JJ II

(1)

volume 6, issue 4, article 92, 2005.

Received 07 April, 2005;

accepted 06 June, 2005.

Communicated by:B. Yang

Abstract Contents

JJ II

J I

Home Page Go Back

Close Quit

Journal of Inequalities in Pure and Applied Mathematics

ON HARDY-HILBERT INTEGRAL INEQUALITIES WITH SOME PARAMETERS

YONG HONG

Department of Mathematics Guangdong Business College Guangzhou 510320

People’s Republic of China EMail:hongyong59@sohu.com

2000c Victoria University ISSN (electronic): 1443-5756 109-05

(2)

On Hardy-Hilbert Integral Inequalities with Some

Parameters Yong Hong

Title Page Contents

JJ II

J I

Go Back Close

Quit Page2of22

J. Ineq. Pure and Appl. Math. 6(4) Art. 92, 2005

http://jipam.vu.edu.au

Abstract

In this paper, we give a new Hardy-Hilbert’s integral inequality with some parameters and a best constant factor. It includes an overwhelming majority of results of many papers.

2000 Mathematics Subject Classification:26D15.

Key words: Hardy-Hilbert’s integral inequality, Weight, Parameter, Best constant fac- tor,β-function,Γ-function.

1. Introduction and Main Result

Let ¹_p + ¹_q = 1(p > 1), f ≥ 0, g ≥ 0, 0 < R∞

0 f^p(x)dx < +∞, 0 <

R∞

0 g^q(x)dx <+∞, then we have the well known Hardy-Hilbert inequality (1.1)

Z ∞ 0

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

< π sin

π p

Z ∞

0

f^p(x)dx

¹_pZ ∞ 0

g^q(x)dx ¹_q

;

and an equivalent form as:

(1.2)

Z ∞ 0

f(x) x+ydx

p

dy <



 π sin

π p





p

Z ∞ 0

f^p(x)dx.

In recent years, many results have been obtained in the research of these two inequalities (see [1] – [13]). Yang [1] and [2] gave:

(1.3) Z ∞

0

Z ∞ 0

f(x)g(y) (x+y)^λdxdy

< B

p+λ−2

p ,p+λ−2 q

× Z ∞

0

x^1−λf^p(x)dx

¹_pZ ∞ 0

x^1−λg^q(x)dx ¹_q

,

(4)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page4of22

whereB(r, s)is theβ-function; and Kuang [3] gave:

(1.4) Z ∞

0

Z ∞ 0

f(x)g(x) x^λ+y^λ dxdy

< π

λsin¹^p π

pλ

sin¹^q

π qλ

Z ∞

0

x^1−λf^p(x)dx

¹_pZ ∞ 0

.

Recently, Hong [4] gave:

(1.5) Z ∞

0

Z ∞ 0

f(x)g(x) px²+y²dxdy

≤ 1 2√

πΓ 1

2p

Γ 1

2q

Z ∞ 0

f^p(x)dx

¹_pZ ∞ 0

g^q(x)dx ¹_q

. And Yang [5] gave:

(1.6) Z ∞

0

Z ∞ 0

f(x)g(x) x^λ+y^λ dxdy

< π λsin

π p

Z ∞

0

x^{(p−1)(1−λ)}f^p(x)dx

¹_pZ ∞ 0

x^{(q−1)(1−λ)}g^q(x)dx ¹_q

;

(1.7) Z ∞

0

y^λ−1 Z ∞

0

f(x) x^λ+y^λdx

p

dy

(5)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page5of22

<



 π λsin

π p





p

Z ∞ 0

x^{(p−1)(1−λ)}f^p(x)dx.

These results generalize and improve (1.1) and (1.2) in a certain degree.

In this paper, by introducing a few parameters, we obtain a new Hardy- Hilbert integral inequality with a best constant factor, which is a more extended inequality, and includes all the results above and the overwhelming majority of results of many recent papers.

Our main result is as follows:

Theorem 1.1. If ¹_p + ¹_q = 1 (p > 1), α > 0, λ > 0, m, n ∈ R, such that 0<1−mp < αλ,0<1−nq < αλ, andf ≥0,g ≥0, satisfy

(1.8) 0<

Z ∞ 0

x(1−αλ)+p(n−m)

f^p(x)dx <∞,

(1.9) 0<

Z ∞ 0

y(1−αλ)+q(m−n)

g^q(y)dy <∞, then

(1.10) Z ∞

0

Z ∞ 0

f(x)g(x) (x^α+y^α)^λdxdy

< Hλ,α(m, n, p, q) Z ∞

0

f^p(x)dx ¹_p

× Z ∞

0

g^q(y)dy ¹_q

;

(6)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page6of22

and

(1.11) Z ∞

0

y(1−αλ)+q(m−n) 1−q

Z ∞ 0

f(x) (x^α+y^α)^λdx

p

dy

<He_λ,α(m, n, p, q) Z ∞

0

f^p(x)dx, where

H_λ,α(m, n, p, q) = 1 αB¹^p

1−mp

α , λ− 1−mp α

B¹^q

1−nq

α , λ− 1−nq α

and

He_λ,α(m, n, p, q) = 1 α^pB

1−mp

α , λ−1−mp α

B^p−1

1−nq

α , λ− 1−nq α

. Theorem 1.2. If p > 1, ¹_p + ¹_q = 1, α > 0, λ > 0, m, n ∈ R, such that 0<1−mp < αλ,mp+nq= 2−αλ, andf(x)≥0,g(y)≥0, satisfy

(1.12) 0<

Z ∞ 0

x^n(p+q)−1f^p(x)dx <∞,

(1.13) 0<

Z ∞ 0

y^m(p+q)−1g^q(y)dy <∞,

(7)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page7of22

then

(1.14) Z ∞

0

Z ∞ 0

f(x)g(x)

(x^α+y^α)^λdxdy < 1 αB

1−mp

α , λ− 1−mp α

× Z ∞

0

x^n(p+q)−1f^p(x)dx

¹_pZ ∞ 0

y^m(p+q)−1g^q(y)dy ¹_q

;

(1.15) Z ∞

0

y^m(p+q)−1^1−q Z ∞

0

p

dy

< 1 α^pB^p

1−mp

α , λ− 1−mp α

Z ∞ 0

x^n(p+q)−1f^p(x)dx, where the constant factors _α¹B ^1−mp_α , λ−^1−mp_α

in (1.14) and

1

α^pB^p ^1−mp_α , λ− ^1−mp_α

in (1.15) are the best possible.

(8)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page8of22

2. Weight Function and Lemmas

The weight function is defined as follows ω_λ,α(m, n, y) =

Z ∞ 0

1

(x^α+y^α)^λ · yⁿ

x^mdx, y∈(0,+∞).

Lemma 2.1. Ifα >0,λ >0,m∈R,0<1−m < αλ, then (2.1) ω_λ,α(m, n, y) = 1

αy(1−αλ)+(n−m)

B

1−m

α , λ−1−m α

. Proof. Settingt = ^x_y^αα, then

ω_λ,α(m, n, y) = 1 α

Z ∞ 0

1

(1 +t)^λy(1−αλ)+(n−m)

t^1−m^α ⁻¹dt

= 1

Z ∞ 0

1

(1 +t)^λt^1−m^α ⁻¹dt

= 1

B

1−m

α , λ−1−m α

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

Lemma 2.2. Ifα >0, λ >0, β <1, a∈R, then (2.2)

Z ∞ 1

1 x^1+ε

Z _xα¹

0

1

(1 +t)^λt^1−β^α ^−1−aεdtdx=O(1), (ε→0⁺).

(9)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page9of22

Proof. Since(1−β)/α >0,forεsmall enough, such that ^1−β_α −aε >0,then Z ∞

1

1 x^1+ε

Z _xα¹

0

1

(1 +t)^λt^1−β^α ^−1−aεdtdx <

Z ∞ 1

1 x

Z _xα¹

0

t(^1−βα −aε)⁻¹dtdx

= 1

1−β−aεα Z ∞

1

x^β+aεα−2dx

= 1

(1−β−aεα)². Hence (2.2) is valid.

(10)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page10of22

3. Proofs of the Theorems

Proof of Theorem1.1. By Hölder’s inequality and Lemma2.1, we have

G= Z ∞

0

Z ∞ 0

f(x)g(y) (x^α+y^α)^λdxdy

= Z ∞

0

Z ∞ 0

f(x) (x^α+y^α)^λ/p

xⁿ y^m

g(y) (x^α+y^α)^λ/q

y^m xⁿ

dxdy

≤ Z ∞

0

Z ∞ 0

f^p(x) (x^α+y^α)^λ

x^np y^mp

¹_pZ ∞ 0

Z ∞ 0

g^q(x) (x^α+y^α)^λ

y^mq x^nqdxdy

¹_q , (3.1)

according to the condition of taking equality in Hölder’s inequality, if (3.1) takes equality, then there exists a constantC, such that

f^p(x) (x^α+y^α)^λ

x^np y^mp

g^q(y) (x^α+y^α)^λ

y^mq x^nq

≡C,a.e. (x, y)∈(0,+∞)×(0,+∞) it follows that

f^p(x)x^n(p+q)≡Cg^q(y)y^m(p+q)≡C₁(constant), a.e. (x, y)∈(0,+∞)×(0,+∞) hence

Z ∞ 0

f^p(x)dx

= Z ∞

0

x(1−αλ)+n(p+q)−nq−mp

f^p(x)dx

=C₁ Z ∞

0

x(1−αλ)−np−mq

dx

(11)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page11of22

=C₁ Z 1

0

dx+C₁ Z ∞

1

dx= +∞, which contradicts (1.8) and (1.9). Hence, by (3.1), we have

G <

Z ∞ 0

1 (x^α+y^α)^λ

x^np y^mpdy

f^p(x)dx ¹_p

× Z ∞

0

Z ∞ 0

1 (x^α+y^α)^λ

y^mq x^nqdy

g^q(y)dy ¹_q

= 1

α Z ∞

0

B

1−mp

α , λ− 1−mp α

f^p(x)dx ¹_p

× 1

α Z ∞

0

B

1−nq

α , λ−1−nq α

g^q(y)dy ¹_q

=H_λ,α(m, n, p, q) Z ∞

0

f^p(x)dx ¹_p

× Z ∞

0

g^q(y)dy ¹_q

. Hence (1.10) is vaild.

Letβ = [(1−αλ) +q(m−n)]/(1−q),and

eg(y) =y^β Z ∞

0

^p_q .

(12)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page12of22

By (1.10), we have Z ∞

0

eg^q(y)dy

= Z ∞

0

y^β(1−q)eg^q(y)dy

= Z ∞

0

y^β Z ∞

0

p

dy

= Z ∞

0

y^β Z ∞

0

^p_q Z ∞ 0

dy

= Z ∞

0

Z ∞ 0

f(x)eg(y) (x^α+y^α)^λdxdy

< H_λ,α(m, n, p, q) Z ∞

0

f^p(x)dx ¹_p

× Z ∞

0

eg^q(y)dy|

¹_q , It follows that

Z ∞ 0

y(1−αλ)+q(m−n) 1−q

Z ∞ 0

p

dy

<He_λ,α(m, n, p, q) Z ∞

0

f^p(x)dx.

Hence, (1.11) is valid.

(13)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page13of22

Proof of Theorem1.2. Since0<1−mp < αλ, mp+nq = 2−αλ,then 0<1−nq < αλ,

(1−αλ) +p(n−m) =n(p+q)−1, (1−αλ) +q(m−n) = m(p+q)−1,

1−mp

α =λ− 1−nq α . By Theorem1.1, (1.14) and (1.15) are valid.

Forε >0,setting f0(x) =

( x[−n(p+q)−ε]/p, x≥1;

0, 0≤x <1,

and

g₀(y) =

( y[−m(p+q)−ε]/q, y ≥1;

0, 0≤y <1.

We have

(3.2) 0<

Z ∞ 0

x^n(p+q)−1f₀^p(x)dx= Z ∞

1

x^−1−εdx= 1 ε <∞,

(3.3) 0<

Z ∞ 0

y^m(p+q)−1g^q₀(y)dy= Z ∞

1

y^−1−εdy = 1 ε <∞.

(14)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page14of22

Z ∞ 0

f₀(x)g₀(y) (x^α+y^α)^λdxdy

= Z ∞

1

Z ∞ 1

1

(x^α+y^α)^λx⁻^n(p+q)+ε^p y⁻^m(p+q)+ε^q dxdy

= Z ∞

1

x⁻^n(p+q)+ε^p Z ∞

1

(x^α+y^α)^λy⁻^m(p+q)+ε^q dydx

= 1 α

Z ∞ 1

1 x^1+ε

Z ∞

1 xα

1

(1 +t)^λt^1−mp^α ⁻¹⁻^qα^ε dtdx

= 1 α

Z ∞ 1

1 x^1+ε

Z ∞ 0

1

(1 +t)^λt^1−mt^α ⁻¹⁻^qα^ε dtdx

− Z ∞

1

1 x^1+ε

Z _xα¹

0

1

(1 +t)^λt^1−mt^α ⁻¹⁻^qα^ε dtdx

# . By Lemma2.2, whenε→0,we have

Z ∞ 1

1 x^1+ε

Z _xα¹

0

1

(1 +t)^λt^1−mp^α ⁻¹⁻^qα^ε dtdx=O(1).

Since Z ∞

0

1

(1 +t)^λt^1−mp^α ⁻¹⁻^qα^ε dt=B

1−mp

α , λ− 1−mp α

+o(1),

(15)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page15of22

we have Z ∞

0

Z ∞ 0

f₀(x)g₀(y)

(x^α+y^α)^λdxdy = 1 α

1 ε

B

1−mp

α , λ− 1−mp α

+o(1)

−O(1)

= 1 ε

1 αB

1−mp

α , λ− 1−mp α

−o(1)

= 1 εαB

1−mp

α , λ− 1−mp α

(1−o(1)).

(3.4)

If the constant _α¹B ^1−mp_α , λ− ^1−mp_α

in (1.14) is not the best possible, then there exists a K < _α¹B ^1−mp_α , λ− ^1−mp_α

,such that (1.14) still is valid when we replace _α¹B ^1−mp_α , λ− ^1−mp_α

byK. By (3.2), (3.3) and (3.4), we find 1

εαB

1−mp

α , λ− 1−mp α

(1−o(1))

< K Z ∞

0

x^n(p+q)−1f₀^p(x)dx

¹_pZ ∞ 0

y^m(p+q)−1g^q₀(y)dy ¹_q

=K1 ε. For ε → 0⁺, we have _α¹B ^1−mp_α , λ− ^1−mp_α

≤ K, which contradicts the fact thatK < _α¹B ^1−mp_α , λ− ^1−mp_α

. It follows that_α¹B ^1−mp_α , λ− ^1−mp_α

in (1.14) is the best possible.

Since (1.14) is equivalent to (1.15), then the constant _α¹pB^p ^1−mp_α , λ− ^1−mp_α in (1.15) is the best possible. The theorem is proved.

(16)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page16of22

4. Some Corollaries

When we take the appropriate parameters, many new inequalities can be obtained as follows:

Corollary 4.1. If ¹_p + ¹_q = 1 (p > 1), α > 0, λ > 0, f ≥ 0, g ≥ 0, and x(1−αλ)(p−1)/pf(x)∈L^p(0,+∞),x(1−αλ)(q−1)/qg(x)∈L^q(0,+∞), then

(4.1) Z ∞

0

Z ∞ 0

f(x)g(y) (x^α+y^α)^λdxdy

<

Γ

λ p

Γ

λ q

αΓ(λ)

Z ∞ 0

x(1−αλ)(p−1)

f^p(x)dx ¹_p

× Z ∞

0

x(1−αλ)(q−1)

g^q(x)dx ¹_q

;

(4.2) Z ∞

0

y^α−1 Z ∞

0

p

dy

<



 Γ

λ p

Γ

λ q

αΓ(λ)





p

Z ∞ 0

x(1−αλ)(p−1)

f^p(x)dx,

where the constantsΓ

λ p

Γ

λ q

.

(αΓ(λ))in (4.1) andh Γ

λ p

Γ

λ q

.

(αΓ(λ))ip

(17)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page17of22

Proof. If we takem= ¹_p −^αλ_p2, n= ¹_q −^αλ_q2 in Theorem1.2, (4.1) and (4.2) can be obtained.

Corollary 4.2. If¹_p+¹_q = 1 (p >1), λ >0, f ≥0, g≥0andx(1−λ)(p−1)/pf(x)∈ L^p(0,+∞), x(1−λ)(q−1)/qg(x)∈L^q(0,+∞), then

(4.3) Z ∞

0

Z ∞ 0

<

Γ

λ p

Γ

λ q

Γ(λ)

Z ∞ 0

x^{(1−λ)(p−1)}f^p(x)dx

¹_pZ ∞ 0

x^{(1−λ)(q−1)}g^q(x)dx ¹_q

,

(4.4)

Z ∞ 0

f(x) (x+y)^λdx

p

dy

<



 Γ

λ p

Γ

λ q

Γ(λ)





p

Z ∞ 0

x^{(1−λ)(p−1)}f^p(x)dx,

where Γ

λ p

Γ

λ q

.

Γ(λ) in (4.3) and h

Γ

λ p

Γ

λ q

.

Γ(λ)ip

in (4.4) are the best possible.

Proof. If we takeα= 1in Corollary4.1, (4.3) and (4.4) can be obtained.

Corollary 4.3. If ¹_p + ¹_q = 1 (p > 1), λ > 0, p+λ−2 > 0, q+λ−2 > 0,

(18)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page18of22

f ≥0, g ≥0, andx^(1−λ)/pf(x)∈L^p(0,+∞), x^(1−λ)/qg(x)∈L^q(0,+∞), then (4.5)

Z ∞ 0

< B

p+λ−2

p ,q+λ−2 q

Z ∞ 0

x^1−λf^p(x)

¹_pZ ∞ 0

,

(4.6) Z ∞

0

y^1−λ^1−q Z ∞

0

f(x) (x+y)^λdx

p

dy

< B^p

p+λ−2

p ,q+λ−2 q

Z ∞ 0

x^1−λf^p(x)dx, where B_p+λ−2

p ,^q+λ−2_q

in (4.5) andB^p_p+λ−2

p ,^q+λ−2_q

in (4.6) are the best possible.

Proof. If we takeα = 1, m =n = ^2−λ_pq in Theorem1.2, (4.5) and (4.6) can be obtained.

Corollary 4.4. If ¹_p +¹_q = 1 (p >1), α >0, f ≥0, g ≥0, andx^(1−α)/pf(x)∈ L^p(0,+∞), x^(1−α)/qg(x)∈L^q(0,+∞), then

(4.7) Z ∞

0

Z ∞ 0

f(x)g(x) x^α+y^α dxdy

< π

αsin¹^p

π pα

sin¹^q

π qα

Z ∞

0

x^1−αf^p(x)dx

pZ ∞ 0

x^1−αg^q(x)dx q

,

(19)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page19of22

(4.8) Z ∞

0

y^1−α^1−q Z ∞

0

f(x) x^α+y^αdx

p

dy

<





π αsin¹^p

π pα

sin¹^q

π qα





p

Z ∞ 0

x^1−αf^p(x)dx.

Proof. If we takeλ = 1, m = n = _pq¹, in Theorem1.1, (4.7) and (4.8) can be obtained.

Corollary 4.5. If ¹_p + ¹_q = 1 (p > 1), α > 0, f ≥ 0, g ≥ 0, and f(x) ∈ L^p(0,+∞), g(x)∈L^q(0,+∞), then

(4.9) Z ∞

0

Z ∞ 0

f(x)g(x) (x^α+y^α)^α¹ dxdy

<

Γ 1

pα

Γ

1 qα

αΓ _α¹

Z ∞ 0

f^p(x)dx

¹_pZ ∞ 0

g^q(x)dx ¹_q

,

(4.10)

Z ∞ 0

f(x) (x^α+y^α)^α¹ dx

!p

dy

<



 Γ

1 pα

Γ

1 qα

αΓ _α¹





p

Z ∞ 0

f^p(x)dx,

where Γ

1 pα

Γ

1 qα

.

αΓ _α¹

in (4.9) andh Γ

1 pα

Γ

1 qα

.

αΓ _α¹ip

(20)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page20of22

Proof. If we takeλ = _α¹, m =n = _pq¹ in Theorem1.2, (4.9) and (4.10) can be obtained.

Remark 1. (4.1) and (4.2) are respectively generalizations of (1.6) and (1.7).

Forα= 1in (4.1) and (4.2), (1.6) and (1.7) can be obtained.

Remark 2. (4.3) and (4.4) are respectively generalizations of (1.1) and (1.2) . Remark 3. (4.5) is the result of [1] and [2]. (4.6) is a new inequality.

Remark 4. (4.7) is the result of [3]. (1.7) is a new inequality.

Remark 5. (4.9) is a generalization of (1.5). (4.10) is a new inequality.

For other appropriate values of parameters taken in Theorems 1.1 and 1.2, many new inequalities and the inequalities of [6] – [13] can yet be obtained.

(21)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page21of22

References

[1] BICHENG YANG, A generalized Hardy-Hilbert’s inequality with a best constant factor, Chin. Ann. of Math. (China), 21A(2000), 401–408.

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

Appl., 261 (2001), 295–306.

[3] JICHANG KUANG, On new extensions of Hilbert’s integtal inequality, Math. Anal. Appl., 235 (1999), 608–614.

[4] YONG HONG, All-sided Generalization about Hardy-Hilbert’s integral inequalities, Acta Math. Sinica (China), 44 (2001), 619–626.

[5] BICHENG YANG, On a generalization of Hardy-Hilbert’s inequality, Ann. of Math. (China), 23 (2002), 247–254.

[6] BICHENG YANG, On a generalization of Hardy-Hilbert’s integral in- eguality, Acta Math. Sinica (China), 41(4) (1998), 839–844.

[7] KE HU, On Hilbert’s inequality, Ann. of Math. (China), 13B (1992), 35–

39

[8] KE HU, On Hilbert’s inequality and it’s application, Adv. in Math.

(China), 22 (1993), 160–163.

[9] BICHENG YANG AND MINGZHE GAO, On a best value of Hardy- Hilbert’s inequality, Adv. in Math. (China), 26 (1999), 159–164.

(22)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page22of22

[10] MINGZHE GAO, An improvement of Hardy-Riesz’s extension of the Hilbert inequality, J. Mathematical Research and Exposition, (China) 14 (1994), 255–359.

[11] MINGZHE GAO AND BICHENG YANG, On the extended Hilbert’s in- equality, Proc. Amer. Math. Soc., 126 (1998), 751–759.

[12] BICHENG YANG AND L. DEBNATH, On a new strengtheaed Hardy- Hilbert’s intequality, Internat. J. Math. & Math. Sci., 21 (1998): 403–

408.

[13] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequal- ity, J. Math. Anal. & Appl., 226 (1998), 166–179.