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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 71049, 14 pages
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
x
α
(x ∈ R) and two parameters α, 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
If p>1, 1/p+1/q
= 1, f ,g ≥ 0, satisfy 0 <

∞
0
f
p
(t)dt < ∞ and 0 <

∞
0

g
q
(t)dt < ∞,then
the well-known Hardy-Hilbert’s integral inequality is given by (see [1, 2])

∞
0
f (x)g(y)
x + y
dxdy <
π
sin(π/p)


∞
0
f
p
(t)dt

1/p


∞
0
g
q
(t)dt

1/q

, (1.1)
where the constant factor π/sin(π/p) is the best possible. Its equivalent form is

∞
0


∞
0
f (x)
x + y
dx

p
dy <

π
sin(π/p)

p

∞
0
f
p
(x) dx, (1.2)
where the constant factor (π/sin(π/p))
p
is still the best possible.
Hardy-Hilbert integral inequality is important in analysis and applications. During

the past few years, many researchers obtained var ious gener alizations, variants, and ex-
tensions of inequality (1.1) (see [3–9] and the references cited therein).
2 Journal of Inequalities and Applications
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)dxdy <

π
sin(π/p)

2


∞
0
f
p
(x) dx

1/p


∞
0
g
q

(x) dx

1/q
, (1.3)
where the constant factor (π/sin(π/p))
2
is the best possible.
Recently, Yang gave a generalization of (1.3) as (see [9])

∞
0
ln(x/y) f (x)g(y)
x
λ
− y
λ
dxdy
<

π
λsin(π/p)

2


∞
0
x
(p−1)(1−λ)
f

p
(x) dx

1/p


∞
0
x
(q−1)(1−λ)
g
q
(x) dx

1/q
,
(1.4)
where the constant factor (π/λsin(π/p))
2
is the best possible. Its equivalent form is

∞
0
y
λ−1


∞
0
ln(x/y) f (x)

x
λ
− y
λ
dx

p
dy <

π
λsin(π/p)

2p

∞
0
x
(p−1)(1−λ)
f
p
(x) dx, (1.5)
where the constant factor (π/λsin(π/p))
2p
is 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
p
i
= 1, p
i
> 1, r
i
=
1
p
i
n

i=1
p
i
, λ>
1
a

n − 1 −
1
r
i

, i = 1,2, ,n, (1.6)
then

∞

α
···

∞
α
1


n
i
=1

x
i
− α

a

λ
n

i=1
f
i

x
i

dx
1

dx
2
···dx
n
≤
Γ
n
(1/α)
α
n−1
Γ(λ)
n

i=1

Γ

1
a

1 −
1
r
i

Γ

λ −
1
a


n −1 −
1
r
i


∞
α
(t − α)
n−1−αλ
f
p
i
i
(t)dt

1/p
i
.
(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 =

x
1
,x
2
, ,x
n

: x
1
,x
2
, ,x
n
> 0

,
x
α
=

x
α
1
+ x
α
2
+ ···+ x

α
n

1/α
, α>0,
(1.8)
and we agree with
x
α
<crepresenting {x ∈ R
+
n
: x
α
<c}.
Baoju Sun 3
2. Lemmas
Firstwegivesomemultipleintegralformulas.
Lemma 2.1 (see [13]). If p
i
> 0, i = 1,2, ,n, f (τ) is a measurable function, then

t
1
,t
2
, ,t
n
>0;t
1

+t
2
+···+t
n
≤1
f

t
1
+ t
2
+ ···+ t
n

t
p
1
−1
1
t
p
2
−1
2
···t
p
n
−1
n
dt

1
dt
2
···dt
n
=
Γ

p
1

Γ

p
2

···
Γ

p
n

Γ

p
1
+ p
2
+ ···+ p
n



1
0
f (τ)τ
p
1
+p
2
+···+p
n
−1
dτ.
(2.1)
Lemma 2.2. If r>0, p
i
> 0, i = 1,2, ,n, f (τ) is a measurable function, then

t
1
,t
2
, ,t
n
>0;t
1
+t
2
+···+t
n

≤r
f

t
1
+ t
2
+ ···+ t
n

t
p
1
−1
1
t
p
2
−1
2
···t
p
n
−1
n
dt
1
dt
2
···dt

n
=
Γ

p
1

Γ

p
2

···
Γ

p
n

Γ

p
1
+ p
2
+ ···+ p
n


r
0

f (τ)τ
p
1
+p
2
+···+p
n
−1
dτ,
(2.2)

t
1
,t
2
, ,t
n
>0
f

t
1
+ t
2
+ ···+ t
n

t
p
1

−1
1
t
p
2
−1
2
···t
p
n
−1
n
dt
1
dt
2
···dt
n
=
Γ

p
1

Γ

p
2

···

Γ

p
n

Γ

p
1
+ p
2
+ ···+ p
n


∞
0
f (τ)τ
p
1
+p
2
+···+p
n
−1
dτ.
(2.3)
Proof. Setting t
i
/r = u

i
(i = 1, 2, ,n) on the left-hand side of (2.2)weobtain(2.2)from
Lemma 2.1.

From (2.1)and(2.3), we have the following lemma.
Lemma 2.3.

t
1
,t
2
, ,t
n
>0;t
1
+t
2
+···+t
n
≥1
f

t
1
+ t
2
+ ···+ t
n

t

p
1
−1
1
t
p
2
−1
2
···t
p
n
−1
n
dt
1
dt
2
···dt
n
=
Γ

p
1

Γ

p
2


···
Γ

p
n

Γ

p
1
+ p
2
+ ···+ p
n


∞
1
f (τ)τ
p
1
+p
2
+···+p
n
−1
dτ.
(2.4)
Setting t

i
= (x
i
/a
i
)
α
i
(i = 1,2, ,n)in(2.1), (2.2), (2.3), (2.4) we have the following
lemma.
4 Journal of Inequalities and Applications
Lemma 2.4. If p
i
> 0, a
i
> 0, α
i
> 0, i = 1,2, ,n, f (τ) is a measurable function, then

x
1
,x
2
, ,x
n
>0;(x
1
/a
1
)

α
1
+(x
2
/a
2
)
α
2
+···+(x
1
/a
n
)
α
n
≤1
f

x
1
a
1

α
1
+

x
2

a
2

α
2
+ ···+

x
1
a
n

α
n

×
x
p
1
−1
1
x
p
2
−1
2
···x
p
n
−1

n
dx
1
dx
2
···dx
n
=
a
p
1
1
a
p
2
2
···a
p
n
n
Γ

p
1
/α
1

Γ

p

2
/α
2

···
Γ

p
n
/α
n

α
1
α
2
···α
n
Γ

p
1
/α
1
+ p
2
/α
2
+ ···+ p
n

/α
n


1
0
f (τ)τ
p
1
/α
1
+p
2
/α
2
+···+p
n
/α
n
−1
dτ,

x
1
,x
2
, ,x
n
>0;(x
1

/a
1
)
α
1
+(x
2
/a
2
)
α
2
+···+(x
1
/a
n
)
α
n
≤r
f

x
1
a
1

α
1
+


x
2
a
2

α
2
+ ···+

x
1
a
n

α
n

×
x
p
1
−1
1
x
p
2
−1
2
···x

p
n
−1
n
dx
1
dx
2
···dx
n
=
a
p
1
1
a
p
2
2
···a
p
n
n
Γ

p
1
/α
1


Γ

p
2
/α
2

···
Γ

p
n
/α
n

α
1
α
2
···α
n
Γ

p
1
/α
1
+ p
2
/α

2
+ ···+ p
n
/α
n


r
0
f (τ)τ
p
1
/α
1
+p
2
/α
2
+···+p
n
/α
n
−1
dτ,

x
1
,x
2
, ,x

n
>0
f

x
1
a
1

α
1
+

x
2
a
2

α
2
+ ···+

x
1
a
n

α
n


x
p
1
−1
1
x
p
2
−1
2
···x
p
n
−1
n
dx
1
dx
2
···dx
n
=
a
p
1
1
a
p
2
2

···a
p
n
n
Γ

p
1
/α
1

Γ

p
2
/α
2

···
Γ

p
n
/α
n

α
1
α
2

···α
n
Γ

p
1
/α
1
+ p
2
/α
2
+ ···+ p
n
/α
n


∞
0
f (τ)τ
p
1
/α
1
+p
2
/α
2
+···+p

n
/α
n
−1
dτ,

x
1
,x
2
, ,x
n
>0;(x
1
/a
1
)
α
1
+(x
2
/a
2
)
α
2
+···+(x
1
/a
n

)
α
n
≥1
f

x
1
a
1

α
1
+

x
2
a
2

α
2
+ ···+

x
1
a
n

α

n

×
x
p
1
−1
1
x
p
2
−1
2
···x
p
n
−1
n
dx
1
dx
2
···dx
n
=
a
p
1
1
a

p
2
2
···a
p
n
n
Γ

p
1
/α
1

Γ

p
2
/α
2

···
Γ

p
n
/α
n

α

1
α
2
···α
n
Γ

p
1
/α
1
+ p
2
/α
2
+ ···+ p
n
/α
n


∞
1
f (τ)τ
p
1
/α
1
+p
2

/α
2
+···+p
n
/α
n
−1
dτ.
(2.5)
In particular , if p>0, α>0, f (τ) is a measurable function, then

x
1
,x
2
, ,x
n
>0;x
α
1
+x
α
2
+···+x
α
n
≤1
f

x

α
1
+ x
α
2
+ ···+ x
α
n

dx
1
dx
2
···dx
n
=
Γ
n
(1/α)
α
n
Γ(n/α)

1
0
f (τ)τ
n/α−1
dτ,
(2.6)


x
1
,x
2
, ,x
n
>0;x
α
1
+x
α
2
+···+x
α
n
≥1
f

x
α
1
+ x
α
2
+ ···+ x
α
n

dx
1

dx
2
···dx
n
=
Γ
n
(1/α)
α
n
Γ(n/α)

∞
1
f (τ)τ
n/α−1
dτ.
(2.7)
The following result holds.
Baoju Sun 5
Lemma 2.5. If p>1, n
∈ Z
+
, α>0, λ>0,definetheweightfunctionw
α,λ
(x, p) as
w
α,λ
(x, p) =


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α

x
α
y
α

n−λ/p
dy. (2.8)
Then
w
α,λ
(x, p) =x
n−λ

α
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2
. (2.9)
Proof. By (2.6)and(2.7), we have
w
α,λ
(x, p) =x
n−λ/p
α

R
n
+
ln


x
α
/y
α


x
λ
α
−y
λ
α
y
λ/p−n
α
dy
=x
n−λ/p
α

y
1
,y
2
, ,y
n
>0
ln

y
α
1
+ y
α
2

+ ···+ y
α
n

1/α
/x
α


y
α
1
+ y
α
2
+ ···+ y
α
n

λ/α
−x
λ
α
×

y
α
1
+ y
α

2
+ ···+ y
α
n

(1/α)(λ/p−n)
dy
1
dy
2
···dy
n
=x
n−λ/p
α
Γ
n
(1/α)
α
n
Γ(n/α)

∞
0
ln

t
1/α
/x
α


t
λ/α
−x
λ
α
t
(1/α)(λ/p−n)
t
n/α−1
dt.
(2.10)
Setting (t
1/α
/x
α
)
λ
= u we have
w
α,λ
(x, p) =x
n−λ
α
Γ
n
(1/α)
α
n−1
Γ(n/α)

1
λ
2

∞
0
lnu
u − 1
u
1/p−1
du. (2.11)
From [1, Theorem 342] we have (1/λ
2
)

∞
0
(lnu/(u − 1))u
1/p−1
du = (π/λsin(π/p))
2
.
So we obtain
w
α,λ
(x, p) =x
n−λ
α
Γ
n

(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2
. (2.12)
Thus Lemma 2.5 is proved.

Lemma 2.6. If λ>0, s>0, then

∞
1
1
x

1/x
λ
0
lnu
u − 1
u
s−1
dudx =
2
λ
∞


n=0
1
(n + s)
3
. (2.13)
Proof. Since
lnu
u − 1
u
s−1
=−lnu
∞

n=0
u
n+s−1
,0<u<1, (2.14)
6 Journal of Inequalities and Applications
then

1/x
λ
0
lnu
u − 1
u
s−1
du =
∞


n=0

1/x
λ
0
(−lnu)u
n+s−1
du
=
∞

n=0

λ
n + s
x
−λ(n+s)
lnx +
1
(n + s)
2
x
−λ(n+s)

,

∞
1
1

x

1/x
λ
0
lnu
u − 1
u
s−1
dudx =

∞
1

∞

n=0

λ
n + s
x
−λ(n+s)−1
+
1
(n + s)
2
x
−λ(n+s)−1



dx
=
∞

n=0


∞
1
λ
n + s
x
−λ(n+s)−1
lnxdx+

∞
1
1
(n + s)
2
x
−λ(n+s)−1
dx

=
2
λ
∞

n=0

1
(n + s)
3
.
(2.15)

We next give a key lemma in this paper.
Lemma 2.7. If p>1, 1/p+1/q
= 1, n ∈ Z
+
, α>0, λ>0, 0 <ε<qλ/2p, then
A :
=

x
α
≥1

y
α
≥1
ln


x
α
/y
α

x

λ
α
−y
λ
α
x
−((n−λ)(p−1)+n+ε)/p
α
y
−((n−λ)(q−1)+n+ε)/q
α
dxdy
≥

Γ
n
(1/α)
α
n−1
Γ(n/α)

2

π
λsin(π/p)

2
1
ε


1+o(1)

, ε −→ 0
+
.
(2.16)
Proof. We have
A
=

x
α
≥1
x
λ/q−n−ε/p
α
dx ×

y
α
1
+y
α
2
+···+y
α
n
≥1
ln


y
α
1
+ y
α
2
+ ···+ y
α
n

1/α
/x
α


y
α
1
+ y
α
2
+ ···+ y
α
n

λ/α
−x
λ
α
×


y
α
1
+ y
α
2
+ ···+ y
α
n

(1/α)(λ/p−n−ε/q)
dy
1
dy
2
···dy
n
=

x
α
≥1
x
λ/q−n−ε/p
α
dx
Γ
n
(1/α)

α
n
Γ(n/α)

∞
1
ln

t
1/α
/x
α

t
λ/α
−x
λ
α
t
(1/α)(λ/p−n−ε/q)
t
n/α−1
dt.
(2.17)
Setting (t
1/α
/x
α
)
λ

= u,wehave
A
=

x
α
≥1
x
−n−ε
α
dx
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2

∞
1/x
λ
α
lnu
u − 1
u
1/p−1−ε/λq
du

=

x
α
≥1
x
−n−ε
α
dx
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2

∞
0
lnu
u − 1
u
1/p−1−ε/λq
du
−

x
α

≥1
x
−n−ε
α
dx
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2

1/x
λ
α
0
lnu
u − 1
u
1/p−1−ε/λq
du.
(2.18)
Baoju Sun 7
Notice

x
α

≥1
x
−n−ε
α
dx =

x
1
,x
2
, ,x
n
>0;x
α
1
+x
α
2
+···+x
α
n
≥1
(x
α
1
+ x
α
2
+ ···+ x
α

n
)
−(n+ε)/α
dx
1
dx
2
···dx
n
=
Γ
n
(1/α)
α
n
Γ(n/α)

∞
1
u
−(n+ε)/α
u
n/α−1
du
=
Γ
n
(1/α)
α
n

Γ(n/α)

∞
1
u
−ε/α−1
du =
Γ
n
(1/α)
α
n−1
Γ(n/α)
·
1
ε
,
1
λ
2

∞
0
lnu
u − 1
u
1/p−1−ε/λq
du =

π

λsin(π/p)

2
+ o(1).
(2.19)
Further , from (2.7)andLemma 2.6 we have
0
≤

x
α
≥1
x
−n−ε
α
dx
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2

1/x
λ
α
0

lnu
u − 1
u
1/p−1−ε/λq
du
≤

x
α
≥1
x
−n
α
dx
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2

1/x
λ
α
0
lnu
u − 1

u
1/2p−1
du
=
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2

∞
1
t
−n/α


1/t
λ/α
0
lnu
u − 1
u
1/2p−1
du

t

n/α−1
dt
=
Γ
n
(1/α)
α
n−1
Γ(n/α)
1
λ
2
2α
λ
∞

n=0
1
(n +1/2p)
3
=
2Γ
n
(1/α)
α
n−2
λ
3
Γ(n/α)
∞


n=0
1
(n +1/2p)
3
.
(2.20)
Then
A
≥

Γ
n
(1/α)
α
n−1
Γ(n/α)

2

π
λsin(π/p)

2
1
ε

1+o(1)

. (2.21)


3. Main results
Our main result is given in the following theorem.
Theorem 3.1. If p>1, 1/p+1/q
= 1, n ∈ Z, α>0, λ>0, f ,g ≥ 0,satisfy
0 <

R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx < ∞,
0 <

R
n
+
y
(n−λ)(q−1)
α
g
q
(y)dy <∞.
(3.1)
8 Journal of Inequalities and Applications
Then

J :
=

R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)g(y)dxdy <
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)


2
×


R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx

1/p


R
n
+
y
(n−λ)(q−1)
α
g
q
(y)dy

1/q
,

(3.2)

R
n
+
y
λ−n
α


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx

p

dy
<

π
λsin(π/p)

2
Γ
n
(1/α)
α
n−1
Γ(n/α)

p

R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx.
(3.3)
The constant factors (Γ
n
(1/α)/α
n−1

Γ(n/α))[π/λsin(π/p)]
2
, [(π/λsin(π/p))
2
(Γ
n
(1/α)/
α
n−1
Γ(n/α))]
p
are the best possible.
Proof. By H
¨
older’s inequalit y, one has
J
=

R
n
+

ln


x
α
/y
α


x
λ
α
−y
λ
α

1/p

x
α
y
α

(n−λ)/p+λ/ pq
x
(1/q−1/p)(n−λ)
α
f (x)
×

ln


x
α
/y
α

x

λ
α
−y
λ
α

1/q


y
α
x
α

(n−λ)/q+λ/qp
y
(1/p−1/q)(n−λ)
α
g(y)dxdy
≤


R
n
+
ln


x
α

/y
α

x
λ
α
−y
λ
α

x
α
y
α

n−λ+λ/q
x
(p/q−1)(n−λ)
α
f
p
(x) dxdy

1/p
×


R
n
+

ln


x
α
/y
α

x
λ
α
−y
λ
α


y
α
x
α

n−λ+λ/p
y
(q/p−1)(n−λ)
α
g
q
(y)dxdy

1/q

=


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α


x
α
y
α

n−λ/p
x
(p−2)(n−λ)

α
f
p
(x) dxdy

1/p
×


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α


y
α

x
α

n−λ/q
y
(q−2)(n−λ)
α
g
q
(y)dxdy

1/q
=


R
n
+
w
α,λ
(x, p)x
(n−λ)(p−2)
α
f
p
(x) dx

1/p



R
n
+
w
α,λ
(y,q)y
(n−λ)(q−2)
α
g
q
(y)dy

1/q
.
(3.4)
According to the condition of taking equality in H
¨
older’s inequality, if this inequality
takes the form of an equality, then there exist constants C
1
and C
2
, such that they are not
Baoju Sun 9
all zero, and
C
1
ln



x
α
/y
α

x
λ
α
−y
λ
α

x
α
y
α

n−λ/p
x
(p−2)(n−λ)
α
f
p
(x)
= C
2
ln


x

α
/y
α

x
λ
α
−y
λ
α


y
α
x
α

n−λ/q
y
(q−2)(n−λ)
α
g
q
(y), a.e. in R
n
+
× R
n
+
.

(3.5)
It follows that
C
1
x
n
α
x
(p−1)(n−λ)
α
f
p
(x) = C
2
y
n
α
y
(q−1)(n−λ)
α
g
q
(y)
= C (constant), a.e. in R
n
+
× R
n
+
,

(3.6)
which contradicts (3.1). Hence we have
J<


R
n
+
w
α,λ
(x, p)x
(n−λ)(p−2)
α
f
p
(x) dx

1/p


R
n
+
w
α,λ
(y,q)y
(n−λ)(q−2)
α
g
q

(y)dy

1/q
.
(3.7)
By Lemma 2.5 and since π/sin(π/p)
= π/sin(π/q), we have
J<
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2


R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx


1/p
×


R
n
+
y
(n−λ)(q−1)
α
g
q
(y)dy

1/q
.
(3.8)
Hence (3.2)isvalid.
For 0 <a<b<
∞, let us define
g
a,b
(y) =
⎧
⎪
⎪
⎨
⎪
⎪

⎩

y
λ−n
α


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx

p−1
, a<y
α
<b,

0, 0 <
y
α
≤ a,ory
α
≥ b,
g(y) =y
λ−n
α


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx


p−1
, y ∈ R
n
+
.
(3.9)
By (3.1), for sufficiently small a>0andsufficiently large b>0, we have
0 <

a<y
α
<b
y
(n−λ)(q−1)
α
g
q
a,b
(y)dy <∞. (3.10)
10 Journal of Inequalities and Applications
Hence by (3.2)wehave

a<y
α
<b
y
(n−λ)(q−1)
α
g
q

(y)dy
=

a<y
α
<b
y
λ−n
α


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx


p
dy
=

a<y
α
<b
y
λ−n
α


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx


p−1
×


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx

dy
=

R
n
+

ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)g
a,b
(y)dxdy
<
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2



R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx

1/p
×


R
n
+
y
(n−λ)(q−1)
α
g
q
a,b
(y)dy

1/q
=
Γ
n

(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2


R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx

1/p
×


a<y
α
<b
y
(n−λ)(q−1)

α
g
q
(y)dy

1/q
.
(3.11)
It follows that

a<y
α
<b
y
(n−λ)(q−1)
α
g
q
(y)dy <

π
λsin(π/p)

2
Γ
n
(1/α)
α
n−1
Γ(n/α)


p

R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx.
(3.12)
For a
→ 0
+
, b → +∞,by(3.1), we have

R
n
+
y
(n−λ)(q−1)
α
g
q
(y)dy ≤

π
λsin(π/p)


2
Γ
n
(1/α)
α
n−1
Γ(n/α)

p

R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx < ∞.
(3.13)
Baoju Sun 11
Hence by (3.2)wehave

R
n
+
y
λ−n
α



R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx

p
dy
=

R
n
+
ln



x
α
/y
α

x
λ
α
−y
λ
α
f (x)g(y)dxdy
<
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2


R
n
+

x
(n−λ)(p−1)
α
f
p
(x) dx

1/p
×


R
n
+
y
(n−λ)(q−1)
α
g
q
(y)dy

1/q
=
Γ
n
(1/α)
α
n−1
Γ(n/α)


π
λsin(π/p)

2


R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx

1/p
×


R
n
+
y
λ−n
α


R
n

+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx

p
dy

1/q
.
(3.14)
It follows that

R
n
+
y
λ−n

α


R
n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f (x)dx

p
dy
<

π
λsin(π/p)

2

Γ
n
(1/α)
α
n−1
Γ(n/α)

p

R
n
+
x
(n−λ)(p−1)
α
f
p
(x) dx,
(3.15)
hence (3.3)isvalid.
If the constant factor C
n,α
(λ, p) = (π/λsin(π/p))
2
(Γ
n
(1/α)/α
n−1
Γ(n/α)) in (3.2)isnot
the best possible, then there exists a positive number k (with k<C

n,α
(λ, p)), such that
(3.2)isstillvalidifonereplacesC
n,α
(λ, p)byk.
For 0 <ε<qλ/2p, by sitting
f
ε
(x) =
⎧
⎪
⎨
⎪
⎩

x
−((n−λ)(p−1)+n+ε)/p
α
, x
α
≥ 1,
0,
x
α
< 1,
g
ε
(y) =
⎧
⎪

⎨
⎪
⎩

y
−((n−λ)(q−1)+n+ε)/q
α
, y
α
≥ 1,
0,
y
α
< 1
(3.16)
12 Journal of Inequalities and Applications
we have

R
n
+
ln


x
α
/y
α

x

λ
α
−y
λ
α
f
ε
(x) g
ε
(y)dxdy
<k


R
n
+
x
(n−λ)(p−1)
α
f
p
ε
(x) dx

1/p


R
n
+

y
(n−λ)(q−1)
α
g
q
ε
(y)dy

1/q
,

x
α
≥1

y
α
≥1
ln


x
α
/y
α

x
λ
α
−y

λ
α
f
ε
(x) g
ε
(y)dxdy
<k


x
α
≥1
x
(n−λ)(p−1)
α
x
−(n−λ)(p−1)−n−ε
α
dx

1/p
×


y
α
≥1
y
(n−λ)(q−1)

α
y
−(n−λ)(q−1)−n−ε
α
dy

1/q
= k

x
α
≥1
x
−n−ε
α
dx = k ·
Γ
n
(1/α)
α
n−1
Γ(n/α)
·
1
ε
.
(3.17)
On the other hand, from Lemma 2.7 we have

R

n
+
ln


x
α
/y
α

x
λ
α
−y
λ
α
f
ε
(x) g
ε
(y)dxdy
≥

Γ
n
(1/α)
α
n−1
Γ(n/α)


2

π
λsin(π/p)

2
1
ε

1+o(1)

, ε −→ 0
+
.
(3.18)
Hence we have

Γ
n
(1/α)
α
n−1
Γ(n/α)

2

π
λsin(π/p)

2

1
ε

1+o(1)

≤
k ·
Γ
n
(1/α)
α
n−1
Γ(n/α)
·
1
ε
,
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2

1+o(1)


≤
k.
(3.19)
By sitting ε
→ 0
+
we have
C
n,α
(λ, p) =
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
λsin(π/p)

2
≤ k. (3.20)
This contradicts the fact that k<C
n,α
(λ, p), hence the constant factor in (3.2)isthebest
possible. Since inequality (3.2)isequivalentto(3.3), the constant factor in (3.3)isalso
the best possible. Thus the theorem is proved.

Remark 3.2. By using (3.3)wecanobtain(3.2), hence inequality (3.2)isequivalentto

(3.3).
Baoju Sun 13
Corollary 3.3. If p>1, 1/p+1/q
= 1, n ∈ Z, α>0, f ,g ≥ 0,satisfy
0 <

R
n
+
f
p
(x) dx < ∞,0<

R
n
+
g
q
(y)dy <∞. (3.21)
Then

R
n
+
ln


x
α
/y

α

x
n
α
−y
n
α
f (x)g(y)dxdy
<
Γ
n
(1/α)
α
n−1
Γ(n/α)

π
nsin(π/p)

2


R
n
+
f
p
(x) dx


1/p


R
n
+
g
q
(y)dy

1/q
,

R
n
+


R
n
+
ln


x
α
/y
α

x

n
α
−y
n
α
f (x)dx

p
dy <

π
nsin(π/p)

2
Γ
n
(1/α)
α
n−1
Γ(n/α)

p

R
n
+
f
p
(x)dx.
(3.22)

The constant factors (Γ
n
(1/α)/α
n−1
Γ(n/α))[π/nsin(π/p)]
2
, [(π/nsin(π/p))
2
(Γ
n
(1/α)/
α
n−1
Γ(n/α))]
p
in (3.22) are all the best possible.
Corollary 3.4. If p>1, 1/p+1/q
= 1, n ∈ Z, f ,g ≥ 0,satisfy
0 <

R
n
+
f
p
(x) dx < ∞,0<

R
n
+

g
q
(y)dy <∞. (3.23)
Then

R
n
+
ln


n
i
=1
x
i
/

n
i
=1
y
i



n
i
=1
x

i

n
−


n
i
=1
y
i

n
f (x)g(y)dxdy
<
1
(n − 1)!

π
nsin(π/p)

2


R
n
+
f
p
(x) dx


1/p


R
n
+
g
q
(y)dy

1/q
,

R
n
+


R
n
+
ln


n
i
=1
x
i

/

n
i
=1
y
i



n
i
=1
x
i

n
−


n
i
=1
y
i

n
f (x)dx

p

dy
<

1
(n − 1)!

π
nsin(π/p)

2

p

R
n
+
f
p
(x)dx,
(3.24)
where the constant factors in (3.24) are all the best possible.
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´
olya, Inequalities, Cambridge University Press, Cam-
bridge, UK, 2nd edition, 1952.
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c, J. E. Pe
ˇ

cari
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c,andA.M.Fink,Inequalities Involving Functions and Their Inte-
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Academic Publishers, Dordrecht, The Netherlands, 1991.
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Baoju Sun: Zhejiang Water Conservancy and Hydropower College, Zhejiang University,
Hangzhou 310018, China
Email address: