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
R∞
R∞
If p > 1, p1 + 1q = 1, f ≥ 0, g ≥ 0, 0 < 0 f p (x)dx < +∞, 0 < 0 g q (x)dx < +∞,
then the well known Hardy-Hilbert’s integral inequality is given by (see [1]):
Z
0
∞Z ∞
0
f (x)g(x)
π
dxdy <
x+y
sin( πp )
where the constant factor
Z
0
π
sin( π
)
p
∞ Z ∞
0
where the constant factor
Z
∞
p
1 Z
p
f (x)dx
0
∞
1
q
g (x)dx
,
q
(1)
0
is the best possible. it’s equivalent form is:
"
#p Z
p
∞
π
f (x)
f p (x)dx,
dx dy <
x+y
sin( πp )
0
π
sin( π
)
p
(2)
p
is also the best possible.
Hardy-Hilbert’t inequality is valuable in harmonic analysis, real analysis and operator theory. In recent years, many valuable results (see [2-5]) have been obtained in
1
generalization and improvement of Hardy-Hilbert’s inequality. In 1999,Kuang [6] gave
a generalization with a parameter λ of (1) as follows:
Z ∞
1 Z ∞
1
Z ∞Z ∞
p
q
f (x)g(y)
1−λ p
1−λ q
dxdy
<
h
(p)
x
f
(x)dx
x
g
(x)dx
,
(3)
λ
xλ + y λ
0
0
0
0
1
1
π
π −1
where max{ p1 , 1q } < λ ≤ 1, hλ (p) = π[λ sin p ( pλ
) sin q ( qλ
)] . 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 ∞Z ∞
f (x)g(y)
π
dxdy <
π
λ + yλ
λ
sin(
x
0
0
p)
1 Z ∞
Z ∞
1
p
q
(1−λ)(q−1) q
(1−λ)(p−1) p
x
f (x)dx
×
x
g (x)dx
, (4)
0
0
it’s equivalent form is:
"
#p Z
Z ∞
p
Z ∞
∞
f
(x)
π
y λ−1
dx
dy
<
x(1−λ)(p−1) f p (x)dx,
π
λ + yλ
λ
sin(
)
x
0
0
0
p
where the constant factors
π
λ sin( π
)
p
(5)
π
p
in (4) and [ λ sin(
π ] in (5) are all the best possible.
)
p
At present, because of the requirement of higher-dimensional harmonic analysis and
higher-dimensional operator theory, multiple
P Hardy-Hilbert’s integral inequality
Q is researched. Hong [8] obtained: If a > 0, ni=1 p1i = 1, pi > 1, fi ≥ 0, ri = p1i nj=1 pj ,
λ > a1 (n − 1 − r1i ), i = 1, 2, · · · , n, then
Z ∞
Z ∞
n
Y
Γn−2 ( α1 )
1
···
f
(x)dx
·
·
·
dx
≤
Pn
i
1
n
λ
αn−1 Γ(λ)
α
α [ i=1 (xi − αi )a ] i=1
Z ∞
1
n Y
pi
1
1
1
1
n−1−αλ pi
(t − α)
fi dt
×
Γ
(1 − ) Γ λ − (n − 1 − )
. (6)
a
ri
a
ri
α
i=1
Afterwards, Bicheng Yang and Kuang Jichang etc. obtained some multiple HardyHilbert’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:
n
R+
= {x = (x1 , · · · , xn ) : x1 , . . . , xn > 0},
1
kxkα = (xα1 + · · · + xαn ) α , α > 0.
2
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 α1
xn αn
Ψ ( ) + ··· + ( )
...
x
a1
an
x1 ,···,xn >0;( 1 )α1 +···+( xn )αn ≤1
a1
an
×xp11 −1 · · · xpnn −1 dx1 · · · dxn
=
ap11
· · · apnn Γ( αp11 ) · · · Γ( αpnn )
α1 · · · αn Γ( αp11 + · · · + αpnn )
1
Z
p1
Ψ(u)u α1
pn
+···+ α
−1
n
du.
(7)
0
where Γ(·) is the Γ-function.
Lemma 2.2 If p > 1,
ωα,λ (x, p, q) as:
1
p
+
1
q
= 1, n ∈ Z+ , α > 0, λ > 0, setting the weight function
1
q

Z
ωα,λ (x, p, q) =
n
R+
1
 kxkα1 
λ
λ
kxkα + kykα
kykαp
then
ωα,λ (x, p, q) = kxk(n−λ)(p−1)
α
Proof
(n−λ)p
kxkα
kykα
λ
q
dy,
πΓn ( α1 )
.
sin( πp )λαn−1 Γ( αn )
By lemma 2.1, we have
ωα,λ (x, p, q) =
(n−λ)(p−1)+ λ
q
kxkα
(n−λ)(p−1)+ λ
q
= kxkα
Z
n
R+
Z
lim
R→+∞
−(n−λ+ λ
)
1
q
kyk
dy
α
kxkλα + kykλα
Z
···
α <r α
y1 ,···,yn >0;y1α +···+yn
λ
i
1 −(n−λ+ q )
r(( yr1 )α + · · · + ( yrn )α ) α
1−1
1−1
×
h
i y1 · · · yn dy1 · · · dyn
1 λ
y
y
n
1
kxkλα + r(( r )α + · · · + ( r )α ) α
h
=
(n−λ)(p−1)+ λ
q
kxkα
rn Γn ( α1 )
lim
n
R→+∞ αn Γ( α )
Z
1
0
1
)
−(n−λ+ λ
q
(ru α )
1
α
kxkλα + (ru )λ
n
u α −1 du
r
Γn ( α1 )
1
λ− λ −1
lim
u q du
n
n−1
λ
λ
α
Γ( α ) R→+∞ 0 kxkα + u
Z ∞
n
λ
(n−λ)(p−1)+ q
Γ ( α1 )
1
λ− λ −1
u q du
= kxkα
n
n−1
λ
λ
α
Γ( α ) 0 kxkα + u
Z ∞
1
n
1
Γ (α)
1
−1
(n−λ)(p−1)
u p du
= kxkα
n
n−1
λα
Γ( α ) 0 1 + u
(n−λ)(p−1)+ λ
q
Z
= kxkα
3
(8)
= kxk(n−λ)(p−1)
α
Γn ( α1 )
1
1
n Γ( )Γ(1 − )
n−1
λα
Γ( α ) p
p
= kxk(n−λ)(p−1)
α
πΓn ( α1 )
,
sin( πp )λαn−1 Γ( αn )
hence (8) is valid.
Lemma 2.3 If p > 1,
ω̃α,λ (x, q, ε) as:
1
p
1
q
+
= 1, n ∈ Z+ , α > 0, λ > 0, 0 < ε < λ(q − 1), setting
Z
ω̃α,λ (x, q, ε) =
n
R+
(n−λ)(q−1)+n+ε
−
1
q
kyk
dy,
α
kxkλα + kykλα
then we have
ω̃α,λ (x, q, ε) =
Proof
3
−λ−ε
kxkα q q
Γn ( α1 )
1
ε
1
ε
Γ( −
)Γ( +
).
λαn−1 Γ( αn ) p λq
q λq
(9)
By the same way of lemma 2.2, lemma 2.3 can be proved.
Main Results
Theorem 3.1 If p > 1,
Z
0<
n
R+
1
p
+
1
q
= 1, n ∈ Z+ , α > 0, λ > 0, f ≥ 0, g ≥ 0, and
kxk(n−λ)(p−1)
f p (x)dx
α
Z
< ∞,
0<
n
R+
kxk(n−λ)(q−1)
g q (x)dx < ∞,
α
(10)
then
Z
πΓn ( α1 )
f (x)g(y)
dxdy
<
n kxkλ + kykλ
sin( πp )λαn−1 Γ( αn )
R+
α
α
!1 Z
Z
Z
n
R+
!1
p
×
n
R+
Z
n
R+
kxk(n−λ)(p−1)
f p (x)dx
α
kykλ−n
α
Z
n
R+
n
R+
q
kxk(n−λ)(q−1)
g q (x)dx
α
!p
"
#p
πΓn ( α1 )
f (x)
dx dy <
sin( πp )λαn−1 Γ( αn )
kxkλα + kykλα
Z
×
kxk(n−λ)(p−1)
f p (x)dx,
α
;
(11)
(12)
n
R+
where the constant factors
1
πΓn ( α
)
n−1 Γ( n )
sin( π
)λα
p
α
πΓn ( 1 )
α
and [ sin( π )λαn−1
]p are all the best possible.
Γ( n )
p
4
α
Proof
By Hölder’s inequality, we have
Z Z
f (x)g(y)
A :=
dxdy
n
n kxkλ + kykλ
R+
R+
α
α
n−λ

1
λ
Z Z
q
kxkα pq
kxkα 
f (x)

=
1
1
n
n
kykα
R+
R+
(kxkλα + kykλα ) p kykαp
n−λ

1
λ
p
pq
kyk
g(y)
kyk
α
α


dxdy
×
1
1
kxkα
(kxkλα + kykλα ) q kxkαq

1

(n−λ)p
p
1
λ
Z Z
q
q
f p (x)
kxk
kxk


α
α


≤ 
dxdy 
1
n
n kxkλ + kykλ
kykα
p
R+
R+
α
α
kykα
1

(n−λ)q

q
1
λ
Z Z
p
q
p
kyk
kyk
g
(x)


α
α

dxdy 
×
1
λ + kykλ
n
n
kxk
kxk
q
α
R+ R+
α
α
kxkα


 1

(n−λ)p
p
1
λ
Z
Z
q
q
1
kxk
kxk




α
α
p


= 
f (x) 
dy  dx
1
n
n kxkλ + kykλ
kykα
p
R+
R+
α
α
kykα
 1


(n−λ)q

q
1
λ
Z
Z
p
p
1
kyk
kyk
 


α
α

dx dy 
×
g q (y) 
1
n
n kxkλ + kykλ
kxk
q
α
R+
R+
α
α
kxkα
1
!
!1
Z
Z
p
=
n
R+
f p (x)ωα,λ (x, p, q)dx
q
n
R+
g q (y)ωα,λ (y, q, p)dy
,
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 all zero, and
=

1
q
(n−λ)p
λ
C1
kxkα q
 kxkα1 
kykα
kxkλα + kykλα
kykαp

(n−λ)q
1
λ
C2 g q (y)  kykαp 
kykα p
,
1
kxkα
kxkλα + kykλα
q
kxkα
f p (x)
n
n
a.e. in R+
× R+
,
it follows that
C1 kxk(n−λ)(p−1)+n
f p (x) = C2 kyk(n−λ)(q−1)+n
g q (y) = C(constant),
α
α
5
n
n
a.e. in R+
× R+
,
which contradicts (10), hence we have
Z
A<
n
R+
!1
p
f p (x)ωα,λ (x, p, q)dx
!1
q
Z
g q (y)ωα,λ (y, q, p)dy
n
R+
.
By lemma 2.2 and sin( πp ) = sin( πq ), we have
#1 Z
!1
p
p
πΓn ( α1 )
(n−λ)(p−1) p
kxk
f
(x)dx
α
n
sin( πp )λαn−1 Γ( αn )
R+
"
#1 Z
!1
q
q
πΓn ( α1 )
(n−λ)(q−1) q
×
kyk
g
(y)dy
α
n
sin( πq )λαn−1 Γ( αn )
R+
!1
!1 Z
Z
p
q
πΓn ( α1 )
(n−λ)(p−1) p
(n−λ)(q−1) q
kxk
f
(x)dx
kxk
g
(x)dx
.
α
α
n
n
sin( πp )λαn−1 Γ( αn )
R+
R+
"
A <
=
Hence (11) is valid.
For 0 < a < b < ∞, setting
(
R
kykλ−n
n
α
R+
ga,b (y) =
f (x)
λ
kxkλ
α +kykα
p−1
dx
,
a < kykα < b
0 < kykα ≤ a
!p−1
0,
g̃(y) = kykλ−n
α
Z
n
R+
f (x)
dx
+ kykλα
or
kykα ≥ b
n
, y ∈ R+
,
kxkλα
by (10), for sufficiently small a > 0 and sufficiently large b > 0, we have
Z
q
0<
kyk(n−λ)(q−1)
ga,b
(y)dy < ∞.
α
a<kykα <b
Hence by (11), we have
Z
Z
kykα(n−λ)(q−1) g̃ q (y)dy =
a<kykα <b
Z
a<kykα <b
kykλ−n
α
=
a<kykα <b
Z
Z
=
n
R+
<
=
n
R+
kykλ−n
α
Z
n
R+
f (x)
dx
λ
kxkα + kykλα
!p−1
Z
n
R+
!p
f (x)
dx dy
n kxkλ + kykλ
R+
α
α
!
f (x)
dx dy
kxkλα + kykλα
Z
f (x)ga,b (y)
dxdy
kxkλα + kykλα
πΓn ( α1 )
sin( πp )λαn−1 Γ( αn )
Z
πΓn ( α1 )
sin( πp )λαn−1 Γ( αn )
Z
n
R+
n
R+
!1
p
kxk(n−λ)(p−1)
f p (x)dx
α
n
R+
!1
p
kxk(n−λ)(p−1)
f p (x)dx
α
Z
!1
q
q
kyk(n−λ)(q−1)
ga,b
(y)dy
α
Z
a<kykα <b
6
!1
q
kykα(n−λ)(q−1) g̃ q (y)dy
,
it follows that
"
Z
kyk(n−λ)(q−1)
g̃ q (y)dy
α
a<kykα <b
πΓn ( α1 )
<
sin( πp )λαn−1 Γ( αn )
#p Z
n
R+
kxk(n−λ)(p−1)
f p (x)dx.
α
For a → 0+ , b → +∞, by (10), we obtain
"
#p Z
Z
n( 1 )
πΓ
α
0<
kyk(n−λ)(q−1)
g̃ q (y)dy ≤
kxk(n−λ)(p−1)
f p (x)dx < ∞,
α
α
π
n−1 Γ( n )
n
n
sin(
)λα
R+
R+
p
α
hence by (11), we have
Z
n
R+
<
=
Z
kykλ−n
α
n
R+
πΓn ( α1 )
sin( πp )λαn−1 Γ( αn )
πΓn ( α1 )
sin( πp )λαn−1 Γ( αn )
!p
Z Z
f (x)
f (x)g̃(y)
dx dy =
dxdy
λ
λ
n
n kxkλ + kykλ
kxkα + kykα
R+
R+
α
α
!1 Z
Z
!1
p
n
R+
Z
n
R+
"Z
×
n
R+
kxk(n−λ)(p−1)
f p (x)dx
α
n
R+
q
kyk(n−λ)(q−1)
g̃ q (y)dy
α
!1
p
kxk(n−λ)(p−1)
f p (x)dx
α
kykλ−n
α
Z
n
R+
f (x)
dx
λ
kxkα + kykλα
#1
!p
q
dy
,
it follows that
!p
f
(x)
kykλ−n
dx dy
α
n
n kxkλ + kykλ
R+
R+
α
α
#p Z
"
πΓn ( α1 )
kxk(n−λ)(p−1)
f p (x)dx.
<
α
n
sin( πp )λαn−1 Γ( αn )
R+
Z
Z
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) :=
1
πΓn ( α
)
n−1 Γ( n )
sin( π
)λα
p
α
in (11) is not the best possible,
then there exists a positive constant K < Cn,α (λ, p), such that
Z Z
f (x)g(y)
dxdy
n
n kxkλ + kykλ
R+
R+
α
α
!1 Z
Z
p
< K
n
R+
kxk(n−λ)(p−1)
f p (x)dx
α
7
n
R+
kyk(n−λ)(q−1)
g q (y)dy
α
!1
q
.
(13)
In particular, setting
−
f (x) = kxkα
(n−λ)(p−1)+n+ε
p
−
g(y) = kykα
,
(n−λ)(q−1)+n+ε
q
,
0 < ε < λ(q − 1),
(13) is still true. By the properties of limit, there exists a sufficiently small a > 0, such
that
Z
Z
(n−λ)(q−1)+n+ε
(n−λ)(p−1)+n+ε
−
−
1
p
q
kxk
kyk
dxdy
α
α
λ + kykλ
n
kxk
kxkα >a R+
α
α
!1
Z
p
< K
kxkα >a
kxk(n−λ)(p−1)
kxk−(n−λ)(p−1)−n−ε
dx
α
α
!1
q
Z
×
kykα >a
Z
= K
kxkα >a
kyk(n−λ)(q−1)
kyk−(n−λ)(q−1)−n−ε
dy
α
α
kxk−n−ε
dx.
α
On the other hand, by lemma 2.3, we have
Z
Z
(n−λ)(p−1)+n+ε
(n−λ)(q−1)+n+ε
−
−
1
p
q
kxk
kyk
dxdy
α
α
n kxkλ + kykλ
kxkα >a R+
α
α
Z
Z
(n−λ)(q−1)+n+ε
−n+ λ − ε
−
1
q
=
kxkα q p
kyk
dydx
α
λ + kykλ
n
kxk
kxkα >a
R+
α
α
Z
−n+ λ − ε
=
kxkα q p ω̃α,λ (x, q, ε)dx
=
kxkα >a
Γn ( α1 )
1
n Γ
n−1
λα
Γ( α )
p
hence we obtain
ε
−
λq
Γn ( α1 )
Γ
λαn−1 Γ( αn )
Z
1
ε
Γ
+
kxk−n−ε
dx,
α
q λq
kxkα >a
ε
1
−
p λq
ε
1
Γ
+
< K.
q λq
for ε → 0+ , we have
Cn,α (λ, p) =
πΓn ( α1 )
≤ K,
sin( πp )λαn−1 Γ( αn )
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, p1 + 1q = 1, n ∈ Z+ , α > 0, f ≥ 0, g ≥ 0, and
Z
Z
p
0<
f (x)dx < ∞,
0<
g q (x)dx < ∞,
n
R+
n
R+
8
(14)
then
Z
Z
n
R+
n
R+
πΓn ( α1 )
f (x)g(y)
dxdy
<
kxknα + kyknα
sin( πp )nαn−1 Γ( αn )
!1 Z
Z
!1
p
Z
n
R+
f (x)
dx
n
kxkα + kyknα
n
R+
!p
;
(15)
n
R+
n
R+
Z
q
g q (x)dx
f p (x)dx
×
"
πΓn ( α1 )
dy <
sin( πp )nαn−1 Γ( αn )
#p Z
f p (x)dx,
(16)
n
R+
where the constant factors in (15) and (16) are all the best possible.
Proof
By taking λ = n in theorem 3.1, (15) and (16) can be obtained.
Corollary 3.2 If p > 1,
1
p
Z
+
1
q
= 1, n ∈ Z+ , f ≥ 0, g ≥ 0, and
Z
p
f (x)dx < ∞,
0<
0<
g q (x)dx < ∞,
(17)
n
R+
n
R+
then
Z
n
R+
Z
n
R+
π
f (x)g(y)
Pn
Pn
n
n dxdy <
n! sin( πp )
( i=1 xi ) + ( i=1 yi )
!1 Z
!1
Z
p
q
p
q
×
f (x)dx
g (x)dx
;
n
R+
Z
n
R+
Z
n
R+
f (x)
Pn
P
dx
n
( i=1 xi ) + ( ni=1 yi )n
!p
(18)
n
R+
"
π
dy <
n! sin( πp )
#p Z
f p (x)dx,
(19)
n
R+
where the constant factors in (18) and (19) are all the best possible.
Proof
By taking α = 1 in corollary 3.1, (18) and (19) can be obtained.
References
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