Volume 2007, Article ID 71049,14pages doi:10.1155/2007/71049
Research Article
A Multiple Hilbert-Type Integral Inequality with
the Best Constant Factor
Baoju Sun
Received 9 February 2007; Accepted 29 April 2007
Recommended by Eugene H. Dshalalow
By introducing the norm xα (x∈R) and two parametersα, λ, we give a multiple
Hilbert-type integral inequality with a best possible constant factor. Also its equivalent form is considered.
Copyright © 2007 Baoju Sun. 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,g≥0, satisfy 0<0∞fp(t)dt <∞and 0< ∞
0 gq(t)dt <∞, then the well-known Hardy-Hilbert’s integral inequality is given by (see [1,2])
∞
0
f(x)g(y) x+y dx d y <
π sin(π/ p)
∞
0 f
p(t)dt1/ p∞
0 g
q(t)dt1/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 still the best possible.
Hardy et al. [1] gave a Hilbert-type integral inequality similar to (1.1) as
∞
0
ln(x/ y)
x−y f(x)g(y)dx d y <
π
sin(π/ p)
2∞
0 f
p(x)dx1/ p∞
0 g
q(x)dx1/q, (1.3)
where the constant factor (π/sin(π/ p))2is the best possible. Recently, Yang gave a generalization of (1.3) as (see [9]) ∞
0
ln(x/ y)f(x)g(y) xλ−yλ dx d y
<
π
λsin(π/ p)
2∞
0 x
(p−1)(1−λ)fp(x)dx
1/ p∞
0 x
(q−1)(1−λ)gq(x)dx
1/q
,
(1.4)
where the constant factor (π/λsin(π/ p))2is the best possible. Its equivalent form is
∞
0 y
λ−1∞ 0
ln(x/ y)f(x) xλ−yλ dx
p
d y <
π
λsin(π/ p) 2p∞
0 x
(p−1)(1−λ)fp(x)dx, (1.5)
where the constant factor (π/λsin(π/ p))2pis the best possible.
At present, because of the requirement of higher-dimensional harmonic analysis and higher-dimensional operator theory, multiple Hilbert-type integral inequalities have been studied. Hong [10] obtained the following. If
a >0,
n
i=1 1
pi=1, pi>1, ri=
1 pi
n
i=1
pi, λ >1a
n−1−r1
i
, i=1, 2,. . .,n, (1.6)
then ∞
α ···
∞
α
1
n
i=1
xi−α
aλ n
i=1 fi
xi
dx1dx2···dxn
≤ Γn(1/α) αn−1Γ(λ)
n
i=1
Γ1a1−1 ri
Γλ−1 a
n−1−1 ri
∞
α (t−α)
n−1−αλfpi
i (t)dt
1/ pi .
(1.7)
Yang and Kuang, and others obtained some multiple Hilbert-type integral inequalities (see [5,11,12]).
The main objective of this paper is to build multiple Hilbert-type integral inequalities with best constant factor of (1.4) and (1.5).
For this reason, we introduce signs as
R+
n=
x=x1,x2,. . .,xn
:x1,x2,. . .,xn>0
,
xα=
xα
1+xα2+···+xnα
1/α
, α >0, (1.8)
2. Lemmas
First we give some multiple integral formulas.
Lemma2.1 (see [13]). Ifpi>0,i=1, 2,. . .,n, f(τ)is a measurable function, then
t1,t2,...,tn>0;t1+t2+···+tn≤1
ft1+t2+···+tnt1p1−1tp 2−1 2 ···tpn−
1
n dt1dt2···dtn
=Γ
p1
Γp2
···Γpn
Γp1+p2+···+pn
1 0 f(τ)τ
p1+p2+···+pn−1dτ.
(2.1)
Lemma2.2. Ifr >0,pi>0,i=1, 2,. . .,n, f(τ)is a measurable function, then
t1,t2,...,tn>0;t1+t2+···+tn≤r
ft1+t2+···+tnt1p1−1tp 2−1 2 ···tpn−
1
n dt1dt2···dtn
=Γ
p1
Γp2
···Γpn
Γp1+p2+···+pn
r 0 f(τ)τ
p1+p2+···+pn−1dτ,
(2.2)
t1,t2,...,tn>0
ft1+t2+···+tn
tp1−1
1 t
p2−1 2 ···t
pn−1
n dt1dt2···dtn
=Γ
p1
Γp2
···Γpn
Γp1+p2+···+pn
∞ 0 f(τ)τ
p1+p2+···+pn−1dτ.
(2.3)
Proof. Settingti/r=ui(i=1, 2,. . .,n) on the left-hand side of (2.2) we obtain (2.2) from
Lemma 2.1.
From (2.1) and (2.3), we have the following lemma.
Lemma2.3.
t1,t2,...,tn>0;t1+t2+···+tn≥1
ft1+t2+···+tn
tp1−1
1 t
p2−1 2 ···t
pn−1
n dt1dt2···dtn
=Γ
p1
Γp2
···Γpn
Γp1+p2+···+pn
∞ 1 f(τ)τ
p1+p2+···+pn−1dτ.
(2.4)
Settingti=(xi/ai)αi (i=1, 2,. . .,n) in (2.1), (2.2), (2.3), (2.4) we have the following
Lemma2.4. Ifpi>0,ai>0,αi>0,i=1, 2,. . .,n,f(τ)is a measurable function, then
x1,x2,...,xn>0; (x1/a1)α1+(x2/a2)α2+···+(x1/an)αn≤1 f x 1 a1 α1 + x 2 a2 α2 +···+ x 1 an αn
×xp1−1 1 x
p2−1 2 ···x
pn−1
n dx1dx2···dxn
=a
p1 1 a
p2 2 ···a
pn
nΓp1/α1Γp2/α2···Γpn/αn
α1α2···αnΓ
p1/α1+p2/α2+···+pn/αn
1
0 f(τ)τ
p1/α1+p2/α2+···+pn/αn−1dτ,
x1,x2,...,xn>0; (x1/a1)α1+(x2/a2)α2+···+(x1/an)αn≤r f x 1 a1 α1 + x 2 a2 α2 +···+ x 1 an αn
×xp1−1 1 x
p2−1 2 ···x
pn−1
n dx1dx2···dxn
=a
p1 1 a
p2 2 ···a
pn
nΓp1/α1
Γp2/α2
···Γpn/αn
α1α2···αnΓ
p1/α1+p2/α2+···+pn/αn
r
0 f(τ)τ
p1/α1+p2/α2+···+pn/αn−1dτ,
x1,x2,...,xn>0 f x 1 a1 α1 + x 2 a2 α2 +···+ x 1 an αn
xp1−1 1 x
p2−1 2 ···x
pn−1
n dx1dx2···dxn
=a
p1 1 a
p2 2 ···a
pn
nΓp1/α1
Γp2/α2
···Γpn/αn
α1α2···αnΓ
p1/α1+p2/α2+···+pn/αn
∞
0 f(τ)τ
p1/α1+p2/α2+···+pn/αn−1dτ,
x1,x2,...,xn>0; (x1/a1)α1+(x2/a2)α2+···+(x1/an)αn≥1 f x 1 a1 α1 + x 2 a2 α2 +···+ x 1 an αn
×xp1−1 1 x
p2−1 2 ···x
pn−1
n dx1dx2···dxn
=a
p1 1 a
p2 2 ···a
pn
nΓp1/α1
Γp2/α2
···Γpn/αn
α1α2···αnΓ
p1/α1+p2/α2+···+pn/αn
∞
1 f(τ)τ
p1/α1+p2/α2+···+pn/αn−1dτ.
(2.5)
In particular, ifp >0,α >0, f(τ)is a measurable function, then
x1,x2,...,xn>0;x1α+xα2+···+xnα≤1 fxα
1+x2α+···+xαn
dx1dx2···dxn
= Γn(1/α) αnΓ(n/α)
1
0 f(τ)τ
n/α−1dτ,
(2.6)
x1,x2,...,xn>0;x1α+xα2+···+xnα≥1
fxα1+x2α+···+xαn
dx1dx2···dxn
= Γn(1/α) αnΓ(n/α)
∞
1 f(τ)τ
n/α−1dτ.
(2.7)
Lemma2.5. Ifp >1,n∈Z+,α >0,λ >0, define the weight functionwα,λ(x,p)as
wα,λ(x,p)=
Rn +
lnxα/yα
xλ α− yλα
x
α
yα
n−λ/ p
d y. (2.8)
Then
wα,λ(x,p)= xnα−λ Γ n(1/α)
αn−1Γ(n/α)
π
λsin(π/ p) 2
. (2.9)
Proof. By (2.6) and (2.7), we have
wα,λ(x,p)= xnα−λ/ p
Rn +
lnxα/yα
xλ α− yλα
yλ/ pα −nd y
= xnα−λ/ p
y1,y2,...,yn>0 lnyα
1+y2α+···+yαn
1/α
/xα
y1α+y2α+···+yαn
λ/α
− xλ α
×yα
1+y2α+···+yαn
(1/α)(λ/ p−n)
d y1d y2···d yn
= xnα−λ/ p Γ n(1/α)
αnΓ(n/α)
∞
0
lnt1/α/x α
tλ/α− xλ α
t(1/α)(λ/ p−n)tn/α−1dt.
(2.10)
Setting (t1/α/x
α)λ=uwe have
wα,λ(x,p)= xnα−λ Γ n(1/α)
αn−1Γ(n/α) 1 λ2
∞
0 lnu u−1u
1/ p−1du. (2.11)
From [1, Theorem 342] we have (1/λ2)∞
0 (lnu/(u−1))u1/ p−1du=(π/λsin(π/ p))2. So we obtain
wα,λ(x,p)= xnα−λ Γ n(1/α)
αn−1Γ(n/α)
π
λsin(π/ p) 2
. (2.12)
ThusLemma 2.5is proved.
Lemma2.6. Ifλ >0,s >0, then
∞
1 1 x
1/xλ
0 lnu u−1u
s−1du dx=2 λ
∞
n=0 1
(n+s)3. (2.13)
Proof. Since
lnu u−1u
s−1= −lnu∞
n=0
then
1/xλ
0 lnu u−1u
s−1du=∞
n=0 1/xλ
0 (−lnu)u
n+s−1du
=
∞
n=0
λ
n+sx
−λ(n+s)lnx+ 1 (n+s)2x−
λ(n+s),
∞
1 1 x
1/xλ
0 lnu u−1u
s−1du dx= ∞
1 ∞
n=0
λ
n+sx−
λ(n+s)−1+ 1 (n+s)2x−
λ(n+s)−1dx
=
∞
n=0
∞
1 λ n+sx
−λ(n+s)−1lnx dx+ ∞
1 1 (n+s)2x−
λ(n+s)−1dx
=2 λ
∞
n=0 1 (n+s)3.
(2.15)
We next give a key lemma in this paper.
Lemma2.7. Ifp >1,1/ p+ 1/q=1,n∈Z+,α >0,λ >0,0< ε < qλ/2p, then
A:=
xα≥1
yα≥1
lnxα/yα
xλ
α− yλα
xα−((n−λ)(p−1)+n+ε)/ pyα−((n−λ)(q−1)+n+ε)/qdx d y
≥ Γn
(1/α) αn−1Γ(n/α)
2 π λsin(π/ p)
2 1 ε
1 +o(1), ε−→0+.
(2.16)
Proof. We have
A=
xα≥1
xλ/qα −n−ε/ pdx×
yα
1+y2α+···+yαn≥1 lnyα
1+y2α+···+yαn
1/α
/xα
yα
1+y2α+···+yαn
λ/α
− xλ α
×y1α+y2α+···+ynα
(1/α)(λ/ p−n−ε/q)
d y1d y2···d yn
=
xα≥1
xλ/qα −n−ε/ pdx Γ n(1/α)
αnΓ(n/α)
∞
1
lnt1/α/x α
tλ/α− xλ α
t(1/α)(λ/ p−n−ε/q)tn/α−1dt.
(2.17)
Setting (t1/α/x
α)λ=u, we have
A=
xα≥1
x−n−ε α dx Γ
n(1/α)
αn−1Γ(n/α) 1 λ2
∞
1/xλ α
lnu u−1u
1/ p−1−ε/λqdu
=
xα≥1
x−n−ε α dx Γ
n(1/α)
αn−1Γ(n/α) 1 λ2
∞
0 lnu u−1u
1/ p−1−ε/λqdu
−
xα≥1
x−n−ε α dx Γ
n(1/α)
αn−1Γ(n/α) 1 λ2
1/xλ α
0
lnu u−1u
1/ p−1−ε/λqdu.
Notice
xα≥1
x−n−ε α dx=
x1,x2,...,xn>0;xα1+x2α+···+xαn≥1 (xα
1+xα2+···+xαn)−(n+ε)/αdx1dx2···dxn
= Γn(1/α) αnΓ(n/α)
∞
1 u
−(n+ε)/αun/α−1du
= Γn(1/α) αnΓ(n/α)
∞
1 u
−ε/α−1du= Γn(1/α) αn−1Γ(n/α)·
1 ε,
1 λ2
∞
0 lnu u−1u
1/ p−1−ε/λqdu=
π λsin(π/ p)
2 +o(1).
(2.19)
Further, from (2.7) andLemma 2.6we have
0≤
xα≥1 x−n−ε
α dx Γ n(1/α)
αn−1Γ(n/α) 1 λ2
1/xλ α
0
lnu u−1u
1/ p−1−ε/λqdu
≤
xα≥1 x−n
α dx Γ n(1/α)
αn−1Γ(n/α) 1 λ2
1/xλ α
0
lnu u−1u
1/2p−1du
= Γn(1/α) αn−1Γ(n/α)
1 λ2
∞
1 t
−n/α1/t
λ/α
0
lnu u−1u
1/2p−1du
tn/α−1dt
= Γn(1/α) αn−1Γ(n/α)
1 λ2
2α λ
∞
n=0 1 (n+ 1/2p)3 =
2Γn(1/α)
αn−2λ3Γ(n/α)
∞
n=0 1 (n+ 1/2p)3.
(2.20)
Then
A≥ Γn(1/α) αn−1Γ(n/α)
2 π
λsin(π/ p) 21
ε
1 +o(1). (2.21)
3. Main results
Our main result is given in the following theorem.
Theorem3.1. Ifp >1,1/ p+ 1/q=1,n∈Z,α >0,λ >0, f,g≥0, satisfy
0<
Rn+x
(n−λ)(p−1)
α fp(x)dx <∞,
0<
Rn +
yα(n−λ)(q−1)gq(y)d y <∞.
Then
J:=
Rn +
lnxα/yα
xλ α− yλα
f(x)g(y)dx d y < Γn(1/α) αn−1Γ(n/α)
π λsin(π/ p)
2
×
Rn +
x(αn−λ)(p−1)fp(x)dx
1/ p
Rn +
y(αn−λ)(q−1)gq(y)d y
1/q
,
(3.2)
Rn+y
λ−n α
Rn+
lnxα/yα
xλ α− yλα
f(x)dx p
d y
<
π
λsin(π/ p)
2 Γn(1/α)
αn−1Γ(n/α) p
Rn +
x(αn−λ)(p−1)fp(x)dx.
(3.3)
The constant factors (Γn(1/α)/αn−1Γ(n/α))[π/λsin(π/ p)]2, [(π/λsin(π/ p))2(Γn(1/α)/
αn−1Γ(n/α))]pare the best possible.
Proof. By H¨older’s inequality, one has
J=
Rn+
lnx
α/yα
xλ α− yλα
1/ px α
yα
(n−λ)/ p+λ/ pq
x(1α/q−1/ p)(n−λ)f(x)
×
lnx
α/yα
xλ α− yλα
1/qy α
xα
(n−λ)/q+λ/qp
y(1α/ p−1/q)(n−λ)g(y)dx d y
≤
Rn +
lnxα/yα
xλ α− yλα
x
α
yα
n−λ+λ/q
x(αp/q−1)(n−λ)fp(x)dx d y
1/ p
×
Rn +
lnxα/yα
xλ α− yλα
y
α
xα
n−λ+λ/ p
y(αq/ p−1)(n−λ)gq(y)dx d y
1/q
=
Rn+
lnxα/yα
xλ α− yλα
x
α
yα
n−λ/ p
x(αp−2)(n−λ)fp(x)dx d y
1/ p
×
Rn +
lnxα/yα
xλ α− yλα
y
α
xα
n−λ/q
y(αq−2)(n−λ)gq(y)dx d y
1/q
=
Rn+wα,λ(x,p)x
(n−λ)(p−2)
α fp(x)dx
1/ p
Rn+wα,λ(y,q)y
(n−λ)(q−2)
α gq(y)d y
1/q
.
(3.4)
all zero, and
C1
lnxα/yα
xλ α− yλα
xα
yα
n−λ/ p
x(αp−2)(n−λ)fp(x)
=C2
lnxα/yα
xλ α− yλα
y
α
xα
n−λ/q
y(αq−2)(n−λ)gq(y), a.e. inRn+×Rn+.
(3.5)
It follows that
C1xn αx
(p−1)(n−λ)
α fp(x)=C2ynαy
(q−1)(n−λ)
α gq(y)
=C(constant), a.e. inRn
+×Rn+,
(3.6)
which contradicts (3.1). Hence we have
J <
Rn+wα,λ(x,p)x
(n−λ)(p−2)
α fp(x)dx
1/ p
Rn+wα,λ(y,q)y
(n−λ)(q−2)
α gq(y)d y
1/q
.
(3.7)
ByLemma 2.5and sinceπ/sin(π/ p)=π/sin(π/q), we have
J < Γn(1/α) αn−1Γ(n/α)
π
λsin(π/ p) 2
Rn+x
(n−λ)(p−1)
α fp(x)dx
1/ p
×
Rn +
y(αn−λ)(q−1)gq(y)d y
1/q
.
(3.8)
Hence (3.2) is valid.
For 0< a < b <∞, let us define
ga,b(y)=
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
yλ−n α
Rn+
lnxα/yα
xλ α− yλα
f(x)dx p−1
, a <yα< b,
0, 0<yα≤a, oryα≥b,
g(y)= yλ−n α
Rn +
lnxα/yα
xλ α− yλα
f(x)dx p−1
, y∈Rn
+.
(3.9)
By (3.1), for sufficiently smalla >0 and sufficiently largeb >0, we have
0<
a<yα<b
Hence by (3.2) we have
a<yα<b
y(αn−λ)(q−1)gq(y)d y
=
a<yα<b yλ−n
α
Rn +
lnxα/yα
xλ α− yλα
f(x)dx p
d y
=
a<yα<b yλ−n
α
Rn +
lnxα/yα
xλ α− yλα
f(x)dx p−1
×
Rn +
lnxα/yα
xλ α− yλα
f(x)dx
d y
=
Rn+
lnxα/yα
xλ α− yλα
f(x)ga,b(y)dx d y
< Γ
n(1/α)
αn−1Γ(n/α)
π
λsin(π/ p) 2
Rn +
x(αn−λ)(p−1)fp(x)dx
1/ p
×
Rn +
yα(n−λ)(q−1)gaq,b(y)d y
1/q
= Γn(1/α) αn−1Γ(n/α)
π
λsin(π/ p) 2
Rn +
x(αn−λ)(p−1)fp(x)dx
1/ p
×
a<yα<b
y(αn−λ)(q−1)gq(y)d y
1/q
.
(3.11)
It follows that
a<yα<b
y(αn−λ)(q−1)gq(y)d y <
π
λsin(π/ p)
2 Γn(1/α)
αn−1Γ(n/α) p
Rn +
x(αn−λ)(p−1)fp(x)dx.
(3.12)
Fora→0+,b→+∞, by (3.1), we have
Rn +
yα(n−λ)(q−1)gq(y)d y≤
π λsin(π/ p)
2 Γn (1/α) αn−1Γ(n/α)
p
Rn +
xα(n−λ)(p−1)fp(x)dx <∞.
Hence by (3.2) we have
Rn +
yλ−n α
Rn +
lnxα/yα
xλ α− yλα
f(x)dx p
d y
=
Rn+
lnxα/yα
xλ α− yλα
f(x)g(y)dx d y
< Γn(1/α) αn−1Γ(n/α)
π
λsin(π/ p) 2
Rn +
x(αn−λ)(p−1)fp(x)dx
1/ p
×
Rn+y
(n−λ)(q−1)
α gq(y)d y
1/q
=αnΓ−n1(1/α)Γ(n/α)
π
λsin(π/ p) 2
Rn +
x(αn−λ)(p−1)fp(x)dx
1/ p
×
Rn +
yλ−n α
Rn +
lnxα/yα
xλ α− yλα
f(x)dx p
d y 1/q
.
(3.14)
It follows that
Rn +
yλ−n α
Rn +
lnxα/yα
xλ α− yλα
f(x)dx p
d y
<
π
λsin(π/ p)
2 Γn(1/α)
αn−1Γ(n/α) p
Rn +
x(αn−λ)(p−1)fp(x)dx,
(3.15)
hence (3.3) is valid.
If the constant factorCn,α(λ,p)=(π/λsin(π/ p))2(Γn(1/α)/αn−1Γ(n/α)) in (3.2) is not
the best possible, then there exists a positive numberk (withk < Cn,α(λ,p)), such that
(3.2) is still valid if one replacesCn,α(λ,p) byk.
For 0< ε < qλ/2p, by sitting
fε(x)=
⎧ ⎪ ⎨ ⎪ ⎩
xα−((n−λ)(p−1)+n+ε)/ p, xα≥1,
0, xα<1,
gε(y)=
⎧ ⎪ ⎨ ⎪ ⎩
yα−((n−λ)(q−1)+n+ε)/q, yα≥1,
0, yα<1
we have
Rn +
lnxα/yα
xλ α− yλα
fε(x)gε(y)dx d y
< k
Rn +
x(αn−λ)(p−1)fεp(x)dx
1/ p
Rn +
y(αn−λ)(q−1)gεq(y)d y
1/q
,
xα≥1
yα≥1
lnxα/yα
xλ α− yλα
fε(x)gε(y)dx d y
< k
xα≥1
x(αn−λ)(p−1)xα−(n−λ)(p−1)−n−εdx
1/ p
×
yα≥1
y(αn−λ)(q−1)yα−(n−λ)(q−1)−n−εd y
1/q
=k
xα≥1
x−n−ε
α dx=k· Γ n(1/α)
αn−1Γ(n/α)· 1 ε.
(3.17)
On the other hand, fromLemma 2.7we have
Rn +
lnxα/yα
xλ α− yλα
fε(x)gε(y)dx d y
≥
Γn(1/α)
αn−1Γ(n/α)
2 π
λsin(π/ p) 21
ε
1 +o(1), ε−→0+.
(3.18)
Hence we have
Γn(1/α)
αn−1Γ(n/α)
2 π
λsin(π/ p) 21
ε
1 +o(1)≤k· Γn(1/α) αn−1Γ(n/α)·
1 ε,
Γn(1/α)
αn−1Γ(n/α)
π
λsin(π/ p) 2
1 +o(1)≤k.
(3.19)
By sittingε→0+we have
Cn,α(λ,p)= Γ n(1/α)
αn−1Γ(n/α)
π
λsin(π/ p) 2
≤k. (3.20)
This contradicts the fact thatk < Cn,α(λ,p), hence the constant factor in (3.2) is the best
possible. Since inequality (3.2) is equivalent to (3.3), the constant factor in (3.3) is also
the best possible. Thus the theorem is proved.
Corollary3.3. Ifp >1,1/ p+ 1/q=1,n∈Z,α >0,f,g≥0, satisfy
0<
Rn +
fp(x)dx <∞, 0<
Rn +
gq(y)d y <∞. (3.21)
Then
Rn +
lnxα/yα
xn α− ynα
f(x)g(y)dx d y
< Γn(1/α) αn−1Γ(n/α)
π
nsin(π/ p) 2
Rn+f
p(x)dx1/ p
Rn+g
q(y)d y1/q,
Rn +
Rn +
lnxα/yα
xn α− ynα
f(x)dx p
d y <
π
nsin(π/ p)
2 Γn(1/α)
αn−1Γ(n/α) p
Rn +
fp(x)dx.
(3.22)
The constant factors (Γn(1/α)/αn−1Γ(n/α))[π/nsin(π/ p)]2, [(π/nsin(π/ p))2(Γn(1/α)/
αn−1Γ(n/α))]pin (3.22) are all the best possible.
Corollary3.4. Ifp >1,1/ p+ 1/q=1,n∈Z,f,g≥0, satisfy
0<
Rn +
fp(x)dx <∞, 0<
Rn +
gq(y)d y <∞. (3.23)
Then
Rn +
lnni=1xi/
n
i=1yi
n
i=1xi
n
−n i=1yi
nf(x)g(y)dx d y
< 1 (n−1)!
π
nsin(π/ p) 2
Rn +
fp(x)dx1/ p
Rn +
gq(y)d y1/q,
Rn +
Rn +
lnni=1xi/
n i=1yi
n
i=1xi
n
−n i=1yi
nf(x)dx
p
d y
<
1
(n−1)!
π
nsin(π/ p) 2p
Rn+f
p(x)dx,
(3.24)
where the constant factors in (3.24) are all the best possible.
References
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[9] B. Yang, “On a generalization of a Hilbert’s type integral inequality and its applications,” Math-ematica Applicata, vol. 16, no. 2, pp. 82–86, 2003.
[10] Y. Hong, “All-sided generalization about Hardy-Hilbert integral inequalities,”Acta Mathematica Sinica. Chinese Series, vol. 44, no. 4, pp. 619–626, 2001.
[11] B. Yang, “A multiple Hardy-Hilbert integral inequality,”Chinese Annals of Mathematics. Series A, vol. 24, no. 6, pp. 743–750, 2003.
[12] H. Yong, “A multiple Hardy-Hilbert integral inequality with the best constant factor,”Journal of Inequalities in Pure and Applied Mathematics, vol. 7, no. 4, article 139, p. 10, 2006.
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Baoju Sun: Zhejiang Water Conservancy and Hydropower College, Zhejiang University, Hangzhou 310018, China
