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
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(pλπ ) sin1q(qλπ)]−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.
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
= 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 q−ε 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.
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+,
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)α g˜q(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 ,
it follows that Z
a<kykα<b
kyk(n−λ)(q−1)α g˜q(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)α g˜q(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)
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 q−ε α p
Z
Rn+
1
kxkλα+kykλαkyk−
(n−λ)(q−1)+n+ε
α q dydx
= Z
kxkα>a
kxk−n+
λ q−ε
α pω˜α,λ(x, q, ε)dx
= Γn(α1) λαn−1Γ(nα)Γ
1 p− ε
λq
Γ 1
q + ε λq
Z
kxkα>a
kxk−n−εα dx, hence we obtain
Γn(α1) λαn−1Γ(nα)Γ
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)
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.
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