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 xα (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 Creative 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)
π
dx d y <
x+y
sin(π/ p)
∞
0
f p (t)dt
1/ p ∞
0
1/q
g q (t)dt
,
(1.1)
where the constant factor π/ sin(π/ p) is the best possible. Its equivalent form is
∞ ∞
0
0
f (x)
dx
x+y
p
π
dy <
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 various generalizations, variants, and extensions 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)
π
f (x)g(y)dx d y <
x−y
sin(π/ p)
2 ∞
0
f p (x)dx
1/ p ∞
0
1/q
g q (x)dx
,
(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)
dx d y
xλ − y λ
<
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
(1.4)
1/q
,
where the constant factor (π/λ sin(π/ p))2 is the best possible. Its equivalent form is
∞
0
y λ −1
∞
0
ln(x/ y) f (x)
dx
xλ − y λ
p
dy <
π
λ sin(π/ p)
2p ∞
0
x(p−1)(1−λ) f p (x)dx,
(1.5)
where the constant factor (π/λ sin(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
n
1
a > 0,
i=1
1
pi ,
p i i =1
n
pi
= 1,
pi > 1,
ri =
λ>
1
1
n−1−
,
a
ri
i = 1,2,...,n, (1.6)
then
∞
α
···
∞
α
n i =1
a λ
xi − α
n
fi xi dx1 dx2 · · · dxn
i =1
Γn (1/α) 1
1
Γ
1−
αn−1 Γ(λ) i=1
a
ri
n
≤
1
Γ λ−
1
1
n −1 −
a
ri
∞
α
p
(t − α)n−1−αλ fi 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 ,
1/α
xα = x1α + x2α + · · · + xnα
, α > 0,
and we agree with xα < c representing {x ∈ R+n : xα < c}.
(1.8)
Baoju Sun 3
2. Lemmas
First we give some multiple integral formulas.
Lemma 2.1 (see [13]). If pi > 0, i = 1,2,...,n, f (τ) is a measurable function, then
t1 ,t2 ,...,tn >0;t1 +t2 +···+tn ≤1
p −1 p −1
f t1 + t2 + · · · + tn t1 1 t2 2
0
dt1 dt2 · · · dtn
(2.1)
1
Γ p1 Γ p2 · · · Γ pn
= Γ p1 + p2 + · · · + pn
p n −1
· · · tn
f (τ)τ
p1 +p2 +···+pn −1
dτ.
Lemma 2.2. If r > 0, pi > 0, i = 1,2,...,n, f (τ) is a measurable function, then
t1 ,t2 ,...,tn >0;t1 +t2 +···+tn ≤r
t1 ,t2 ,...,tn >0
0
f (τ)τ
p −1 p −1
f t1 + t2 + · · · + tn t1 1 t2 2
p1 +p2 +···+pn −1
0
p n −1
· · · tn
dt1 dt2 · · · dtn
dτ,
dt1 dt2 · · · dtn
(2.3)
∞
Γ p1 Γ p2 · · · Γ pn
= Γ p1 + p2 + · · · + pn
p n −1
· · · tn
(2.2)
r
Γ p1 Γ p2 · · · Γ pn
= Γ p1 + p2 + · · · + pn
p −1 p −1
f t1 + t2 + · · · + tn t1 1 t2 2
f (τ)τ
p1 +p2 +···+pn −1
dτ.
Proof. Setting ti /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.
Lemma 2.3.
t1 ,t2 ,...,tn >0;t1 +t2 +···+tn ≥1
p −1 p −1
f t1 + t2 + · · · + tn t1 1 t2 2
1
dt1 dt2 · · · dtn
(2.4)
∞
Γ p1 Γ p2 · · · Γ pn
= Γ p1 + p2 + · · · + pn
p n −1
· · · tn
f (τ)τ
p1 +p2 +···+pn −1
dτ.
Setting ti = (xi /ai )αi (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 pi > 0, ai > 0, αi > 0, i = 1,2,...,n, f (τ) is a measurable function, then
f
x1 ,x2 ,...,xn >0;(x1 /a1 )α1 +(x2 /a2 )α2 +···+(x1 /an )αn ≤1
x1
a1
α1
+
x2
a2
α2
+ ··· +
x1
an
αn p 1 −1 p 2 −1
p −1
x2 · · · xn n dx1 dx2 · · · dxn
×x1
p
p
p
a1 1 a2 2 · · · ann Γ p1 /α1 Γ p2 /α2 · · · Γ pn /αn
=
α1 α2 · · · αn Γ p1 /α1 + p2 /α2 + · · · + pn /αn
f
x1 ,x2 ,...,xn >0;(x1 /a1 )α1 +(x2 /a2 )α2 +···+(x1 /an )αn ≤r
x1
a1
α1
1
f (τ)τ p1 /α1 +p2 /α2 +···+pn /αn −1 dτ,
0
+
x2
a2
α2
+ ··· +
x1
an
αn p 1 −1 p 2 −1
p −1
x2 · · · xn n dx1 dx2 · · · dxn
×x1
p
=
p
p
x1 ,x2 ,...,xn >0
p
=
a1 1 a2 2 · · · ann Γ p1 /α1 Γ p2 /α2 · · · Γ pn /αn
α1 α2 · · · αn Γ p1 /α1 + p2 /α2 + · · · + pn /αn
f
p
x1
a1
α1
+
p
x2
a2
α2
+ ··· +
x1
an
αn p −1 p −1
x1 ,x2 ,...,xn >0;(x1 /a1 )α1 +(x2 /a2 )α2 +···+(x1 /an )αn ≥1
f
f (τ)τ p1 /α1 +p2 /α2 +···+pn /αn −1 dτ,
0
x1 1 x2 2
a1 1 a2 2 · · · ann Γ p1 /α1 Γ p2 /α2 · · · Γ pn /αn
α1 α2 · · · αn Γ p1 /α1 + p2 /α2 + · · · + pn /αn
r
x1
a1
α1
∞
0
x
+ 2
a2
p n −1
· · · xn
dx1 dx2 · · · dxn
f (τ)τ p1 /α1 +p2 /α2 +···+pn /αn −1 dτ,
α2
x
+ ··· + 1
an
αn p 1 −1 p 2 −1
p −1
x2 · · · xn n dx1 dx2 · · · dxn
×x1
p
p
p
a1 1 a2 2 · · · ann Γ p1 /α1 Γ p2 /α2 · · · Γ pn /αn
=
α1 α2 · · · αn Γ p1 /α1 + p2 /α2 + · · · + pn /αn
∞
1
f (τ)τ p1 /α1 +p2 /α2 +···+pn /αn −1 dτ.
(2.5)
In particular, if p > 0, α > 0, f (τ) is a measurable function, then
x1 ,x2 ,...,xn >0;x1α +x2α +···+xnα ≤1
Γn (1/α)
= n
α Γ(n/α)
(2.6)
1
0
f (τ)τ
f x1α + x2α + · · · + xnα dx1 dx2 · · · dxn
n/α−1
dτ,
x1 ,x2 ,...,xn >0;x1α +x2α +···+xnα ≥1
Γn (1/α)
= n
α Γ(n/α)
The following result holds.
∞
1
f x1α + x2α + · · · + xnα dx1 dx2 · · · dxn
f (τ)τ
(2.7)
n/α−1
dτ.
Baoju Sun 5
Lemma 2.5. If p > 1, n ∈ Z+ , α > 0, λ > 0, define the weight function wα,λ (x, p) as
wα,λ (x, p) =
Rn+
ln xα / y α
xλα − y λα
x α
y α
n−λ/ p
d y.
(2.8)
Then
2
n
π
n−λ Γ (1/α)
.
wα,λ (x, p) = xα
αn−1 Γ(n/α) λ sin(π/ p)
(2.9)
Proof. By (2.6) and (2.7), we have
n−λ/ p
wα,λ (x, p) = xα
n−λ/ p
= x α
Rn+
ln xα / y α
λ/ p−n
y α
dy
xλα − y λα
ln
y1 ,y2 ,...,yn >0
y1α + y2α + · · · + ynα
y1α + y2α + · · · + ynα
1/α
λ/α
/ x α
− xλα
(1/α)(λ/ p−n)
× y1α + y2α + · · · + ynα
d y1 d y2 · · · d yn
n−λ/ p
= x α
Γn (1/α)
αn Γ(n/α)
∞
0
ln t 1/α / xα (1/α)(λ/ p−n) n/α−1
t
t
dt.
t λ/α − xλα
(2.10)
Setting (t 1/α / xα )λ = u we have
wα,λ (x, p) = xαn−λ
Γn (1/α) 1
n
α −1 Γ(n/α) λ2
From [1, Theorem 342] we have (1/λ2 )
So we obtain
∞
0
wα,λ (x, p) = xαn−λ
∞
0
lnu 1/ p−1
du.
u
u−1
(2.11)
(lnu/(u − 1))u1/ p−1 du = (π/λsin(π/ p))2 .
Γn (1/α)
π
αn−1 Γ(n/α) λ sin(π/ p)
2
.
(2.12)
Thus Lemma 2.5 is proved.
Lemma 2.6. If λ > 0, s > 0, then
∞
1
1
x
1/xλ
0
∞
ln u s−1
2
1
.
u dudx =
u−1
λ n=0 (n + s)3
(2.13)
Proof. Since
∞
ln u s−1
u = − ln u un+s−1 ,
u−1
n =0
0 < u < 1,
(2.14)
6
Journal of Inequalities and Applications
then
1/xλ
∞
ln u s−1
u du =
u−1
n =0
0
1/xλ
∞ λ
=
n =0
∞
1
1
x
1/xλ
0
lnu s−1
u dudx =
u−1
(− lnu)un+s−1 du
0
n+s
x
∞ ∞ 1
n =0
∞ ∞
=
n =0
1
1
lnx +
x−λ(n+s) ,
(n + s)2
−λ(n+s)
1
λ −λ(n+s)−1
+
x−λ(n+s)−1
x
n+s
(n + s)2
λ −λ(n+s)−1
lnx dx +
x
n+s
∞
1
dx
1
x−λ(n+s)−1 dx
(n + s)2
∞
2
1
.
λ n=0 (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:=
xα ≥1
≥
ln xα / y α
−((n−λ)(p−1)+n+ε)/ p
−((n−λ)(q−1)+n+ε)/q
x α
y α
dx d y
λ
λ
y α ≥1 x α − y α
Γn (1/α)
n
α −1 Γ(n/α)
2 π
λ sin(π/ p)
2
1
1 + o(1) ,
ε
ε −→ 0+ .
(2.16)
Proof. We have
A=
λ/q−n−ε/ p
xα ≥1
x α
dx ×
ln
y1α +y2α +···+ynα ≥1
y1α + y2α + · · · + ynα
y1α + y2α + · · · + ynα
1/α
λ/α
/ x α
− xλα
(1/α)(λ/ p−n−ε/q)
× y1α + y2α + · · · + ynα
d y1 d y2 · · · d yn
Γn (1/α) ∞ ln t 1/α / xα (1/α)(λ/ p−n−ε/q) n/α−1
λ/q−n−ε/ p
=
x α
dx n
t
t
dt.
α Γ(n/α) 1 t λ/α − xλα
xα ≥1
(2.17)
Setting (t 1/α / xα )λ = u, we have
A=
=
xα ≥1
xα ≥1
xα−n−ε dx
Γn (1/α) 1
n
α −1 Γ(n/α) λ2
xα−n−ε dx
Γn (1/α) 1
αn−1 Γ(n/α) λ2
−
xα ≥1
xα−n−ε dx
∞
1/ xλα
∞
0
Γn (1/α) 1
n
α −1 Γ(n/α) λ2
lnu 1/ p−1−ε/λq
du
u
u−1
lnu 1/ p−1−ε/λq
u
du
u−1
1/xλα
0
lnu 1/ p−1−ε/λq
du.
u
u−1
(2.18)
Baoju Sun 7
Notice
−n −ε
xα ≥1
x α
dx =
x1 ,x2 ,...,xn >0;x1α +x2α +···+xnα ≥1
=
Γn (1/α)
αn Γ(n/α)
=
Γn (1/α)
αn Γ(n/α)
1
λ2
∞
0
∞
1
∞
1
(x1α + x2α + · · · + xnα )−(n+ε)/α dx1 dx2 · · · dxn
u−(n+ε)/α un/α−1 du
u−ε/α−1 du =
Γn (1/α)
αn−1 Γ(n/α)
lnu 1/ p−1−ε/λq
π
du =
u
u−1
λsin(π/ p)
1
ε
· ,
2
+ o(1).
(2.19)
Further, from (2.7) and Lemma 2.6 we have
0≤
xα ≥1
≤
xα ≥1
xα−n−ε dx
n
x −
α dx
Γn (1/α) 1
= n −1
α Γ(n/α) λ2
Γn (1/α) 1
n
α −1 Γ(n/α) λ2
Γn (1/α) 1
n
α −1 Γ(n/α) λ2
∞
1
t −n/α
1/tλ/α
0
1/xλα
0
1/xλα
0
lnu 1/ p−1−ε/λq
du
u
u−1
lnu 1/2p−1
du
u
u−1
∞
=
(2.20)
lnu 1/2p−1
du t n/α−1 dt
u
u−1
∞
Γn (1/α) 1 2α 1
2Γn (1/α) 1
=
.
αn−1 Γ(n/α) λ2 λ n=0 (n + 1/2p)3 αn−2 λ3 Γ(n/α) n=0 (n + 1/2p)3
Then
Γn (1/α)
A ≥ n −1
α Γ(n/α)
2 π
λ sin(π/ p)
2
1
1 + o(1) .
ε
(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<
(n−λ)(p−1)
Rn+
0<
Rn+
x α
f p (x)dx < ∞,
(3.1)
(n−λ)(q−1) q
y α
g (y)d y
< ∞.
8
Journal of Inequalities and Applications
Then
J :=
Rn+
×
Rn+
Rn+
(n−λ)(p−1) p
x α
f (x)dx
y αλ−n
<
ln xα / y α
Γn (1/α)
π
f
(x)g(y)dx
d
y
<
xλα − y λα
αn−1 Γ(n/α) λ sin(π/ p)
1/ p Rn+
ln xα / y α
f (x)dx
xλα − y λα
Rn+
π
λ sin(π/ p)
2
Γn (1/α)
n
α −1 Γ(n/α)
(n−λ)(q−1) q
y α
g (y)d y
2
1/q
(3.2)
,
p
dy
(3.3)
p (n−λ)(p−1)
Rn+
x α
f p (x)dx.
The constant factors (Γn (1/α)/αn−1 Γ(n/α))[π/λsin(π/ p)]2 , [(π/λ sin(π/ p))2 (Γn (1/α)/
αn−1 Γ(n/α))] p are the best possible.
Proof. By Hölder’s inequality, one has
J=
Rn+
ln xα / y α
xλα − y λα
×
ln xα / y α
xλα − y λα
ln xα / y α
xλα − y λα
Rn+
Rn+
ln xα / y α
xλα − y λα
Rn+
×
Rn+
=
ln xα / y α
xλα − y λα
Rn+
×
=
1/q ≤
1/ p ln xα / y α
xλα − y λα
y α
x α
x α
y α
x α
y α
(1/ p−1/q)(n−λ)
y α
n−λ+λ/q
(p/q−1)(n−λ)
xα
n−λ+λ/ p
g(y)dx d y
1/ p
f p (x)dx d y
1/q
g (y)dx d y
(p−2)(n−λ) p
xα
f (x)dx d y
n−λ/q
(n−λ)(p−2) p
wα,λ (x, p)xα
f (x)dx
f (x)
(q/ p−1)(n−λ) q
y α
n−λ/ p
y α
x α
(1/q−1/ p)(n−λ)
x α
(n−λ)/q+λ/qp
y α
x α
x α
y α
(n−λ)/ p+λ/ pq
1/ p
(q−2)(n−λ) q
y α
g (y)dx d y
1/ p 1/q
(n−λ)(q−2) q
wα,λ (y, q) y α
n
R+
g (y)d y
1/q
.
(3.4)
According to the condition of taking equality in Hölder’s inequality, if this inequality
takes the form of an equality, then there exist constants C1 and C2 , such that they are not
Baoju Sun 9
all zero, and
ln xα / y α
C1
xλα − y λα
x α
y α
ln xα / y α
= C2
xλα − y λα
n−λ/ p
(p−2)(n−λ)
x α
y α
x α
n−λ/q
f p (x)
(3.5)
(q−2)(n−λ) q
y α
g (y),
a.e. in Rn+ × Rn+ .
It follows that
(p−1)(n−λ)
C1 xnα xα
(q−1)(n−λ) q
f p (x) = C2 y nα y α
g (y)
(3.6)
a.e. in Rn+ × Rn+ ,
= C (constant),
which contradicts (3.1). Hence we have
J<
(n−λ)(p−2)
wα,λ (x, p)xα
n
R+
1/ p f p (x)dx
(n−λ)(q−2) q
wα,λ (y, q) y α
n
1/q
g (y)d y
R+
.
(3.7)
By Lemma 2.5 and since π/sin(π/ p) = π/sin(π/q), we have
J<
Γn (1/α)
π
n
α −1 Γ(n/α) λ sin(π/ p)
×
Rn+
2 (n−λ)(p−1)
Rn+
(n−λ)(q−1) q
y α
g (y)d y
x α
1/ p
f p (x)dx
(3.8)
1/q
.
Hence (3.2) is valid.
For 0 < a < b < ∞, let us define
⎧
⎪
⎪
⎨ y αλ−n
ga,b (y) = ⎪
⎪
⎩0,
g(y) = y αλ−n
Rn+
ln xα / y α
f (x)dx
xλα − y λα
p −1
,
a < y α < b,
0 < y α ≤ a, or y α ≥ b,
Rn+
ln xα / y α
f (x)dx
xλα − y λα
p −1
,
y ∈ Rn+ .
(3.9)
By (3.1), for sufficiently small a > 0 and sufficiently large b > 0, we have
0<
a< y α <b
(n−λ)(q−1) q
ga,b (y)d y
y α
< ∞.
(3.10)
10
Journal of Inequalities and Applications
Hence by (3.2) we have
(n−λ)(q−1) q
y α
a< y α <b
g (y)d y
=
a< y α <b
=
a< y α <b
y αλ−n
y αλ−n
=
Rn+
Rn+
Rn+
ln xα / y α
f (x)dx
xλα − y λα
Rn+
Rn+
p −1
2 Rn+
(n−λ)(q−1) q
ga,b (y)d y
(3.11)
1/ p
(n−λ)(p−1)
f p (x)dx
(n−λ)(p−1)
f p (x)dx
x α
1/q
y α
Γn (1/α)
π
n
α −1 Γ(n/α) λ sin(π/ p)
×
=
dy
ln xα / y α
f (x)ga,b (y)dx d y
xλα − y λα
Γn (1/α)
π
αn−1 Γ(n/α) λ sin(π/ p)
×
p
ln xα / y α
f (x)dx d y
xλα − y λα
<
ln xα / y α
f (x)dx
xλα − y λα
×
2 Rn+
x α
1/q
(n−λ)(q−1) q
a< y α <b
y α
1/ p
g (y)d y
.
It follows that
(n−λ)(q−1) q
a< y α <b
y α
g (y)d y <
π
λ sin(π/ p)
2
Γn (1/α)
n
α −1 Γ(n/α)
p (n−λ)(p−1)
Rn+
x α
f p (x)dx.
(3.12)
For a → 0+ , b → +∞, by (3.1), we have
(n−λ)(q−1) q
Rn+
y α
g (y)d y ≤
π
λ sin(π/ p)
2
Γn (1/α)
αn−1 Γ(n/α)
p Rn+
(n−λ)(p−1)
x α
f p (x)dx < ∞.
(3.13)
Baoju Sun 11
Hence by (3.2) we have
Rn+
y αλ−n
Rn+
ln xα / y α
f (x)dx
xλα − y λα
=
Rn+
Γn (1/α)
π
n
−
1
α Γ(n/α) λ sin(π/ p)
×
Rn+
2 Γn (1/α)
π
n
α −1 Γ(n/α) λ sin(π/ p)
×
Rn+
y αλ−n
(n−λ)(p−1)
Rn+
1/ p
f p (x)dx
(3.14)
1/q
2 (n−λ)(p−1)
Rn+
x α
Rn+
(n−λ)(q−1) q
y α
g (y)d y
=
dy
ln xα / y α
f (x)g(y)dx d y
xλα − y λα
<
p
x α
ln xα / y α
f (x)dx
xλα − y λα
1/ p
f p (x)dx
p
1/q
dy
.
It follows that
Rn+
y αλ−n
<
Rn+
ln xα / y α
f (x)dx
xλα − y λα
π
λ sin(π/ p)
2
Γn (1/α)
n
α −1 Γ(n/α)
p
dy
(3.15)
p Rn+
(n−λ)(p−1) p
x α
f (x)dx,
hence (3.3) is valid.
If the constant factor Cn,α (λ, p) = (π/λ sin(π/ p))2 (Γn (1/α)/αn−1 Γ(n/α)) in (3.2) is not
the best possible, then there exists a positive number k (with k < Cn,α (λ, p)), such that
(3.2) is still valid if one replaces Cn,α (λ, p) by k.
For 0 < ε < qλ/2p, by sitting
⎧
⎪
⎨xα−((n−λ)(p−1)+n+ε)/ p ,
fε (x) = ⎪
⎩0,
⎧
((n−λ)(q−1)+n+ε)/q
⎪
⎨ y −
,
α
gε (y) = ⎪
⎩0,
xα ≥ 1,
xα < 1,
(3.16)
y α ≥ 1,
y α < 1
12
Journal of Inequalities and Applications
we have
Rn+
ln xα / y α
fε (x)gε (y)dx d y
xλα − y λα
<k
xα ≥1
Rn+
(n−λ)(p−1) p
x α
fε (x)dx
1/ p Rn+
(n−λ)(q−1) q
gε (y)d y
1/q
y α
,
ln xα / y α
fε (x)gε (y)dx d y
λ
λ
y α ≥1 x α − y α
<k
(n−λ)(p−1)
xα ≥1
x α
×
(n−λ)(q−1)
y α ≥1
=k
xα ≥1
y α
−(n−λ)(p−1)−n−ε
x α
dx
−(n−λ)(q−1)−n−ε
y α
xα−n−ε dx = k ·
(3.17)
1/ p
1/q
dy
Γn (1/α)
1
· .
αn−1 Γ(n/α) ε
On the other hand, from Lemma 2.7 we have
Rn+
ln xα / y α
fε (x)gε (y)dx d y
xλα − y λα
Γn (1/α)
≥
n
α −1 Γ(n/α)
2 π
λ sin(π/ p)
2
1
1 + o(1) ,
ε
(3.18)
ε −→ 0+ .
Hence we have
Γn (1/α)
αn−1 Γ(n/α)
2 π
λ sin(π/ p)
Γn (1/α)
2
1
Γn (1/α)
1
1 + o(1) ≤ k · n−1
· ,
ε
α Γ(n/α) ε
π
αn−1 Γ(n/α) λ sin(π/ p)
2
(3.19)
1 + o(1) ≤ k.
By sitting ε → 0+ we have
Cn,α (λ, p) =
Γn (1/α)
π
αn−1 Γ(n/α) λ sin(π/ p)
2
≤ k.
(3.20)
This contradicts the fact that k < 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.
Remark 3.2. By using (3.3) we can obtain (3.2), hence inequality (3.2) is equivalent to
(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<
Rn+
f p (x)dx < ∞,
0<
Rn+
g q (y)d y < ∞.
(3.21)
Then
Rn+
ln xα / y α
f (x)g(y)dx d y
xnα − y nα
<
Γn (1/α)
π
n
−
1
α Γ(n/α) nsin(π/ p)
Rn+
Rn+
2 ln xα / y α
f (x)dx
xnα − y nα
1/ p f p (x)dx
Rn+
p
dy <
1/q
g q (y)d y
n
,
R+
π
nsin(π/ p)
2
Γn (1/α)
n
α −1 Γ(n/α)
p Rn+
f p (x)dx.
(3.22)
The constant factors (Γn (1/α)/αn−1 Γ(n/α))[π/nsin(π/ p)]2 , [(π/nsin(π/ p))2 (Γn (1/α)/
αn−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<
Rn+
f p (x)dx < ∞,
0<
Rn+
g q (y)d y < ∞.
(3.23)
Then
n
i =1 x i / i =1 y i
n
n n
n
−
Rn+
i=1 xi
i=1 yi
ln
<
π
(n − 1)! nsin(π/ p)
n
f (x)g(y)dx d y
2 Rn+
1/ p f p (x)dx
xi / n y i
n i=1n i=n1 n f (x)dx
−
i =1 x i
i =1 y i
ln
Rn+
<
1
Rn+
n
π
(n − 1)! nsin(π/ p)
1
1/q
g q (y)d y
n
R+
,
(3.24)
p
dy
2 p Rn+
f p (x)dx,
where the constant factors in (3.24) are all the best possible.
References
[1] G. H. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, Cambridge University Press, Cambridge, UK, 2nd edition, 1952.
[2] D. S. Mitrinović, J. E. Pečarić, and A. M. Fink, Inequalities Involving Functions and Their Integrals and Derivatives, vol. 53 of Mathematics and Its Applications (East European Series), Kluwer
Academic Publishers, Dordrecht, The Netherlands, 1991.
14
Journal of Inequalities and Applications
[3] M. Gao and B. Yang, “On the extended Hilbert’s inequality,” Proceedings of the American Mathematical Society, vol. 126, no. 3, pp. 751–759, 1998.
[4] K. Jichang and L. Debnath, “On new generalizations of Hilbert’s inequality and their applications,” Journal of Mathematical Analysis and Applications, vol. 245, no. 1, pp. 248–265, 2000.
[5] K. Jichang, Applied Inequalities, Shandong Science and Technology Press, Jinan, China, 2004.
[6] B. G. Pachpatte, “On some new inequalities similar to Hilbert’s inequality,” Journal of Mathematical Analysis and Applications, vol. 226, no. 1, pp. 166–179, 1998.
[7] B. Yang and L. Debnath, “On new strengthened Hardy-Hilbert’s inequality,” International Journal of Mathematics and Mathematical Sciences, vol. 21, no. 2, pp. 403–408, 1998.
[8] B. Yang, “On a general Hardy-Hilbert’s integral inequality with a best constant,” Chinese Annals
of Mathematics. Series A, vol. 21, no. 4, pp. 401–408, 2000.
[9] B. Yang, “On a generalization of a Hilbert’s type integral inequality and its applications,” Mathematica 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.
[13] L. Hua, An Introduction to Advanced Mathematics (Remaining Sections), Science Publishers, Beijing, China, 1984.
Baoju Sun: Zhejiang Water Conservancy and Hydropower College, Zhejiang University,
Hangzhou 310018, China
Email address: sunbj@mail.zjwchc.com