INTRODUCTION Iff, g ≥0, p >1, 1p + 1q = 1,0<R∞ 0 fp(x)dx <∞and0<R∞ 0 gq(x)dx &lt

http://jipam.vu.edu.au/

Volume 7, Issue 2, Article 61, 2006

SOME EXTENSIONS OF HILBERT’S TYPE INEQUALITY AND ITS APPLICATIONS

YONGJIN LI AND BING HE DEPARTMENT OFMATHEMATICS

SUNYAT-SENUNIVERSITY

GUANGZHOU, 510275 P. R. CHINA

[email protected] [email protected]

Received 28 November, 2005; accepted 19 January, 2006 Communicated by B. Yang

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

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

2000 Mathematics Subject Classification. 26D15.

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

Z ∞ 0

f(x)g(y)

x+y dxdy < π sin(π/p)

Z ∞ 0

f^p(x)dx

¹_p Z ∞ 0

g^q(x)dx ¹_q

,

(1.2)

Z ∞ 0

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

,

ISSN (electronic): 1443-5756 c

2006 Victoria University. All rights reserved.

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.

342-05

(1.4)

Z ∞ 0

Z ∞ 0

f(x) max{x, y}dx

p

dy <(pq)^p Z ∞

0

f^p(x)dx,

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

(1.5)

∞

X

n=1

∞

X

m=1

ambn

max{m, n} < pq

∞

X

n=1

a^p_n

!¹_{p ∞} X

n=1

b^q_n

!¹_q ,

where the constant factorpq is 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

and (1.9)

Z ∞ 0

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_pZ ∞ 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 in- equalities.

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

Z ∞ 0

Z ∞ 0

f(x^λ)g(y^µ)

max{x, y}dxdy < pq λ^1pµ^1q

Z ∞ 0

x^λ1−1f^p(x)dx

¹_pZ ∞ 0

x^µ1−1g^q(x)dx ¹_q

,

whereλ >0andµ >0.

2. MAINRESULTS

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

(2.1)

Z ∞ 1

x^{−ε(1s+λrµ)−1} Z x^−λ1

0

1

max{1, t}t^−1−µεr dtdx=O(1) (ε→o⁺).

Proof. There existn ∈ Nwhich is large enough, such that1 + ^−1−µε_r > 0forε ∈ (0,_µn¹ ]and x≥1, we have

Z x^−λ1 0

1

max{1, t}t^−1−µεr dt= Z x^−λ1

0

t^−1−µεr 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^−1λ1+^−1−µε_r

≤ 1

1 + ^−1−1/n_r

x^−λ11+^−1−1/n_r

,

so

0<

Z ∞ 1

x^{−ε(1s+λrµ)−1} Z x^−1λ

0

1

max{1, t}t^−1−µr 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.

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

x^µ1−1g^q(x)dx <∞.

Then (2.2)

Z ∞ 0

Z ∞ 0

f(x^λ)g(y^µ)

max{x, y}dxdy < pq λ^1pµ^1q

Z ∞ 0

x^λ1−1f^p(x)dx

¹_p Z ∞ 0

x^µ1−1g^q(x)dx ¹_q

,

where the constant factor ^pq

λ^1pµ^1q

is the best possible forλ =µ.

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

0

Z ∞ 0

f(x^λ)g(y^µ) max{x, y}dxdy

= 1 λ

1 µ

Z ∞ 0

Z ∞ 0

h

x^1λ−1f(x) i h

y^µ1−1g(y) i

maxn

x^1λ, y^1µo dxdy

= 1 λ

1 µ

Z ∞ 0

Z ∞ 0







x^λ1−1f(x)

maxn

x^λ1, y^µ1o¹_p

x^(1−λ1)p/q y^1−µ1

!¹_p x^1λ y^µ1

!_pq¹







×







y^µ1−1g(y)

maxn

x^1λ, y^1µo¹_q

y^(1−µ1) x^(1−λ1)p/q

!¹_p y^1µ x^λ1

!_pq¹





dxdy

≤ 1 λ

1 µ



 Z ∞

0

Z ∞ 0

x^p(1λ−1) f^p(x) maxn

x^λ1, y^µ1o

x^(1−λ1)p/q y^1−µ1

x^1λ y^µ1

!¹_q dxdy





1 p

×



 Z ∞

0

Z ∞ 0

y^q(1µ−1) g^q(y) maxn

x^λ1, y^µ1o

y^(1−µ1)q/p x^1−λ1

y^µ1 x^1λ

!¹_p dxdy





1 q

.

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

Z ∞ 0

1 maxn

x^λ1, y^µ1o · x^(1−1λ)p/q y^1−1µ

x^λ1 y^µ1

!¹_q

dy, x∈(0,∞),

ψ(y) :=

Z ∞ 0

1 maxn

x^λ1, y^µ1o · y^(1−µ1)q/p x^1−λ1

y^µ1 x^1λ

!¹_p

dx, y∈(0,∞),

then above inequality yields Z ∞

0

Z ∞ 0

f(x^λ)g(y^µ) max{x, y}dxdy

≤ 1 λ

1 µ

Z ∞ 0

ϕ(x)x^p(1λ−1)f^p(x)dx

¹_pZ ∞ 0

ψ(y)y^q(1µ−1)g^q(y)dy ¹_q

.

For fixedx, lety^1µ =x^λ1t, we have

ϕ(x) := µx^{(p−1)(1−1λ)} Z ∞

0

1

max{1, t}t^p1−1dt

=µpqx^{(p−1)(1−λ1)}

=µpqx^{(p−1)(1−λ1)}.

By the same token,ψ(y) =λpqy^{(q−1)(1−µ1)}, thus (2.3)

Z ∞ 0

Z ∞ 0

f(x^λ)g(y^µ)

max{x, y}dxdy ≤ pq λ^1pµ^1q

Z ∞ 0

x^λ1−1f^p(x)dx

_p¹ Z ∞ 0

x^µ1−1g^q(x)dx ¹_q

.

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{x^λ1, y^µ1} · x^(1−1λ)p/q y^1−1µ

x^λ1 y^1µ

!¹_q

=d y^q(1µ−1)g^q(y) maxn

x^λ1, y^µ1o · y^(1−µ1)q/p x^1−λ1

y^µ1 x^1λ

!¹_p

a.e. on(0,∞)×(0,∞).

Then we have

cx^1λf^p(x) = dy^µ1g^q(y) a.e. on(0,∞)×(0,∞).

Hence we have

cx^λ1f^p(x) =dy^µ1g^q(y) = constant a.e. on(0,∞)×(0,∞), which contradicts the facts that

0<

Z ∞ 0

x^1λf^p(x)dx <∞ and 0<

Z ∞ 0

y^1µg^q(x)dx <∞.

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

For 0 < ε < ¹₂, settingf_ε(x) = x

−ε−1/λ

p , forx ∈ [1,∞); f_ε(x) = 0, forx ∈ (0,1), and gε(y) = y

−ε−1/µ

q , for y ∈ [1,∞);gε(y) = 0, for y ∈ (0,1). Assume that the constant factor

pq

λ^1pµ^1q in (2.2) is not the best possible, then there exists a positive numberK with K < ^pq

λ^1pµ^1q, such that (2.2) is valid by changing ^pq

λ^1pµ^1q

toK. We have Z ∞

0

Z ∞ 0

f(x^λ)g(y^µ)

max{x, y}dxdy < K Z ∞

0

x^1λ−1f^p(x)dx

¹_pZ ∞ 0

x^1µ−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⁺),

settingy^µ1 =x^λ1t, by (2.1), we find Z ∞

0

Z ∞ 0

f(x^λ)g(y^µ) max{x, y}dxdy

= 1 λ

1 µ

Z ∞ 0

Z ∞ 0

h

x^1λ−1f(x)i h

y^µ1−1g(y)i maxn

x^1λ, y^1µo dxdy

= 1 λ

1 µ

Z ∞ 1

Z ∞ 1

x^(λ1−1)+

−ε−1 λ

p y^(1µ−1)+

−ε−1 µ q

max n

x^1λ, y^1µ

o dxdy

= 1 λ

1 µ

Z ∞ 1

Z ∞ x^−1λ

x^(λ1−1)+

−ε−1 λ

p (t^µx^µλ)^(µ1−1)+

−ε−1 µ q

maxn

x^λ1, tx^λ1o x^µλµt^µ−1dxdt

= 1 λ

Z ∞ 1

x^{−ε(1p+λqµ)−1} Z ∞

x^−λ1

1

max{1, t}t^−1−εµq dtdx

= 1 λ

Z ∞ 1

x^{−ε(1p+λqµ)−1}



 Z ∞

0

1

max{1, t}t^−1−εµq dtdx− Z x^−λ1

0

1

max{1, t}t^−1−εµq dtdx





= 1 λε

"

pq

1

p + _λq^µ +o(1)

# .

Since forε >0small enough, we have 1 λ

"

pq

1

p +_λq^µ +o(1)

#

< K.

It is obvious thatλ^1pµ^1q ≤ ^λ_p+^µ_q (i.e. _λ¹ 1 ¹

p+_λq^µ ≤ ¹

λ^1pµ^1q

)by Young’s inequality. Consider the case of taking the form of the equality for Young’s inequality, we getµ^1q =λ

p−1

p ,i.e. λ=µ, Then 1

λ

"

pq

1

p +_λq^µ +o(1)

#

= pq λ^1pµ^1q

+o(1) < K.

Thus we get ^pq

λ^1pµ^1q

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

λ^1pµ^1q in

(2.2) is best possible forλ =µ.

Remark 2.3. 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.

Theorem 2.4. Supposef ≥0, p >1, λ > 0and0<R∞

0 x^1λ−1f^p(x)dx <∞. Then (2.5)

Z ∞ 0

Z ∞ 0

f(x^λ) max{x, y}dx

p

dy < 1 λ(pq)^p

Z ∞ 0

x^1λ−1f(x)^pdx,

where the constant factor _λ¹(pq)^pis the best possible. Inequality (2.5) is equivalent to (2.4).

Proof. Settingg(y)as



 Z ∞

0

x^λ1−1f(x) maxn

x^λ1, y^λ1odx





p−1

>0, y∈(0,∞).

then by (2.4), we find λ⁻²

Z ∞ 0

y^1λ−1g^q(y)dy=λ^p−1 Z ∞

0

Z ∞ 0

f(x^λ) max{x, y}dx

p

dy

= Z ∞

0

Z ∞ 0

f(x^λ)g(y^λ) max{x, y}dxdy

≤ pq λ

Z ∞ 0

x^λ1−1f^p(x)dx

¹_pZ ∞ 0

y^1λ−1g^q(y)dy ¹_q

. (2.6)

Hence we obtain

(2.7) 0<

Z ∞ 0

y^λ1−1g^q(y)dy≤λ^p(pq)^p Z ∞

0

x^1λ−1f^p(x)dx <∞.

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

f(x^λ)g(y^λ) max{x, y}dxdy

= 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 2.5.

(i) For λ = µ = 1, (2.2) and (2.5) reduce respectively to (1.3) and (1.4). It follows 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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