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volume 6, issue 2, article 39, 2005.

Received 04 June, 2004;

accepted 18 March, 2005.

Communicated by:B.G. Pachpatte

Abstract Contents

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Journal of Inequalities in Pure and Applied Mathematics

ON A NEW MULTIPLE EXTENSION OF HILBERT’S INTEGRAL INEQUALITY

BICHENG YANG

Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, People’s Republic Of China.

EMail:bcyang@pub.guangzhou.gd.cn URL:http://page.gdei.edu.cn/~yangbicheng

c

2000Victoria University ISSN (electronic): 1443-5756 114-04

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On a New Multiple Extension of Hilbert’s Integral Inequality

Bicheng Yang

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Abstract

This paper gives a new multiple extension of Hilbert’s integral inequality with a best constant factor, by introducing a parameterλand theΓfunction. Some particular results are obtained.

2000 Mathematics Subject Classification:26D15.

Key words: Hilbert’s integral inequality; Weight coefficient ,Γfunction.

Contents

1 Introduction. . . 3 2 Some Lemmas. . . 7 3 Proof of the Theorem and Remarks. . . 13

References

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On a New Multiple Extension of Hilbert’s Integral Inequality

Bicheng Yang

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1. Introduction

Iff, g≥0satisfy 0<

Z 0

f2(x)dx <∞ and 0<

Z 0

g2(x)dx <∞,

then (1.1)

Z 0

Z 0

f(x)g(y)

x+y dxdy < π Z

0

f2(x)dx Z

0

g2(x)dx 12

, where the constant factorπis the best possible (cf. Hardy et al. [2]). Inequality (1.1) is well known as Hilbert’s integral inequality, which had been extended by Hardy [1] as:

Ifp > 1, 1p +1q = 1, f, g ≥0satisfy 0<

Z 0

fp(x)dx <∞ and 0<

Z 0

gq(x)dx <∞,

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(x)dx 1q

,

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On a New Multiple Extension of Hilbert’s Integral Inequality

Bicheng Yang

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where the constant factor sin(π/p)π is the best possible. Inequality (1.2) is called Hardy- Hilbert’s integral inequality, and is important in analysis and its appli- cations (cf. Mitrinovi´c et al.[6]).

Recently, by introducing a parameterλ,Yang [9] gave an extension of (1.2) as:

Ifλ >2−min{p, q}, f, g ≥0satisfy 0<

Z 0

x1−λfp(x)dx <∞ and 0<

Z 0

x1−λgq(x)dx <∞,

then (1.3)

Z 0

Z 0

f(x)g(y) (x+y)λdxdy

< kλ(p) Z

0

x1−λfp(x)dx

1p Z 0

x1−λgq(x)dx 1q

,

where the constant factorkλ(p) = B

p+λ−2

p ,q+λ−2q

is the best possible (B(u, v) is theβ function). Forλ= 1,inequality (1.3) reduces to (1.2).

On the problem for multiple extension of (1.1), [3,4] gave some new results and Yang [8] gave an improvement of their works as:

Ifn ∈N\{1}, pi >1,Pn i=1

1

pi = 1, λ > n− min

1≤i≤n{pi}, fi ≥0,satisfy 0<

Z 0

xn−1−λfipi(x)dx <∞ (i= 1,2, . . . , n),

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On a New Multiple Extension of Hilbert’s Integral Inequality

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then (1.4)

Z 0

· · · Z

0

1 Pn

j=1xjλ n

Y

i=1

fi(xi)dx1. . . dxn

< 1 Γ(λ)

n

Y

i=1

Γ

pi+λ−n pi

Z 0

xn−1−λfipi(x)dx 1

pi ,

where the constant factor Γ(λ)1 Qn i=1Γ

pi+λ−n pi

is the best possible. Forn= 2, inequality (1.4) reduces to (1.3). It follows that (1.4) is a multiple extension of (1.3), (1.2) and (1.1).

In 2003, Yang et. al [11] provided an extensive account of the above results.

The main objective of this paper is to build a new extension of (1.1) with a best constant factor other than (1.4), and give some new particular results. That is

Theorem 1.1. Ifn ∈N\{1}, pi >1,Pn i=1

1

pi = 1, λ > 0, fi ≥0,satisfy 0<

Z 0

xpi−1−λfipi(x)dx <∞ (i= 1,2, . . . , n),

then (1.5)

Z 0

· · · Z

0

1 Pn

j=1xjλ n

Y

i=1

fi(xi)dx1. . . dxn

< 1 Γ(λ)

n

Y

i=1

Γ λ

pi

Z 0

xpi−1−λfipi(x)dx 1

pi ,

(6)

On a New Multiple Extension of Hilbert’s Integral Inequality

Bicheng Yang

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where the constant factor Γ(λ)1 Qn i=1Γ

λ pi

is the best possible. In particular, (a) forλ= 1,we have

(1.6) Z

0

· · · Z

0

Qn

i=1fi(xi) Pn

j=1xj

dx1. . . dxn

<

n

Y

i=1

Γ 1

pi

Z 0

xpi−2fipi(x)dx 1

pi ;

(b) for n = 2, using the symbol of (1.3) and setting ekλ(p) = B λ

p,λq

,we have

(1.7) Z

0

Z 0

f(x)g(y) (x+y)λdxdy

<ekλ(p) Z

0

xp−1−λfp(x)dx

1p Z 0

xq−1−λgq(x)dx 1q

, where the constant factors in (1.6) and (1.7) are still the best possible.

In order to prove the theorem, we introduce some lemmas.

(7)

On a New Multiple Extension of Hilbert’s Integral Inequality

Bicheng Yang

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2. Some Lemmas

Lemma 2.1. Ifk ∈N, ri >1 (i= 1,2, . . . , k+ 1),andPk+1

i=1 ri =λ(k),then (2.1)

Z 0

· · · Z

0

1

1 +Pk

j=1ujλ(k) k

Y

j=1

urjj−1du1. . . duk = Qk+1

i=1 Γ(ri) Γ(λ(k)) .

Proof. We establish (2.1) by mathematical induction. Fork= 1, sincer1+r2 = λ(1),and (see [7])

(2.2) B(p, q) = Z

0

up−1

(1 +u)p+qdu=B(q, p) (p, q >0),

we have (2.1). Suppose fork ∈ N,that (2.1) is valid. Then for k+ 1, since r1+Pk+1

i=2 ri =λ(k+ 1), by settingv =u1.

1 +Pk+1 j=2 uj

, we obtain Z

0

· · · Z

0

1

1 +Pk+1 j=1uj

λ(k+1) k+1

Y

j=1

urjj−1du1. . . duk+1 (2.3)

= Z

0

· · · Z

0

vr1−1

1 +Pk+1 j=2ujr1

Qk+1 j=2urjj−1

1 +Pk+1

j=2 ujλ(k+1)

(1 +v)λ(k+1)

dvdu2. . . duk+1

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On a New Multiple Extension of Hilbert’s Integral Inequality

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= Z

0

· · · Z

0

Qk+1 j=2urjj−1

1 +Pk+1

j=2ujλ(k+1)−r1du2. . . duk+1

× Z

0

vr1−1

(1 +v)λ(k+1)dv.

In view of (2.2) and the assumption ofk, we have (2.4)

Z 0

vr1−1

(1 +v)λ(k+1)dv= 1

Γ(λ(k+ 1))Γ

k+1

X

i=2

ri

! Γ(r1);

(2.5) Z

0

· · · Z

0

Qk+1 j=2urjj−1

1 +Pk+1

j=2ujλ(k+1)−r1du2. . . duk+1 =

Qk+2 i=2 Γ(ri) Γ

Pk+1 i=2 ri. Then, by (2.5), (2.4) and (2.3), we find

Z 0

· · · Z

0

Qk+1 j=1 urjj−1

1 +Pk+1 j=1uj

λ(k+1)du1. . . duk+1 = Qk+2

i=1 Γ(ri) Γ(λ(k+ 1)).

Hence (2.1) is valid fork ∈Nby induction. The lemma is proved.

Lemma 2.2. Ifn ∈ N\{1}, pi >1 (i = 1,2, . . . , n),Pn i=1

1

pi = 1andλ > 0, set the weight coefficientω(xi)as

(2.6) ω(xi) :=x

λ pj

i

Z 0

· · · Z

0

Qn

j=1(j6=i)x(λ−pj j)/pj Pn

j=1xjλ dx1. . . dxi−1dxi+1. . . dxn.

(9)

On a New Multiple Extension of Hilbert’s Integral Inequality

Bicheng Yang

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Then, eachω(xi)is constant, that is (2.7) ω(xi) = 1

Γ(λ)

n

Y

j=1

Γ λ

pj

, (i= 1,2, . . . , n).

Proof. Fixi. Settingpen =pi,andpej =pj, uj =xj/xi, forj = 1,2, . . . , i−1;

pej =pj+1, uj =xj+1/xi, forj =i, i+ 1, . . . , n−1in (2.6), by simplification, we have

(2.8) ω(xi) = Z

0

· · · Z

0

1

1 +Pn−1 j=1 ujλ

n−1

Y

j=1

u

−1+λ

pj

j du1. . . dun−1.

Substitution of n −1 fork, λ forλ(k)and λ/pej for rj (j = 1,2, . . . , n)into (2.1), in view of (2.8), we have

ω(xi) = 1 Γ(λ)

n

Y

j=1

Γ λ

pej

= 1

Γ(λ)

n

Y

j=1

Γ λ

pj

.

Hence, (2.7) is valid. The lemma is proved.

Lemma 2.3. As in the assumption of Lemma2.2, for0< ε < λ, we have

I :=ε Z

1

· · · Z

1

Qn

i=1x(λ−pi i−ε)/pi Pn

j=1xjλ dx1. . . dxn

≥ 1

Γ(λ)

n

Y

i=1

Γ λ

pi

(ε →0+).

(2.9)

(10)

On a New Multiple Extension of Hilbert’s Integral Inequality

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Proof. Settingui =xi/xn(i= 1,2, . . . , n−1)in the following, we find

I =ε Z

1

x−1−εn

 Z

x−1n

· · · Z

x−1n

Qn−1

i=1 u(λ−pi i−ε)/pi

1 +Pn−1

j=1 ujλdu1. . . dun−1

dxn

≥ε Z

1

x−1−εn

 Z

0

· · · Z

0

Qn−1

i=1 u(λ−pi i−ε)/pi

1 +Pn−1

j=1 ujλdu1. . . dun−1

dxn

−ε Z

1

x−1n

n−1

X

j=1

Aj(xn)dxn, (2.10)

where, forj = 1,2, . . . , n−1, Aj(xn)is defined by (2.11) Aj(xn) :=

Z

· · · Z

Dj

Qn−1

i=1 u(λ−pi i−ε)/pi (1 +Pn−1

j=1 uj)λ du1. . . dun−1,

satisfyingDj ={(u1, u2, . . . , un−1)|0< uj ≤x−1n , 0< uk <∞(k 6=j)}.

Without loss of generality, we estimate the integralAj(xn)forj = 1.

(a) Forn= 2, we have A1(xn) =

Z x−1n

0

1

(1 +u1)λu(λ−p1 1−ε)/p1du1

≤ Z x−1n

0

u(λ−p1 1−ε)/p1du1 = p1

λ−εx−(λ−ε)/pn 1;

(11)

On a New Multiple Extension of Hilbert’s Integral Inequality

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(b) Forn∈N\{1,2},by (2.1), we have

A1(xn)≤ Z

0

· · · Z

0

Qn−1 i=2 u−1+

λ−ε pi

i

1 +Pn−1

j=2 ujλdu1. . . dun−1

Z x−1n

0

u

λ−p1−ε p1

1 du1

≤ p1x

λ−ε p1

n

λ−ε Z

0

· · · Z

0

Qn−1

i=2 u−1+(λ−ε)/pi i

1 +Pn−1

j=2uj(λ−ε)(1−p−11 )du1. . . dun−1

= p1x

λ−ε p1

n

λ−ε ·

Qn

i=2Γ(λ−εp

i ) Γ((λ−ε)(1−p−11 )).

By virtue of the results of (a) and (b), forj = 1,2, . . . , n−1,we have (2.12) Aj(xn)≤ pj

λ−εx−(λ−ε)/pn jOj(1) (ε→0+, n∈N\{1}).

By (2.11), since forε→0+, Z

0

· · · Z

0

Qn−1

i=1 u−1+(λ−ε)/pi i

1 +Pn−1 j=1 uj

λ du1. . . dun−1 = Qn

i=1Γ

λ pi

Γ(λ) +o(1);

(12)

On a New Multiple Extension of Hilbert’s Integral Inequality

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Z 1

x−1n

n−1

X

j=1

Aj(xn)dxn =

n−1

X

j=1

Z 1

x−1n Aj(xn)dxn

n−1

X

j=1

pj

λ−εOj(1) Z

1

x−1−(λ−ε)/pn jdxn

=

n−1

X

j=1

pj λ−ε

2

Oj(1),

then by (2.10) , we find

I ≥

 Qn

i=1Γ

λ pi

Γ(λ) +o(1)

−ε

n−1

X

j=1

pj λ−ε

2

Oj(1)

→ Qn

i=1Γ

λ pi

Γ(λ) (ε→0+).

Thereby, (2.9) is valid and the lemma is proved.

(13)

On a New Multiple Extension of Hilbert’s Integral Inequality

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3. Proof of the Theorem and Remarks

Proof of Theorem1.1. By Hölder’s inequality, we have

(3.1) J :=

Z 0

· · · Z

0

1 Pn

j=1xjλ n

Y

i=1

fi(xi)dx1. . . dxn

= Z

0

· · · Z

0









n

Y

i=1

fi(xi) Pn

j=1xjλ/pi

x(pi−λ)(1−p

−1 i ) i

n

Y

j=1

(j6=i)

x

λ−pj pj

j

1 pi









dx1. . . dxn

n

Y

i=1





 Z

0

· · · Z

0

fipi(xi) Pn

j=1xjλx(pi−λ)(1−p

−1 i ) i

n

Y

j=1

(j6=i)

x

λ−pj pj

j dx1. . . dxn





1 pi

.

If (3.1) takes the form of equality, then there exists constantsC1, C2, . . . , Cn, such that they are not all zero and for anyi6=k ∈ {1,2, . . . , n}(see [5]),

Cifipi(xi)x(pi−λ)(1−p

−1 i ) i

Pn

j=1xjλ

n

Y

j=1

(j6=i)

x

λ−pj pj

j =Ckfkpk(xk)x(pk−λ)(1−p

−1 k ) k

Pn

j=1xjλ

n

Y

j=1

(j6=k)

x

λ−pj pj

j ,

a.e. in(0,∞)× · · · ×(0,∞).

(3.2)

(14)

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Assume thatCi 6= 0.By simplification of (3.2), we find xpii−λfipi(xi) =F(x1, . . . , xi−1, xi+1, . . . , xn)

=constant a.e. in(0,∞)× · · · ×(0,∞), which contradicts the fact that0 <R

0 xpii−λ−1fipi(x)dx < ∞.Hence by (2.6) and (3.1), we conclude

(3.3) J <

n

Y

i=1

Z 0

ω(xi)xpii−1−λfipi(xi)dxi pi1

. Then by (2.7), we have (1.5).

For0< ε < λ,settingfei(xi)as:fei(xi) = 0,forxi ∈(0,1);

fei(xi) =x(λ−pi i−ε)/pi, forxi ∈[1,∞) (i= 1,2, . . . , n), then we find

(3.4) ε

n

Y

i=1

Z 0

xpii−1−λfeipi(xi)dxi pi1

= 1.

By (2.9), we find (3.5) ε

Z 0

· · · Z

0

1 Pn

j=1xjλ n

Y

i=1

fei(xi)dx1. . . dxn

=I ≥ 1 Γ(λ)

n

Y

i=1

Γ λ

pi

(ε→0+).

(15)

On a New Multiple Extension of Hilbert’s Integral Inequality

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If the constant factor Γ(λ)1 Qn i=1Γ

λ pi

in (1.5) is not the best possible, then there exists a positive constant K < Γ(λ)1 Qn

i=1Γ

λ pi

, such that (1.5) is still valid if one replaces Γ(λ)1 Qn

i=1Γ

λ pi

byK.In particular, one has

ε Z

0

· · · Z

0

1 Pn

j=1xjλ n

Y

i=1

fei(xi)dx1. . . dxn

< εK

n

Y

i=1

Z 0

xpi−1−λfeipi(x)dx pi1

,

and in view of (3.4) and (3.5), it follows that Γ(λ)1 Qn i=1Γ

λ pi

≤ K (ε →

0+).This contradicts the factK < Γ(λ)1 Qn i=1Γ

λ pi

.Hence the constant factor

1 Γ(λ)

Qn i=1Γ

λ pi

in (1.5) is the best possible.

The theorem is proved.

Remark 1. Forλ= 1, inequality (1.7) reduces to (see [10])

(3.6) Z

0

Z 0

f(x)g(y) x+y dxdy

< π sin

π p

Z

0

xp−2fp(x)dx

1pZ 0

xq−2gq(x)dx 1q

.

(16)

On a New Multiple Extension of Hilbert’s Integral Inequality

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Forp =q = 2, both (3.6) and (1.2) reduce to (1.1). It follows that inequalities (3.6) and (1.2) are different extensions of (1.1). Hence, inequality (1.5) is a new multiple extension of (1.1). Since all the constant factors in the obtained inequalities are the best possible, we have obtained new results.

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On a New Multiple Extension of Hilbert’s Integral Inequality

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References

[1] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive terms, Proc. Math. Soc., 23(2) (1925), Records of Proc. XLV-XLVI.

[2] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cam- bridge University Press, Cambridge, 1952.

[3] YONG HONG, All-sided generalization about Hardy-Hilbert’s integral in- equalities, Acta Math. Sinica, 44(4) (2001), 619–626.

[4] LEPING HE, JIANGMING YU AND MINGZHE GAO, An extension of Hilbert’s integral inequality, J. Shaoguan University (Natural Science), 23(3) (YEAR), 25–29.

[5] JICHANG KUANG, Applied Inequalities, Hunan Education Press, Changsha, 2004.

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

[7] ZHUXI WANG AND DUNRIN GUO, An Introduction to Special Func- tions, Science Press, Beijing, 1979.

[8] BICHENG YANG, On a multiple Hardy-Hilbert’s integral inequality, Chi- nese Annals of Math., 24A(6) (2003), 743–750.

[9] BICHENG YANG, On Hardy-Hilbert’s integral inequality, J. Math. Anal.

Appl., 261 (2001), 295–306.

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On a New Multiple Extension of Hilbert’s Integral Inequality

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[10] BICHENG YANG, On the extended Hilbert’s integral inequality, J. In- equal. Pure and Appl. Math., 5(4) (2004), Art. 96. [ONLINE: http:

//jipam.vu.edu.au/article.php?sid=451]

[11] BICHENG YANG AND Th. M. RASSIAS, On the way of weight coeffi- cient and research for the Hilbert-type inequalities, Math. Ineq. Appl., 6(4) (2003), 625–658.

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In this paper, by introducing the norm kxk α (x ∈ R n ), we give a multiple Hardy- Hilbert’s integral inequality with a best constant factor and two parameters α, λ.. Key words

By introducing some parameters and the β function and improving the weight function, we obtain a generalization of Hilbert’s integral inequality with the best constant factorJ. As

By introducing some parameters and the β function and improving the weight func- tion, we obtain a generalization of Hilbert’s integral inequality with the best constant factor.. As

This paper gives a new multiple extension of Hilbert’s integral inequality with a best constant factor, by introducing a parameter λ and the Γ function.. Some particular results

This paper deals with some extensions of Hardy-Hilbert’s inequality with the best constant factors by introducing two parameters λ and α and using the Beta functionJ. The

In 1998, Yang [6] first introduced an indepen- dent parameter λ and the Beta function to build an extension of Hilbert’s integral inequality.. Recently, by introducing a parameter

Key words: Hardy-Hilbert’s integral inequality, Weight, Parameter, Best constant fac- tor, β-function,

Key words and phrases: Hardy-Hilbert’s integral inequality, Weight, Parameter, Best constant factor, β-function, Γ-function.. 2000 Mathematics