volume 4, issue 2, article 33, 2003.
Received 25 May, 2002;
accepted 7 April, 2003.
Communicated by:S.S. Dragomir
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Journal of Inequalities in Pure and Applied Mathematics
SOME NEW INEQUALITIES SIMILAR TO HILBERT-PACHPATTE TYPE INEQUALITIES
ZHONGXUE LÜ
Department of Basic Science of Technology College, Xuzhou Normal University, 221011,
People’s Republic of China.
E-Mail:lvzx1@163.net
2000c Victoria University ISSN (electronic): 1443-5756 059-02
Some New Inequalities Similar to Hilbert-Pachpatte Type
Inequalities Zhongxue Lü
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Abstract
In this paper, some new inequalities similar to Hilbert-Pachpatte type inequali- ties are given.
2000 Mathematics Subject Classification:26D15.
Key words: Inequalities, Hilbert-Pachpatte inequalities, Hölder inequality.
Contents
1 Introduction. . . 3 2 Main Results . . . 5 3 Discrete Analogues. . . 13
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1. Introduction
In [1, Chap. 9], the well-known Hardy-Hilbert inequality is given as follows.
Theorem 1.1. Let p >1, q >1, 1p + 1q = 1, am, bn≥0, 0 < P∞
n=1apn<∞, 0<P∞
n=1bqn<∞.Then
(1.1)
∞
X
m=1
∞
X
n=1
ambn
(m+n)λ ≤ π sin(π/p)
∞
X
m=1
apm
!1p ∞ X
n=1
bqn
!1q
where sin(π/p)π is best possible.
The integral analogue of the Hardy-Hilbert inequality can be stated as fol- lows
Theorem 1.2. Letp >1, q >1, 1p + 1q= 1, f(x), g(y)≥0, 0<R∞
0 fp(x)dx <
∞, 0<R∞
0 gq(y)dy <∞.Then (1.2)
Z ∞
0
Z ∞
0
f(x)g(y)
x+y dxdy≤ π sin(π/p)
Z ∞
0
fp(x)dx
1pZ ∞
0
gq(y)dy 1q
,
where sin(π/p)π is best possible.
In [1, Chap. 9] the following extension of Hardy-Hilbert’s double-series theorem is given.
Theorem 1.3. Letp >1, q >1, 1p +1q ≥1, 0< λ= 2−1p− 1q = 1p +1q ≤1.
Then
∞
X
m=1
∞
X
n=1
ambn
(m+n)λ ≤K
∞
X
m=1
apm
!p1 ∞ X
n=1
bqn
!1q ,
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whereK =K(p, q)depends onpandqonly.
The following integral analogue of Theorem 1.3 is also given in [1, Chap.
9].
Theorem 1.4. Under the same conditions as in Theorem1.1we have Z ∞
0
Z ∞
0
f(x)g(y)
(x+y)λdxdy ≤K Z ∞
0
fpdx
1pZ ∞
0
gqdy 1q
,
whereK =K(p, q)depends onpandqonly.
The inequalities in Theorems1.1 and1.2were studied by Yang and Kuang (see [2,3]). In [4,5], some new inequalities similar to the inequalities given in Theorems1.1,1.2,1.3and1.4were established.
In this paper, we establish some new inequalities similar to the Hilbert- Pachpatte inequality.
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2. Main Results
In what follows we denote by Rthe set of real numbers. Let N = {1,2, . . .}, N0 ={0,1,2, . . .}. We define the operator∇by∇u(t) = u(t)−u(t−1)for any functionudefined on N. For any functionu(t) : [0,∞) → R, we denote byu0the derivatives ofu.
First we introduce some Lemmas.
Lemma 2.1. (see [2]). Letp > 1, q >1, 1p +1q = 1,λ >2−min{p, q}, define the weight functionω1(q, x)as
ω1(q, x) :=
Z ∞
0
1 (x+y)λ
x y
2−λq
dy, x∈[0,∞).
Then
(2.1) ω1(q, x) =B
q+λ−2
q ,p+λ−2 p
x1−λ,
whereB(p, q)isβ-function.
Lemma 2.2. (see [3]). Letp > 1, q >1, 1p +1q = 1,λ >2−min{p, q}, define the weight functionω2(q, x)as
ω2(q, x) :=
Z ∞
0
1 xλ +yλ
x y
2−λq
dy, x∈[0,∞).
Then
(2.2) ω2(q, x) = 1 λB
q+λ−2
qλ ,p+λ−2 pλ
x1−λ.
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Lemma 2.3. Let p > 1, q > 1, 1p + 1q = 1, λ > 2−min{p, q}, define the weight functionω3(q, m)as
ω3(q, m) :=
∞
X
n=1
1 (m+n)λ
m n
2−λq
, m∈ {1,2, . . .}.
Then
(2.3) ω3(q, m)< B
q+λ−2
q ,p+λ−2 p
m1−λ,
whereB(p, q)isβ-function.
Proof. By Lemma2.1,we have
ω3(q, m)<
Z ∞
0
1 (m+y)λ
m y
2−λq dy
=B
q+λ−2
q ,p+λ−2 p
m1−λ.
The proof is completed.
Lemma 2.4. Let p > 1, q > 1, 1p + 1q = 1, λ > 2−min{p, q}, define the weight functionω4(q, m)as
ω4(q, m) :=
∞
X
n=1
1 mλ +nλ
m n
2−λq
, m∈ {1,2, . . .}.
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Then
(2.4) ω4(q, m)< 1 λB
q+λ−2
qλ ,p+λ−2 pλ
m1−λ. Proof. By Lemma2.2, we have
ω4(q, m)<
Z ∞
0
1 mλ+yλ
m y
2−λq dy
= 1 λB
q+λ−2
qλ ,p+λ−2 pλ
m1−λ.
The proof is completed.
Our main result is given in the following theorem.
Theorem 2.5. Letp >1, 1p +1q = 1, andf(x), g(y)be real-valued continuous functions defined on[0,∞), respectively, and letf(0) =g(0) = 0, and
0<
Z ∞
0
Z x
0
|f0(τ)|pdτ dx <∞, 0<
Z ∞
0
Z y
0
|g0(δ)|qdδdy <∞.
Then
(2.5) Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(qxp−1+pyq−1)(x+y)dxdy
≤ π
sin(π/p)pq Z ∞
0
Z x
0
|f0(τ)|pdτ dx
1p Z ∞
0
Z y
0
|g0(δ)|qdδdy 1q
.
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In particular, whenp=q = 2, we have
(2.6) Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(x+y)2 dxdy
≤ π 2
Z ∞
0
Z x
0
|f0(τ)|2dτ dx
12 Z ∞
0
Z y
0
|g0(δ)|2dδdy 12
.
Proof. From the hypotheses, we have the following identities
(2.7) f(x) =
Z x
0
f0(τ)dτ,
and
(2.8) g(y) =
Z y
0
g0(δ)dδ
forx, y ∈ (0,∞). From (2.7) and (2.8) and using Hölder’s integral inequality, respectively, we have
(2.9) |f(x)| ≤x1q
Z x
0
|f0(τ)|pdτ 1p
and
(2.10) |g(y)| ≤y1p
Z y
0
|g0(δ)|qdδ 1q
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forx, y ∈(0,∞). From (2.9) and (2.10) and using the elementary inequality (2.11) z1z2 ≤ z1p
p +z2q
q , z1 ≥0, z2 ≥0, 1 p +1
q = 1, p > 1,
we observe that
|f(x)| |g(y)| ≤x1qy1p Z x
0
|f0(τ)|pdτ
1pZ y
0
|g0(δ)|qdδ 1q
≤
xp−1
p + yq−1 q
Z x
0
|f0(τ)|pdτ
1pZ y
0
|g0(δ)|qdδ 1q (2.12)
forx, y ∈(0,∞). From (2.12) we observe that (2.13) |f(x)| |g(y)|
qxp−1+pyq−1 ≤ 1 pq
Z x
0
|f0(τ)|pdτ
1pZ y
0
|g0(δ)|qdδ 1q
.
Hence (2.14)
Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(qxp−1+pyq−1)(x+y)dxdy
≤ 1 pq
Z ∞
0
Z ∞
0
Rx
0 |f0(τ)|pdτ1p Ry
0 |g0(δ)|qdδ1q
x+y dxdy.
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By Hölder’s integral inequality and (2.1), we have Z ∞
0
Z ∞
0
Rx
0 |f0(τ)|pdτ1p Ry
0 |g0(δ)|qdδ1q
x+y dxdy
= Z ∞
0
Z ∞
0
Rx
0 |f0(τ)|pdτ1p (x+y)p1
x y
pq1 Ry
0 |g0(δ)|qdδ1q (x+y)1q
y x
pq1 dxdy
≤ Z ∞
0
Z ∞
0
Rx
0 |f0(τ)|pdτ x+y
x y
1q dxdy
!p1
× Z ∞
0
Z ∞
0
Ry
0 |g0(δ)|qdδ x+y
y x
1p dxdy
1q
≤ π
sin(π/p) Z ∞
0
Z x
0
|f0(τ)|pdτ dx
p1 Z ∞
0
Z y
0
|g0(δ)|qdδdy 1q (2.15)
by (2.14) and (2.15), we get (2.5). The proof of Theorem2.5is complete.
In a similar way to the proof of Theorem 2.5, we can prove the following theorems.
Theorem 2.6. Letp > 1, 1p + 1q = 1, λ > 2−min{p, q}, andf(x), g(y)be real-valued continuous functions defined on[0,∞),respectively, and letf(0) = g(0) = 0, and
0<
Z ∞
0
Z x
0
x1−λ|f0(τ)|pdτ dx <∞, 0<
Z ∞
0
Z y
0
y1−λ|g0(δ)|qdδdy <∞,
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then
(2.16) Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(qxp−1+pyq−1)(x+y)λdxdy
≤ B
q+λ−2
q ,p+λ−2p pq
Z ∞
0
Z x
0
x1−λ|f0(τ)|pdτ dx 1p
× Z ∞
0
Z y
0
y1−λ|g0(δ)|qdδdy 1q
. In particular, whenp=q = 2,
(2.17) Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(x+y)1+λ dxdy
≤ B λ2,λ2 2
Z ∞
0
Z x
0
x1−λ|f0(τ)|2dτ dx 12
× Z ∞
0
Z y
0
y1−λ|g0(δ)|2dδdy 12
.
Theorem 2.7. Letp > 1, 1p + 1q = 1, λ > 2−min{p, q}, andf(x), g(y)be real-valued continuous functions defined on[0,∞), respectively, and letf(0) = g(0) = 0, and
0<
Z ∞
0
Z x
0
x1−λ|f0(τ)|pdτ dx <∞, 0<
Z ∞
0
Z y
0
y1−λ|g0(δ)|qdδdy <∞.
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Then
(2.18) Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(qxp−1+pyq−1)(xλ+yλ)dxdy
≤ B
q+λ−2
qλ ,p+λ−2pλ λpq
Z ∞
0
Z x
0
x1−λ|f0(τ)|pdτ dx 1p
× Z ∞
0
Z y
0
y1−λ|g0(δ)|qdδdy 1q
. In particular, whenp=q = 2,
(2.19) Z ∞
0
Z ∞
0
|f(x)| |g(y)|
(xλ+yλ)(x+y)dxdy
≤ π 2λ
Z ∞
0
Z x
0
x1−λ|f0(τ)|2dτ dx 12
× Z ∞
0
Z y
0
y1−λ|g0(δ)|2dδdy 12
.
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3. Discrete Analogues
Theorem 3.1. Let p > 1, 1p + 1q = 1, and {a(m)} and {b(n)} be two se- quences of real numbers where m, n ∈ N0, and a(0) = b(0) = 0, and 0 <
P∞ m=1
Pm
τ=1|∇a(τ)|p <∞,0<P∞ n=1
Pn
δ=1|∇b(δ)|q <∞, then (3.1)
∞
X
m=1
∞
X
n=1
|am| |bn|
(qmp−1+pnq−1)(m+n)
≤ π
sin(π/p)pq
∞
X
m=1 m
X
k=1
apk
!1p ∞ X
n=1 n
X
r=1
bqr
!1q .
In particular, whenp=q = 2, we have
(3.2)
∞
X
m=1
∞
X
n=1
|am| |bn| (m+n)2 ≤ π
2
∞
X
m=1 m
X
k=1
a2k
!12 ∞ X
n=1 n
X
r=1
b2r
!12 .
Proof. From the hypotheses, it is easy to observe that the following identities hold
(3.3) am =
m
X
τ=1
∇a(τ),
and
(3.4) bn=
n
X
δ=1
∇b(δ)
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form, n∈N. From (3.3) and (3.4) and using Hölder’s inequality, we have
(3.5) |am| ≤m1q
m
X
τ=1
|∇a(τ)|p
!1p ,
and
(3.6) |bn| ≤n1p
n
X
δ=1
|∇b(δ)|q
!1q
form, n∈N. From (3.5) and (3.6) and using the elementary inequality (2.11), we observe that
|am| |bn| ≤m1qnp1
m
X
τ=1
|∇a(τ)|p
!1p n X
δ=1
|∇b(δ)|q
!1q
≤
mp−1
p + nq−1 q
m
X
τ=1
|∇a(τ)|p
!1p n X
δ=1
|∇b(δ)|q
!1q (3.7)
form, n∈N. From (3.7), we observe that (3.8) |am| |bn|
qmp−1+pnq−1 ≤ 1 pq
m
X
τ=1
|∇a(τ)|p
!1p n X
δ=1
|∇b(δ)|q
!1q .
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Hence (3.9)
∞
X
m=1
∞
X
n=1
|am| |bn|
(qmp−1+pnq−1)(m+n)
≤ 1 pq
∞
X
m=1
∞
X
n=1
(Pm
τ=1|∇a(τ)|p)1p(Pn
δ=1|∇b(δ)|q)1q
m+n .
By the Hölder inequality and (2.3)
∞
X
m=1
∞
X
n=1
(Pm
τ=1|∇a(τ)|p)1p(Pn
δ=1|∇b(δ)|q)1q m+n
=
∞
X
m=1
∞
X
n=1
(Pm
τ=1|∇a(τ)|p)1p (m+n)1p
m n
pq1 (Pn
δ=1|∇b(δ)|q)1q (m+n)1q
n m
pq1
≤
∞
X
m=1
∞
X
n=1
Pm
τ=1|∇a(τ)|p m+n
m n
1q
!1p
×
∞
X
m=1
∞
X
n=1
Pn
δ=1|∇b(δ)|q m+n
n m
1p
!1q
< π sin(π/p)
∞
X
m=1 m
X
τ=1
|∇a(τ)|p
!1p ∞ X
n=1 n
X
δ=1
|∇b(δ)|q
!1q (3.10)
by (3.9) and (3.10), we get (3.1). The proof of Theorem3.1is complete.
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In a similar manner to the proof of Theorem3.1, we can prove the following theorems.
Theorem 3.2. Letp > 1, 1p+1q = 1,λ >2−min{p, q}, and{a(m)}and{b(n)}
be two sequences of real numbers wherem, n∈N0,anda(0) =b(0) = 0, and
0<
∞
X
m=1 m
X
τ=1
m1−λ|∇a(τ)|p <∞,
0<
∞
X
n=1 n
X
δ=1
n1−λ|∇b(δ)|q <∞,
then
(3.11)
∞
X
m=1
∞
X
n=1
|am| |bn|
(qmp−1+pnq−1)(m+n)λ
≤
Bq+λ−2
q ,p+λ−2p pq
∞
X
m=1 m
X
τ=1
m1−λ|∇a(τ)|p
!1p
×
∞
X
n=1 n
X
δ=1
n1−λ|∇b(δ)|q
!1q .
In particular, whenp=q = 2, we have
(3.12)
∞
X
m=1
∞
X
n=1
|am| |bn| (m+n)1+λ
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≤ B λ2,λ2 2
∞
X
m=1 m
X
τ=1
m1−λ|∇a(τ)|2
!12 ∞ X
n=1 n
X
δ=1
n1−λ|∇b(δ)|2
!12 .
Theorem 3.3. Letp > 1, 1p+1q = 1,λ >2−min{p, q}, and{a(m)}and{b(n)}
be two sequences of real numbers wherem, n∈N0, anda(0) = b(0) = 0, and
0<
∞
X
m=1 m
X
τ=1
m1−λ|∇a(τ)|p <∞,
0<
∞
X
n=1 n
X
δ=1
n1−λ|∇b(δ)|q <∞,
then
(3.13)
∞
X
m=1
∞
X
n=1
|am| |bn|
(qmp−1+pnq−1)(mλ+nλ)
≤ B
q+λ−2
qλ ,p+λ−2pλ λpq
∞
X
m=1 m
X
τ=1
m1−λ|∇a(τ)|p
!1p
×
∞
X
n=1 n
X
δ=1
n1−λ|∇b(δ)|q
!1q .
In particular, whenp=q = 2, we have
(3.14)
∞
X
m=1
∞
X
n=1
|am| |bn| (m+n)(mλ+nλ)
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≤ π 2λ
∞
X
m=1 m
X
τ=1
m1−λ|∇a(τ)|2
!12 ∞ X
n=1 n
X
δ=1
n1−λ|∇b(δ)|2
!12 .
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References
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