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volume 7, issue 2, article 61, 2006.

Received 28 November, 2005;

accepted 19 January, 2006.

Communicated by:B. Yang

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Journal of Inequalities in Pure and Applied Mathematics

SOME EXTENSIONS OF HILBERT’S TYPE INEQUALITY AND ITS APPLICATIONS

YONGJIN LI AND BING HE

Department of Mathematics Sun Yat-sen University

Guangzhou, 510275 P. R. China.

EMail:stslyj@mail.sysu.edu.cn EMail:hzs314@163.com

c

2000Victoria University ISSN (electronic): 1443-5756 342-05

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Some Extensions of Hilbert’s Type Inequality and its

Applications Yongjin Li and Bing He

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J. Ineq. Pure and Appl. Math. 7(2) Art. 61, 2006

http://jipam.vu.edu.au

Abstract

By introducing parametersλandµ, we give a generalization of the Hilbert’s type integral inequality. As applications, we give its equivalent form.

2000 Mathematics Subject Classification:26D15.

Key words: Hilbert’s integral inequality, Weight function.

The authors are grateful to the referee for the careful reading of the manuscript and for several useful remarks. The work was partially supported by the Foundation of Sun Yat-sen University Advanced Research Centre.

1. Introduction

Iff, g ≥ 0, p > 1, ¹_p + ¹_q = 1,0< R∞

0 f^p(x)dx < ∞and0 <R∞

0 g^q(x)dx <

∞, then (1.1)

Z ∞ 0

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

< π sin(π/p)

Z ∞ 0

f^p(x)dx

¹_pZ ∞ 0

g^q(x)dx ¹_q

,

(1.2)

Z ∞ 0

f(x) x+ydx

p

dy <

π sin(π/p)

pZ ∞ 0

f^p(x)dx,

where the constant factor _sin(π/p)^π is the best possible. Inequality (1.1) is known as Hardy-Hilbert’s integral inequality (see [1]); it is important in analysis and its applications (see [4]). Under the same condition of (1.1), we have the Hardy- Hilbert type inequality similar to (1.1):

(1.3) Z ∞

0

Z ∞ 0

f(x)g(y)

max{x, y}dxdy < pq Z ∞

0

f^p(x)dx

¹_pZ ∞ 0

g^q(x)dx ¹_q

,

(1.4)

Z ∞ 0

f(x) max{x, y}dx

p

dy <(pq)^p Z ∞

0

f^p(x)dx,

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where the constant factor pqis the best possible. The corresponding inequality for series is:

(1.5)

∞

X

n=1

∞

X

m=1

a_mb_n

max{m, n} < pq

∞

X

n=1

a^p_n

!¹_p _∞ X

n=1

b^q_n

!¹_q ,

where the constant factor pqis also the best possible. In particular, whenp = q = 2, we have the well-known Hilbert type inequality:

(1.6) Z ∞

0

Z ∞ 0

f(x)g(y)

max{x, y}dxdy <4 Z ∞

0

f²(x)dx

¹₂ Z ∞ 0

g²(x)dx ¹₂

.

In recent years, Kuang (see [3]) gave a strengthened form as:

(1.7)

∞

X

n=1

∞

X

m=1

a_mb_n max{m, n}

<

( _∞ X

n=1

[pq−G(p, n)]a^p_n

)¹_p( _∞ X

n=1

[pq−G(q, n)]b^q_n )¹_q

,

whereG(r, n) = ^r+1/3r−4/3_(2n+1)1/r >0 (r=p, q).

Yang (see [5,8]) gave: forλ >2−min{p, q}

(1.8) Z ∞

0

Z ∞ 0

f(x)g(y)

max{x^λ, y^λ}dxdy < pqλ

(p+λ−2)(q+λ−2)

× Z ∞

0

x(p−1)(2−λ)−1

f^p(x)dx

¹_pZ ∞ 0

x(q−1)(2−λ)−1

g^q(x)dx ¹_q

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and (1.9)

Z ∞ 0

f(x)g(y)

max{x^λ, y^λ}dxdy < pqλ

(p+λ−2)(q+λ−2)

× Z ∞

0

x^1−λf^p(x)dx

¹_p Z ∞ 0

x^1−λg^q(x)dx ¹_q

.

At the same time, Yang (see [6,7]) considered the refinement of other types of Hilbert’s inequalities.

In this paper, we give a generalization of Hilbert’s type inequality and an improvement as:

Z ∞ 0

f(x^λ)g(y^µ) max{x, y}dxdy

< pq λ¹^pµ¹^q

Z ∞ 0

x^λ¹⁻¹f^p(x)dx

¹_p Z ∞ 0

x^µ¹⁻¹g^q(x)dx ¹_q

,

whereλ >0andµ >0.

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2. Main Results

Lemma 2.1. Supposer >1,¹_s +¹_r = 1, λ, µ, ε >0.Then

(2.1)

Z ∞ 1

x^−ε(¹^s⁺^λr^µ⁾⁻¹ Z x⁻^λ¹

0

1

max{1, t}t^−1−µε^r dtdx=O(1) (ε→o⁺).

Proof. There exist n ∈ Nwhich is large enough, such that1 + ^−1−µε_r > 0for ε ∈(0,_µn¹ ]andx≥1, we have

Z x⁻^λ¹ 0

1

max{1, t}t^−1−µε^r dt= Z x⁻^λ¹

0

t^−1−µε^r dt= 1 1 + ^−1−µε_r

x⁻^λ¹1+^−1−µε_r

.

Since fora≥1the functiong(y) = _ya¹y (y∈(0,∞))is decreasing, we find 1

1 + ^−1−µε_r

x⁻^λ¹1+^−1−µε_r

≤ 1

1 + ^−1−1/n_r

x⁻¹^λ1+^−1−1/n_r

,

so

0<

Z ∞ 1

x^−ε(¹^s⁺^λr^µ⁾⁻¹ Z x⁻^λ¹

0

1

max{1, t}t^−1−µ^r dtdx

<

Z ∞ 1

x⁻¹ 1 1 + ^−1−1/n_r

x⁻^λ¹1+^−1−1/n_r

= 1 λ

1 1 + ^−1−1/n_r

!2

.

Hence relation (2.1) is valid. The lemma is proved.

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Now we study the following inequality:

Theorem 2.2. Supposef(x), g(x)≥0, p >1, ¹_p +¹_q = 1,λ >0, µ >0and

0<

Z ∞ 0

x^λ¹⁻¹f^p(x)dx <∞, 0<

Z ∞ 0

x¹^µ⁻¹g^q(x)dx <∞.

Then (2.2)

Z ∞ 0

< pq λ¹^pµ¹^q

Z ∞ 0

x^λ¹⁻¹f^p(x)dx

¹_p Z ∞ 0

,

where the constant factor ^pq

λ

1 pµ

1

q is the best possible forλ =µ.

Proof. By Hölder’s inequality, we have Z ∞

0

Z ∞ 0

= 1 λ

1 µ

Z ∞ 0

h

x^λ¹⁻¹f(x)i h

y¹^µ⁻¹g(y)i maxn

x^λ¹, y^µ¹o dxdy

= 1 λ

1 µ

Z ∞ 0







x^λ¹⁻¹f(x)

maxn

x^λ¹, y^µ¹o¹_p

x⁽¹⁻¹^λ^)p/q y¹⁻¹^µ

!¹_p x¹^λ y^µ¹

!_pq¹







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×







y^µ¹⁻¹g(y)

max n

x^λ¹, y^µ¹ o¹_q

y⁽¹⁻^µ¹⁾ x⁽¹⁻¹^λ^)p/q

!¹_p y^µ¹ x¹^λ

!_pq¹





dxdy

≤ 1 λ

1 µ



 Z ∞

0

Z ∞ 0

x^p(¹^λ⁻¹⁾ f^p(x) maxn

x^λ¹, y^µ¹o

x⁽¹⁻¹^λ^)p/q y¹⁻^µ¹

x^λ¹ y^µ¹

!¹_q dxdy





1 p

×



 Z ∞

0

Z ∞ 0

y^q(^µ¹⁻¹⁾ g^q(y) maxn

x^λ¹, y^µ¹o

y⁽¹⁻^µ¹^)q/p x¹⁻¹^λ

y^µ¹ x¹^λ

!_p¹ dxdy





1 q

.

Define the weight functionϕ(x),ψ(y)as ϕ(x) :=

Z ∞ 0

1 maxn

x¹^λ, y^µ¹o · x⁽¹⁻¹^λ^)p/q y¹⁻¹^µ

x^λ¹ y¹^µ

!¹_q

dy, x∈(0,∞),

ψ(y) :=

Z ∞ 0

1 maxn

x¹^λ, y^µ¹o · y⁽¹⁻¹^µ^)q/p x¹⁻¹^λ

y^µ¹ x^λ¹

!¹_p

dx, y ∈(0,∞),

then above inequality yields Z ∞

0

Z ∞ 0

≤ 1 λ

1 µ

Z ∞ 0

ϕ(x)x^p(¹^λ⁻¹⁾f^p(x)dx

¹_pZ ∞ 0

ψ(y)y^q(¹^µ⁻¹⁾g^q(y)dy ¹_q

.

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For fixedx, lety^µ¹ =x¹^λt, we have ϕ(x) :=µx^(p−1)(1−^λ¹⁾

Z ∞ 0

1

max{1, t}t¹^p⁻¹dt

=µpqx^(p−1)(1−¹^λ⁾

=µpqx^(p−1)(1−¹^λ⁾.

By the same token,ψ(y) = λpqy^(q−1)(1−¹^µ⁾, thus (2.3)

Z ∞ 0

≤ pq λ¹^pµ¹^q

Z ∞ 0

x^λ¹⁻¹f^p(x)dx

¹_p Z ∞ 0

.

If (2.3) takes the form of the equality, then there exist constantscandd, such that Kuang (see [2])

cx^p(^λ¹⁻¹⁾f^p(x)

max{x¹^λ, y^µ¹} · x⁽¹⁻¹^λ^)p/q y¹⁻¹^µ

x^λ¹ y¹^µ

!¹_q

=d y^q(^µ¹⁻¹⁾g^q(y) max

n x¹^λ, y^µ¹

o · y⁽¹⁻¹^µ^)q/p x¹⁻¹^λ

y^µ¹ x^λ¹

!_p¹

a.e. on(0,∞)×(0,∞).

Then we have

cx^λ¹f^p(x) = dy¹^µg^q(y) a.e. on(0,∞)×(0,∞).

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Hence we have

cx¹^λf^p(x) = dy^µ¹g^q(y) = constant a.e. on(0,∞)×(0,∞), which contradicts the facts that

0<

Z ∞ 0

x¹^λf^p(x)dx <∞ and 0<

Z ∞ 0

y^µ¹g^q(x)dx <∞.

Hence (2.3) takes the form of strict inequality. So we have (2.2).

For0 < ε < ¹₂, setting f_ε(x) = x^−ε−1/λ^p , for x ∈ [1,∞); f_ε(x) = 0, for x ∈ (0,1), and g_ε(y) = y^−ε−1/µ^q , for y ∈ [1,∞); g_ε(y) = 0, for y ∈ (0,1).

Assume that the constant factor ^pq

λ¹^pµ¹^q

in (2.2) is not the best possible, then there exists a positive numberKwithK < ^pq

λ^p¹µ¹^q, such that (2.2) is valid by changing

pq λ^p¹µ¹^q

toK. We have Z ∞

0

Z ∞ 0

< K Z ∞

0

x¹^λ⁻¹f^p(x)dx

¹_pZ ∞ 0

x¹^µ⁻¹g^q(x)dx ¹_q

= K ε .

Since

Z ∞ 0

1

max{1, t}t^−1−εµ^q dt=pq+o(1) (ε→0⁺),

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settingy¹^µ =x^λ¹t, by (2.1), we find Z ∞

0

Z ∞ 0

= 1 λ

1 µ

Z ∞ 0

h

x^λ¹⁻¹f(x) i h

y¹^µ⁻¹g(y) i

maxn

= 1 λ

1 µ

Z ∞ 1

x⁽¹^λ⁻¹⁾⁺

−ε−1 λ

p y⁽^µ¹⁻¹⁾⁺

−ε−1 µ q

maxn

= 1 λ

1 µ

Z ∞ 1

Z ∞ x⁻^λ¹

x⁽¹^λ⁻¹⁾⁺

−ε−1 λ

p (t^µx^µ^λ)⁽^µ¹⁻¹⁾⁺

−ε−1 µ q

maxn

x¹^λ, tx¹^λo x^µ^λµt^µ−1dxdt

= 1 λ

Z ∞ 1

x^−ε(¹^p⁺^λq^µ⁾⁻¹ Z ∞

x⁻¹^λ

1

max{1, t}t^−1−εµ^q dtdx

= 1 λ

Z ∞ 1

x^−ε(¹^p⁺^λq^µ⁾⁻¹ Z ∞

0

1

− Z x⁻¹^λ

0

1





= 1 λε

"

pq

1

p +_λq^µ +o(1)

# .

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Since forε >0small enough, we have 1

λ

"

pq

1

p +_λq^µ +o(1)

#

< K.

It is obvious thatλ¹^pµ¹^q ≤ ^λ_p +^µ_q (i.e. _λ¹ 1 ¹

p+_λq^µ ≤ ¹

λ¹^pµ¹^q

)by Young’s inequality.

Consider the case of taking the form of the equality for Young’s inequality, we getµ¹^q =λ^p−1^p ,i.e.λ=µ, Then

1 λ

"

pq

1

p +_λq^µ +o(1)

#

= pq λ¹^pµ¹^q

+o(1)< K.

Thus we get ^pq

λ

1 pµ

1

q ≤ K, which contradicts the hypothesis. Hence the constant factor ^pq

λ¹^pµ¹^q in (2.2) is best possible forλ =µ.

Remark 1. Forλ=µ, inequality (2.2) becomes

(2.4) Z ∞

0

Z ∞ 0

f(x^λ)g(y^λ) max{x, y}dxdy

< pq λ

Z ∞ 0

x^λ¹⁻¹f^p(x)dx

_p¹ Z ∞ 0

x^λ¹⁻¹g^q(x)dx ¹_q

,

where the constant factor ^pq_λ is the best possible.

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Theorem 2.3. Supposef ≥0, p > 1, λ > 0and0<R∞

0 x^λ¹⁻¹f^p(x)dx < ∞.

Then (2.5)

Z ∞ 0

f(x^λ) max{x, y}dx

p

dy < 1 λ(pq)^p

Z ∞ 0

x¹^λ⁻¹f(x)^pdx,

where the constant factor _λ¹(pq)^p is the best possible. Inequality (2.5) is equiv- alent to (2.4).

Proof. Settingg(y)as



 Z ∞

0

x¹^λ⁻¹f(x) maxn

x¹^λ, y¹^λodx





p−1

>0, y∈(0,∞).

then by (2.4), we find λ⁻²

Z ∞ 0

y^λ¹⁻¹g^q(y)dy=λ^p−1 Z ∞

0

Z ∞ 0

f(x^λ) max{x, y}dx

^p dy

= Z ∞

0

Z ∞ 0

≤ pq λ

Z ∞ 0

x^λ¹⁻¹f^p(x)dx

¹_p Z ∞ 0

y^λ¹⁻¹g^q(y)dy ¹_q

. (2.6)

Hence we obtain (2.7) 0<

Z ∞ 0

y¹^λ⁻¹g^q(y)dy≤λ^p(pq)^p Z ∞

0

x¹^λ⁻¹f^p(x)dx <∞.

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By (2.4), both (2.6) and (2.7) take the form of strict inequality, so we have (2.5).

On the other hand, suppose that (2.5) is valid. By Hölder’s inequality, we find

Z ∞ 0

= Z ∞

0

Z ∞ 0

f(x^λ) max{x, y}dx

g(y^λ)dy

≤ Z ∞

0

Z ∞ 0

f(x^λ) max{x, y}dx

p

dy

¹_pZ ∞ 0

g^q(y^λ)dy ¹_q

. (2.8)

Then by (2.5), we have (2.4). Thus (2.4) and (2.5) are equivalent.

If the constant _λ¹(pq)^p in (2.5) is not the best possible, by (2.8), we may get a contradiction that the constant factor in (2.4) is not the best possible. Thus we complete the proof of the theorem.

Remark 2.

(i) Forλ =µ= 1, (2.2) and (2.5) reduce respectively to (1.3) and (1.4). It fol- lows that (2.2) is a new extension of (1.6) and (1.3) with some parameters and the equivalent form (2.4) is a new extension of (1.4).

(ii) It is amazing that (2.4) and (1.9) are different, although both of them are the extensions of (1.6) with(p, q)−parameter and the best constant factor.

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References

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

[2] J. KUANG, Applied Inequalities (Chinese), Hunan Jiaoyu Chubanshe, Changsha, 1993.

[3] J. KUANG AND L. DEBNATH, On new generalizations of Hilbert’s in- equality and their applications, J. Math. Anal. Appl., 245(1) (2000), 248–

265.

[4] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´CANDA.M. FINK, Inequalities Involv- ing Functions and their Integrals and Derivatives, Mathematics and its Ap- plications, 53. Kluwer Academic Publishers Group, Dordrecht, 1991.

[5] B. YANG, Generalization of a Hilbert type inequality and its applications (Chinese), Gongcheng Shuxue Xuebao, 21(5) (2004), 821–824.

[6] B. YANG, On a Hilbert’s type inequality and its applications, Math. Appl.

(Wuhan), 15(1) (2002), 52–56.

[7] B. YANG, A new Hardy-Hilbert type inequality and its applications (Chi- nese), Acta Math. Sinica, 46(6) (2003), 1079–1086.

[8] B. YANG, Best generalization of a Hilbert-type inequality (Chinese), J. Jilin Univ. Sci., 42(1) (2004), 30–34.