WITH SOME PARAMETERS

WITH SOME PARAMETERS

HONG YONG

Received 19 April 2006; Revised 30 May 2006; Accepted 5 June 2006

By introducing some parameters and norm xα(x∈Rⁿ), we give multiple Hardy- Hilbert integral inequalities, and prove that their constant factors are the best possible when parameters satisfy appropriate conditions.

Copyright © 2006 Hong Yong. This is an open access article distributed under the Cre- ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Ifp >1, 1/ p+ 1/q₌1, f _≥0,g_≥0, 0<₀^∞f^p(x)dx <+∞, 0<₀^∞g^q(x)dx <+∞, then we have the well-known Hardy-Hilbert inequality (see [4]):

_+∞

0

f(x)g(x)

x+y dx d y < π sin(π/ p)

+∞

0 f^p(x)dx

1/ p+∞ 0 g^q(x)dx

1/q

, (1.1)

where the constant factorπ/sin(π/ p) is the best possible. Its equivalent form is +_∞

0

+_∞ 0

f(x) x+ydx

p

d y <

π sin(π/ p)

p+_∞

0 f^p(x)dx, (1.2)

where the constant factor [π/sin(π/ p)]^pis also the best possible.

Hardy-Hilbert inequalities are important in analysis and in their applications (see [7]).

In recent years, many results (see [1,3,8–10]) have been obtained in the research of Hardy-Hilbert inequality. At present, because of the requirement of higher-dimensional harmonic analysis and higher-dimensional operator theory, multiple Hardy-Hilbert in- tegral inequalities are researched (see [5, 6, 11]). Yang [11] obtains the following: if α∈R, n≥2, p_i>1 (i=1, 2,. . .,n), ⁿ_i=1(1/ p_i)=1, λ > n−min1≤i≤n{p_i}, f_i≥0, and

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 94960, Pages1–11 DOI 10.1155/JIA/2006/94960

0<_α^+∞(t−α)^n−1−λf_i^pi(t)dt <+∞, (i=1, 2,. . .,n), then _+∞

α ···

_+∞

α

1

ni=1xi−nα^λ n i=1

fi xi

dx1. . . dxn

< 1 Γ(λ)

n i=1

Γ1−n−λ p_i

+_∞

α (t−α)^n−1−λf_i^pi(t)dt 1/ pi

,

(1.3)

where the constant factor (1/Γ(λ))ⁿ_i=1Γ(1−(n−λ)/ pi) is the best possible.

In this paper, by introducing some parameters and normxα(x∈Rⁿ), we give mul- tiple Hardy-Hilbert integral inequalities, and discuss the problem of the best constant factor. For this reason, we introduce the notation

Rⁿ₊=

x= x1,. . .,x_n:x1,. . .,x_n>0, xα= x^α₁+···+x_n^α1/α, (α >0),

(1.4)

and we agree onxα< crepresenting{x∈Rⁿ+:xα< c}. 2. Some lemmas

Lemma 2.1 (see [2]). Ifpi>0,ai>0,αi>0, (i=1, 2,. . .,n),Ψ(u) is a measurable function, then

···

x1,...,xn>0; (x1/a1)^α1+···+(xn/an)^αn≤1Ψx1

a1

α1

+···+ xn

an

αn

×x₁^p1−1. . . xn^pn−1dx1. . . dx_n

= a₁^p1. . . a^pnⁿΓ p1/α1

. . .Γ pn/αn α1. . . αnΓ p1/α1+···+pn/αn1

0Ψ(u)u^{p1/α1+···+pn/αn−1}du,

(2.1)

where theΓ(·) isΓ-function.

Lemma 2.2. Ifn∈Z+,α >0,β >0,λ >0,m∈R, 0< n−m < βλ, and setting weight func- tionω_α,β,λ(m,n,y) as

ωα,β,λ(m,n,y)=

Rⁿ+

1

x^βα+y^βαλx⁻α^mdx, (2.2)

then

ωα,β,λ(m,n,y)= yⁿα^−βλ−m Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−m

β ,λ−n−m β

, (2.3)

where theB(·,·) isβ-function.

Proof. ByLemma 2.1, we have

ω_α,β,λ(m,n,y)=

Rⁿ+

1

x^βα+y^βαλx⁻_α^md y

= lim

r→+∞

···

x1,...,xn>0;x^α1+···+x^α_n<r^α

×

r x1/r^α+···+ x_n/r^α1/α−m

r^β x1/r^α+···+ x_n/r^αβ/α+y^βαλx¹₁⁻¹. . . x_n¹⁻¹dx1. . . dx_n

= lim

r→+_∞

rⁿΓⁿ(1/α) αⁿΓ(n/α)

1 0

ru^1/α−m

y^βα+r^βu^β/αλu^n/α−1du

= Γⁿ(1/α) αⁿ⁻¹Γ(n/α) lim

r→+∞

_r

0

1

y^βα+t^βλt^n−m−1dt

= Γⁿ(1/α) αⁿ⁻¹Γ(n/α)

+∞ 0

1

y^βα+t^βλt^n−m−1dt

= yⁿα^−βλ−m Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)

1 0

1

(1 +u)^λu^{(n−m)/β−1}du

= yⁿα^−βλ−m Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−m

β ,λ−n−m β

.

(2.4)

Hence (2.3) is valid.

3. Main results

Theorem 3.1. If p >1, 1/ p+ 1/q=1,n∈Z+,α >0,β >0,λ >0,a∈R,b∈R, 0< n− ap < βλ, 0< n−bq < βλ, f ≥0,g≥0, and

0<

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx <+∞, (3.1) 0<

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y <+∞, (3.2)

then

Rⁿ+

f(x)g(y)

x^βα+y^βαλdx d y

< C_α,β,λ(a,b,p,q)×

R+ⁿ

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx 1/ p

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y 1/q

, (3.3)

Rⁿ+

y⁽⁽ⁿα ^{−βλ)+q(a−b))/(1−q)}

Rⁿ+

f(x) x^βα+y^βαλdx

p

d y

< C_α,β,λ^p (a,b,p,q)×

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx,

(3.4)

whereCα,β,λ(a,b,p,q)=(Γⁿ(1/α)/βαⁿ⁻¹Γ(n/α))B^{1/ p}((n−ap)/β,λ−(n−ap)/β)B^1/q((n− bq)/β,λ−(n−bq)/β).

Proof. By H ¨older’s inequality, we have

G:=

Rⁿ+

f(x)g(y)

x^βα+y^βαλdx d y

=

Rⁿ+

f(x) x^βα+y^βαλ/ p

x^b_α y^a_α

g(y) x^βα+y^βαλ/q

y^a_α x^b_α

dx d y

≤

Rⁿ+

f^p(x) x^βα+y^βαλ

x^bpα

y^apα

dx d y 1/ p

×

Rⁿ+

g^q(y) x^βα+y^βαλ

y^aqα

x^bqα

dx d y 1/q

, (3.5) according to the condition of taking equality in H ¨older’s inequality, if this inequality takes the form of an equality, then there exist constantsC1 andC2, such that they are not all zero, and

C1f^p(x) x^βα+y^βαλ

x^bpα

y^apα = C2g^q(y) x^βα+y^βαλ

y^aqα

x^bqα

, a.e. (x,y)∈Rⁿ₊×Rⁿ₊. (3.6)

Without losing generality, we suppose thatC1=0, we may get x^b(p+q)α f^p(x)=C2

C1y^a(p+q)α g^q(y), a.e. (x,y)∈Rⁿ₊×Rⁿ₊, (3.7) hence, we obtain

x^b(p+q)α f^p(x)=C(constant), a.e.x∈Rⁿ₊, (3.8)

hence, we have

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx=

Rⁿ+

x⁽ⁿα^{−βλ)−bq−ap+b(p+q)}f^p(x)dx

=C

Rⁿ+

x⁽ⁿα^{−βλ)−bq−ap}dx= ∞,

(3.9)

which contradicts (3.1). Hence, and byLemma 2.2, we obtain G <

Rⁿ+

Rⁿ+

1 x^βα+y^βαλ

1 y^apα

d y

x^bpα f^p(x)dx 1/ p

×

Rⁿ+

Rⁿ+

1 x^βα+y^βαλ

1 x^bqα

dx

y^aqα g^q(y)d y 1/q

=

Rⁿ+

ω_α,β,λ,(ap,n,x)x^bpα f^p(x)dx 1/ p

Rⁿ+

ω_α,β,λ,(bq,n,y)y^aqα g^q(y)d y 1/q

=

Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−ap

β ,λ−n−ap β

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx 1/ p

×

Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−bq

β ,λ−n−bq β

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y 1/q

=Cα,β,λ,(a,b,p,q)

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx 1/ p

×

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y 1/q

. (3.10) Hence, (3.3) is valid.

Letk=((n−βλ) +q(a−b))/(1−q), for 0< h < l <+∞, setting

gh,l(y)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ y^k_α

Rⁿ+

f(x) x^βα+y^βα_λdx

p/q

, h <yα< l,

0, 0<yα≤horyα≥l,

g(y)= y^kα

Rⁿ+

f(x) x^βα+y^βαλdx

p/q

, y∈Rⁿ₊,

(3.11)

by (3.1), for sufficiently smallh >0 and sufficiently largel >0, we have 0<

h<yα<ly⁽ⁿα^{−βλ)+q(a−b)}g_h,l^q (y)d y <+∞. (3.12)

Hence, by (3.3), we have

h<yα<ly⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y

=

h<yα<ly^k(1α ^−q)g^q(y)d y=

h<yα<ly^k_α

Rⁿ+

f(x) x^βα+x^βαλdx

p

d y

=

h<yα<ly^k_α

Rⁿ+

f(x) x^βα+y^βαλdx

p/q

Rⁿ+

f(x) x^βα+y^βαλdx

d y

=

Rⁿ+

f(x)gh,l(y)

x^βα+y^βαλdx d y < Cα,β,λ,(a,b,p,q)

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx 1/ p

×

R+ⁿ

y⁽ⁿα^{−βλ)+q(a−b)}g_h,l^q(y)d y 1/q

=Cα,β,λ,(a,b,p,q)

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx 1/ p

×

h<yα<ly⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y 1/q

,

(3.13) it follows that

h<yα<ly⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y < C_α,β,λ,^p (a,b,p,q)

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx. (3.14) Forh→0⁺,l→+∞, we obtain

0<

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y

≤C_α,β,λ,^p (a,b,p,q)

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx <+∞,

(3.15)

hence, by (3.3), we obtain

Rⁿ+

y⁽⁽ⁿα ^{−βλ)+q(a−b))/(1−q)}

Rⁿ+

f(x) x^βα+y^βαλdx

p

d y

=

Rⁿ+

f(x)g(y)

x^βα+y^βαλdx d y < C_α,β,λ,(a,b,p,q)

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx 1/ p

×

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y 1/q

=C_α,β,λ,(a,b,p,q)

Rⁿ+

x⁽α^{n−βλ)+p(b−a)}f^p(x)dx 1/ p

×

Rⁿ+

y⁽⁽ⁿα ^{−βλ)+q(a−b))/(1−q)}

Rⁿ+

f(x) x^βα+y^βαλdx

p

d y 1/q

.

(3.16)

Hence, we can obtain (3.4).

Remark 3.2. If f andgdo not satisfy (3.1) and (3.2), by the proof ofTheorem 3.1, we can obtain

Rⁿ+

f(x)g(y)

x^βα+y^βαλdx d y

≤Cα,β,λ(a,b,p,q)×

Rⁿ+

x⁽α^{n−βλ)+p(b−a)}f^p(x)dx 1/ p

Rⁿ+

y⁽ⁿα^{−βλ)+q(a−b)}g^q(y)d y 1/q

, (3.17)

Rⁿ+

y⁽⁽ⁿα ^{−βλ)+q(a−b))/(1−q)}

Rⁿ+

f(x) x^βα+y^βαλdx

p

d y

≤C_α,β,λ^p (a,b,p,q)×

Rⁿ+

x⁽ⁿα^{−βλ)+p(b−a)}f^p(x)dx.

(3.18)

Remark 3.3. By (3.4), we can also obtain (3.3), hence (3.4) and (3.3) are equivalent.

Theorem 3.4. If p >1, 1/ p+ 1/q=1,n∈Z+,α >0,β >0,λ >0,a∈R,b∈R, 0< n− ap < βλ,ap+bq=2n−βλ, f ≥0,g≥0, and

0<

Rⁿ+

x^b(p+q)α ⁻ⁿf^p(x)dx <+∞, 0<

Rⁿ+

y^a(p+q)α ⁻ⁿg^q(y)d y <+∞,

(3.19)

then

Rⁿ+

f(x)g(y)

x^βα+y^βα_λdx d y

< Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−ap

β ,λ−n−ap β

×

Rⁿ+

x^b(p+q)α ⁻ⁿf^p(x)dx 1/ p

Rⁿ+

y^a(p+q)α ⁻ⁿg^q(y)d y 1/q

,

(3.20)

Rⁿ+

y^(a(p+q)α ^{−n)/(1−q)}

Rⁿ+

f(x) x^βα+y^βαλdx

p

d y

<

Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−ap

β ,λ−n−ap β

p

Rⁿ+

x^b(p+q)α ⁻ⁿf^p(x)dx,

(3.21)

where the constant factors (Γⁿ(1/α)/βαⁿ⁻¹Γ(n/α))B((n−ap)/β,λ−(n−ap)/β) and [(Γⁿ(1/α)/βαⁿ⁻¹Γ(n/α))B((n−ap)/β,λ−(n−ap)/β)]^pare all the best possible.

Proof. Sinceap+bq=2n−βλ, we have

n−bq=n−(2n−βλ−ap)=βλ−(n−ap), (3.22)

hence, by 0< n−ap < βλ, we obtain 0< n−bq < βλ, and

(n−βλ) +p(b−a)=b(p+q)−n, (n−βλ) +q(a−b)=a(p+q)−n, n−ap

β ⁼λ−n−bq

β , λ−n−ap

β ⁼

n−bq

β . (3.23)

ByTheorem 3.1, (3.20) and (3.21) are valid.

If the constant factor K1:=(Γⁿ(1/α)/βαⁿ⁻¹Γ(n/α))B((n−ap)/β,λ−(n−ap)/β) in (3.20) is not the best possible, then there exists a positive constant K < K1, such that (3.20) is still valid when we replaceK1byK.

In particular, for 0< ε < q(n−ap), we take

f_ε(x)= x⁻α^{bq−ε/ p}, g_ε(y)= y⁻α^ap−ε/q, (3.24) by (3.17) and the properties of limit, whenδ >0 is sufficiently small, we have

xα>δ

Rⁿ+

fε(x)gε(y) x^βα+y^βαλdx d y

≤K

xα>δx^b(p+q)α ⁻ⁿfε^p(x)dx 1/ p

yα>δy^a(p+q)α ⁻ⁿgε^q(y)d y 1/q

=K

xα>δx⁻_α^n−ε1/ p

yα>δy⁻_α^n−εd y 1/q

=K

xα>δx⁻_α^n−εdx.

(3.25)

On the other hand, byLemma 2.2, we have

xα>δ

Rⁿ+

f_ε(x)g_ε(y) x^βα+y^βαλdx d y

=

xα>δx⁻α^{bq−ε/ p}

Rⁿ+

1

x^βα+y^βαλy⁻α^ap−ε/qd y dx

=

xα>δx⁻α^{bq−ε/ p}ω_α,β,λ

ap+ε q,n,x

dx

= Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

1 β

n−ap−ε q

,λ−1 β

n−ap−ε q

xα>δx⁻_α^n−εdx.

(3.26) Hence, we obtain

Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

1 β

n−ap−ε q

,λ−1 β

n−ap−ε q

≤K, (3.27)

forε→0⁺, we have

K1= Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

n−ap

β ,λ−n−ap β

≤K, (3.28)

which contradicts the fact thatK < K1. Hence the constant factor in (3.20) is the best possible.

Since (3.21) and (3.20) are equivalent, the constant factor in (3.21) is also the best

possible.

4. Some corollaries

Corollary 4.1. Ifp >1, 1/ p+ 1/q=1,n∈Z+,α >0,β >0,λ >0, f ≥0,g≥0, and 0<

Rⁿ+

x⁽ⁿα^{−βλ)(p−1)}f^p(x)dx <+∞, 0<

Rⁿ+

y⁽ⁿα^{−βλ)(q−1)}g^q(y)d y <+∞,

(4.1)

then

Rⁿ+

f(x)g(y)

x^βα+y^βαλdx d y

< Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

λ p,λ

q

Rⁿ+

x⁽ⁿα^{−βλ)(p−1)}f^p(x)dx 1/ p

Rⁿ+

y⁽ⁿα^{−βλ)(q−1)}g^q(y)d y 1/q

,

Rⁿ+

y^βλα⁻ⁿ

Rⁿ+

f(x) x^βα+y^βαλdx

p

d y

<

Γⁿ(1/α) βαⁿ⁻¹Γ(n/α)B

λ p,λ

q p

Rⁿ+

x⁽ⁿα^{−βλ)(p−1)}f^p(x)dx,

(4.2) where the constant factors in (4.2) are all the best possible.

Proof. If we takea=n/ p−βλ/ p²,b=n/q−βλ/q²inTheorem 3.4, (4.2) can be obtained.

Remark 4.2. If we taken=λ=1 in (4.2), we can obtain the results of [10]:

+∞ 0

f(x)g(y) x^β+y^β dx d y

< π βsin(π/ p)

+_∞

0 x^{(p−1)(1−β)}f^p(x)dx

1/ p+_∞

0 y^{(q−1)(1−β)}g^q(y)d y 1/q

, _+∞

0 y^β−1 +∞

0

f(x) x^β+y^βdx

p

d y <

π βsin(π/ p)

p_+∞

0 x^{(p−1)(1−β)}f^p(x)dx,

(4.3)

where the constant factors in (4.3) are all the best possible.