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Hilbert’s Integral Inequality Bicheng Yang vol. 9, iss. 2, art. 59, 2008

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A RELATION TO HILBERT’S INTEGRAL INEQUALITY AND SOME BASE HILBERT-TYPE

INEQUALITIES

BICHENG YANG

Department of Mathematics Guangdong Education Institute

Guangzhou, Guangdong 510303, P.R. China EMail:bcyang@pub.guangzhou.gd.cn

Received: 17 April, 2008

Accepted: 18 May, 2008

Communicated by: J. Peˇcari´c 2000 AMS Sub. Class.: 26D15.

Key words: Base Hilbert-type integral inequality; Parameter; Weight function.

Abstract: In this paper, by using the way of weight function and real analysis techniques, a new integral inequality with some parameters and a best constant factor is given, which is a relation to Hilbert’s integral inequality and some base Hilbert-type integral inequalities. The equivalent form and the reverse forms are considered.

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Hilbert’s Integral Inequality Bicheng Yang vol. 9, iss. 2, art. 59, 2008

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Contents

1 Introduction 3

2 Some Lemmas 6

3 Main Results 10

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1. Introduction

Iff, g ≥ 0, 0 < R

0 f2(x)dx < ∞and 0 < R

0 g2(x)dx < ∞ then we have the following Hilbert’s integral inequality [1]:

(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 factorπ is the best possible. Under the same condition of (1.1), we also have the following basic Hilbert-type integral inequalities [1,2,3]:

(1.2)

Z 0

Z 0

f(x)g(y)

max{x, y}dxdy <4 Z

0

f2(x)dx Z

0

g2(x)dx 12

;

(1.3)

Z 0

Z 0

|ln(x/y)|f(x)g(y)

x+y dxdy < c0

Z 0

f2(x)dx Z

0

g2(x)dx 12

;

(1.4)

Z 0

Z 0

|ln(x/y)|f(x)g(y)

max{x, y} dxdy <8 Z

0

f2(x)dx Z

0

g2(x)dx 12

, where the constant factors 4, c0

=P k=1

8(−1)k−1

(2k−1)2 = 7.3277+

and 8 are the best possible. In 2005, Hardy-Riesz gave a best extension of (1.1) by introducing one pair of conjugate exponents(p, q) (p >1,1p +1q = 1)as [4]

(1.5) Z

0

Z 0

f(x)g(y)

x+y dxdy < π sin

π p

Z

0

fp(x)dx

p1 Z 0

gq(x)dx 1q

,

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where the constant factor sin(π/p)π is the best possible. Inequality (1.5) is referred to as Hardy-Hilbert’s integral inequality, which is important in analysis and its applica- tions [5]. In 1998, Yang gave a best extension of (1.1) by introducing an independent parameterλ >0as [6,7]

(1.6) Z

0

Z 0

f(x)g(y) (x+y)λdxdy

< B λ

2,λ 2

Z 0

x1−λf2(x)dx Z

0

x1−λg2(x)dx 12

, where the constant factorB λ2,λ2

is the best possible and the Beta functionB(u, v) is defined by [8]:

(1.7) B(u, v) :=

Z 0

1

(1 +t)u+vtu−1dt (u, v >0).

In 2004-2005, by introducing two pairs of conjugate exponents and an independent parameter, Yang et al. [9,10] gave two different extensions of (1.1) and (1.5) as: If p, r > 1, p1 + 1q = 1, 1r + 1s = 1, λ > 0, φ(x) = xp(1−λr)−1, ψ(x) = xq(1−λs)−1, f, g≥0,

0<||f||p,φ :=

Z 0

xp(1−λr)−1fp(x)dx 1p

<∞ and

0<||g||q,ψ :=

Z 0

xq(1−λs)−1gq(x)dx 1q

<∞, then

(1.8)

Z 0

Z 0

f(x)g(y)

xλ+yλ dxdy < π

λsin(πr)||f||p,φ||g||q,ψ;

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(1.9)

Z 0

Z 0

f(x)g(y)

(x+y)λdxdy < B λ

r,λ s

||f||p,φ||g||q,ψ, where the constant factors λsin(π π

r) andB λr,λs

are the best possible. Yang [11] also considered the reverse of (1.8) and (1.9).

In this paper, by using weight functions and real analysis techniques, a new inte- gral inequality with the homogeneous kernel of−λdegree

kλ(x, y) = |ln(x/y)|β

(x+y)λ−α(max{x, y})α (λ >0, α∈R, β >−1)

is given, which is a relation to (1.1) and the above basic Hilbert-type integral in- equalities (1.2), (1.3) and (1.4). The equivalent and reverse forms are considered.

All the new inequalities possess the best constant factors.

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2. Some Lemmas

We introduce the following Gamma function [8]:

(2.1) Γ(s) =

Z 0

e−tts−1dt (s >0).

Lemma 2.1. Fora, b >0,it follows that (2.2)

Z 1 0

xa−1(−lnx)b−1dx= 1

abΓ(b) = Z

1

y−a−1(lny)b−1dy.

Proof. Settingx=e−t/ain first integral of (2.2), by (2.1), we find the first equation of (2.2). Settingy= 1/xin the first integral of (2.2), we obtain the second equation of (2.2). The lemma is hence proved.

Lemma 2.2. If r > 1, 1r + 1s = 1, λ > 0, α ∈ Randβ > −1,define the weight function as

(2.3) $λ(s, x) := xλr Z

0

ln

x y

β

yλs−1

(x+y)λ−α(max{x, y})αdy (x∈(0,∞)).

Then we have

(2.4) $λ(s, x) =kλ(r) :=

Z 0

|lnu|βuλr−1

(u+ 1)λ−α(max{u,1})αdu, wherekλ(r)is a positive number and

(2.5) kλ(r) = Γ(β+ 1)

X

k=0

α−λ k

"

1

k+λrβ+1 + 1 k+λsβ+1

# .

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Proof. Settingu=x/yin (2.3), by simplification, we obtain (2.4). In view of (2.2), we obtain

0< kλ(r) = Z 1

0

(−lnu)βuλr−1 (u+ 1)λ−α du+

Z 1

(lnu)βuλr−α−1 (u+ 1)λ−α du

≤2|α−λ|

Z 1 0

(−lnu)(β+1)−1uλr−1du+ Z

1

(lnu)(β+1)−1u−λs −1du

= 2|α−λ|

r λ

β+1

+s λ

β+1

Γ(β+ 1) <∞.

Hencekλ(r)is a positive number. Using the property of power series, we find kλ(r) =

Z 1 0

(−lnu)βuλr−1 (u+ 1)λ−α du+

Z 1

(lnu)βu−λs −1 (1 +u−1)λ−αdu

= Z 1

0

X

k=0

α−λ k

(−lnu)βuλr+k−1du+ Z

1

X

k=0

α−λ k

(lnu)βu−λs −k−1du

=

X

k=0

α−λ k

Z 1 0

(−lnu)βuλr+k−1du+ Z

1

(lnu)βu−λs −k−1du

. Then in view of (2.2), we have (2.5). The lemma is proved.

Lemma 2.3. Ifp >0 (p6= 1), r >1, 1p+1q = 1, 1r+1s = 1, λ >0, α ∈R, β >−1, n∈N, n > |q|λr ,then forn→ ∞,we have

(2.6) In:= 1 n

Z 1

Z 1

ln

x y

β

xλrnp1 −1yλsnq1 −1

(x+y)λ−α(max{x, y})α dxdy=kλ(r) +o(1).

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Proof. Settingu=y/x, by Fubini’s theorem [12], we obtain

In = 1 n

Z 1

 Z

1

ln

x y

β

xλrnp1 −1yλsnq1−1 (x+y)λ−α(max{x, y})α dx

dy

= 1 n

Z 1

yn1−1

"

Z y 0

|lnu|βuλs+np1 −1

(1 +u)λ−α(max{1, u})αdu

# dy

= 1 n

Z 1

yn1−1

"

Z 1 0

(−lnu)βuλs+np1−1 (1 +u)λ−α du+

Z y 1

(lnu)βuλs+np1 −1 (1 +u)λ−αuα du

# dy

= Z 1

0

(−lnu)βuλs+np1 −1

(1 +u)λ−α du+ 1 n

Z 1

yn1−1

"

Z y 1

(lnu)βuλs+np1 −1 (1 +u)λ−αuα du

# dy

= Z 1

0

(−lnu)βuλs+np1 −1

(1 +u)λ−α du+ 1 n

Z 1

Z u

yn1−1dy

(lnu)βuλs+np1−1 (1 +u)λ−αuα du

= Z 1

0

(−lnu)βuλs+np1 −1 (1 +u)λ−α du+

Z 1

(lnu)βuλsnq1−1 (1 +u)λ−αuα du.

(2.7)

(i) Ifp >0 (p6= 1)andq >0,then by Levi’s theorem [12], we find Z 1

0

(−lnu)βuλs+np1 −1 (1 +u)λ−α du=

Z 1 0

(−lnu)βuλs−1

(1 +u)λ−α du+o1(1), Z

1

(lnu)βuλsnq1 −1 (1 +u)λ−αuα du=

Z 1

(lnu)βuλs−1

(1 +u)λ−αuαdu+o2(1) (n→ ∞);

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(ii) If q < 0, settingn0 ∈ N, n0 > |q|λr , s10 = 1sn1

0, r10 = 1r + n1

0, then for n≥n0,we find

Z 1

(lnu)βuλsnq1 −1 (1 +u)λ−αuα du≤

Z 1

(lnu)βuλsn10q−1

(1 +u)λ−αuα du≤kλ(r0), and by Lebesgue’s control convergence theorem, we have

Z 1

(lnu)βuλsnq1 −1 (1 +u)λ−αuα du=

Z 1

(lnu)βuλs−1

(1 +u)λ−αuαdu+o3(1) (n → ∞).

Hence by the above results and (2.7), we obtain (2.6). The lemma is proved.

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3. Main Results

Theorem 3.1. Assume that p > 0 (p 6= 1), r > 1, 1p + 1q = 1, 1r + 1s = 1, λ > 0, α∈R, β >−1, φ(x) = xp(1−λr)−1, ψ(x) = xq(1−λs)−1 (x∈(0,∞)), f, g≥0,

0<||f||p,φ = Z

0

xp(1−λr)−1fp(x)dx 1p

<∞, 0<||g||q,ψ <∞.

(i) Forp >1,we have the following inequality:

(3.1) I :=

Z 0

Z 0

ln

x y

β

f(x)g(y)

(x+y)λ−α(max{x, y})αdxdy < kλ(r)||f||p,φ||g||q,ψ; (ii) For 0 < p < 1,we have the reverse of (3.1), where the constant factorkλ(r)

expressed by (2.5) in (3.1) and its reverse is the best possible.

Proof. (i) By Hölder’s inequality with weight [13], in view of (2.3), we find

I = Z

0

Z 0

ln

x y

β

(x+y)λ−α(max{x, y})α

"

x(1−λr)/q y(1−λs)/pf(x)

# "

y(1−λs)/p x(1−λr)/qg(y)

# dxdy



 Z

0

Z 0

ln

x y

β

(x+y)λ−α(max{x, y})α · x(1−λr)(p−1)

y1−λs fp(x)dxdy





1 p

×



 Z

0

Z 0

ln

x y

β

(x+y)λ−α(max{x, y})α · y(1−λs)(q−1)

x1−λr gq(y)dxdy





1 q

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(3.2) =

Z 0

$λ(s, x)φ(x)fp(x)dx

1pZ 0

$λ(r, y)ψ(y)gq(y)dy 1q

. We confirm that the middle of (3.2) keeps the form of strict inequality. Otherwise, there exist constantsAandB,such that they are not all zero and [13]

Ax(1−λr)(p−1)

y1−λs fp(x) = By(1−λs)(q−1)

x1−λr gq(y) a.e. in (0,∞)×(0,∞).

It follows thatAxp(1−λr)fp(x) =Byq(1−λs)gq(y)a.e.in (0,∞)×(0,∞).Assuming thatA6= 0,there existsy >0,such thatxp(1−λr)−1fp(x) =

h

Byq(1−λs)gq(y) i 1

Ax a.e.

inx∈(0,∞).This contradicts the fact that0<||f||p,φ <∞.Then inequality (3.1) is valid by using (2.4) and (2.5).

Forn ∈N, n > |q|λr ,settingfn, gnas

fn(x) :=

( 0, 0< x≤1;

xλrnp1 −1, x >1; gn(x) :=

( 0, 0< x≤1;

xλrnq1 −1, x >1;

if there exists a constant factor 0 < k ≤ kλ(r), such that (3.1) is still valid if we replacekλ(r)byk,then by (2.6), we have

kλ(r) +o(1) =In= 1 n

Z 0

Z 0

ln

x y

β

fn(x)gn(y)

(x+y)λ−α(max{x, y})αdxdy

< 1

nk||fn||p,φ||gn||q,ψ =k,

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andkλ(r)≤k (n → ∞).Hencek =kλ(r)is the best constant factor of (3.1).

(ii) For0 < p < 1,by the reverse Hölder’s inequality with weight [13], in view of (2.3), we find the reverse of (3.2), which still keeps the strict form. Then by (2.4) and (2.5), we have the reverse of (3.1). By using (2.6) and the same manner as mentioned above , we can show that the constant factor in the reverse of (3.1) is still the best possible. The theorem is proved.

Theorem 3.2. Assume that p > 0 (p 6= 1), r > 1, 1p + 1q = 1, 1r + 1s = 1, λ > 0, α ∈ R, β > −1, φ(x) = xp(1−λr)−1, ψ(x) = xq(1−λs)−1 (x ∈ (0,∞)), f ≥ 0, 0<||f||p,φ <∞.

(i) For p > 1, we have the following inequality, which is equivalent to (3.1) and with the best constant factorkλp(r):

J :=

Z 0

ys−1

 Z

0

ln

x y

β

f(x)

(x+y)λ−α(max{x, y})αdx

p

dy (3.3)

< kλp(r)||f||pp,φ;

(ii) For0< p <1,we have the reverse of (3.3), which is equivalent to the reverse of (3.1), with the best constant factorkpλ(r).

Proof. (i) Forp >1, x >0,setting a bounded measurable function as [f(x)]n:= min{f(x), n}=

( f(x), forf(x)< n;

n, forf(x)≥n,

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since||f||p,φ > 0,there existsn0 ∈ N,such thatRn

1 n

φ(x)[f(x)]pndx > 0 (n ≥ n0).

Settingegn(y) (y∈(n1, n);n ≥n0)as

(3.4) egn(y) :=ys−1

 Z n

1 n

ln

x y

β

(x+y)λ−α(max{x, y})α[f(x)]ndx

p−1

,

then by (3.1), we find 0<

Z n

1 n

ψ(y)egqn(y)dy

= Z n

1 n

ys −1

 Z n

1 n

ln

x y

β

[f(x)]ndx (x+y)λ−α(max{x, y})α

p

dy

= Z n

1 n

Z n

1 n

ln

x y

β

[f(x)]negn(y) (x+y)λ−α(max{x, y})αdxdy

< kλ(r) (Z n

1 n

φ(x)[f(x)]pndx )1p(

Z n

1 n

ψ(y)egnq(y)dy )1q

<∞;

(3.5)

(3.6) 0<

Z n

1 n

ψ(y)egnq(y)dy < kpλ(r) Z

0

φ(x)fp(x)dx <∞.

It follows0<||g||q,ψ <∞.Forn→ ∞,by (3.1), both (3.5) and (3.6) still keep the forms of strict inequality. Hence we have (3.3). On the other-hand, suppose (3.3) is

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valid. By Hölder’s inequality, we have

I = Z

0

y−1p +λs Z

0

ln

x y

β

f(x)dx (x+y)λ−α(max{x, y})α

 h

y1pλsg(y) i

dy (3.7)

≤J1p||g||q,ψ.

In view of (3.3), we have (3.1), which is equivalent to (3.3). We confirm that the constant factor in (3.3) is the best possible. Otherwise, we may get a contradiction by (3.7) that the constant factor in (3.1) is not the best possible.

(ii) For0 < p < 1, since||f||p,φ > 0,we confirm thatJ > 0.If J = ∞, then the reverse of (3.3) is naturally valid. Suppose0< J <∞.Setting

g(y) :=ys−1

 Z

0

ln

x y

β

(x+y)λ−α(max{x, y})αf(x)dx

p−1

,

by the reverse of (3.1), we obtain

∞>||g||qq,ψ =J =I > kλ(r)||f||p,φ||g||q,ψ >0;

J1p =||g||q−1q,ψ > kλ(r)||f||p,φ.

Hence we have the reverse of (3.3). On the other-hand, suppose the reverse of (3.3) is valid. By the reverse Hölder’s inequality, we can get the reverse of (3.7). Hence in view of the reverse of (3.3), we obtain the reverse of (3.1), which is equivalent to the reverse of (3.3). We confirm that the constant factor in the reverse of (3.3) is the best possible. Otherwise, we may get a contradiction by the reverse of (3.7) that

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the constant factor in the reverse of (3.1) is not the best possible. The theorem is proved.

Remark 1. Forp=r= 2in (3.1), settingα =β = 0, λ= 1, we obtain (1.1); setting α= 0, β =λ= 1, we obtain (1.3). Forα=λ >0, β > −1in (3.1), we have

(3.8) Z

0

Z 0

ln

x y

β

f(x)g(y)

(max{x, y})λ dxdy < rβ+1+sβ+1

λβ+1 Γ(β+ 1)||f||p,φ||g||q,ψ, where the constant factor λβ+11 (rβ+1+sβ+1)Γ(β+ 1)is the best possible. Forp = r= 2in (3.8), settingλ= 1, β = 0, we obtain (1.2); settingλ= 1, β = 1, we obtain (1.4). Hence inequality (3.1) is a relation to (1.1), (1.2), (1.3) and (1.4).

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References

[1] G.H. HARDY, J.E. LITTLEWOOD ANDG. PÓLYA, Inequalities, Cambridge Univ. Press, Cambridge, 1952.

[2] BICHENG YANG, On a base Hilbert-type inequality, Journal of Guangdong Education Institute, 26(3) (2006), 1–5.

[3] BICHENG YANG, On a base Hilbert-type integral inequality and extensions, College Mathematics, 24(2) (2008), 87–92.

[4] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive terms, Proc. London Math. Soc., 23(2) (1925), Records of Proc. xlv-xlvi.

[5] D.S. MINTRINOVI ´C, J.E. PE ˇCARI ´CANDA.M. FINK, Inequalities Involving Functions and their Integrals and Derivatives, Kluwer Academic Publishers, Boston, 1991.

[6] BICHENG YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl., 220 (1998), 778–785.

[7] BICHENG YANG, A note on Hilbert’s integral inequality, Chin. Quart. J.

Math., 13(4) (1998), 83–86.

[8] ZHUXI WANGANDDUNREN GUO, Introduction to Special Functions, Sci- ence Press, Beijing, 1979.

[9] BICHENG YANG, On an extension of Hilbert’s integral inequality with some parameters, Austral. J. Math. Anal. Applics., 1(1) (2004), Art. 11. [ONLINE:

http://ajmaa.org/].

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[10] BICHENG YANG, ILKO BRNETIC, MARIO KRNIC AND J. PE ˇCARI ´C, Generalization of Hilbert and Hardy-Hilbert integral inequalities, Math. Ineq.

and Applics., 8(2) (2005), 259–272.

[11] BICHENG YANG, On a reverse Hardy-Hilbert’s inequality, Kyungpook Math.

J., 47 (2007), 411–423.

[12] JICHANG KUANG, Introduction to Real Analysis, Hunan Education Press, Changsha, 1996.

[13] JICHANG KUANG, Applied Inequalities, Shangdong Science Press, Jinan, 2004.

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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

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

This paper deals with a relation between Hardy-Hilbert’s integral inequality and Mulholland’s integral inequality with a best constant factor, by using the Beta function and

This paper deals with a relation between Hardy-Hilbert’s integral inequality and Mulholland’s integral inequality with a best constant factor, by using the Beta function and

In this paper we use a new approach to obtain a class of multivariable integral inequalities of Hilbert type from which we can recover as special cases integral inequalities

In this paper we use a new approach to obtain a class of multivariable integral inequalities of Hilbert type from which we can recover as special cases integral inequalities