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volume 4, issue 2, article 33, 2003.

Received 25 May, 2002;

accepted 7 April, 2003.

Communicated by:S.S. Dragomir

Abstract Contents

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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:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 059-02

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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= 2−1p1q = 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)|p1p

and

(2.10) |g(y)| ≤y1p

Z y

0

|g0(δ)|q1q

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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)|p

1pZ y

0

|g0(δ)|q1q

xp−1

p + yq−1 q

Z x

0

|f0(τ)|p

1pZ y

0

|g0(δ)|q1q (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)|p

1pZ y

0

|g0(δ)|q1q

.

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)|p1p Ry

0 |g0(δ)|q1q

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)|p1p Ry

0 |g0(δ)|q1q

x+y dxdy

= Z

0

Z

0

Rx

0 |f0(τ)|p1p (x+y)p1

x y

pq1 Ry

0 |g0(δ)|q1q (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

,p+λ−2 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

,p+λ−2 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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[2] BICHENG YANG, A generalized Hilbert’s integral inequality with the best constant, Ann. of Math.(Chinese), 21A:4 (2000), 401–408.

[3] JICHANG KUANG,On new extensions of Hilbert’s integral inequality, J.

Math. Anal. Appl., 235 (1999), 608–614.

[4] B.G. PACHPATTE, Inequalities similar to certain extensions of Hilbert’s inequality, J.Math. Anal. Appl., 243 (2000), 217–227.

[5] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequal- ity, J. Math. Anal. Appl., 226 (1998), 166–179.

[6] D.S. MITRINOVI ´C, Analytic Inequalities, Springer-Verlag, Berlin/New York,1970.

[7] E.F. BECKENBACH AND R. BELLMAN, Inequalities, Springer-Verlag, Berlin, 1983.

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