• 検索結果がありません。

JJ II

N/A
N/A
Protected

Academic year: 2022

シェア "JJ II"

Copied!
14
0
0

読み込み中.... (全文を見る)

全文

(1)

volume 1, issue 2, article 22, 2000.

Received 8 May, 2000;

accepted 10 June 2000.

Communicated by:L. Debnath

Abstract Contents

JJ II

J I

Home Page Go Back

Close Quit

Journal of Inequalities in Pure and Applied Mathematics

ON A STRENGTHENED HARDY-HILBERT INEQUALITY

BICHENG YANG

Department of Mathematics Guangdong Education College Guangzhou

Guangdong 510303

PEOPLE’S REPUBLIC OF CHINA EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 012-00

(2)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page2of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

Abstract

In this paper, a new inequality for the weight coefficientW(n, r)of the form

W(n, r) =

X

m=0

1 m+n+ 1

n+12 m+12

!1r

< π

sin πr− 1

13 (n+ 1) (2n+ 1)1−1r

(r >1, n∈N0=N∪ {0})

is proved. This is followed by a strengthened version of the more accurate Hardy-Hilbert inequality.

2000 Mathematics Subject Classification:26D15

Key words: Hardy-Hilbert inequality, Weight Coefficient, Hölder’s inequality.

Contents

1 Introduction. . . 3 2 Some Lemmas. . . 6 3 Theorem and Remarks . . . 10

References

(3)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page3of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

1. Introduction

Ifp >1,p1+1q = 1,an, bn≥0, and0<P

n=1−λapn <∞,0<P

n=1−λbqn <∞ (λ= 0,1), then the Hardy-Hilbert inequality is

(1.1)

X

m=1−λ

X

n=1−λ

ambn

m+n+λ < π sin

π p

X

n=1−λ

apn

!1p X

n=1−λ

bqn

!1q ,

where the constant π/sin(π/p) is best possible (see [3, Chapter 9]). In- equality (1.1) is important in analysis and it’s applications (see [4, Chapter 5]).

In recent years, Yang and Gao [2,7], have given a strengthened version of (1.1) forλ= 0as

(1.2)

X

m=1

X

n=1

ambn m+n <

X

n=1

 π sin

π p

− 1−γ np1

apn

1 p

×

X

n=1

 π sin

π p

−1−γ n1q

bqn

1 q

,

whereγ = 0.5772+, is Euler’s constant. Later, Yang and Debnath [6] proved a

(4)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page4of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

distinctly strengthened version of (1.1) forλ = 0as

(1.3)

X

m=1

X

n=1

ambn m+n <

X

n=1

 π sin

π p

− 1 2n1p +n1q

apn

1 p

×

X

n=1

 π sin

π p

− 1 2n1q +n1p

bqn

1 q

,

which is not comparable with (1.2).

Inequality (1.1) for λ = 1 is called the more accurate Hardy-Hilbert’s in- equality, which has been strengthened as

(1.4)

X

m=0

X

n=0

ambn m+n+ 1

<

X

n=0

 π sin

π p

− ln 2−γ (2n+ 1)1+1p

apn

1 p

×

X

n=0

 π sin

π p

− ln 2−γ (2n+ 1)1+1q

bqn

1 q

,

by introducing an inequality of the weight coefficientW(n, r)in the form (see

(5)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page5of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

[5, equation (2.9)]):

W(n, r) =

X

m=0

1 m+n+ 1

n+12 m+ 12

1 r

(1.5)

< π

sin πr − ln 2−γ (2n+ 1)2−1r

(r >1, n∈N0).

In this paper we will give another strengthened version of (1.1) for λ = 1, which is not comparable with (1.4). We need some preparatory works.

(6)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page6of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

2. Some Lemmas

Lemma 2.1. If f ∈ C6[0,∞), f(2q)(x) > 0 (q= 0,1,2,3), f(i−1)(∞) = 0 (i= 1,2, . . . ,6), andP

n=0f(n)<∞, then we have

X

k=2

(−1)kf(k)> 1

2f(2) (see [1, equation (4.4)]), (2.1)

X

m=0

f(m) = Z

0

f(t)dt+ 1

2f(0) + Z

0

1f0(t)dt, (2.2)

Z 0

1f0(t)dt =− 1

12f0(0) +δ2, δ2 = 1 6

Z 0

3f000(t)dt <0, (2.3)

wherei(t) (i= 1,3)are Bernoulli functions (see [5, equations (1.7)-(1.9)]).

Setting the weight coefficientW(n, r)in the form:

W(n, r) =

X

m=0

1 m+n+ 1

n+12 m+ 12

1r (2.4)

= π

sin πr − θ(n, r) (2n+ 1)2−1r

(r >1, n∈N0), then we find

(2.5)

θ(n, r) = π

sin πr(2n+ 1)2−1r −(2n+ 1)2

X

m=0

1 m+n+ 1

1 2m+ 1

1r .

(7)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page7of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

If we define the functionf(x)asf(x) = (x+n+1)1 2x+11 1r

,x∈[0,∞), then we havef(0) = (n+1)1 ,

f0(x) =− 1 (x+n+ 1)2

1 2x+ 1

1r

− 2

r(x+n+ 1) 1

2x+ 1 1+1r

, f0(0) =− 1

(n+ 1)2 − 2 r(n+ 1), and

Z 0

f(x)dx= 1 (2n+ 1)1r

Z

1 2n+1

1 (y+ 1)

1 y

1r dy

= 1

(2n+ 1)1r

"

π sin πr

X

ν=0

(−1)ν 1− 1r

(2n+ 1)ν+1−1r

# . By (2.2) and (2.5), we have

θ(n, r) =−(2n+ 1)2 2 (n+ 1) +

X

ν=0

(−1)ν 1− 1r

(2n+ 1)ν−1 (2.6)

+ Z

0

1(t)

"

(2n+ 1)2 (t+n+ 1)2(2t+ 1)1r

+ 2 (2n+ 1)2 r(t+n+ 1) (2t+ 1)1+1r

# dt.

By Lemma 4 of [5, p. 1106] and (2.4), we have

(8)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page8of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

Lemma 2.2. Forr >1,n ∈N0, we haveθ(n, r)> θ(n,∞), and (2.7) W(n, r)< π

sin πr − θ(n,∞) (2n+ 1)2−1r

(r >1, n∈N0), where

(2.8)

θ(n,∞) = −(2n+ 1)2 2 (n+ 1)+

X

ν=0

(−1)ν

(1 +ν) (2n+ 1)ν−1+ Z

0

1(t)

"

(2n+ 1)2 (t+n+ 1)2

# dt.

Since by (2.3) and (2.1),we have Z

0

1(t) 1

(t+n+ 1)2dt =− 1

12 (n+ 1)2 − 1 3!

Z 0

3(t)

1 (t+n+ 1)

000

dt

>− 1 12 (n+ 1)2 and

X

ν=0

(−1)ν

(1 +ν) (2n+ 1)ν−1 = (2n+ 1)− 1 2 +

X

ν=2

(−1)ν

(1 +ν) (2n+ 1)ν−1

>(2n+ 1)− 1

2 + 1

6 (2n+ 1). Then by (2.8), we find

(2.9) θ(n,∞)> 1

6− 1

6 (n+ 1) − 1

12 (n+ 1)2 + 1 6 (2n+ 1).

(9)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page9of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

Lemma 2.3. Forr >1,n ∈N0, we have (2.10) W(n, r)< π

sin πr − 1

13 (n+ 1) (2n+ 1)1−1r .

Proof. Define the functiong(x)as g(x) = 1

12− 1

6 (2x+ 1) + 1

12 (x+ 1) + 1

12 (2x+ 1)2, x∈[0,∞).

Then by (2.8), we have θ(n,∞) > 2n+1n+1g(n). Sinceg(1) >0.0787> 131, and forx∈[1,∞),

g0(x) = 1

3 (2x+ 1)2− 1

12 (x+ 1)2− 1

3 (2x+ 1)3 = 4x2+ 2x−1

12 (x+ 1)2(2x+ 1)3 >0, then for n ≥ 1, we have θ(n,∞) > (n+1)2n+1g(1) > 13(n+1)2n+1 . Hence by (2.7), inequality (2.10) is valid forn ≥ 1. Sinceln 2−γ = 0.1159+ > 131, then by (1.5), we find

(2.11)

W(0, r)< π

sin πr− ln 2−γ (2×0 + 1)2−1r

< π

sin πr− 1

13 (0 + 1) (2×0 + 1)1−1r . It follows that (2.10) is valid forr >1,andn∈N0. This proves the lemma.

(10)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page10of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

3. Theorem and Remarks

Theorem 3.1. If p > 1,1p + 1q = 1, an, bn ≥ 0, 0 < P

n=0apn < ∞, and 0<P

n=0bqn<∞, then (3.1)

X

m=0

X

n=0

ambn m+n+ 1 <

X

n=0

 π sin

π p

− 1

13 (n+ 1) (2n+ 1)1p

apn

1 p

×

X

n=0

 π sin

π p

− 1

13 (n+ 1) (2n+ 1)1q

bqn

1 q

,

(3.2)

X

m=0

X

n=0

an

m+n+ 1

!p

<

 π sin

π p

p−1

X

n=0

 π sin

π p

− 1

13 (n+ 1) (2n+ 1)1p

apn. Proof. By Hölder’s inequality, we have

(11)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page11of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

X

m=0

X

n=0

ambn

m+n+ 1

=

X

m=0

X

n=0

"

am (m+n+ 1)1p

m+ 12 n+12

pq1 # "

bn (m+n+ 1)1q

n+ 12 m+12

pq1 #

≤ (

X

m=0

X

n=0

"

1 (m+n+ 1)

m+12 n+12

1 q#

apm )

1 p

× (

X

m=0

X

n=0

"

1 (m+n+ 1)

n+ 12 m+12

1 p#

bqn )1q

= (

X

m=0

X

n=0

"

1 (m+n+ 1)

m+12 n+ 12

1q# apm

)

1 p

× (

X

n=0

X

m=0

"

1 (m+n+ 1)

n+ 12 m+12

1p# bqn

)

1 q

= (

X

m=0

W(m, q)apm

)1p( X

n=0

W(n, p)bqn )1q

.

Sincesin(π/q) = sin(π/p), by (2.10) forr =p, q, we have (3.1).

By (2.10), we have W(n, p) < π/sin(π/p). Then by Hölder’s inequality,

(12)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page12of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

we obtain

X

n=0

an m+n+ 1

=

X

n=0

"

an (m+n+ 1)p1

n+ 12 m+12

1 pq# "

1 (m+n+ 1)1q

m+ 12 n+12

1 pq#

≤ (

X

n=0

"

1 (m+n+ 1)

n+ 12 m+12

1 q#

apn

)1p ( X

n=0

1 (m+n+ 1)

m+ 12 n+12

1 p)1q

= (

X

n=0

"

1 (m+n+ 1)

n+12 m+ 12

1q# apn

)

1 p

{W(m, p)}1q

<

( X

n=0

"

1 (m+n+ 1)

n+12 m+ 12

1q# apn

)

1

p

 π sin

π p

1 q

.

Then we find

X

m=0

X

n=0

an m+n+ 1

!p

<

X

m=0

X

n=0

"

1 (m+n+ 1)

n+12 m+ 12

1 q#

apn

 π sin

π p

p q

(13)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page13of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

=

 π sin

π p

p−1

X

n=0

X

m=0

"

1 (m+n+ 1)

n+ 12 m+12

1q# apn

=

 π sin

π p

p−1

X

n=0

W(n, q)apn.

Hence by (2.10) forr=q, we have (3.2). The theorem is proved.

Remark 3.1. Inequality (3.1) is a definite improvement over (1.1) forλ= 1.

Remark 3.2. Since forn≥2,13 (2−γ)<2− n+11 , then (3.3)

π

sin πr− ln 2−γ (2n+ 1)2−1r

> π

sin πr− 1

13 (n+ 1) (2n+ 1)1−1r

(r >1, n≥2). In view of (2.11) and (3.3), it follows that (3.1) and (1.4) represent two dis- tinct versions of strengthened inequalities. However, they are not comparable.

Remark 3.3. Inequality (3.2) reduces to

(3.4)

X

m=0

X

n=0

an m+n+ 1

!p

<

 π sin

π p

p

X

n=0

apn.

This is an equivalent form of the more accurate Hardy-Hilbert’s inequality (see [3, Chapter 9]).

(14)

On a Strengthened Hardy-Hilbert Inequality

Bicheng Yang

Title Page Contents

JJ II

J I

Go Back Close

Quit Page14of14

J. Ineq. Pure and Appl. Math. 1(2) Art. 22, 2000

http://jipam.vu.edu.au

References

[1] R.P. BOAS, Estimating remainders, Math. Mag., 51(2) (1978), 83–89.

[2] M. GAO ANDB. YANG, On the extended Hilbert’s inequality, Proc. Amer.

Math. Soc., 126(3) (1998), 751–759.

[3] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cam- bridge University Press, 1934.

[4] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´CANDA.M. FINK, Inequalities Involv- ing Functions and their Integrals and Derivatives, Kluwer Academic Pub- lishers, 1991.

[5] B. YANG, On a strengthened version of the more accurate Hardy-Hilbert’s inequality, Acta Math. Sinica, 42(6) (1999), 1103–1110.

[6] B. YANG AND L. DEBNATH, On new strengthened Hardy-Hilbert’s in- equality, Int. J. Math. Math. Sci., 21(2) (1998), 403–408.

[7] B. YANGANDM. GAO, On a best value of Hardy-Hilbert’s inequality, Adv.

Math., 26(2) (1997), 156–164.

参照

関連したドキュメント

Hilbert’s 12th problem conjectures that one might be able to generate all abelian extensions of a given algebraic number field in a way that would generalize the so-called theorem

In order to implement the explicit finite difference method, we will replace the space derivative by a finite difference formula at the j − th time step and the time derivative by

The question of the connectedness of the Hilbert schemes H d,g of locally Cohen–Macaulay curves C ⊂ P 3 of degree d and arithmetic genus g arose naturally after Hartshorne proved in

In Section 3, we shall state (without proof) Y. Chinwarakorn’s F q [t]- analogue of Lerch’s formula together with the properties of the associated Fermat quotient operator. In

Remark 3 Because the Bernstein’s bivariate operator B m,n conserve only the lineares functions in x and respectively y, it follows that the degree of exactness for the cubature

In this article we propose to study a dynamical system, namely the geodesic flow, over more structurally complex manifolds, the tangent bundles of a family of Hilbert

Debnath, “A new generalized and sharp version of Jordan’s inequality and its applications to the improvement of the Yang Le inequality,” Applied Mathematics Letters, vol.. Debnath,

Abstract: Based on a modified iterative algorithm, fixed points of the operators of the form S = T + U on nonempty closed convex subsets of Hilbert spaces are ap- proximated... To