ON MULTIPLE HARDY-HILBERT INTEGRAL INEQUALITIES
WITH SOME PARAMETERS
HONG YONG
Received 19 April 2006; Revised 30 May 2006; Accepted 5 June 2006
By introducing some parameters and norm xα (x ∈ Rn ), we give multiple HardyHilbert integral inequalities, and prove that their constant factors are the best possible
when parameters satisfy appropriate conditions.
Copyright © 2006 Hong Yong. 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 ≥ 0, g ≥ 0, 0 < 0 f p (x)dx < +∞, 0 <
have the well-known Hardy-Hilbert inequality (see [4]):
+∞
0
f (x)g(x)
π
dx d y <
x+y
sin(π/ p)
+∞
p
0
1/ p +∞
f (x)dx
0
∞
0
g q (x)dx < +∞, then we
1/q
g q (x)dx
,
(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
dy <
π
sin(π/ p)
p +∞
0
f p (x)dx,
(1.2)
where the constant factor [π/ sin(π/ p)] p is also the best possible.
Hardy-Hilbert inequalities are important in analysis and in their applications (see [7]).
In recent years, many results (see [1, 3, 8–10]) have been obtained in the research of
Hardy-Hilbert inequality. At present, because of the requirement of higher-dimensional
harmonic analysis and higher-dimensional operator theory, multiple Hardy-Hilbert integral inequalities are researched (see[5, 6, 11]). Yang [11] obtains the following: if
α ∈ R, n ≥ 2, pi > 1 (i = 1,2,...,n), ni=1 (1/ pi ) = 1, λ > n − min1≤i≤n { pi }, fi ≥ 0, and
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 94960, Pages 1–11
DOI 10.1155/JIA/2006/94960
2
0<
Multiple Hardy-Hilbert integral inequalities
+∞
α
p
(t − α)n−1−λ fi i (t)dt < +∞, (i = 1,2,...,n), then
+∞
α
···
+∞
n
1
i=1 xi − nα
α
λ
1 n−λ
<
Γ 1−
Γ(λ) i=1
pi
n
n
fi xi dx1 ...dxn
i=1
+∞
α
pi
(t − α)n−1−λ fi (t)dt
(1.3)
1/ pi
,
where the constant factor (1/Γ(λ)) ni=1 Γ(1 − (n − λ)/ pi ) is the best possible.
In this paper, by introducing some parameters and norm xα (x ∈ Rn ), we give multiple Hardy-Hilbert integral inequalities, and discuss the problem of the best constant
factor. For this reason, we introduce the notation
Rn+ = x = x1 ,...,xn : x1 ,...,xn > 0 ,
1/α
xα = x1α + · · · + xnα
,
(1.4)
(α > 0),
and we agree on xα < c representing {x ∈ Rn+ : xα < c}.
2. Some lemmas
Lemma 2.1 (see [2]). If pi > 0, ai > 0, αi > 0, (i = 1,2,...,n), Ψ(u) is a measurable function,
then
···
x1 ,...,xn >0; (x1 /a1 )α1 +···+(xn /an )αn ≤1
p 1 −1
× x1
p
=
p
Ψ
x1
a1
α1
x
+ ··· + n
an
αn p −1
...xn n dx1 ...dxn
(2.1)
1
a1 1 ...ann Γ p1 /α1 ...Γ pn /αn
α1 ...αn Γ p1 /α1 + · · · + pn /αn
0
Ψ(u)u p1 /α1 +···+pn /αn −1 du,
where the Γ(·) is Γ-function.
Lemma 2.2. If n ∈ Z+ , α > 0, β > 0, λ > 0, m ∈ R, 0 < n − m < βλ, and setting weight function ωα,β,λ (m,n, y) as
ωα,β,λ (m,n, y) =
1
Rn+
−m
β
β λ x α dx,
x α + y α
(2.2)
Hong Yong 3
then
n−βλ−m
ωα,β,λ (m,n, y) = y α
Γn (1/α)
n−m
n−m
,
B
,λ −
βαn−1 Γ(n/α)
β
β
(2.3)
where the B(·, ·) is β-function.
Proof. By Lemma 2.1, we have
1
−m
β
β λ x α d y
x α + y α
= lim · · ·
ωα,β,λ (m,n, y) =
Rn+
r →+∞
x1 ,...,xn >0; x1α +···+xnα <r α
r x1 /r
× β
r
x1 /r
α
α
1
Γn (1/α)
= n −1
lim
α Γ(n/α) r →+∞
n−βλ−m
= y α
n−βλ−m
= y α
0
α β/α
−m
ru1/α
1−1
1−1
β λ x1 ...xn dx1 ...dxn
+ y α
λ u
β
y α + r β uβ/α
r
0
+∞
0
α 1/α −m
+ · · · + xn /r
r n Γn (1/α)
= lim n
r →+∞ α Γ(n/α)
Γn (1/α)
= n −1
α Γ(n/α)
+ · · · + xn /r
1
λ t
β
y α + t β
1
λ t
β
y α + t β
Γn (1/α)
βαn−1 Γ(n/α)
1
0
n/α−1
n−m−1
n−m−1
du
(2.4)
dt
dt
1
u(n−m)/β−1 du
(1 + u)λ
Γn (1/α)
n−m
n−m
.
B
,λ −
βαn−1 Γ(n/α)
β
β
Hence (2.3) is valid.
3. Main results
Theorem 3.1. If p > 1, 1/ p + 1/q = 1, n ∈ Z+ , α > 0, β > 0, λ > 0, a ∈ R, b ∈ R, 0 < n −
ap < βλ, 0 < n − bq < βλ, f ≥ 0, g ≥ 0, and
0<
(n−βλ)+p(b−a)
Rn+
0<
x α
f p (x)dx < +∞,
(n−βλ)+q(a−b) q
Rn+
y α
g (y)d y < +∞,
(3.1)
(3.2)
4
Multiple Hardy-Hilbert integral inequalities
then
f (x)g(y)
Rn+
β
β λ dx d y
x α + y α
1/ p
1/q
(n−βλ)+p(b−a) p
(n−βλ)+q(a−b) q
< Cα,β,λ (a,b, p, q)×
x α
f (x)dx
y α
g (y)d y
,
R+n
((n−βλ)+q(a−b))/(1−q)
Rn+
Rn+
y α
f (x)
Rn+
p
< Cα,β,λ (a,b, p, q) ×
Rn+
(3.3)
p
β
β λ dx
x α + y α
dy
(3.4)
(n−βλ)+p(b−a) p
x α
f (x)dx,
where Cα,β,λ (a,b, p, q) = (Γn (1/α)/βαn−1 Γ(n/α))B1/ p ((n − ap)/β,λ − (n − ap)/β)B1/q ((n −
bq)/β,λ − (n − bq)/β).
Proof. By Hölder’s inequality, we have
G :=
f (x)g(y)
Rn+
=
β
β λ dx d y
x α + y α
f (x)
β
β λ/ p
x α + y α
Rn+
≤
x α
ap dx d y
β
β λ
x α + y α y α
g(y)
β
β λ/q
x α + y α
bp
f p (x)
Rn+
xbα
y aα
1/ p
×
y aα
dx d y
xbα
g q (y)
Rn+
aq
y α
bq dx d y
β
β λ
x α + y α x α
1/q
,
(3.5)
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
aq
bp
C1 f p (x)
C2 g q (y)
y α
x α
=
ap
β
β λ
β
β λ
bq ,
x α + y α y α
x α + y α x α
a.e. (x, y) ∈ Rn+ × Rn+ .
(3.6)
Without losing generality, we suppose that C1 = 0, we may get
b(p+q)
x α
f p (x) =
C2
a(p+q) q
y α
g (y),
C1
a.e. (x, y) ∈ Rn+ × Rn+ ,
(3.7)
hence, we obtain
b(p+q)
x α
f p (x) = C(constant),
a.e. x ∈ Rn+ ,
(3.8)
Hong Yong 5
hence, we have
(n−βλ)+p(b−a)
x α
Rn+
f p (x)dx =
(n−βλ)−bq−ap+b(p+q)
x α
Rn+
(3.9)
=C
f p (x)dx
Rn+
(n−βλ)−bq−ap
x α
dx
= ∞,
which contradicts (3.1). Hence, and by Lemma 2.2, we obtain
G<
1
bp p
ap d y x α f (x)dx
β
β λ
y
α
x α + y α
Rn+
Rn+
×
Rn+
1
1
Rn+
1
1/q
aq
y α g q (y)d y
β
β λ
bq dx
x α + y α x α
1/ p bp
ωα,β,λ, (ap,n,x)xα f p (x)dx
n
=
R+
n − ap
n − ap
Γn (1/α)
=
B
,λ −
βαn−1 Γ(n/α)
β
β
×
Rn+
n − bq
n − bq
Γn (1/α)
B
,λ −
βαn−1 Γ(n/α)
β
β
= Cα,β,λ, (a,b, p, q)
(n−βλ)+p(b−a)
Rn+
1/q
aq
ωα,β,λ, (bq,n, y) y α g q (y)d y
n
R+
1/ p
x α
(n−βλ)+p(b−a) p
x α
f (x)dx
1/ p
(n−βλ)+q(a−b) q
Rn+
y α
1/ p
f p (x)dx
1/q
g (y)d y
×
(n−βλ)+q(a−b) q
Rn+
y α
1/q
g (y)d y
.
(3.10)
Hence, (3.3) is valid.
Let k = ((n − βλ) + q(a − b))/(1 − q), for 0 < h < l < +∞, setting
⎧
p/q
⎪
⎪
f (x)
⎪
⎨ y k
,
α
β
β λ dx
gh,l (y) = ⎪
Rn+ x α + y α
⎪
⎪
⎩0,
g(y) = y kα
Rn+
f (x)
β
β λ dx
x α + y α
h < y α < l,
0 < y α ≤ h or y α ≥ l,
(3.11)
p/q
,
y ∈ Rn+ ,
by (3.1), for sufficiently small h > 0 and sufficiently large l > 0, we have
0<
h< y α <l
(n−βλ)+q(a−b) q
gh,l (y)d y
y α
< +∞.
(3.12)
6
Multiple Hardy-Hilbert integral inequalities
Hence, by (3.3), we have
(n−βλ)+q(a−b) q
h< y α <l
y α
g (y)d y
=
h< y α <l
k(1−q) q
y α
g (y)d y
=
h< y α <l
=
y kα
=
h< y α <l
y kα
β
x α + y α
dy
β
β λ dx
x α + x α
f (x)
Rn+
β
β λ dx d y
x α + y α
f (x)gh,l (y)
Rn+
Rn+
β λ dx
p
f (x)
p/q f (x)
Rn+
β λ dx d y < Cα,β,λ, (a,b, p, q)
β
x α + y α
(n−βλ)+p(b−a)
Rn+
x α
1/ p
f p (x)dx
1/q
1/ p
(n−βλ)+q(a−b) q
(n−βλ)+p(b−a) p
×
yα
gh,l (y)d y
= Cα,β,λ, (a,b, p, q)
x α
f (x)dx
R+n
R+n
×
1/q
(n−βλ)+q(a−b) q
h< y α <l
y α
g (y)d y
,
(3.13)
it follows that
h< y α <l
(n−βλ)+q(a−b) q
y α
g (y)d y
p
< Cα,β,λ, (a,b, p, q)
(n−βλ)+p(b−a)
Rn+
x α
f p (x)dx. (3.14)
For h → 0+ , l → +∞, we obtain
0<
(n−βλ)+q(a−b) q
Rn+
y α
g (y)d y
p
≤ Cα,β,λ, (a,b, p, q)
Rn+
(n−βλ)+p(b−a) p
x α
f (x)dx
(3.15)
< +∞,
hence, by (3.3), we obtain
Rn+
((n−βλ)+q(a−b))/(1−q)
y α
=
Rn+
f (x)
p
β
β λ dx
x α + y α
dy
f (x)g(y)
(n−βλ)+p(b−a) p
x α
f (x)dx
β
β λ dx d y < Cα,β,λ, (a,b, p, q)
Rn+
x α + y α
1/q
(n−βλ)+q(a−b) q
×
yα
g (y)d y
=Cα,β,λ, (a,b, p, q)
Rn+
Rn+
×
Rn+
((n−βλ)+q(a−b))/(1−q)
y α
f (x)
Rn+
(n−βλ)+p(b−a)
Rn+
β
β λ dx
x α + y α
x α
p
1/ p
1/ p
f p (x)dx
1/q
dy
.
(3.16)
Hence, we can obtain (3.4).
Hong Yong 7
Remark 3.2. If f and g do not satisfy (3.1) and (3.2), by the proof of Theorem 3.1, we can
obtain
f (x)g(y)
Rn+
β
β λ dx d y
x α + y α
≤ Cα,β,λ (a,b, p, q) ×
Rn+
(n−βλ)+p(b−a)
Rn+
((n−βλ)+q(a−b))/(1−q)
y α
p
≤ Cα,β,λ (a,b, p, q) ×
x α
Rn+
1/q
(n−βλ)+q(a−b) q
yα
g (y)d y
,
(3.17)
p
f (x)
β
β λ dx
x α + y α
Rn+
1/ p f p (x)dx
dy
(3.18)
(n−βλ)+p(b−a) p
x α
f (x)dx.
Rn+
Remark 3.3. By (3.4), we can also obtain (3.3), hence (3.4) and (3.3) are equivalent.
Theorem 3.4. If p > 1, 1/ p + 1/q = 1, n ∈ Z+ , α > 0, β > 0, λ > 0, a ∈ R, b ∈ R, 0 < n −
ap < βλ, ap + bq = 2n − βλ, f ≥ 0, g ≥ 0, and
0<
b(p+q)−n
Rn+
0<
Rn+
x α
f p (x)dx < +∞,
(3.19)
a(p+q)−n q
y α
g (y)d y
< +∞,
then
Rn+
f (x)g(y)
β λ dx d y
β
x α + y α
<
n − ap
n − ap
Γn (1/α)
B
,λ −
βαn−1 Γ(n/α)
β
β
×
Rn+
b(p+q)−n
Rn+
x α
(a(p+q)−n)/(1−q)
y α
(3.20)
1/ p f p (x)dx
a(p+q)−n q
Rn+
f (x)
Rn+
y α
1/q
,
p
β
β λ dx
x α + y α
n − ap
n − ap
Γn (1/α)
B
,λ −
<
βαn−1 Γ(n/α)
β
β
g (y)d y
dy
p Rn+
(3.21)
b(p+q)−n p
x α
f (x)dx,
where the constant factors (Γn (1/α)/βαn−1 Γ(n/α))B((n − ap)/β,λ − (n − ap)/β) and
[(Γn (1/α)/βαn−1 Γ(n/α))B((n − ap)/β,λ − (n − ap)/β)] p are all the best possible.
Proof. Since ap + bq = 2n − βλ, we have
n − bq = n − (2n − βλ − ap) = βλ − (n − ap),
(3.22)
8
Multiple Hardy-Hilbert integral inequalities
hence, by 0 < n − ap < βλ, we obtain 0 < n − bq < βλ, and
(n − βλ) + p(b − a) = b(p + q) − n,
(n − βλ) + q(a − b) = a(p + q) − n,
n − bq
n − ap
= λ−
,
β
β
(3.23)
n − ap n − bq
=
.
β
β
λ−
By Theorem 3.1, (3.20) and (3.21) are valid.
If the constant factor K1 := (Γn (1/α)/βαn−1 Γ(n/α))B((n − ap)/β,λ − (n − ap)/β) in
(3.20) is not the best possible, then there exists a positive constant K < K1 , such that
(3.20) is still valid when we replace K1 by K.
In particular, for 0 < ε < q(n − ap), we take
−bq−ε/ p
fε (x) = xα
−ap−ε/q
gε (y) = y α
,
,
(3.24)
by (3.17) and the properties of limit, when δ > 0 is sufficiently small, we have
xα >δ
fε (x)gε (y)
β
β λ dx d y
x α + y α
Rn+
≤K
xα >δ
b(p+q)−n p
x α
fε (x)dx
=K
−n −ε
xα >δ
1/ p y α >δ
1/ p x α
−n −ε
y α >δ
y α
a(p+q)−n q
y α
gε (y)d y
1/q
=K
dy
xα >δ
1/q
(3.25)
xα−n−ε dx.
On the other hand, by Lemma 2.2, we have
fε (x)gε (y)
β
β λ dx d y
x α + y α
1
−bq−ε/ p
−ap−ε/q
=
x α
d y dx
β
β λ y α
xα >δ
Rn+ x α + y α
xα >δ
Rn+
=
−bq−ε/ p
xα >δ
x α
ω
α,β,λ
ε
ap + ,n,x dx
q
=
Γn (1/α)
1
ε
1
ε
n − ap − ,λ − n − ap −
B
n
−
1
βα Γ(n/α) β
q
β
q
xα >δ
xα−n−ε dx.
(3.26)
Hence, we obtain
Γn (1/α)
1
ε
1
ε
n − ap − ,λ − n − ap −
B
βαn−1 Γ(n/α) β
q
β
q
for ε → 0+ , we have
K1 =
≤ K,
(3.27)
n − ap
n − ap
Γn (1/α)
B
,λ −
≤ K,
n
−
1
βα Γ(n/α)
β
β
(3.28)
Hong Yong 9
which contradicts the fact that K < K1 . Hence the constant factor in (3.20) is the best
possible.
Since (3.21) and (3.20) are equivalent, the constant factor in (3.21) is also the best
possible.
4. Some corollaries
Corollary 4.1. If p > 1, 1/ p + 1/q = 1, n ∈ Z+ , α > 0, β > 0, λ > 0, f ≥ 0, g ≥ 0, and
0<
(n−βλ)(p−1)
Rn+
0<
Rn+
x α
f p (x)dx < +∞,
(4.1)
(n−βλ)(q−1) q
y α
g (y)d y
< +∞,
then
f (x)g(y)
β
β λ dx d y
x α + y α
Rn+
<
Rn+
λ λ
Γn (1/α)
B ,
βαn−1 Γ(n/α) p q
βλ−n
y α
<
f (x)dx
Rn+
β
β λ dx
x α + y α
1/q
(n−βλ)(q−1) q
Rn+
yα
g (y)d y
,
p
f (x)
Rn+
1/ p
(n−βλ)(p−1) p
x α
Γn (1/α)
λ λ
B ,
βαn−1 Γ(n/α)
p q
dy
p (n−βλ)(p−1)
Rn+
x α
f p (x)dx,
(4.2)
where the constant factors in (4.2) are all the best possible.
Proof. If we take a = n/ p − βλ/ p2 , b = n/q − βλ/q2 in Theorem 3.4, (4.2) can be obtained.
Remark 4.2. If we take n = λ = 1 in (4.2), we can obtain the results of [10]:
+∞
0
f (x)g(y)
dx d y
xβ + y β
<
+∞
0
π
β sin(π/ p)
y β −1
+∞
0
+∞
0
x(p−1)(1−β) f p (x)dx
f (x)
dx
xβ + y β
p
dy <
1/ p +∞
π
β sin(π/ p)
0
p +∞
0
y (q−1)(1−β) g q (y)d y
1/q
x(p−1)(1−β) f p (x)dx,
where the constant factors in (4.3) are all the best possible.
,
(4.3)
10
Multiple Hardy-Hilbert integral inequalities
If we take n = β = 1 in (4.2), we can obtain
+∞
f (x)g(y)
dx d y
(x + y)λ
0
<B
+∞
0
y λ −1
λ λ
,
p q
+∞
+∞
0
0
x(1−λ)(p−1) f p (x)dx
f (x)
dx
(x + y)λ
p
dy < Bp
1/ p +∞
0
λ λ
,
p q
+∞
0
y (1−λ)(q−1) g q (y)d y
1/q
(4.4)
,
x(1−λ)(p−1) f p (x)dx,
where the constant factors in (4.4) are all the best possible.
Corollary 4.3. If p > 1, 1/ p + 1/q = 1, n ∈ Z+ , λ > 0, np + λ − 2n > 0, nq + λ − 2n > 0
f ≥ 0, g ≥ 0, and
0<
Rn+
0<
Rn+
xαn−λ f p (x)dx < +∞,
(4.5)
y αn−λ g q (y)d y
< +∞,
then
f (x)g(y)
λ dx d y
x α + y α
Rn+
<B
np + λ − 2n nq + λ − 2n
,
p
q
(n−λ)/(1−q)
yα
Rn+
p
f (x)
Rn+
Rn+
xαn−λ f p (x)dx
1/ p Rn+
y αn−λ g q (y)d y
p
λ dx d y < B
xα +yα
np+λ −2n nq+λ − 2n
,
p
q
1/q
,
xαn−λ f p (x)dx,
Rn+
(4.6)
where the constant factors in (4.6) are all the best possible.
Proof. If we take β = 1, a = b = (2n − λ)/ pq in Theorem 3.4, (4.6) can be obtained.
Remark 4.4. If we take n = 1 in (4.6), we can obtain the results of [1]:
+∞
0
f (x)g(y)
dx d y
(x + y)λ
<B
+∞
0
p+λ−2 q+λ−2
,
p
q
y (1−λ)/(1−q)
+∞
0
+∞
f (x)
dx
(x + y)λ
0
x1−λ f p (x)dx
p
dy < Bp
1/ p +∞
0
y 1−λ g q (y)d y
p+λ−2 q+λ−2
,
p
q
+∞
0
1/q
,
x1−λ f p (x)dx,
(4.7)
where the constant factors in (4.7) are all the best possible.
If we take other appropriate parameters, we can obtain many new inequalities.
Hong Yong 11
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Hong Yong: Department of Mathematics, Guangdong University of Business Study,
Guangzhou 510320, China
E-mail address: hongyong59@sohu.com