφp
Z
a
b
f (x)
dx
(u(x) + u(y))λ
p
dy
p
Z
< [B(φp , φq )]
a
b
(u(x))p(1−φq )−1 p
f (x)dx;
(u0 (x))p−1
(ii) If 0 < p < 1, one obtains the reverse of (3.5) equivalent to the reverse of (3.1);
(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
Z b
p−1
0
f (x)
u (y)
g(y) :=
dx
,
(u(y))1−pφp a (u(x) + u(y))λ
and use (3.1) to obtain
Z b
(u(y))q(1−φp )−1 q
0<
g (y)dy
(u0 (y))q−1
a
Z b
p
Z b
0
u (y)
f (x)
dx dy
=
1−pφp
λ
a (u(x) + u(y))
a (u(y))
Z bZ b
f (x)g(y)
=
dxdy ≤ B(φp , φq )
λ
a
a (u(x) + u(y))
Z b
p1 Z b
1q
(u(x))p(1−φq )−1 p
(u(y))q(1−φp )−1 q
(3.6)
f (x)dx
g (y)dy ;
×
(u0 (x))p−1
(u0 (y))q−1
a
a
1− 1q
(u(y))q(1−φp )−1 q
0<
g (y)dy
(u0 (y))q−1
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
≤ B(φp , φq )
f (x)dx
< ∞.
(u0 (x))p−1
a
Z
(3.7)
b
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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A R ELATION TO H ARDY-H ILBERT ’ S I NTEGRAL I NEQUALITY
AND
M ULHOLLAND ’ S I NEQUALITY
7
It follows that (3.6) takes the form of strict inequality by using (3.1); so does (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
f (x)
(u(y)) p −φp
=
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
f (x)
u (y)
≤
dx dy
1−pφp
λ
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.
4. S OME PARTICULAR R ESULTS
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 φr = (1 − 1r )(λ − 2) +
R b (u(x))1−λ p
R b (u(x))1−λ q
1(r = p, q), 0 < a (u
g (x)dx < ∞, setting
0
p−1 f (x)dx < ∞ and 0 <
a (u0 (x))q−1
(x))
kλ (p) = B p+λ−2
, q+λ−2
,
p
q
(i) If p > 1, p1 + 1q = 1, λ > 2 − min{p, q} , then we have the following two equivalent
inequalities:
Z bZ b
f (x)g(y)
(4.1)
dxdy
λ
a
a (u(x) + u(y))
Z b
1q
p1 Z b
(u(x))1−λ p
(u(x))1−λ q
g (x)dx
< kλ (p)
f (x)dx
0
0
q−1
p−1
a (u (x))
a (u (x))
and
Z b
p
Z b
Z b
0
u (y)
f (x)
(u(x))1−λ p
p
(4.2)
dx
dy
<
[k
(p)]
f (x)dx.
λ
0
(p−1)(1−λ)
λ
p−1
a (u(y))
a (u(x) + u(y))
a (u (x))
(ii) If 0 < p < 1 and 2 − p < λ < 2 − q, one obtains two equivalent reverses of (4.1) and
(4.2),
(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),
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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8
(4.3)
(4.4)
B ICHENG YANG
(a) setting u(x) = xα (α > 0, x ∈ (0, ∞)), one has
Z ∞Z ∞
f (x)g(y)
dxdy
(xα + y α )λ
0
0
1q
Z ∞
p1 Z ∞
1
q−1+α(2−λ−q) q
p−1+α(2−λ−p) p
x
g (x)dx ;
< kλ (p)
x
f (x)dx
α
0
0
(b) setting u(x) = ln x, x ∈ (1, ∞), one has
Z ∞Z ∞
f (x)g(y)
dxdy
(ln xy)λ
1
1
Z ∞
p1 Z
< kλ (p)
xp−1 (ln x)1−λ f p (x)dx
1
(4.5)
(4.7)
xq−1 (ln x)1−λ g q (x)dx
1q
;
1
(c) setting u(x) = ex , x ∈ (−∞, ∞), one has
Z ∞Z ∞
f (x)g(y)
dxdy
x
y λ
−∞ −∞ (e + e )
Z ∞
p1 Z
(2−p−λ)x p
< kλ (p)
e
f (x)dx
−∞
(0, π2 ),
(4.6)
∞
∞
(2−q−λ)x q
e
g (x)dx
1q
;
−∞
(d) setting u(x) = tan x, x ∈
one has
Z πZ π
2
2
f (x)g(y)
dxdy
λ
0
0 (tan x + tan y)
) p1 (Z π
) 1q
(Z π
2
2
tan1−λ x p
tan1−λ x q
f (x)dx
g (x)dx
< kλ (p)
;
2(p−1) x
2(q−1) x
0 sec
0 sec
(e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has
Z πZ π
2
2
f (x)g(y)
dxdy
λ
0
0 (sec x + sec y − 2)
(Z π
) p1 (Z π
) 1q
2 (sec x − 1)1−λ
2 (sec x − 1)1−λ
f p (x)dx
g q (x)dx
.
< kλ (p)
p−1
q−1
0 (sec x tan x)
0 (sec x tan x)
4.2. The second reversible form.
Corollary 4.2. Let the assumptions of Theorems 3.1 – 3.2 hold. For φr = λ−1
+ 1r (r =
2
Rb
R b (u(x))q 1−λ
p 1−λ
2
2
p
p, q), 0 < a (u(x))
f
(x)dx
<
∞
and
0
<
g q (x)dx < ∞, setting keλ (p) =
0
p−1
a (u0 (x))q−1
(u
(x))
B pλ−p+2
, qλ−q+2
,
2p
2q
(i) If p > 1, p1 + 1q = 1, λ > 1 − 2 min{ p1 , 1q } , then one can get two equivalent inequalities
as follows:
Z bZ b
f (x)g(y)
(4.8)
dxdy
λ
a
a (u(x) + u(y))
(Z
) p1 (Z
) 1q
b
b
p 1−λ
q 1−λ
2
2
(u(x))
(u(x))
< keλ (p)
f p (x)dx
g q (x)dx
;
0
0
p−1
q−1
(u
(x))
(u
(x))
a
a
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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A R ELATION TO H ARDY-H ILBERT ’ S I NTEGRAL I NEQUALITY
0
b
Z
u (y)
(4.9)
a
p 1−λ
2
(u(y))
Z
b
a
f (x)
dx
(u(x) + u(y))λ
p
AND
M ULHOLLAND ’ S I NEQUALITY
b
ip Z
e
dy < kλ (p)
h
a
9
1−λ
(u(x))p 2 p
f (x)dx,
(u0 (x))p−1
(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),
(a) setting u(x) = xα (α > 0, x ∈ (0, ∞)), one has
Z ∞Z ∞
f (x)g(y)
dxdy
(4.10)
(xα + y α )λ
0
0
1q
Z ∞
p1 Z ∞
1+λ
1e
)
q
p−1+α(1−p 1+λ
)
p
q−1+α(1−q
2
2
g (x)dx ;
< kλ (p)
x
f (x)dx
x
α
0
0
(b) setting u(x) = ln x, x ∈ (1, ∞), one has
Z ∞Z ∞
f (x)g(y)
(4.11)
dxdy
(ln xy)λ
1
1
Z ∞
p1 Z
1−λ
p−1
p
p
< keλ (p)
x (ln x) 2 f (x)dx
1
(4.12)
∞
x
q−1
q 1−λ
q
2
(ln x)
1q
g (x)dx
;
1
(c) setting u(x) = ex , x ∈ (−∞, ∞), one has
Z ∞Z ∞
f (x)g(y)
dxdy
x
y λ
−∞ −∞ (e + e )
Z ∞
p1 Z
1+λ
< keλ (p)
e(1−p 2 )x f p (x)dx
−∞
∞
)x
(1−q 1+λ
2
e
g q (x)dx
1q
;
−∞
(d) setting u(x) = tan x, x ∈ (0, π2 ), one has
Z
π
2
Z
π
2
(4.13)
0
0
f (x)g(y)
dxdy
(tan x + tan y)λ
(Z π
) p1 (Z π
) 1q
1−λ
1−λ
2 tanp 2 x
2 tanq 2 x
f p (x)dx
g q (x)dx
< keλ (p)
;
2(p−1) x
2(q−1) x
sec
sec
0
0
(e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has
Z
(4.14)
0
π
2
Z
π
2
f (x)g(y)
dxdy
λ
0 (sec x + sec y − 2)
(Z π
) p1 (Z π
) 1q
1−λ
1−λ
2 (sec x − 1)p 2
2 (sec x − 1)q 2
f p (x)dx
g q (x)dx
< keλ (p)
.
p−1
q−1
0 (sec x tan x)
0 (sec x tan x)
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 = λr (r = p, q), if p >
Rb
Rb
(p−1)(1−λ)
(q−1)(1−λ)
1, p1 + 1q = 1, λ > 0 , 0 < a (u(x))
f p (x)dx < ∞ and 0 < a (u(x))
g q (x)dx <
(u0 (x))p−1
(u0 (x))q−1
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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10
B ICHENG YANG
∞, then one can get two equivalent inequalities as:
Z bZ
(4.15)
b
f (x)g(y)
dxdy
λ
a
a (u(x) + u(y))
1q
p1 Z b
Z b
(u(x))(p−1)(1−λ) p
(u(x))(q−1)(1−λ) q
λ λ
g (x)dx ;
,
f (x)dx
<B
(u0 (x))q−1
p q
(u0 (x))p−1
a
a
b
Z
(4.16)
a
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
,
f (x)dx.
< B
p q
(u0 (x))p−1
a
In particular, by (4.15),
(a) setting u(x) = xα (α > 0; x ∈ (0, ∞)), one has
Z
∞
∞
Z
f (x)g(y)
dxdy
(xα + y α )λ
0
Z ∞
p1 Z ∞
1q
λ λ
1
(p−1)(1−αλ) p
(q−1)(1−αλ) q
,
< B
x
f (x)dx
x
g (x)dx ;
α
p q
0
0
(4.17)
0
(b) setting u(x) = ln x, x ∈ (1, ∞), one has
Z
∞
∞
Z
(4.18)
1
1
f (x)g(y)
dxdy < B
(ln xy)λ
λ λ
,
p q
Z
p1
∞
x
p−1
(p−1)(1−λ) p
(ln x)
f (x)dx
1
Z
×
∞
x
q−1
(q−1)(1−λ) q
(ln x)
g (x)dx
1q
;
1
(c) setting u(x) = ex , x ∈ (−∞, ∞), one has
Z
∞
Z
∞
(4.19)
−∞
−∞
f (x)g(y)
dxdy
(ex + ey )λ
Z ∞
p1 Z ∞
1q
λ λ
(1−p)λx p
(1−q)λx q
<B
,
e
f (x)dx
e
g (x)dx ;
p q
−∞
−∞
(d) setting u(x) = tan x, x ∈ (0, π2 ), one has
Z
(4.20)
π
2
Z
π
2
f (x)g(y)
dxdy
λ
0
0 (tan x + tan y)
) p1 (Z π
) 1q
(Z π
2 tan(p−1)(1−λ) x
2 tan(q−1)(1−λ) x
λ λ
<B
,
f p (x)dx
g q (x)dx
;
2(p−1) x
2(q−1) x
p q
sec
sec
0
0
J. Inequal. Pure and Appl. Math., 6(4) Art. 112, 2005
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A R ELATION TO H ARDY-H ILBERT ’ S I NTEGRAL I NEQUALITY AND M ULHOLLAND ’ S I NEQUALITY
11
(e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has
Z
π
2
Z
(4.21)
0
0
π
2
f (x)g(y)
dxdy
(sec x + sec y − 2)λ
) p1
(Z π
2 (sec x − 1)(p−1)(1−λ)
λ λ
f p (x)dx
<B
,
p−1
(sec
x
tan
x)
p q
0
(Z π
) 1q
2 (sec x − 1)(q−1)(1−λ)
.
×
g q (x)dx
q−1
(sec
x
tan
x)
0
Remark 4.4. 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−p)x p
(1−q)x q
<
e
f (x)dx
e
g (x)dx ,
−∞
−∞
sin πp
(4.6), (4.13) and (4.20) reduce to
Z πZ π
2
2
f (x)g(y)
(4.23)
dxdy
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
2
π
cos x
cos x
<
f p (x)dx
g q (x)dx
.
sin x
sin x
0
0
sin πp
R EFERENCES
[1] G. H. HARDY, J.E. LITTLEWOOD
Cambridge, 1952.
AND
G. POLYA, Inequalities, Cambridge University Press,
[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.
[4] JICHANG KUANG, Applied Inequalities, Shangdong Science and Technology Press, Jinan, 2004.
[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]
[7] ZHUXI WANG AND DUNRIN GUO, An Introduction to Special Functions, Science Press, Beijing,
1979.
[8] BICHENG YANG, On a general Hardy-Hilbert’s inequality with a best value, Chinese Annals of
Math., 21A(4) (2000), 401–408.
[9] BICHENG YANG, On an extension of Hardy-Hilbert’s inequality, Chinese Annals of Math.,
23A(2) (2002), 247–254.
[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.
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