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A MULTIPLE HARDY-HILBERT’S INTEGRAL INEQUALITY WITH THE BEST CONSTANT FACTOR

HONG Yong

Department of Mathematics,Guangdong University of Business Study , Guangzhou 510320,People’s Republic of China

November 14, 2006

Abstract

In this paper, by introducing the norm kxkα(x Rn), we give a multiple Hardy-Hilbert’s integral inequality with a best constant factor and two parameters α,λ.

Key words: multiple Hardy-Hilbert’s integral inequality, the Γ-function, the best constant factor.

Mathematics subject classification (2000): 26D15

1 Introduction

If p >1, 1p +1q = 1,f ≥0, g ≥0, 0<R

0 fp(x)dx <+∞, 0<R

0 gq(x)dx <+∞, then the well known Hardy-Hilbert’s integral inequality is given by (see [1]):

Z 0

Z 0

f(x)g(x)

x+y dxdy < π sin(πp)

Z 0

fp(x)dx

1pZ 0

gq(x)dx 1q

, (1)

where the constant factor sin(ππ

p) is the best possible. it’s equivalent form is:

Z 0

Z 0

f(x) x+ydx

p

dy <

"

π sin(πp)

#p

Z 0

fp(x)dx, (2)

where the constant factor

π sin(πp)

p

is also the best possible.

Hardy-Hilbert’t inequality is valuable in harmonic analysis, real analysis and oper- ator theory. In recent years, many valuable results (see [2-5]) have been obtained in

(2)

generalization and improvement of Hardy-Hilbert’s inequality. In 1999,Kuang [6] gave a generalization with a parameterλof (1) as follows:

Z 0

Z 0

f(x)g(y)

xλ+yλ dxdy < hλ(p) Z

0

x1−λfp(x)dx

1pZ 0

x1−λgq(x)dx 1q

, (3) where max{1p,1q} < λ≤ 1, hλ(p) =π[λsin1p(π ) sin1q(π)]−1. Because of the constant factor hλ(p) being not the best possible, Yang [7] gave a new generalization of (1) in 2002 as follows:

Z 0

Z 0

f(x)g(y)

xλ+yλ dxdy < π λsin(πp)

× Z

0

x(1−λ)(p−1)fp(x)dx

p1Z 0

x(1−λ)(q−1)gq(x)dx 1q

, (4) it’s equivalent form is:

Z 0

yλ−1 Z

0

f(x) xλ+yλdx

p

dy <

"

π λsin(πp)

#p

Z 0

x(1−λ)(p−1)fp(x)dx, (5) where the constant factors λsin(π π

p) in (4) and [λsin(π π

p)]p in (5) are all the best possible.

At present, because of the requirement of higher-dimensional harmonic analysis and higher-dimensional operator theory, multiple Hardy-Hilbert’s integral inequality is re- searched. Hong [8] obtained: If a > 0, Pn

i=1 1

pi = 1, pi > 1, fi ≥ 0, ri = p1

i

Qn j=1pj, λ > 1a(n−1− r1

i),i= 1,2,· · ·, n, then Z

α

· · · Z

α

1 [Pn

i=1(xi−αi)a]λ

n

Y

i=1

fi(x)dx1· · ·dxn≤ Γn−2(α1) αn−1Γ(λ)

×

n

Y

i=1

Γ

1 a(1− 1

ri

)

Γ

λ− 1

a(n−1− 1 ri

) Z

α

(t−α)n−1−αλfipidt 1

pi .(6)

Afterwards, Bicheng Yang and Kuang Jichang etc. obtained some multiple Hardy- Hilbert’s integral inequalities (see [9-10]).

In this paper, by introducing the Γ-function, we generalize (3) and (4) into multiple Hardy-Hilbert’s integral inequalities with the best constant factors.

2 Some lemmas

First of all, we introduce the signs as:

Rn+={x= (x1,· · ·, xn) :x1, . . . , xn>0}, kxkα = (xα1 +· · ·+xαn)α1, α >0.

(3)

Lemma 2.1 (see [11]) If pi >0, ai >0, αi >0, i= 1,2,· · ·, n, Ψ(u) is a measurable function, then

Z . . .

Z

x1,···,xn>0;(xa1

1)α1+···+(xn

an)αn≤1

Ψ

(x1

a1)α1 +· · ·+ (xn

an)αn

×xp11−1· · ·xpnn−1dx1· · ·dxn

= ap11· · ·apnnΓ(αp1

1)· · ·Γ(αpn

n) α1· · ·αnΓ(αp1

1 +· · ·+αpn

n) Z 1

0

Ψ(u)u

p1 α1+···+pn

αn−1

du. (7)

where Γ(·) is the Γ-function.

Lemma 2.2 If p > 1, 1p + 1q = 1, n∈ Z+, α >0, λ > 0, setting the weight function ωα,λ(x, p, q) as:

ωα,λ(x, p, q) = Z

Rn+

1 kxkλα+kykλα

 kxk

1 q

α

kyk

1 p

α

(n−λ)p

kxkα kykα

λq dy,

then

ωα,λ(x, p, q) =kxk(n−λ)(p−1)α πΓn(α1)

sin(πp)λαn−1Γ(αn). (8)

Proof By lemma 2.1, we have

ωα,λ(x, p, q) =kxk(n−λ)(p−1)+λ α q

Z

Rn+

1

kxkλα+kykλαkyk−(n−λ+

λ q)

α dy

= kxk(n−λ)(p−1)+λ

α q lim

R→+∞

Z

· · · Z

y1,···,yn>0;yα1+···+yαn<rα

× h

r((yr1)α+· · ·+ (yrn)α)α1

i−(n−λ+λ

q)

kxkλα+h

r((yr1)α+· · ·+ (yrn)α)α1iλy1−11 · · ·yn1−1dy1· · ·dyn

= kxk(n−λ)(p−1)+λq

α lim

R→+∞

rnΓn(α1) αnΓ(nα)

Z 1 0

(ru1α)−(n−λ+λq) kxkλα+ (ruα1)λ

unα−1du

= kxk(n−λ)(p−1)+λ α q

Γn(α1)

αn−1Γ(nα) lim

R→+∞

Z r

0

1

kxkλα+uλuλ−λq−1du

= kxk(n−λ)(p−1)+λ α q

Γn(α1) αn−1Γ(nα)

Z 0

1

kxkλα+uλuλ−λq−1du

= kxk(n−λ)(p−1)α Γn(α1) λαn−1Γ(nα)

Z 0

1

1 +uu1p−1du

(4)

= kxk(n−λ)(p−1)α Γn(α1) λαn−1Γ(nα)Γ(1

p)Γ(1−1 p)

= kxk(n−λ)(p−1)α πΓn(α1) sin(πp)λαn−1Γ(nα), hence (8) is valid.

Lemma 2.3 If p > 1, 1p + 1q = 1, n ∈ Z+, α > 0, λ > 0, 0 < ε < λ(q−1), setting

˜

ωα,λ(x, q, ε) as:

˜

ωα,λ(x, q, ε) = Z

Rn+

1

kxkλα+kykλαkyk

(n−λ)(q−1)+n+ε q

α dy,

then we have

˜

ωα,λ(x, q, ε) =kxk

λ qε α q

Γn(α1) λαn−1Γ(nα)Γ(1

p − ε λq)Γ(1

q + ε

λq). (9)

Proof By the same way of lemma 2.2, lemma 2.3 can be proved.

3 Main Results

Theorem 3.1 If p >1, 1p +1q = 1, n∈Z+,α >0, λ >0, f ≥0, g≥0, and 0<

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx <∞, 0<

Z

Rn+

kxk(n−λ)(q−1)α gq(x)dx <∞, (10) then

Z

Rn+

Z

Rn+

f(x)g(y)

kxkλα+kykλαdxdy < πΓn(α1) sin(πp)λαn−1Γ(nα)

× Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!1p Z

Rn+

kxk(n−λ)(q−1)α gq(x)dx

!1q

; (11)

Z

Rn+

kykλ−nα Z

Rn+

f(x) kxkλα+kykλαdx

!p

dy <

"

πΓn(1α) sin(πp)λαn−1Γ(nα)

#p

× Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx, (12)

where the constant factors πΓ

n(α1)

sin(πp)λαn−1Γ(αn) and[ πΓ

n(α1)

sin(πp)λαn−1Γ(nα)]p are all the best possible.

(5)

Proof By H¨older’s inequality, we have A:=

Z

Rn+

Z

Rn+

f(x)g(y) kxkλα+kykλαdxdy

= Z

Rn+

Z

Rn+

f(x) (kxkλα+kykλα)1p

 kxk

1 q

α

kyk

1 p

α

n−λ

kxkα kykα

pqλ

× g(y)

(kxkλα+kykλα)1q

 kyk

1

αp

kxk

1

αq

n−λ

kykα kxkα

pqλ dxdy

 Z

Rn+

Z

Rn+

fp(x) kxkλα+kykλα

 kxk

1

αq

kyk

1

αp

(n−λ)p

kxkα kykα

λ

q

dxdy

1 p

×

 Z

R+n

Z

Rn+

gq(x) kxkλα+kykλα

 kyk

1 p

α

kxk

1 q

α

(n−λ)q

kykα kxkα

λp dxdy

1 q

=

 Z

Rn+

fp(x)

 Z

Rn+

1 kxkλα+kykλα

 kxk

1

αq

kyk

1

αp

(n−λ)p

kxkα kykα

λq dy

dx

1 p

×

 Z

Rn+

gq(y)

 Z

Rn+

1 kxkλα+kykλα

 kyk

1 p

α

kxk

1 q

α

(n−λ)q

kykα kxkα

λ

p

dx

dy

1 q

= Z

Rn+

fp(x)ωα,λ(x, p, q)dx

!1

p Z

R+n

gq(y)ωα,λ(y, q, p)dy

!1

q

,

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 and C2, such that they are not all zero, and

C1fp(x) kxkλα+kykλα

 kxk

1

αq

kyk

1

αp

(n−λ)p

kxkα kykα

λq

= C2gq(y) kxkλα+kykλα

 kyk

1 p

α

kxk

1 q

α

(n−λ)q

kykα kxkα

λp

, a.e. in Rn+×Rn+,

it follows that C1kxk(n−λ)(p−1)+n

α fp(x) =C2kyk(n−λ)(q−1)+n

α gq(y) =C(constant), a.e. in Rn+×Rn+,

(6)

which contradicts (10), hence we have A <

Z

Rn+

fp(x)ωα,λ(x, p, q)dx

!1p Z

R+n

gq(y)ωα,λ(y, q, p)dy

!1q .

By lemma 2.2 and sin(πp) = sin(πq), we have

A <

"

πΓn(α1) sin(πp)λαn−1Γ(nα)

#1

p Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!1

p

×

"

πΓn(α1) sin(πq)λαn−1Γ(nα)

#1

q Z

Rn+

kyk(n−λ)(q−1)α gq(y)dy

!1

q

= πΓn(α1) sin(πp)λαn−1Γ(nα)

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!1

p Z

Rn+

kxk(n−λ)(q−1)α gq(x)dx

!1

q

.

Hence (11) is valid.

For 0< a < b <∞, setting ga,b(y) =

(

kykλ−nα R

Rn+

f(x)

kxkλα+kykλαdxp−1

, a <kykα< b

0, 0<kykα ≤a or kykα≥b

˜

g(y) =kykλ−nα Z

Rn+

f(x) kxkλα+kykλαdx

!p−1

, y ∈Rn+,

by (10), for sufficiently smalla >0 and sufficiently largeb >0, we have 0<

Z

a<kykα<b

kyk(n−λ)(q−1)α ga,bq (y)dy <∞.

Hence by (11), we have Z

a<kykα<b

kyk(n−λ)(q−1)αq(y)dy= Z

a<kykα<b

kykλ−nα Z

R+n

f(x) kxkλα+kykλαdx

!p

dy

= Z

a<kykα<b

kykλ−nα Z

Rn+

f(x) kxkλα+kykλαdx

!p−1

Z

R+n

f(x) kxkλα+kykλαdx

! dy

= Z

Rn+

Z

Rn+

f(x)ga,b(y) kxkλα+kykλαdxdy

< πΓn(α1) sin(πp)λαn−1Γ(αn)

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!1p Z

Rn+

kyk(n−λ)(q−1)α gqa,b(y)dy

!1q

= πΓn(α1) sin(πp)λαn−1Γ(αn)

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!1p Z

a<kykα<b

kyk(n−λ)(q−1)α ˜gq(y)dy

!1q ,

(7)

it follows that Z

a<kykα<b

kyk(n−λ)(q−1)αq(y)dy <

"

πΓn(α1) sin(πp)λαn−1Γ(αn)

#p

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx.

Fora→0+,b→+∞, by (10), we obtain 0<

Z

Rn+

kyk(n−λ)(q−1)α ˜gq(y)dy≤

"

πΓn(α1) sin(πp)λαn−1Γ(nα)

#p

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx <∞, hence by (11), we have

Z

Rn+

kykλ−nα Z

Rn+

f(x) kxkλα+kykλαdx

!p

dy= Z

Rn+

Z

R+n

f(x)˜g(y) kxkλα+kykλαdxdy

< πΓn(α1) sin(πp)λαn−1Γ(nα)

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!p1 Z

Rn+

kyk(n−λ)(q−1)αq(y)dy

!1q

= πΓn(α1) sin(πp)λαn−1Γ(nα)

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!p1

×

"

Z

Rn+

kykλ−nα Z

Rn+

f(x) kxkλα+kykλαdx

!p

dy

#1q ,

it follows that

Z

Rn+

kykλ−nα Z

Rn+

f(x) kxkλα+kykλαdx

!p

dy

<

"

πΓn(α1) sin(πp)λαn−1Γ(nα)

#p

Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx.

Hence (12) is valid.

Remark: By (12), we can also obtain (11), hence (12) and (11) are equivalent.

If the constant factor Cn,α(λ, p) := πΓ

n(1α)

sin(πp)λαn−1Γ(αn) in (11) is not the best possible, then there exists a positive constantK < Cn,α(λ, p), such that

Z

Rn+

Z

Rn+

f(x)g(y) kxkλα+kykλαdxdy

< K Z

Rn+

kxk(n−λ)(p−1)α fp(x)dx

!1p Z

Rn+

kyk(n−λ)(q−1)α gq(y)dy

!1q

. (13)

(8)

In particular, setting f(x) =kxk

(n−λ)(p−1)+n+ε

α p , g(y) =kyk

(n−λ)(q−1)+n+ε

α q , 0< ε < λ(q−1), (13) is still true. By the properties of limit, there exists a sufficiently smalla >0, such that

Z

kxkα>a

Z

R+n

1

kxkλα+kykλαkxk

(n−λ)(p−1)+n+ε

α p kyk

(n−λ)(q−1)+n+ε

α q dxdy

< K Z

kxkα>a

kxk(n−λ)(p−1)α kxk−(n−λ)(p−1)−n−ε

α dx

!1p

× Z

kykα>a

kyk(n−λ)(q−1)α kyk−(n−λ)(q−1)−n−ε

α dy

!1q

= K

Z

kxkα>a

kxk−n−εα dx.

On the other hand, by lemma 2.3, we have Z

kxkα>a

Z

Rn+

1

kxkλα+kykλαkxk

(n−λ)(p−1)+n+ε

α p kyk

(n−λ)(q−1)+n+ε

α q dxdy

= Z

kxkα>a

kxk−n+

λ qε α p

Z

Rn+

1

kxkλα+kykλαkyk

(n−λ)(q−1)+n+ε

α q dydx

= Z

kxkα>a

kxk−n+

λ qε

α pω˜α,λ(x, q, ε)dx

= Γn(α1) λαn−1Γ(nα

1 p− ε

λq

Γ 1

q + ε λq

Z

kxkα>a

kxk−n−εα dx, hence we obtain

Γn(α1) λαn−1Γ(nα

1 p− ε

λq

Γ 1

q + ε λq

< K.

forε→0+, we have

Cn,α(λ, p) = πΓn(α1)

sin(πp)λαn−1Γ(nα) ≤K,

this contradicts the fact that K < Cn,α(λ, p).Hence the constant factor in (11) is the best possible.

Since (12) and (11) are equivalent, the constant factor in (12) is also the best possible.

Corollary 3.1 If p >1, 1p +1q = 1, n∈Z+, α >0, f ≥0, g≥0, and 0<

Z

Rn+

fp(x)dx <∞, 0<

Z

Rn+

gq(x)dx <∞, (14)

(9)

then

Z

Rn+

Z

Rn+

f(x)g(y)

kxknα+kyknαdxdy < πΓn(α1) sin(πp)nαn−1Γ(nα)

× Z

Rn+

fp(x)dx

!1p Z

Rn+

gq(x)dx

!1q

; (15)

Z

Rn+

Z

Rn+

f(x) kxknα+kyknαdx

!p

dy <

"

πΓn(1α) sin(πp)nαn−1Γ(nα)

#p

Z

Rn+

fp(x)dx, (16) where the constant factors in (15) and (16) are all the best possible.

Proof By takingλ=n in theorem 3.1, (15) and (16) can be obtained.

Corollary 3.2 If p >1, 1p +1q = 1, n∈Z+, f ≥0, g≥0, and 0<

Z

Rn+

fp(x)dx <∞, 0<

Z

Rn+

gq(x)dx <∞, (17) then

Z

Rn+

Z

Rn+

f(x)g(y) (Pn

i=1xi)n+ (Pn

i=1yi)ndxdy < π n! sin(πp)

× Z

Rn+

fp(x)dx

!1p Z

Rn+

gq(x)dx

!1q

; (18)

Z

Rn+

Z

Rn+

f(x) (Pn

i=1xi)n+ (Pn

i=1yi)ndx

!p

dy <

"

π n! sin(πp)

#p

Z

Rn+

fp(x)dx, (19) where the constant factors in (18) and (19) are all the best possible.

Proof By takingα= 1 in corollary 3.1, (18) and (19) can be obtained.

References

[1] G. H. Hardy, J. E. Littewood and G. Polya. Inequalities. Gambridge Unie. press, London, 1952.

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[2] A. E. Ingham. A note on Hilbert’s inequality. J. London Math. Soc., 11(1936), 237-240.

[3] Gao Mingzhe and Tan Li. Some improvements on Hilbert’s integral inequality. J.

Math. Anal. Appl. 229(1999), 682-689.

[4] B. G. pachpatte.On some new inequalities similar to Hilbert’s inequality. J. Math.

Anal. Appl. 226(1998), 166-179.

[5] Bicheng Yang and Themistocles M. Rassias. On the way of weight coefficient and research for the Hilbert-type inequalities. Math. Ineq. appl., 6: 4(2003), 625-658.

[6] Kuang Jichang. On new extensions of Hilbert’s integtal inequality. Math. Anal.

Appl, 235(1999): 608-614.

[7] Bicheng Yang. On a generalization of Hardy-Hilbert’s inequality. Chin. Ann. of Math., 23(2002): 247-254.

[8] Hong Yong. All-sided Generalization about Hardy-Hilbert’s integral inequalities.

Acta Math. sinica ,(China) 44(2001): 619-626.

[9] Bichebg Yang. On a multiple Hardy-Hilbert’s integral inequality. Chin. Anna. of Math. 24(A): 6(2003), 743-750.

[10] Kuang Jichang.Applied Inequalities. Shandong Science and Technology press, Ji- nan,2004.

[11] Fichtingoloz G. M..A course in differential and integral calculus. Renmin Jiaoyu publisgers, Beijing, 1957.

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