JIPAM - Journal of Inequalities in Pure and Applied Mathematics

A REVERSE HARDY-HILBERT-TYPE INTEGRAL INEQUALITY

GAOWEN XI

DEPARTMENT OFMATHEMATICS

LUOYANGTEACHERS’ COLLEGE

LUOYANG471022, P. R. CHINA

xigaowen@163.com

Received 07 December, 2007; accepted 25 May, 2008 Communicated by B. Opi´c

ABSTRACT. By estimating a weight function, a reverse Hardy-Hilbert-type integral inequality with a best constant factor is obtained. As an application, some equivalent forms and some particular results have been established.

Key words and phrases: Hardy-Hilbert-type integral inequality; weight function;βfunction; Hölder’s inequality.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

Letp >1, ¹_p + ¹_q = 1,f(x) ≥0,g(x)≥ 0, and0<R∞

0 f^p(x)dx <∞, 0<R∞

0 g^q(x)dx <

∞. Then the Hardy-Hilbert’s integral inequality is as follows:

(1.1)

Z ∞

0

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

, where the constant factor _sin(^ππ

p) is the best possible (see [1]). For (1.1), Yang et al. [2], [3], [4], [8] and [9] gave some strengthened versions and extensions. In particular, in [7], Yang obtained (1.2)

Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)^λdxdy < B

p+λ−2

p ,q+λ−2 q

× Z ∞

α

(x−α)^1−λf^p(x)dx

¹_p Z ∞

α

(x−α)^1−λg^q(x)dx ¹_q

, whereα ∈R, λ > 2−min{p, q},

The author wishes to thank the referee for several helpful comments. This research is supported by the Natural Science Foundation of Henan Province (Grant No. 2008B110011).

360-07

Recently, in [5], Xi gave a reverse Hardy-Hilbert-type inequality

(1.3)

∞

X

n=0

∞

X

m=0

a_mb_n (m+n+ 1)^λ

> 2 λ−1

( _∞ X

n=0

(n+ 1)^2−λ 2n+ 3−λ

1− (λ−1)² 4(n+ 1)²

a^p_n

)¹_p ( _∞ X

n=0

(n+ 1)^2−λ 2n+ 3−λb^q_n

)¹_q , where0< p <1,¹_p+¹_q = 1,1.5≤λ <3andan≥0, bn >0,such that0<P∞

n=0

(n+1)^2−λa^pn

2n+3−λ <

∞,0<P∞ n=0

(n+1)^2−λb^qn

2n+3−λ <∞.

In this paper, by estimating the weight function, a reverse Hardy-Hilbert-type integral in- equality with a best constant factor is obtained. As an application, some equivalent forms and some particular results have been established.

2. SOME LEMMAS

Lemma 2.1.

(2.1) B(p, q) =B(q, p) = Z ∞

0

1

(1 +u)^p+qu^p−1du (p, q >0), whereB(p, q)is theβfunction .

Proof. See [6].

Lemma 2.2. Letp < 0or0< p <1,and ¹_p+¹_q = 1, f(x), g(x)≥0, f ∈L^p(E), g∈L^q(E), x∈R^kand the setE is Borel measurable inR^k, wherek is a positive integer. Then

(2.2)

Z

E

f(x)g(x)dx≥ Z

E

f^p(x)dx ¹_p Z

E

g^q(x)dx ¹_q

,

where the equality holds if and only if there exist non-negative real numbersaandb, such that they are not all zero andaf^p(x) =bg^q(x),a. e. inE.

Proof. See [1, Section 6.9, Theorem 189].

Lemma 2.3. Letp < 0or0< p <1,and ¹_p+¹_q = 1,2−max{p, q}< λ <2−min{p, q}, y >

α,The weight functionω_λ(y, p)is defined by (2.3) ω_λ(y, p) =

Z ∞

α

1 (x+y−2α)^λ

y−α x−α

^2−λ_p

dx, y∈(0, ∞).

Then we have ^p+λ−2_p >0, ^q+λ−2_q >0,and (2.4) ω_λ(y, p) = (y−α)^1−λB

p+λ−2

p ,q+λ−2 q

.

Proof. Ifp <0,then0< q <1.Since1< λ <2−p, ^p+λ−2_p >0and ^q+λ−2_q >0.If0< p <1, we haveq <0,1< λ <2−q, ^p+λ−2_p >0, ^q+λ−2_q >0.

Settingt = ^x−α_y−α,by ^p+λ−2_p + ^q+λ−2_q =λ, we have

Z ∞

α

1 (x+y−2α)^λ

y−α x−α

^2−λ_p

dx = (y−α)^1−λ Z ∞

0

1

(1 +t)^λt^−2−λp dt

= (y−α)^1−λ Z ∞

0

1

(1 +t)^λt^{p+λ−2p −1}dt

= (y−α)^1−λB

p+λ−2

p ,q+λ−2 q

.

The lemma is thus proved.

Lemma 2.4. Letp <0, ¹_p +¹_q = 1,2−q < λ <2−p,0< ε < ^q+λ−2₂ ,then we have

I = Z ∞

α+1

Z ∞

α+1

1

(x+y−2α)^λ(x−α)^λ−2−εp (y−α)^λ−2−εq dxdy (2.5)

= 1 εB

q+λ−2

q −ε

q,p+λ−2

p + ε

q

−O_ε(1).

Proof. Settingx−α=s, y−α=t,then

I = Z ∞

α+1

Z ∞

α+1

1

(x+y−2α)^λ(x−α)^λ−2−εp (y−α)^λ−2−εq dxdy

= Z ∞

1

Z ∞

1

1

(s+t)^λs^λ−2−εp t^λ−2−εq dsdt

= Z ∞

1

s^λ−2−εp Z ∞

1

1

(s+t)^λt^λ−2−εq dt

ds.

Letu= ^t_s, we have

I = Z ∞

1

s^−1−ε

"

Z ∞

1 s

1

(1 +u)^λu^{q+λ−2−εq −1}du

# ds

= Z ∞

1

s^−1−ε Z ∞

0

1

(1 +u)^λu^{q+λ−2−εq −1}du

ds

− Z ∞

1

s^−1−ε

"

Z ¹_s

0

1

(1 +u)^λu^{q+λ−2−εq −1}du

# ds

= 1 εB

q+λ−2

q − ε

q,p+λ−2

p + ε

q

− Z ∞

1

s^−1−ε

"

Z ¹_s

0

1

(1 +u)^λu^{q+λ−2−εq −1}du

# ds.

By1<2−q < λ, 0< ε < ^q+λ−2₂ .Hence 0<

Z ∞

1

s^−1−ε

"

Z ¹_s

0

1

(1 +u)^λu^{q+λ−2−εq −1}du

# ds

<

Z ∞

1

s^−1−ε

"

Z ¹_s

0

u^{q+λ−2−εq −1}du

# ds

<

Z ∞

1

s⁻¹

"

Z ¹_s

0

u^{q+λ−2q −1}du

# ds

=

q q+λ−2

2

.

The lemma is proved.

3. MAINRESULTS

Theorem 3.1. Let p < 0 or0 < p < 1, ¹_p + ¹_q = 1,2−max{p, q} < λ < 2−min{p, q}, f(x), g(x) ≥ 0, and 0 < R∞

α (x−α)^1−λf^p(x)dx < ∞,0 < R∞

α (x−α)^1−λg^q(x)dx < ∞, α∈R.Then

(3.1) Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)^λdxdy > B

p+λ−2

p ,q+λ−2 q

× Z ∞

α

(x−α)^1−λf^p(x)dx

¹_p Z ∞

α

(x−α)^1−λg^q(x)dx ¹_q

, where the constant factorB_p+λ−2

p ,^q+λ−2_q

is the best possible.

Proof. By (2.2), we have Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)^λdxdy = Z ∞

α

Z ∞

α

"

f(x) (x+y−2α)^λp

x−α y−α

^2−λ_pq # (3.2)

×

"

g(y) (x+y−2α)^λq

x−α y−α

^2−λ_pq # dxdy

≥ Z ∞

α

Z ∞

α

"

f^p(x) (x+y−2α)^λ

x−α y−α

^2−λ_q dx

#_p¹

× Z ∞

α

Z ∞

α

"

g^q(y) (x+y−2α)^λ

y−α x−α

^2−λ_p dy

#¹_q . If (3.2) takes the form of equality, then by (2.2), there exist non-negative numbersaandbsuch that they are not all zero and

a f^p(x) (x+y−2α)^λ

x−α y−α

^2−λ_q

=b g^q(y) (x+y−2α)^λ

y−α x−α

^2−λ_p

, a.e. in(α,∞)×(α,∞).

It follows that

a(x−α)^2−λf^p(x) =b(y−α)^2−λg^q(y) = c,

a. e. in(0, ∞)×(0, ∞).Without loss of generality, supposea6= 0.One has (x−α)^1−λf^p(x) = c

a(x−α)⁻¹, a. e. in(0, ∞)×(0, ∞), which contradicts0<R∞

α (x−α)^1−λf^P(x)dx <∞.Therefore, by (2.3), we have

Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)^λdxdy >

Z ∞

α

ωλ(x, q)(x−α)^1−λf^p(x)dx ¹_p

× Z ∞

α

ω_λ(y, p)(y−α)^1−λg^q(y)dy ¹_q

. By (2.4), we obtain (3.1).

Without loss of generality, supposep <0,which implies that0< q <1. For0< ε < ^q+λ−2₂ set

f_ε(x) = 0, x∈(α, α+ 1); f_ε(x) = (x−α)^λ−2−εp , x∈[α+ 1,∞), g_ε(x) = 0, x∈(α, α+ 1); g_ε(x) = (x−α)^λ−2−εq , x∈[α+ 1,∞).

If the constant factor B

p+λ−2

p , ^q+λ−2_q

in (3.1) is not the best possible, then, there exists a positive constantK > B

p+λ−2

p , ^q+λ−2_q

, such that (3.1) is still valid ifB

p+λ−2

p , ^q+λ−2_q is replaced byK. By (2.5), we have

B

q+λ−2

q − ε

q,p+λ−2

p +ε

q

−εOε(1)

=ε Z ∞

α

Z ∞

α

f_ε(x)g_ε(y)

(x+y−2α)^λdxdy

> εK Z ∞

α+1

(x−α)^1−λf_ε^p(x)dx

¹_pZ ∞

α+1

(y−α)^1−λg^q_ε(y)dy ¹_q

=K.

Letting ε −→ 0⁺, we obtain B

p+λ−2

p ,^q+λ−2_q

≥ K. Hence the constant factor B

p+λ−2

p ,^q+λ−2_q

in (3.1) is the best possible when2−q < λ <2−p,0< ε < ^q+λ−2₂ .

The theorem is proved.

In (3.1), whenα=−¹₂, we have:

Corollary 3.2. Letp < 0,or0 < p < 1, ¹_p + ¹_q = 1,2−max{p, q} < λ < 2−min{p, q}, f(x), g(x)≥0,0<R∞

−¹₂ x+¹₂1−λ

f^p(x)dx <∞,0<R∞

−¹₂ x+¹₂1−λ

g^q(x)dx <∞.Then (3.3)

Z ∞

−¹₂

Z ∞

−¹₂

f(x)g(y)

(x+y+ 1)^λdxdy > B

p+λ−2

p ,q+λ−2 q

×

"

Z ∞

−¹₂

x+ 1

2 ^1−λ

f^p(x)dx

#¹_p"

Z ∞

−¹₂

x+1

2 ^1−λ

g^q(x)dx

#¹_q , where the constant factorB

p+λ−2

p ,^q+λ−2_q

is the best possible.

In (3.1), whenλ= 2, we have:

Corollary 3.3. Letp <0,or0< p <1, ¹_p+¹_q = 1, f(x), g(x)≥0,0<R∞

α (x−α)⁻¹f^p(x)dx

<∞,0<R∞

α (x−α)⁻¹g^q(x)dx <∞, α∈R.Then (3.4)

Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)²dxdy >

Z ∞

α

f^p(x) x−αdx

¹_pZ ∞

α

g^q(x) x−αdx

¹_q . In (3.1), whenα=−¹₂,λ= 2, we have:

Corollary 3.4. Letp <0,or0< p <1,¹_p+¹_q = 1, f(x), g(x)≥0,0<R∞

−¹

2

(x+¹₂)⁻¹f^p(x)dx

<∞,0<R∞

−¹₂ x+¹₂⁻¹

g^q(x)dx <∞.Then (3.5)

Z ∞

−¹₂

Z ∞

−¹₂

f(x)g(y)

(x+y+ 1)²dxdy >

"

Z ∞

−¹₂

f^p(x) x+¹₂dx

#¹_p "

Z ∞

−¹₂

g^q(x) x+ ¹₂dx

#¹_q . 4. AN EQUIVALENTFORM

Theorem 4.1. Letp <0,¹_p+¹_q = 1,2−q < λ <2−p, f(x)≥0,and0<R∞

α (x−α)^1−λf^p(x)dx

<∞, α∈R.Then (4.1)

Z ∞

α

(y−α)^{(p−1)(λ−1)} Z ∞

α

f(x)

(x+y−2α)^λdx p

dy

<

B

p+λ−2

p ,q+λ−2 q

pZ ∞

α

(x−α)^1−λf^p(x)dx.

where the constant factor B

p+λ−2

p ,^q+λ−2_q

is the best possible. Inequalities (4.1) and (3.1) are equivalent asp < 0.

Proof. Let

g(y) = (y−α)^{(p−1)(λ−1)} Z ∞

α

f(x)

(x+y−2α)^λdx p−1

, y∈(α,∞).

Then by (3.1) andp <0,we have 0<

Z ∞

α

(y−α)^1−λg^q(y)dy (4.2)

= Z ∞

α

(y−α)^{(p−1)(λ−1)} Z ∞

α

f(x)

(x+y−2α)^λdx p

dy

= Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)^λdxdy

≥B

p+λ−2

p ,q+λ−2 q

Z ∞

α

(x−α)^1−λf^p(x)dx ¹_p

× Z ∞

α

(y−α)^1−λg^q(y)dy ¹_q

. Sincep < 0, we obtain that

(4.3)

Z ∞

α

(y−α)^1−λg^q(y)dy

≤

B

p+λ−2

p ,q+λ−2 q

pZ ∞

α

(x−α)^1−λf^p(x)dx <∞.

By (4.3) and (4.2), we obtain (4.1).

Whereas, assume that (4.1) is true, by (2.2), we have Z ∞

α

Z ∞

α

f(x)g(y)

(x+y−2α)^λdxdy

= Z ∞

α

(y−α)^λ−1q Z ∞

α

f(x)

(x+y−2α)^λdx h

(y−α)^1−λq g(y)i dy

≥ Z ∞

α

(y−α)^{(p−1)(λ−1)} Z ∞

α

f(x)

(x+y−2α)^λdx p

dy

¹_pZ ∞

α

(y−α)^1−λg^q(y)dy ¹_q

. Sincep <0, by (4.1), then (3.1) is proved. Hence inequalities (4.1) and (3.1) are equivalent as

p <0.

Corollary 4.2. Letp <0,¹_p+¹_q = 1,2−q < λ <2−p, f(x)≥0,0<R∞

−¹₂ x+¹₂1−λ

f^p(x)dx

<∞. Then (4.4)

Z ∞

−¹

2

y+1

2

(p−1)(λ−1)"

Z ∞

−¹

2

f(x) (x+y+ 1)^λdx

#p

dy

<

B

p+λ−2

p ,q+λ−2 q

pZ ∞

−¹

2

x+ 1

2 1−λ

f^p(x)dx,

where the constant factorB_p+λ−2

p ,^q+λ−2_q

is the best possible.

Corollary 4.3. Letp < 0, ¹_p + ¹_q = 1, f(x) ≥ 0,0 < R∞

α (x−α)⁻¹f^p(x)dx < ∞, α ∈ R. Then

(4.5)

Z ∞

α

(y−α)^(p−1) Z ∞

α

f(x)

(x+y−2α)²dx p

dy <

Z ∞

α

f^p(x) x−αdx.

Corollary 4.4. Letp <0, ¹_p + ¹_q = 1, f(x)≥0,0<R∞

−¹₂ x+¹₂−1

f^p(x)dx <∞.Then

(4.6)

Z ∞

−¹

2

y+1

2

(p−1)"

Z ∞

−¹

2

f(x) (x+y+ 1)²dx

#p

dy <

Z ∞

−¹

2

f^p(x) x+¹₂dx.

Theorem 4.5. Let 0 < p < 1, ¹_p + ¹_q = 1, 2 − p < λ < 2− q, f(x) ≥ 0, and 0 <

R∞

α (x−α)^1−λf^p(x)dx <∞, α∈R.Then (4.7)

Z ∞

α

(y−α)^{(p−1)(λ−1)} Z ∞

α

f(x)

(x+y−2α)^λdx p

dy

>

B

p+λ−2

p ,q+λ−2 q

pZ ∞

α

(x−α)^1−λf^p(x)dx, where the constant factor B_p+λ−2

p ,^q+λ−2_q

is the best possible. Inequalities (4.7) and (3.1) are equivalent as0< p <1.

Proof. Since0< p <1,by (2.2) and (2.4),we have Z ∞

α

f(x)

(x+y−2α)^λdx p

= (Z ∞

α

"

f(x) (x+y−2α)^λp

x−α y−α

^2−λ_pq # (4.8)

×

"

1 (x+y−2α)^λq

y−α x−α

^2−λ_pq # dx

)^p

≥ Z ∞

α

f^p(x) (x+y−2α)^λ

x−α y−α

^2−λ_q dx

×

"

Z ∞

α

1 (x+y−2α)^λ

y−α x−α

^2−λ_p dx

#^p−1

=

B

p+λ−2

p ,q+λ−2 q

p−1

(y−α)^{(p−1)(1−λ)}

× Z ∞

α

f^p(x) (x+y−2α)^λ

x−α y−α

^2−λ_q dx.

If (4.8) takes the form of equality, then by (2.2), there exist non-negative numbersaandb, such that they are not all zero and

a f^p(x) (x+y−2α)^λ

x−α y−α

^2−λ_q

=b 1

(x+y−2α)^λp

y−α x−α

^2−λ_p

, a.e.(α,∞).

It follows that

a(x−α)^2−λf^p(x) =b(y−α)^2−λ, a.e. in(α,∞).

Obviouslya6= 0,(otherwisea=b= 0), one has (x−α)^1−λf^p(x) = b

a(y−α)^2−λx⁻¹, a.e.(α,∞) which contradicts0<R∞

α (x−α)^1−λf^P(x)dx <∞. Hence Z ∞

α

(y−α)^{(p−1)(λ−1)} Z ∞

α

f(x)

(x+y−2α)^λdx p

dy

>

B

p+λ−2

p ,q+λ−2 q

p−1Z ∞

α

"

Z ∞

α

f^p(x) (x+y−2α)^λ

x−α y−α

^2−λ_q dx

# dy

=

B

p+λ−2

p ,q+λ−2 q

p−1

× Z ∞

α

"

Z ∞

α

1 (x+y−2α)^λ

x−α y−α

^2−λ_q dy

#

f^p(x)dx.

By (2.3) and (2.4), we obtain (4.7).

Obviously, inequalities (4.7) and (3.1) are equivalent as 0 < p < 1 and the constant factor B

p+λ−2

p ,^q+λ−2_q

is the best possible.

Corollary 4.6. Let0< p <1,¹_p+¹_q = 1,2−p < λ <2−q,0<R∞

−¹₂ x+ ¹₂1−λ

f^p(x)dx <∞, f(x)≥0.Then

Z ∞

−¹₂

y+1

2

(p−1)(λ−1)"

Z ∞

−¹₂

f(x) (x+y+ 1)^λdx

#p

dy

>

B

p+λ−2

p ,q+λ−2 q

pZ ∞

−¹₂

x+ 1

2 1−λ

f^p(x)dx, where the constant factorB

p+λ−2

p ,^q+λ−2_q

is the best possible.

Corollary 4.7. Let0< p < 1, ¹_p +¹_q = 1, f(x)≥0,0<R∞

α (x−α)⁻¹f^p(x)dx <∞, α∈R. Then

(4.9)

Z ∞

α

(y−α)^(p−1) Z ∞

α

f(x)

(x+y−2α)²dx p

dy >

Z ∞

α

f^p(x) x−αdx.

Corollary 4.8. Let0< p <1, ¹_p +¹_q = 1, f(x)≥0,0<R∞

−¹₂ x+¹₂−1

f^p(x)dx <∞.Then (4.10)

Z ∞

−¹

2

y+1

2

(p−1)"

Z ∞

−¹

2

f(x) (x+y+ 1)²dx

#p

dy >

Z ∞

−¹

2

f^p(x) x+¹₂dx.

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