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volume 4, issue 5, article 94, 2003.

Received 01 August, 2003;

accepted 28 August, 2003.

Communicated by:L. Debnath

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Journal of Inequalities in Pure and Applied Mathematics

ON HARDY-HILBERT’S INTEGRAL INEQUALITY WITH PARAMETERS

LEPING HE, MINGZHE GAO AND WEIJIAN JIA

Department of Mathematics and Computer Science, Normal College, Jishou University,

Jishou Hunan, 416000 People’s Republic of China.

EMail:[email protected] EMail:[email protected] EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 108-03

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On Hardy-Hilbert’s Integral Inequality with Parameters

Leping He, Mingzhe Gao and Weijian Jia

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J. Ineq. Pure and Appl. Math. 4(5) Art. 94, 2003

http://jipam.vu.edu.au

Abstract

In this paper, by means of a sharpening of Hölder’s inequality, Hardy-Hilbert’s integral inequality with parameters is improved. Some new inequalities are established.

2000 Mathematics Subject Classification:26D15, 46C99

Key words: Hardy-Hilbert integral inequality, Hölder’s inequality, Weight function, Beta function.

The authors thank the referee for his help and patience in improving the paper.

1. Introduction

Letp > 1,¹_p + ¹_q = 1, f, g > 0. If 0< R∞

0 f^p(t)dt < +∞, 0 <R∞

0 g^q(t)dt <

+∞, then (1.1)

Z ∞ 0

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

< π sin (π/p)

Z ∞ 0

f^p(t)dt

_p¹ Z ∞ 0

g^q(t)dt ¹_q

, where the constant_sin(π/p)^π is best possible. The inequality (1.1) is well known as Hardy-Hilbert’s integral inequality. In recent years, some improvements and extensions of Hilbert’s inequality and Hardy-Hilbert’s inequality have been given in [2] – [6], Yang [2] gave a generalization of (1.1) as follows:

Ifλ >2−min{p, q}, α < T 6∞then (1.2)

Z T α

f(x)g(y)

(x+y−2α)^λdxdy

<

(Z T

α

"

kλ(p)−θλ(p)

t−α T −α

^p+λ−2_p #

(t−α)^1−λf^p(t)dt )¹p

× (Z T

α

"

k_λ(p)−θ_λ(q)

t−α T −α

^q+λ−2_q #

(t−α)^1−λg^q(t)dt )¹_q

(T < ∞)

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and (1.3)

Z ∞ α

f(x)g(y)

(x+y−2α)^λdxdy

< k_λ(p) Z ∞

α

(t−α)^1−λf^p(t)dt

¹_p Z ∞ α

(t−α)^1−λg^q(t)dt ¹_q

, where

k_λ(p) =B

p+λ−2

p ,q+λ−2 q

, θ_λ(r) =

Z 1 0

1 (1 +u)^λ

1 u

(2−λ)/r

du (r =p, q).

The main purpose of this paper is to build a few new inequalities which include improvements of the inequalities (1.2) and (1.3), and extensions of cor- responding results in [3] – [5].

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2. Lemmas and their Proofs

For convenience, we firstly introduce some notations:

(f^r, g^s) = Z T

α

f^r(x)g^s(x)dx, kfk_p = Z T

α

f^p(x)dx

1 p

, kfk₂ =kfk. We next introduce a function defined by

S_r(H, x) = H^r/2, x

kHk^−r/2_r ,

where xis a parametric variable vector which is a variable unit vector. Under the general case, it is properly chosen such that the specific problems discussed are simplified.

Clearly, S_r(H, x) = 0 when the vector x selected is orthogonal to H^p/2. Throughout this paper, the exponent m indicatesm = minn

1 p,¹_qo

, α < T 6

∞.

In order to verify our assertions, we need to build the following lemmas.

Lemma 2.1. Let f(x), g(x) >0, ¹_p + ¹_q = 1 andp >1. If0 <kfk_p < +∞

and0<kgk_q <+∞,then

(2.1) (f, g)<kfk_pkgk_q(1−R)^m,

where R = (S_p(f, h)−S_q(g, h))², khk = 1, f^p/2(x), g^q/2(x)and h(x) are linearly independent.

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Proof. First of all, we discuss the case of p 6= q. Without loss of generality, suppose thatp > q >1, since ¹_p + ¹_q = 1, we havep > 2. LetR = ^p₂,Q= _p−2^p . Then _R¹ +_Q¹ = 1. By Hölder’s inequality we obtain,

(f, g) = Z T

a

f(x)g(x)dx (2.2)

= Z T

a

f·g^q/p

g^1−(q/p)dx 6

Z T a

f ·g^q/pR

dx

1

R Z T

a

g^1−(q/p)Q

dx

1 Q

= f^p/2, g^q/2²_p

kgk^q(¹⁻²p)

q .

And the equality in (2.2) holds if and only if f^p/2 andg^q/2 are linearly dependent. In fact, the equality in (2.2) holds if and only if, there exists ac₁such that

f·g^q/pR

=c₁ g^1−(q/p)Q

. It is easy to deduce thatf^p/2 =c₁g^q/2.

In our previous paper [3], with the help of the positive definiteness of the Gram matrix, we established an important inequality of the form

(2.3) (α, β)² 6kαk²kβk²−(kαkx− kβky)² =kαk²kβk²(1−γ) whereγ =

y

kαk − _kβk^x 2

, x = (β, γ), y = (α, γ)withkγk = 1andxy > 0.

The equality in (2.3) holds if and only ifαandβ are linearly dependent; or the vectorγ is a linear combination ofαandβ, andxy= 0butx6=y. Ifα, β and γ in (2.3) are replaced byf^p/2,g^q/2 andhrespectively, then we get

(2.4) f^p/2, g^q/2²

6kfk^p_pkgk^q_q(1−R),

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whereR = (S_p(f, h)−S_q(g, h))² with||h|| = 1. The equality in (2.4) holds if and only if f^p/2andg^q/2 are linearly dependent, orh is a linear combination of f^p/2 andg^q/2, and f^p/2, h

g^q/2, h

= 0, but f^p/2, h

6= g^q/2, h

. Since f^p/2andg^q/2are linearly independent, it is impossible to have equality in (2.4).

Substituting (2.4) into (2.2), we obtain after simplifications (2.5) (f, g)<kfk_pkgk_q(1−R)¹^p .

Provided that h(x)is properly chosen, thenR 6= 0 is achieved. (The choice of h(x) is quite flexible, as long as condition khk = 1is satisfied, on which we can refer to [3,4], etc.). Noticing the symmetry ofpandq, the inequality (2.1) follows from (2.5).

Next, we discuss the case ofp=q. According to the hypothesis: whenf, g andhare linearly independent, we immediately obtain from (2.3) the following result:

(f, g)<kfk kgk(1−r)¯ ¹² , where¯r=

(f,h)

kfk − ^(g,h)_kgk 2

, andkhk= 1. Thus the lemma is proved.

Lemma 2.2. Let p > 1, ¹_p + ¹_q = 1, λ > 2−min{p, q}, α < T < ∞. Define the weight functionω_λ as follows:

(2.6) ω_λ(α, T, r, x) = Z T

α

1 (x+y−2α)^λ

x−α y−α

2−λ r

dy x∈(α, T].

Settingω_λ(α,∞, r, x) = lim_T→∞ω_λ(α, T, r, x)andk_λ(p) = B

p+λ−2

p ,^q+λ−2_q ,

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(2.7) θλ(r) = Z 1

0

1 (1 +u)^λ

1 u

(2−λ)(1−1/r)

du, (r=p, q), then we have

(2.8) ω_λ(α,∞, r, x) =k_λ(p)(x−α)^1−λ, x∈(α,∞) and

(2.9) ω_λ(α, T, r, x)

< k_λ(p)−θ(r)

x−α T −α

1+(λ−2)(1−1/r)!

(x−α)^1−λ, x∈(α, T), whereB(m, n)is the beta function.

The proof of this lemma is given in the paper [2]; it is omitted here.

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3. Main Results

In order to state it conveniently, we need again to define the functions and introduce some notations

F = f(x)

(x+y−2α)^λ/p

x−α y−α

^2−λ_pq

, G= g(y)

(x+y−2α)^λ/q

y−α x−α

^2−λ_pq ,

Sp(F, hT) = Z T

α

Z T α

F^p/2hTdxdy

Z T α

F^pdxdy ⁻¹2

, S_q(G, h_T) =

Z T α

G^q/2h_Tdxdy

Z T α

G^qdxdy ⁻¹2

, wherehT =hT(x, y)is a unit vector with two variants, namely

||hT||= Z T

α

Z T α

h²_Tdxdy ¹2

= 1, α < T 6∞, andF^p/2, G^q/2, h_T are linearly independent.

Theorem 3.1. Let p > 1, ¹_p + ¹_q = 1, λ > 2− min{p, q}, α < T 6 ∞, f(t), g(t)>0. If

0<

Z ∞ α

(t−α)^1−λf^p(t)dt <+∞ and 0<

Z ∞ α

(t−α)^1−λg^q(t)dt <+∞, then

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(i) ForT < ∞, we have (3.1)

Z T α

f(x)g(y)

(x+y−2α)^λdxdy

<

(Z T

α

k_λ(p)−θ_λ(p)

t−α T −α

(p+λ−2)/p!

(t−α)^1−λf^p(t)dt )¹_p

× (Z T

α

k_λ(p)−θ_λ(q)

t−α T −α

(q+λ+2)/q!

(t−α)^1−λg^q(t)dt )¹_q

(1−R_T)^m, where

k_λ(p) =B

p+λ−2

p ,q+λ−2 q

, θ_λ(r) =

Z 1 0

1 (1 +u)^λ

1 u

^2−λ_r

du (r =p, q).

(ii) ForT =∞, we have (3.2)

Z ∞ α

f(x)g(y)

(x+y−2α)^λdxdy

< kλ(p) Z ∞

α

(t−α)^1−λf^p(t)dt ¹_p

× Z ∞

α

(t−α)^1−λg^q(t)dt ¹_q

(1−R∞)^m,

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whereR_T = (S_p(F, h_T)−S_q(G, h_T))²,

(3.3) hT (x, y) =









 2

π ¹₂

e^a−x (x+y−2a)¹²

x−a y−a

¹₄

, T =∞;

T −α

(x−α)(y−α)e(¹⁻2(x−α)^T−α −_2(y−α)^T−α ), T <∞.

Proof. By Lemma2.1, we get

Z T α

f(x)g(y)

(x+y−2α)^λdxdy (3.4)

= Z T

α

Z T α

F Gdxdy 6

Z T α

F^pdxdy

1

pZ T

α

Z T α

G^qdxdy

1 q

(1−R_T)^m

= Z T

α

ω_λ(α, β, q, t)f^p(t)dt ¹_p

× Z T

α

ω_λ(α, β, p, t)g^q(t)dt ¹_q

(1−R_T)^m, whereωλ(α, T, r, t) (r=p, q)is the function defined by (2.6) .

Now notice that θ_λ(p) = θ_λ(q), θ_λ(q) = θ_λ(p) and substituting (2.9) and (2.8) into (3.4) respectively, the inequalities (3.1) and (3.2) follow.

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It remains to discuss the expression ofR_T. We may choose the functionh_T indicated by (3.3).

WhenT =∞, settings=x−α, t =y−α, then

||h∞||= Z ∞

α

Z ∞ α

h²_∞(x, y)dxdy ¹₂

= 2

π Z ∞

0

e^−2sds Z ∞

0

1 s+t

s t

¹₂ dt

¹₂

= 1.

WhenT <∞, settingξ = ^T_x−α^−α, η= ^T_y−α^−α, then we have kh_Tk=

Z T α

h²_Tdxdy ¹₂

= Z T

α

T −α

(x−α)²e(¹⁻^T−αx−α)dx· Z T

α

T −α

(y−α)²e(¹⁻^T−αy−α)dy ¹2

= Z ∞

1

e^1−ξdξ· Z ∞

1

e^1−ηdη ¹₂

= 1.

According to Lemma 2.1 and the given h_T, we have R_T = (S_p(F, h_T)− S_q(G, h_T))². It is obvious that F^p/2, G^q/2 and h_T are linearly independent, so it is impossible for equality to hold in (3.4). Thus the proof of theorem is completed.

Remark 3.1. Clearly, the inequalities (3.1) and (3.2) are the improvements of (1.2) and (1.3) respectively .

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Owing top, q > 1, when λ = 1,2,3, the condition λ > 2−min(p, q)is satisfied, then we have

θ1(r) = Z 1

0

1 1 +u

1 u

_γ¹ du >

Z 1 0

1

1 +udu= ln 2, k1(p) = B

1 q,1

p

= π

sin(π/p), θ₂(r) =

Z 1 0

1

(1 +u)²du= 1 2, k₂(p) = B

p+ 2−2

p ,q+ 2−2 q

=B(1,1) = 1, θ₃(r) =

Z 1 0

1 (1 +u)³

1 u

−¹

γ

du >

Z 1 0

u

(1 +u)³du= 1 8, k₃(p) = 1

2pqB 1

q,1 p

= (p−1)π 2p²sin(π/p).

By Theorem3.1, some corollaries are established as follows:

Corollary 3.2. If p > 1, ¹_p + ¹_q = 1, λ = 1, α < T 6 ∞andf(t), g(t) > 0, 0<RT

α f^p(t)dt <+∞and0<RT

α g^q(t)dt <+∞, then we have (3.5)

Z T α

f(x)g(y) x+y−2αdxdy

<

(Z T

α

π sin(π/p)−

t−α T −α

¹_q

·ln 2

!

f^p(t)dt )¹_p

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× (Z T

α

π sin(π/p)−

t−α T −α

¹_p

·ln 2

!

·g^q(t)dt )¹_q

(1−r₁)^m, forT < ∞, and

(3.6) Z ∞

α

Z ∞ α

f(x)g(y) x+y−2αdxdy

< π sin(π/p)

Z ∞ α

f^p(t)dt

¹_pZ ∞ α

g^q(t)dt ¹_q

(1−r₁)^m. Remark 3.2. When α = 0andp =q = 2, the inequality (3.6) is reduced to a result which is equivalent to inequality (3.1) in [3] after simple computations.

As a result, the inequalities (3.1), (3.2) and (3.5) – (3.6) are all extensions of (3.1) in [3].

Corollary 3.3. Letp >1, ¹_p + ¹_q = 1,α < T 6∞andf(t), g(t)>0. If 0<

Z T α

1

t−αf^p(t)dt <+∞

and

0<

Z T α

1

t−αg^q(t)dt <+∞,

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then we obtain (3.7)

Z T α

f(x)g(y)

(x+y−2α)²dxdy

<

Z T α

1− t−α 2 (T −α)

1

t−α ·f^p(t)dt ¹p

× Z T

α

1− t−α 2 (T −α)

1

t−α ·g^q(t)dt ¹q

(1−r₂)^m,

for T < ∞, and

(3.8) Z ∞

α

Z ∞ α

f(x)g(y)

(x+y−2α)²dxdy

<

Z ∞ α

1

t−αf^p(t)dt

_p¹ Z ∞ α

1

t−αg^q(t)dt ¹_q

(1−r₂)^m. Corollary 3.4. Ifp > 1, ¹_p + ¹_q = 1,λ= 3,α < T 6∞andf(t), g(t)>0,

0<

Z T α

1

(t−α)²f^p(t)dt <+∞, 0<

Z T α

1

(t−α)²g^q(t)dt <+∞,

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then we get

(3.9) Z T

α

Z T α

f(x)g(y)

(x+y−2α)³dxdy

<

(Z T

α

(p−1)π 2p²sin(π/p)− 1

8

t−α T −α

1+¹_p! 1

(t−α)²f^p(t)dt )¹_p

× (Z T

α

(p−1)π 2p²sin(π/p)− 1

8

t−α T −α

1+¹_q! 1

(t−α)²g^q(t)dt )¹_q

(1−r₃)^m T < ∞, and

(3.10) Z ∞

α

Z ∞ α

f(x)g(y)

(x+y−2α)³dxdy

< (p−1)π 2p²sin(π/p)

Z ∞ α

1

(t−α)²f^p(t)dt ¹_p

× Z ∞

α

1

(t−α)²g^q(t)dt ¹_q

(1−r₃)^m. Sincek_λ(2) = B ^λ₂,^λ₂

, θ_λ(2) = ¹₂B ^λ₂,^λ₂

,andλ > 2−min(2,2) = 0,we also have

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Corollary 3.5. Ifp=q= 2,λ >0,α < T 6∞andf(t), g(t)>0, 0<

Z T α

(t−α)^1−λf²(t)dt <+∞, 0<

Z T α

(t−α)^1−λg²(t)dt <+∞, then we have

(3.11) Z T

α

Z T α

f(x)g(y)

(x+y−2α)^λdxdy

< B λ

2,λ 2

( Z T

α

"

1− 1 2

t−α T −α

λ/2#

(t−α)^1−λf²(t)dt )¹₂

× (Z T

α

"

1− 1 2

t−α T −α

λ/2#

(t−α)^1−λg²(t)dt )¹₂

(1−R)^m, for T < ∞ and

(3.12) Z ∞

α

Z ∞ α

f(x)g(y)

(x+y−2α)^λdxdy

< B λ

2,λ 2

Z ∞ α

(t−α)^1−λf²(t)dt ¹₂

× Z ∞

α

(t−α)^1−λg²(t)dt ¹₂

(1−R)e ^m.

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Remark 3.3. The inequalities (3.11), (3.12) are new generalizations of (20) in [4] and improvements of the inequalities (4) and (12) in [6] respectively.

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References

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

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

Appl., 261 (2001), 295–306.

[3] MINGZHE GAO, LI TAN AND L. DEBNATH, Some improvements on Hilbert’s integral inequality, J. Math. Anal. Appl., 229 (1999), 682–689.

[4] MINGZHE GAO, On the Hilbert inequality, Zeitschrift für Analysis und ihre Anwendungen, 18(4) (1999), 1117–1122.

[5] MINGZHE GAO, SHANG RONG WEI ANDLEPING HE, On the Hilbert inequality with weights, Zeitschrift für Analysis und ihre Anwendungen, 21(1) (2002), 257–263.

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