WITH SOME PARAMETERS
HONG YONGReceived 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∞fp(x)dx <+∞, 0<∞
0 gq(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)dx1/ p+∞
0 g
q(x)dx1/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, pi>1 (i=1, 2,. . .,n),
n
i=1(1/ pi)=1, λ > n−min1≤i≤n{pi}, fi≥0, and
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 94960, Pages1–11
0<α+∞(t−α)n−1−λfpi
i (t)dt <+∞, (i=1, 2,. . .,n), then
+∞
α ···
+∞
α
1 n
i=1xi−nα
λ n
i=1
fi xidx1. . . dxn
< 1
Γ(λ) n
i=1
Γ1−n−p λ i
+∞
α (t−α) n−1−λfpi
i (t)dt
1/ pi
,
(1.3)
where the constant factor (1/Γ(λ))ni=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,. . .,xn
:x1,. . .,xn>0
,
xα= xα
1+···+xnα
1/α
, (α >0),
(1.4)
and we agree onxα< crepresenting{x∈Rn
+:xα< c}.
2. Some lemmas
Lemma2.1 (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≤1
Ψx1 a1
α1 +···+
xn
an
αn
×x1p1−1. . . x pn−1
n dx1. . . dxn
= a p1
1 . . . a pn
nΓ p1/α1
. . .Γ pn/αn
α1. . . αnΓ p1/α1+···+pn/αn
1
0Ψ(u)u
p1/α1+···+pn/αn−1du,
(2.1)
where theΓ(·)isΓ-function.
Lemma2.2. Ifn∈Z+,α >0,β >0,λ >0,m∈R,0< n−m < βλ, and setting weight func-tionωα,β,λ(m,n,y)as
ωα,β,λ(m,n,y)=
Rn
+
1 xβα+yβαλ
x−m
then
ωα,β,λ(m,n,y)= ynα−βλ−m Γ n(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
ωα,β,λ(m,n,y)=
Rn
+
1 xβα+yβαλ
x−m α d y
= lim r→+∞
···
x1,...,xn>0;xα1+···+xαn<rα
×
r x1/r
α
+···+ xn/r
α1/α−m
rβ x
1/r α
+···+ xn/r
αβ/α
+yβαλ
x1−1
1 . . . xn1−1dx1. . . dxn
= lim r→+∞
rnΓn(1/α)
αnΓ(n/α)
1
0
ru1/α−m yβα+rβuβ/αλ
un/α−1du
= Γn(1/α)
αn−1Γ(n/α)rlim→+∞ r
0
1 yβα+tβλ
tn−m−1dt
= Γn(1/α)
αn−1Γ(n/α) +∞
0
1 yβα+tβλ
tn−m−1dt
= ynα−βλ−m Γ n(1/α)
βαn−1Γ(n/α) 1
0
1 (1 +u)λu
(n−m)/β−1du
= ynα−βλ−m Γ n(1/α)
βαn−1Γ(n/α)B n−m
β ,λ− n−m
β
.
(2.4)
Hence (2.3) is valid.
3. Main results
Theorem3.1. 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
+
xα(n−βλ)+p(b−a)fp(x)dx <+∞, (3.1)
0<
Rn
+
then
Rn
+
f(x)g(y) xβα+yβαλ
dx d y
< Cα,β,λ(a,b,p,q)×
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx
1/ p
Rn
+
yα(n−βλ)+q(a−b)gq(y)d y
1/q ,
(3.3)
Rn
+
yα((n−βλ)+q(a−b))/(1−q)
Rn
+
f(x) xβα+yβαλ
dx p
d y
< Cαp,β,λ(a,b,p,q)×
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx,
(3.4)
whereCα,β,λ(a,b,p,q)=(Γn(1/α)/βαn−1Γ(n/α))B1/ p((n−ap)/β,λ−(n−ap)/β)B1/q((n−
bq)/β,λ−(n−bq)/β).
Proof. By H ¨older’s inequality, we have
G:=
Rn
+
f(x)g(y) xβα+yβαλ
dx d y
=
Rn
+
f(x) xβα+yβαλ/ p
xb α ya α
g(y) xβα+yβαλ/q
ya α xb α
dx d y
≤
Rn
+
fp(x) xβα+yβαλ
xbpα yapα
dx d y 1/ p
×
Rn
+
gq(y) xβα+yβαλ
yaqα xbqα
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
C1fp(x) xβα+yβαλ
xbpα yapα =
C2gq(y) xβα+yβαλ
yaqα xbqα
, a.e. (x,y)∈Rn
+×Rn+. (3.6)
Without losing generality, we suppose thatC1=0, we may get
xαb(p+q)fp(x)=C2
C1
yaα(p+q)gq(y), a.e. (x,y)∈Rn+×Rn+, (3.7)
hence, we obtain
hence, we have
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx=
Rn
+
xα(n−βλ)−bq−ap+b(p+q)fp(x)dx
=C
Rn
+
x(αn−βλ)−bq−apdx= ∞,
(3.9)
which contradicts (3.1). Hence, and byLemma 2.2, we obtain
G <
Rn+
Rn+
1 xβα+yβαλ
1 yapα
d y
xbpα fp(x)dx
1/ p
×
Rn
+
Rn
+
1 xβα+yβαλ
1 xbqα
dx
yaqα gq(y)d y
1/q
=
Rn
+
ωα,β,λ,(ap,n,x)xbpα fp(x)dx
1/ p
Rn
+
ωα,β,λ,(bq,n,y)yaqα gq(y)d y
1/q
=
Γn(1/α)
βαn−1Γ(n/α)B
n−ap
β ,λ−
n−ap β
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx
1/ p
×
Γn
(1/α)
βαn−1Γ(n/α)B n−bq
β ,λ−
n−bq β
Rn
+
yα(n−βλ)+q(a−b)gq(y)d y
1/q
=Cα,β,λ,(a,b,p,q)
Rn+x
(n−βλ)+p(b−a)
α fp(x)dx
1/ p ×
Rn+y
(n−βλ)+q(a−b)
α gq(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)=
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
yk α
Rn
+
f(x) xβα+yβαλ
dx p/q
, h <yα< l,
0, 0<yα≤horyα≥l,
g(y)= yk α
Rn
+
f(x) xβα+yβαλ
dx p/q
, y∈Rn+,
(3.11)
by (3.1), for sufficiently smallh >0 and sufficiently largel >0, we have
0<
h<yα<l
Hence, by (3.3), we have
h<yα<l
yα(n−βλ)+q(a−b)gq(y)d y
=
h<yα<l
yαk(1−q)gq(y)d y=
h<yα<l
yk α
Rn
+
f(x) xβα+xβαλ
dx p
d y
=
h<yα<l
yk α
Rn
+
f(x) xβα+yβαλ
dx p/q
Rn
+
f(x) xβα+yβαλ
dx
d y
=
Rn+
f(x)gh,l(y) xβα+yβαλ
dx d y < Cα,β,λ,(a,b,p,q)
Rn+x
(n−βλ)+p(b−a)
α fp(x)dx
1/ p
×
Rn
+
yα(n−βλ)+q(a−b)ghq,l(y)d y
1/q
=Cα,β,λ,(a,b,p,q)
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx
1/ p
×
h<yα<l
yα(n−βλ)+q(a−b)gq(y)d y
1/q ,
(3.13)
it follows that
h<yα<l
yα(n−βλ)+q(a−b)gq(y)d y < Cαp,β,λ,(a,b,p,q)
Rn+x
(n−βλ)+p(b−a)
α fp(x)dx. (3.14)
Forh→0+,l→+∞, we obtain
0<
Rn
+
yα(n−βλ)+q(a−b)gq(y)d y
≤Cαp,β,λ,(a,b,p,q)
Rn+x
(n−βλ)+p(b−a)
α fp(x)dx <+∞,
(3.15)
hence, by (3.3), we obtain
Rn+y
((n−βλ)+q(a−b))/(1−q) α
Rn+
f(x) xβα+yβαλ
dx p
d y
=
Rn
+
f(x)g(y) xβα+yβαλ
dx d y < Cα,β,λ,(a,b,p,q)
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx
1/ p
×
Rn
+
yα(n−βλ)+q(a−b)gq(y)d y
1/q
=Cα,β,λ,(a,b,p,q)
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx
1/ p
×
Rn
+
yα((n−βλ)+q(a−b))/(1−q)
Rn
+
f(x) xβα+yβαλ
dx p
d y 1/q
.
(3.16)
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) xβα+yβαλ
dx d y
≤Cα,β,λ(a,b,p,q)×
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx
1/ p
Rn
+
yα(n−βλ)+q(a−b)gq(y)d y
1/q ,
(3.17)
Rn
+
yα((n−βλ)+q(a−b))/(1−q)
Rn
+
f(x) xβα+yβαλ
dx p
d y
≤Cαp,β,λ(a,b,p,q)×
Rn
+
xα(n−βλ)+p(b−a)fp(x)dx.
(3.18)
Remark 3.3. By (3.4), we can also obtain (3.3), hence (3.4) and (3.3) are equivalent.
Theorem3.4. 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
+
xαb(p+q)−nfp(x)dx <+∞,
0<
Rn
+
yαa(p+q)−ngq(y)d y <+∞,
(3.19)
then
Rn+
f(x)g(y) xβα+yβαλ
dx d y
< Γ
n(1/α)
βαn−1Γ(n/α)B
n−ap
β ,λ−
n−ap β
×
Rn+x
b(p+q)−n
α fp(x)dx
1/ p
Rn+y
a(p+q)−n α gq(y)d y
1/q ,
(3.20)
Rn
+
yα(a(p+q)−n)/(1−q)
Rn
+
f(x) xβα+yβαλ
dx p
d y
< Γn
(1/α)
βαn−1Γ(n/α)B n−ap
β ,λ−
n−ap β
p
Rn
+
xαb(p+q)−nfp(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
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
fε(x)= x−αbq−ε/ p, gε(y)= y−αap−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) xβα+yβαλ
dx d y
≤K
xα>δ
xαb(p+q)−nfεp(x)dx
1/ p
yα>δ
yαa(p+q)−ngεq(y)d y
1/q
=K
xα>δ
x−n−ε α
1/ p
yα>δ
y−n−ε α d y
1/q =K
xα>δ
x−n−ε
α dx.
(3.25)
On the other hand, byLemma 2.2, we have
xα>δ
Rn
+
fε(x)gε(y) xβα+yβαλ
dx d y
=
xα>δ
x−αbq−ε/ p
Rn
+
1 xβα+yβαλ
y−αap−ε/qd y dx
=
xα>δ
x−αbq−ε/ pωα,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>δ
x−n−ε
α dx.
(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= Γ
n(1/α)
βαn−1Γ(n/α)B
n−ap
β ,λ−
n−ap β
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
Corollary4.1. Ifp >1,1/ p+ 1/q=1,n∈Z+,α >0,β >0,λ >0, f ≥0,g≥0, and
0<
Rn+x
(n−βλ)(p−1)
α fp(x)dx <+∞,
0<
Rn
+
yα(n−βλ)(q−1)gq(y)d y <+∞,
(4.1)
then
Rn
+
f(x)g(y) xβα+yβαλ
dx d y
< Γn(1/α) βαn−1Γ(n/α)B
λ
p, λ q
Rn
+
xα(n−βλ)(p−1)fp(x)dx
1/ p
Rn
+
yα(n−βλ)(q−1)gq(y)d y
1/q ,
Rn
+
yβλα−n
Rn
+
f(x) xβα+yβαλ
dx p
d y
<
Γn(1/α)
βαn−1Γ(n/α)B λ
p, λ q
p
Rn
+
xα(n−βλ)(p−1)fp(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)
xβ+yβ dx d y
< π
βsin(π/ p)
+∞
0 x
(p−1)(1−β)fp(x)dx
1/ p+∞
0 y
(q−1)(1−β)gq(y)d y
1/q ,
+∞
0 y
β−1+∞ 0
f(x)
xβ+yβdx
p
d y <
π
βsin(π/ p)
p+∞
0 x
(p−1)(1−β)fp(x)dx,
(4.3)
If we taken=β=1 in (4.2), we can obtain
+∞
0
f(x)g(y) (x+y)λ dx d y
< B λ
p, λ q
+∞
0 x
(1−λ)(p−1)fp(x)dx
1/ p+∞
0 y
(1−λ)(q−1)gq(y)d y
1/q ,
+∞
0 y
λ−1 +∞
0
f(x) (x+y)λdx
p
d y < Bp
λ
p, λ q
+∞
0 x
(1−λ)(p−1)fp(x)dx,
(4.4)
where the constant factors in (4.4) are all the best possible.
Corollary4.3. If p >1,1/ p+ 1/q=1,n∈Z+,λ >0,np+λ−2n >0,nq+λ−2n >0 f ≥0,g≥0, and
0<
Rn
+ xn−λ
α fp(x)dx <+∞,
0<
Rn
+ yn−λ
α gq(y)d y <+∞,
(4.5)
then
Rn+
f(x)g(y)
xα+yαλdx d y
< B
np+λ−2n
p ,
nq+λ−2n q
Rn
+ xn−λ
α fp(x)dx
1/ p
Rn
+ yn−λ
α gq(y)d y
1/q ,
Rn
+
y(αn−λ)/(1−q)
Rn
+
f(x) xα+yαλdx
p
d y < Bp
np+λ−2n
p ,
nq+λ−2n q
Rn
+ xn−λ
α fp(x)dx, (4.6)
where the constant factors in (4.6) are all the best possible.
Proof. If we takeβ=1,a=b=(2n−λ)/ pqinTheorem 3.4, (4.6) can be obtained.
Remark 4.4. If we taken=1 in (4.6), we can obtain the results of [1]:
+∞
0
f(x)g(y) (x+y)λ dx d y
< B
p+λ−2
p ,
q+λ−2
q
+∞
0 x
1−λfp(x)dx1/ p+∞
0 y
1−λgq(y)d y1/q,
+∞
0 y
(1−λ)/(1−q)+∞ 0
f(x) (x+y)λdx
p
d y < Bp
p+λ−2
p ,
q+λ−2
q
+∞
0 x
1−λfp(x)dx,
(4.7)
where the constant factors in (4.7) are all the best possible.
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Hong Yong: Department of Mathematics, Guangdong University of Business Study, Guangzhou 510320, China