Journal of Inequalities in Pure and
Applied Mathematics
ON HARDY-HILBERT’S INTEGRAL INEQUALITY WITH
PARAMETERS
volume 4, issue 5, article 94,
2003.
LEPING HE, MINGZHE GAO AND WEIJIAN JIA
Department of Mathematics and Computer Science,
Normal College, Jishou University,
Jishou Hunan, 416000
People’s Republic of China.
EMail: lianheping@163.com
EMail: mingzhegao@163.com
EMail: jwj1959@163.com
Received 01 August, 2003;
accepted 28 August, 2003.
Communicated by: L. Debnath
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
108-03
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Abstract
In this paper, by means of a sharpening of Hölder’s inequality, Hardy-Hilbert’s
integral inequality with parameters is improved. Some new inequalities are
established.
2000 Mathematics Subject Classification: 26D15, 46C99
Key words: Hardy-Hilbert integral inequality, Hölder’s inequality, Weight function,
Beta function.
On Hardy-Hilbert’s Integral
Inequality with Parameters
The authors thank the referee for his help and patience in improving the paper.
Leping He, Mingzhe Gao and
Weijian Jia
Contents
Title Page
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Lemmas and their Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
5
9
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1.
Introduction
Let p > 1, p1 +
+∞, then
1
q
∞
Z
Z
(1.1)
0
0
= 1, f, g > 0. If 0 <
∞
R∞
0
f p (t)dt < +∞, 0 <
R∞
0
g q (t)dt <
f (x)g(y)
dxdy
x+y
π
<
sin (π/p)
Z
0
∞
p1 Z
f (t) dt
p
∞
1q
g (t) dt ,
q
0
π
where the constant sin(π/p)
is best possible. The inequality (1.1) is well known as
Hardy-Hilbert’s integral inequality. In recent years, some improvements and extensions of Hilbert’s inequality and Hardy-Hilbert’s inequality have been given
in [2] – [6], Yang [2] gave a generalization of (1.1) as follows:
If λ > 2 − min{p, q}, α < T 6 ∞ then
Z
(1.2)
T
Z
T
f (x)g(y)
dxdy
λ
α
α (x + y − 2α)
(Z "
#
) p1
p+λ−2
T
p
t−α
<
kλ (p) − θλ (p)
(t − α)1−λ f p (t)dt
T
−
α
α
(Z "
#
) 1q
q+λ−2
T
q
t−α
×
kλ (p) − θλ (q)
(t − α)1−λ g q (t)dt
T
−
α
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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(T < ∞)
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and
Z
∞
(1.3)
α
Z
∞
f (x)g(y)
dxdy
(x + y − 2α)λ
α
Z ∞
p1 Z
1−λ p
< kλ (p)
(t − α) f (t)dt
α
∞
1q
g (t)dt ,
1−λ q
(t − α)
α
where
p+λ−2 q+λ−2
kλ (p) = B
,
,
p
q
(2−λ)/r
Z 1
1
1
θλ (r) =
du
λ
u
0 (1 + u)
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
(r = p, q).
Title Page
The main purpose of this paper is to build a few new inequalities which
include improvements of the inequalities (1.2) and (1.3), and extensions of corresponding results in [3] – [5].
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2.
Lemmas and their Proofs
For convenience, we firstly introduce some notations:
r
s
Z
T
(f , g ) =
r
s
Z
f (x)g (x) dx,
kf kp =
α
p1
T
p
f (x) dx
,
kf k2 = kf k .
α
We next introduce a function defined by
Sr (H, x) = H r/2 , x kHk−r/2
,
r
On Hardy-Hilbert’s Integral
Inequality with Parameters
where x is a parametric variable vector which is a variable unit vector. Under
the general case, it is properly chosen such that the specific problems discussed
are simplified.
p/2
Clearly, Sr (H, x) = 0 when the vector x selected is orthogonal
n
o to H .
Throughout this paper, the exponent m indicates m = min p1 , 1q , α < T 6
∞.
In order to verify our assertions, we need to build the following lemmas.
Leping He, Mingzhe Gao and
Weijian Jia
Lemma 2.1. Let f (x) , g (x) > 0,
and 0 < kgkq < +∞, then
(2.1)
1
p
+
1
q
= 1 and p > 1. If 0 < kf kp < +∞
where R = (Sp (f, h) − Sq (g, h)) , khk = 1, f
linearly independent.
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(f, g) < kf kp kgkq (1 − R)m ,
2
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p/2
(x), g
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q/2
(x)and h(x) are
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Proof. First of all, we discuss the case of p 6= q. Without loss of generality,
p
suppose that p > q > 1, since p1 + 1q = 1, we have p > 2. Let R = p2 , Q = p−2
.
1
1
Then R + Q = 1. By Hölder’s inequality we obtain,
Z T
(2.2)
(f, g) =
f (x)g (x) dx
a
Z T
f · g q/p g 1−(q/p) dx
=
a
Z
T
f ·g
6
q/p R
R1 Z
dx
g
a
= f p/2 , g
T
1−(q/p) Q
Q1
dx
a
2
q/2 p
q (1− 2 )
kgkq p
Leping He, Mingzhe Gao and
Weijian Jia
.
And the equality in (2.2) holds if and only if f p/2 and g q/2 are linearly dependent. In fact, the equality in (2.2) holds if and only if, there exists a c1 such that
R
Q
f · g q/p = c1 g 1−(q/p) . It is easy to deduce that f p/2 = c1 g q/2 .
In our previous paper [3], with the help of the positive definiteness of the
Gram matrix, we established an important inequality of the form
(α, β)2 6 kαk2 kβk2 − (kαk x − kβk y)2 = kαk2 kβk2 (1 − γ)
2
y
x
where γ = kαk
− kβk
, x = (β, γ) , y = (α, γ) with kγk = 1 and xy > 0.
The equality in (2.3) holds if and only if α and β are linearly dependent; or the
vector γ is a linear combination of α and β, and xy = 0 but x 6= y. If α, β and
γ in (2.3) are replaced by f p/2 , g q/2 and h respectively, then we get
2
(2.4)
f p/2 , g q/2 6 kf kpp kgkqq (1 − R) ,
(2.3)
On Hardy-Hilbert’s Integral
Inequality with Parameters
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where R = (Sp (f, h) − Sq (g, h))2 with ||h|| = 1. The equality in (2.4) holds
if and only if f p/2 and g q/2 are linearly dependent, or h is a linear combination
of f p/2 and g q/2 , and f p/2 , h g q/2 , h = 0, but f p/2 , h 6= g q/2 , h . Since
f p/2 and g q/2 are linearly independent, it is impossible to have equality in (2.4).
Substituting (2.4) into (2.2), we obtain after simplifications
1
(f, g) < kf kp kgkq (1 − R) p .
(2.5)
Provided that h(x) is properly chosen, then R 6= 0 is achieved. (The choice of
h(x) is quite flexible, as long as condition khk = 1 is satisfied, on which we
can refer to [3, 4], etc.). Noticing the symmetry of p and q, the inequality (2.1)
follows from (2.5).
Next, we discuss the case of p = q. According to the hypothesis: when f, g
and h are linearly independent, we immediately obtain from (2.3) the following
result:
1
(f, g) < kf k kgk (1 − r̄) 2 ,
2
(f,h)
(g,h)
where r̄ = kf k − kgk , and khk = 1. Thus the lemma is proved.
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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Lemma 2.2. Let p > 1, p1 + 1q = 1, λ > 2 − min{p, q}, α < T < ∞. Define
the weight function ωλ as follows:
Z
(2.6)
T
ωλ (α, T, r, x) =
α
1
(x + y − 2α)λ
x−α
y−α
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2−λ
r
dy
Setting ωλ (α, ∞, r, x) = limT →∞ ωλ (α, T, r, x) and kλ (p) = B
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x ∈ (α, T ].
Page 7 of 19
p+λ−2 q+λ−2
, q
p
,
J. Ineq. Pure and Appl. Math. 4(5) Art. 94, 2003
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Z
θλ (r) =
(2.7)
0
1
1
(1 + u)λ
(2−λ)(1−1/r)
1
du,
u
(r = p, q),
then we have
ωλ (α, ∞, r, x) = kλ (p)(x − α)1−λ ,
(2.8)
x ∈ (α, ∞)
and
On Hardy-Hilbert’s Integral
Inequality with Parameters
(2.9) ωλ (α, T, r, x)
<
kλ (p) − θ(r)
x−α
T −α
1+(λ−2)(1−1/r)
Leping He, Mingzhe Gao and
Weijian Jia
!
(x − α)1−λ , x ∈ (α, T ),
where B(m, n) is the beta function.
The proof of this lemma is given in the paper [2]; it is omitted here.
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3.
Main Results
In order to state it conveniently, we need again to define the functions and introduce some notations
2−λ
2−λ
f (x)
x − α pq
g(y)
y − α pq
, G=
,
F =
(x + y − 2α)λ/p y − α
(x + y − 2α)λ/q x − α
Z
T
Z
T
F p/2 hT dxdy
Sp (F, hT ) =
α
Z
T
α
Z
G
α
q/2
Z
T
T
Z
α
T
Sq (G, hT ) =
T
Z
F p dxdy
,
α
Z
T
hT dxdy
α
− 21
q
− 21
G dxdy
α
,
Leping He, Mingzhe Gao and
Weijian Jia
α
where hT = hT (x, y) is a unit vector with two variants, namely
Z T Z T
12
2
= 1, α < T 6 ∞,
||hT || =
hT dxdy
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
α
and F p/2 , Gq/2 , hT are linearly independent.
Theorem 3.1. Let p > 1, p1 + 1q = 1, λ > 2 − min {p, q}, α < T 6 ∞,
f (t), g(t) > 0. If
Z ∞
0<
(t − α)1−λ f p (t) dt < +∞ and
Zα∞
0<
(t − α)1−λ g q (t) dt < +∞,
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α
then
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(i) For T < ∞, we have
Z TZ T
f (x)g(y)
(3.1)
dxdy
λ
α
α (x + y − 2α)
(Z
) p1
(p+λ−2)/p !
T
t−α
<
kλ (p) − θλ (p)
(t − α)1−λ f p (t)dt
T −α
α
) 1q
(Z
(q+λ+2)/q !
T
t−α
(t − α)1−λ g q (t)dt
(1−RT )m ,
×
kλ (p) − θλ (q)
T
−
α
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
where
p+λ−2 q+λ−2
kλ (p) = B
,
,
p
q
2−λ
Z 1
1
1 r
θλ (r) =
du (r = p, q).
λ
u
0 (1 + u)
(ii) For T = ∞, we have
Z ∞Z ∞
f (x)g(y)
dxdy
(3.2)
(x + y − 2α)λ
α
α
Z ∞
p1
1−λ p
< kλ (p)
(t − α) f (t)dt
α
Z
∞
×
α
1q
1−λ q
(t − α) g (t)dt (1 − R∞ )m ,
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where RT = (Sp (F, hT ) − Sq (G, hT ))2 ,
(3.3)
 1
14
a−x
2

e
x
−
a
2


, T = ∞;
1

 π
(x + y − 2a) 2 y − a
hT (x, y) =


T −α
T −α

T −α


e(1− 2(x−α) − 2(y−α) ) , T < ∞.
(x − α)(y − α)
Proof. By Lemma 2.1, we get
Z TZ T
f (x)g(y)
(3.4)
dxdy
λ
α
α (x + y − 2α)
Z TZ T
=
F Gdxdy
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
Title Page
α
T
Z
Z
1
p
T
Z
p
T
Z
T
G dxdy
F dxdy
6
α
Z
T
=
α
α
q
1
q
Contents
(1 − RT )m
α
p1
p
ωλ (α, β, q, t)f (t) dt
α
Z
T
×
1q
q
ωλ (α, β, p, t)g (t) dt (1 − RT )m ,
α
where ωλ (α, T, r, t) (r = p, q) is the function defined by (2.6) .
Now notice that θλ (p) = θλ (q), θλ (q) = θλ (p) and substituting (2.9) and
(2.8) into (3.4) respectively, the inequalities (3.1) and (3.2) follow.
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It remains to discuss the expression of RT . We may choose the function hT
indicated by (3.3).
When T = ∞, setting s = x − α, t = y − α, then
Z
∞
Z
∞
h2∞ (x, y)dxdy
||h∞ || =
α
12
α
Z ∞
Z ∞
s 12 12
2
1
dt
= 1.
=
e−2s ds
π 0
s+t t
0
When T < ∞, setting ξ =
T
Z
Z
α
η=
T
khT k =
Z
T −α
,
x−α
h2T dxdy
T −α
,
y−α
On Hardy-Hilbert’s Integral
Inequality with Parameters
then we have
Leping He, Mingzhe Gao and
Weijian Jia
12
α
T
T
−α
−α
T − α (1− Tx−α
T − α (1− Ty−α
) dy
) dx ·
e
e
2
2
(x
−
α)
(y
−
α)
α
α
Z ∞
21
Z ∞
=
e1−ξ dξ ·
e1−η dη
= 1.
Z
12
=
1
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1
According to Lemma 2.1 and the given hT , we have RT = (Sp (F, hT ) −
Sq (G, hT ))2 . It is obvious that F p/2 , Gq/2 and hT are linearly independent,
so it is impossible for equality to hold in (3.4). Thus the proof of theorem is
completed.
Remark 3.1. Clearly, the inequalities (3.1) and (3.2) are the improvements of
(1.2) and (1.3) respectively .
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Owing to p, q > 1, when λ = 1, 2, 3, the condition λ > 2 − min(p, q) is
satisfied, then we have
γ1
Z 1
Z 1
1
1
1
θ1 (r) =
du >
du = ln 2,
u
0 1+u
0 1+u
1 1
π
k1 (p) = B
,
=
,
q p
sin(π/p)
Z 1
1
1
θ2 (r) =
2 du = ,
2
0 (1 + u)
p+2−2 q+2−2
k2 (p) = B
,
= B(1, 1) = 1,
p
q
− γ1
Z 1
Z 1
1
1
u
1
θ3 (r) =
du >
du = ,
3
3
u
8
0 (1 + u)
0 (1 + u)
1 1
(p − 1)π
1
k3 (p) =
B
,
= 2
.
2pq
q p
2p sin(π/p)
By Theorem 3.1, some corollaries are established as follows:
Corollary 3.2. If p > 1, p1 + 1q = 1, λ = 1, α < T 6 ∞ and f (t), g(t) > 0,
RT
RT
0 < α f p (t)dt < +∞ and 0 < α g q (t)dt < +∞ , then we have
Z TZ T
f (x)g(y)
(3.5)
dxdy
α
α x + y − 2α
(Z
!
) p1
1q
T
π
t−α
<
−
· ln 2 f p (t)dt
sin(π/p)
T −α
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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(Z
T
×
α
π
−
sin(π/p)
t−α
T −α
p1
) 1q
!
· ln 2
· g q (t)dt
(1 − r1 )m ,
for T < ∞,
and
Z
∞
Z
∞
(3.6)
α
α
f (x)g(y)
dxdy
x + y − 2α
Z ∞
p1 Z ∞
1q
π
p
q
<
f (t)dt
g (t)dt (1 − r1 )m .
sin(π/p)
α
α
Remark 3.2. When α = 0 and p = q = 2, the inequality (3.6) is reduced to a
result which is equivalent to inequality (3.1) in [3] after simple computations.
As a result, the inequalities (3.1), (3.2) and (3.5) – (3.6) are all extensions of
(3.1) in [3].
Corollary 3.3. Let p > 1,
1
p
+
1
q
= 1, α < T 6 ∞ and f (t), g(t) > 0. If
Z
T
1
f p (t) dt < +∞
t−α
0<
α
and
Z
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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T
0<
α
1 q
g (t) dt < +∞,
t−α
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then we obtain
Z
(3.7)
T
T
Z
f (x)g(y)
2 dxdy
α
α (x + y − 2α)
Z T p1
t−α
1
p
<
1−
· f (t)dt
2 (T − α) t − α
α
1q
Z T 1
t−α
×
1−
· g q (t)dt (1 − r2 )m ,
2
(T
−
α)
t
−
α
α
for T < ∞,
and
Z
∞
Z
∞
(3.8)
α
α
f (x)g(y)
dxdy
(x + y − 2α)2
Z ∞
p1 Z ∞
1q
1
1 q
p
<
f (t)dt
g (t)dt (1 − r2 )m .
t−α
t−α
α
α
Corollary 3.4. If p > 1,
1
p
+
1
q
= 1, λ = 3, α < T 6 ∞ and f (t), g(t) > 0,
Z
T
1
f p (t)dt < +∞,
2
(t − α)
1
g q (t)dt < +∞,
(t − α)2
0<
α
T
Z
0<
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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then we get
T
Z
T
Z
(3.9)
α
α
(Z
f (x)g(y)
dxdy
(x + y − 2α)3
T
(p − 1)π
1
−
2p2 sin(π/p) 8
<
α
(Z
T
(p − 1)π
1
−
2p2 sin(π/p) 8
×
α
t−α
T −α
t−α
T −α
1+ p1 !
1+ 1q !
) p1
1
f p (t)dt
(t − α)2
) 1q
1
g q (t)dt
(t − α)2
(1 − r3 )m
On Hardy-Hilbert’s Integral
Inequality with Parameters
T < ∞,
Leping He, Mingzhe Gao and
Weijian Jia
and
Z
∞
Z
∞
f (x)g(y)
(3.10)
dxdy
(x + y − 2α)3
α
α
Z ∞
p1
(p − 1)π
1
p
f (t)dt
< 2
2p sin(π/p)
(t − α)2
α
1q
Z ∞
1
q
×
g (t)dt (1 − r3 )m .
2
(t
−
α)
α
Since kλ (2) = B λ2 , λ2 , θλ (2) = 12 B λ2 , λ2 , and λ > 2 − min(2, 2) =
0, we also have
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Corollary 3.5. If p = q = 2, λ > 0, α < T 6 ∞ and f (t), g(t) > 0,
Z T
0<
(t − α)1−λ f 2 (t) dt < +∞,
α
Z T
0<
(t − α)1−λ g 2 (t) dt < +∞,
α
then we have
Z TZ
(3.11)
T
f (x)g(y)
dxdy
(x + y − 2α)λ
) 12
(Z T "
λ/2 #
1 t−α
λ λ
1−
(t − α)1−λ f 2 (t)dt
<B
,
2 2
2 T −α
α
(Z "
) 12
λ/2 #
T
1 t−α
1−
(1 − R)m ,
×
(t − α)1−λ g 2 (t)dt
2 T −α
α
α
α
for T < ∞
and
Z
∞
Z
∞
(3.12)
α
α
f (x)g(y)
dxdy
(x + y − 2α)λ
Z ∞
12
λ λ
1−λ 2
,
<B
(t − α)
f (t)dt
2 2
α
Z ∞
12
1−λ 2
e m.
×
(t − α)
g (t)dt (1 − R)
α
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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Remark 3.3. The inequalities (3.11), (3.12) are new generalizations of (20) in
[4] and improvements of the inequalities (4) and (12) in [6] respectively.
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ. Press, Cambridge, UK, 1952.
[2] BICHENG YANG, On Hardy-Hilbert’s integral inequality, J. Math. Anal.
Appl., 261 (2001), 295–306.
[3] MINGZHE GAO, LI TAN AND L. DEBNATH, Some improvements on
Hilbert’s integral inequality, J. Math. Anal. Appl., 229 (1999), 682–689.
[4] MINGZHE GAO, On the Hilbert inequality, Zeitschrift für Analysis und
ihre Anwendungen, 18(4) (1999), 1117–1122.
[5] MINGZHE GAO, SHANG RONG WEI AND LEPING HE, On the Hilbert
inequality with weights, Zeitschrift für Analysis und ihre Anwendungen,
21(1) (2002), 257–263.
On Hardy-Hilbert’s Integral
Inequality with Parameters
Leping He, Mingzhe Gao and
Weijian Jia
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Contents
[6] BICHENG YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl.,
220 (1998), 778–785.
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