General Mathematics Vol. 13, No. 1 (2005), 73–82
Multidimensional Integral Inequalities with
Homogeneous Weights
Sorina Bârză
Dedicated to Professor Emil C. Popa on the occasion of his 60th birthday
Abstract
We give upper and lower bounds estimates for the best constant
which appears in a multidimensional integral inequality, in the particular case of homogeneous weights.
2000 Mathematics Subject Classification: 26D15, 26B99
Keywords and phrases: Multidimensional embedding, decreasing
functions, integral inequalities, best constant.
1
Introduction
Let Rn+ := {(x1 , ..., xn ) : xi ≥ 0, i = 1, 2, ..., n}. Assume that f : Rn+ −→ R+
is decreasing which means that it is decreasing with respect to each variable.
We denote f ↓ when f is decreasing (=nonicreasing). Throughout this paper
u and v are positive, locally integrable functions defined on Rn+ , n ≥ 1. χD
denotes the characteristic function of a set D. In order to motivate the
results of this paper and put them into a frame we use Section 2 to remind
the characterization of the inequality
ÃZ
!1/q
ÃZ
!1/p
(1)
R
n
+
f qu
≤C
R
f pv
n
+
73
, 0 < p ≤ q < ∞,
74
Sorina Bârză
for all f ↓ and some of the cases when the best constant in the above
inequality can be easily computed or estimated. In Section 3 we present
some new facts about the best constant in the inequality (1) in the case of
homogeneous weights. More precisely we give upper and lower estimates
for the best constant. Section 4 is reserved for particular cases and open
questions.
2
Weighted inequalities for decreasing functions of several variables
In the one-dimensional case the inequality (1) was characterized in [7] as
follows:
If n = 1, 0 < p ≤ q < ∞, then (1) is valid for all f ↓ if and only if
µZ
¶−1/p
t
v
u
C1 = sup
t>0
¶1/q µZ
t
<∞
0
0
and the constant C = C1 is sharp.
The multidimensional case was treated in [2] as follows:
If 0 < p ≤ q < ∞, then (1) is valid for all f ↓ if and only if
¡R
(2)
¢1/q
u
Cn = sup ¡R ¢1/p < ∞
D∈D
v
D
D
and the constant C = Cn is sharp. Here D is the set of all ”decreasing
domains” , i.e. for which the characteristic function is a decreasing function
in each variable. It is not always easy to calculate the constant defined in
(2). It was shown that in the case of weights of product type the supremum
in (2) can be reduced to supremum over a much simpler class of decreasing
sets (the rectangles), which is something much simpler to compute, as the
following theorem shows. For it, s complete proof see [2].
75
Multidimensional Integral Inequalities with Homogeneous Weights
Theorem 1.If 0 < p ≤ q < ∞, u(x1 , . . . , xn ) = u1 (x1 ) . . . un (xn ) and
v(x1 , . . . , xn ) = v1 (x1 ) . . . vn (xn ) then
¡R ¢1/q
¡R a1
¢1/q
Ra
u
· · · 0 n u(x1 , · · · , xn )dx1 · · · dxn
D
0
(3)
sup ¡R ¢1/p = sup ¡R a
¢1/p .
R an
1
ai >0
D∈D
v
·
·
·
v(x
,
·
·
·
,
x
)dx
·
·
·
dx
1
n
1
n
D
0
0
In [3] it was given an example which shows that the equality (3) does not
hold for arbitrary weights. The natural and important question is now
whether the constants Cn and
¡R a1
¢1/q
Ra
· · · 0 n u(x1 , · · · , xn )dx1 · · · dxn
0
An = sup ¡R a
¢1/p
Ra
1
ai >0
· · · 0 n v(x1 , · · · , xn )dx1 · · · dxn
0
are comparable in the general case. Clearly An ≤ Cn and we may ask if
the converse inequality Cn ≤ cAn holds with a constant c independent on
the weights. An answer to this question was given in [3] with the help of
the following result of independent interest where explicit lower and upper
estimates are given for the constant Cn when the weights are of the form:
u(xy), v(xy).
Theorem 2.Let 0 < p ≤ q < ∞ and u(s) ≥ 0 and v(s) ≥ 0 be two
Rt
Rt
measurable functions on R+ such that U (t) = 0 u < ∞, V (t) = 0 v < ∞
for all t > 0. Then the inequality
ÃZ
!1/q
ÃZ
!1/p
R
f q (x, y)u(xy)dxdy
2
+
≤C
R
f p (x, y)v(xy)dxdy
.
2
+
holds for all f ≥ 0 decreasing in x and y with a finite constant C > 0
independent on f if and only if
µ
¶1/q µZ t
¶1/q−1/p
U (t)
dx
A = Ap,q := sup
V (x)
<∞
V (t)
x
t>0
0
Moreover, C = Ap,p and
−1/p
2
µ ¶1/q
p
A≤C≤
A, if p < q.
q
76
Sorina Bârză
Using the above theorem it was proved that An and Cn are comparable in
the sense that either both are finite or both are infinite, but the estimate
Cn ≤ cAn is no longer uniform. For more details see [3].
3
Upper and lower estimates of the best constant in the case of homogeneous weights
In this section we study the same type of questions as above but for the
case when the weights are homogeneous functions. The next Lemma gives
important information about the best constant in the inequality (1) when
the weights are homogeneous functions.
Lemma 1. Let 0 < p ≤ q < ∞, α > −n, β > −n, u(εx) = εα u(x),
v(εx) = εβ v(x), x ∈ Rn+ and ε > 0. If Cn < ∞ then
β+n
α+n
=
.
q
p
Proof. Let S = {σ ∈ Rn+ : |σ| = 1} and suppose that
α+n
β+n
6=
.
q
p
Then
¡R
Cn
¢1/q
u
= sup ¡R ¢1/p
D∈D
v
D
¡R r R
¢1/q
n−1
u(ρσ)ρ
dρdσ
≥ sup ¡R0r RS
¢
n−1 dρdσ 1/p
r>0
v(ρσ)ρ
0 S
¢1/q
¡R r R α+n−1
ρ
u(σ)dρdσ
0 S
= sup ¡R r R
¢1/p
r>0
ρβ+n−1 v(σ)dρdσ
0 S
D
= sup r(α+n)/q−(β+n)/p C(u, v, p, q) = ∞.
r>0
Multidimensional Integral Inequalities with Homogeneous Weights
77
where C(u, v, p, q) < ∞ and depends only on u, v, p and q. Hence in the
case of homogeneous weights the only interesting case is when
α+n
β+n
=
.
q
p
In fact, for the case n = 2 we have the following result:
Theorem 3. Let 0 < p ≤ q, α > −2, β > −2, u(εx, εy) = εα u(x, y),
v(εx, εy) = εβ v(x, y), (x, y) ∈ R2+ , ε > 0.
Rs
Rs
Denote by U (s) = 0 u(1, y)dy < ∞ and V (s) = 0 v(1, y)dy < ∞ for
all s > 0. Then the inequality (1) holds for all f ↓ if and only if
¶1/q µZ ∞
µ
¶1/q−1/p
ds
(β−α) U (t)
V (s) β+3
A := sup t
< ∞.
V (t)
s
t>0
t
Moreover,
A ≤ C2 ≤ (p/q)1/q A, if p < q.
and
A = C2 if p = q.
Proof. By Lemma 1 we may suppose that α+2
= β+2
. We will first prove
q
p
the upper bound. We know from (2) that
³R
´1/q
R h(x)
t
dx
u(x,
y)dy
0
0
(4)
C2 = sup Ih (t) = sup ³R
´1/p
R h(x)
t
t>0,h↓
t>0,h↓
dx
v(x,
y)dy
0
0
and by using the homogeneity and changing variables, we find that
´ ´1/q
³R
³
t α+1
h(x)
dx
x
U
x
0
Ih (t) = ³R
³
´ ´1/p .
t β+1
h(x)
x
V
dx
x
0
We observe that
µZ
∞
q
U (t) ≤ A
t
ds
V (s) β+3
s
¶q/p−1
V (t)t(β+2)(q/p−1) , ∀t > 0.
78
Sorina Bârză
By substituting now t by
ÃZ
∞
q
≤A
h(x)
x
h(x)
x
and multiplying by xα+1 we get
¶
µ
h(x)
α+1
x U
x
!q/p−1
ds
V (s) β+3
s
µ
V
h(x)
x
¶
xβ+1 h(β+2)(q/p−1) (x)
i.e.
µ
x
α+1
U
h(x)
x
ÃZ
¶
q
≤A
µ
∞
h(x)
x
V (s)
h(x)
s
¶(β+2)
ds
s
!q/p−1
µ
V
h(x)
x
¶
xβ+1 .
By integrating now with respect to x, from 0 to t, changing the variables
h(x)
ξ =
and using the facts that h and V are decreasing, respectively
s
increasing we obtain
Z
µ
t
x
α+1
U
0
h(x)
x
¶
(5)
This implies that
¶
¶q/p−1 µ
¶
h(x) β+1
h(x)
dx ≤ A
V
ξ dξ
V
xβ+1 dx
ξ
x
0
0
¶
¶q/p−1 µ
¶
Z t µZ x µ
h(ξ) β+1
h(x)
≤
V
ξ dξ
V
xβ+1 dx
ξ
x
0
0
µZ t µ
¶
¶q/p
h(x)
p q
β+1
V
=
A
x dx
.
q
x
0
q
Z t µZ
x
µ
µ ¶1/q
p
Ih ≤
A
q
for all t > 0 and h ↓ and we got the desired upper bound.
For the lower bound suppose first that p = q, thus also that α = β. Consider
a strictly increasing sequence (an )n≥1 , a1 = 0 which converges to 1 and take
h(x) ≡ 1. For a fix t > 0, by using the fact that U is increasing and y −(α+3)
is decreasing we get
µ ¶
µ ¶
Z an t
n Z ak t
X
1
1
α+1
α+1
x U
dx =
x U
dx
x
x
0
a
t
k−1
k=2
79
Multidimensional Integral Inequalities with Homogeneous Weights
=
n Z
X
k=2
n
X
1
ak−1 t
y −(α+3) U (y)dy
1
ak t
¶
1
(ak−1 t)α+3
≥
U
ak t
k=2
µ ¶X
n
1
≥ tα+3 U
ak−1 α+3 .
t k=2
(6)
µ
In the same way since V is also increasing we get
µ ¶
µ ¶
1
Z an t
Z a2 t
n Z
X
ak−1 t
1
1
α+1
α+1
dx =
x V
dx +
y −(α+3) V (y)dy
x V
1
x
x
0
0
k=3 ak t
µ ¶
µ
¶
Z a2 t
n
X
1
1
α+1
α+3
(7)
≤
x V
dx + t
V
ak α+3 .
x
a
t
k−1
0
k=3
By (6) and (7) we have
R an t
R0an t
0
x
α+1
U
xα+1 V
¡1¢
¡ x1 ¢
x
dx
dx
tα+3 U
≥
n
¡1¢ X
t
ak−1 α+3
k=2
R a2 t
0
xα+1 V
¡1¢
x
dx + tα+3
n
X
µ
V
k=3
1
ak−1 t
.
¶
ak α+3
P
Dividing now by tα+3 nk=2 ak−1 α+3 we get
¡ ¢
R an t α+1 ¡ 1 ¢
U 1t
x U x dx
0
≥
(8) R an t
.
¡ ¢
µ
¶
n
X
xα+1 V x1 dx
1
α+3
0
V
ak
R a2 t α+1 ¡ 1 ¢
a
k−1 t
dx
x
V
x
0
+ k=3 n
n
X
X
tα+3
ak−1 α+3
ak−1 α+3
k=2
k=2
Since V is continuous, we get by a theorem of Stolz-Cesaro (see e.g Theorem
2.10.2 in [5]) that
³
´
Pn
1
µ ¶
ak α+3
k=3 V
ak−1 t
1
Pn
=V
lim
.
α+3
n→∞
t
k=2 ak−1
80
Sorina Bârză
Since t was arbitrarily chosen, letting now n → ∞ in (8) we finally get
µ
C2 ≥ sup
t>0
U (t)
V (t)
¶1/q
.
If now p < q, i.e. α 6= β one can prove exactly as above that for any t > 0,
³R
´1/q
µ
1 ¶1/q
dx
x
α−β U ( t )
.
³R
´1/q ≥ t
V ( 1t )
t β+1 ¡ 1 ¢
x V x dx
0
t
0
xα+1 U
¡1¢
Hence
³R
t
0
U
¡1¢
´1/q
dx
¡ 1 ¢ ´1/p
β+1
x V x dx
0
³R
´1/q
t α+1 ¡ 1 ¢
µ ¶ ¶1/p−1/q
µZ t
x U x dx
0
1
β+1
= ³R
x V
dx
´1/q
¢
¡
x
t β+1
1
0
x V x dx
0
ÃZ
!1/p−1/q
µ
1 ¶1/q
∞
U
(
)
ds
≥
tα−β 1t
V (s) β+3
.
1
s
V (t)
t
C2 ≥ ³R
t
x
α+1
x
Since t > 0 was arbitrarily chosen, the above inequality implies the upper
bound for the constant and this completes the proof.
Remark. Let
³R R
a b
0
0
u(x, y)dxdy
0
0
v(x, y)dxdy
A2 := sup ³R R
a b
a,b>0
´1/q
´1/p .
it is clear from (4), with h(x) ≡ b and changing variables that we obtain
³R
t
´1/q
dx
x
0
A2 = sup ³R
¡
¢ ´1/p
t β+1
t,b>0
b
x
V
dx
x
0
xα+1 U
¡b¢
Multidimensional Integral Inequalities with Homogeneous Weights
e where
and A2 ≥ A
³R
81
´1/q
dx
x
e = sup ³
A
¡ 1 ¢ ´1/p .
Rt
t>0
β+1
x V x dx
0
t
0
xα+1 U
¡1¢
Obviously, Theorem 3 yields
e ≤ A2 ≤ C2 ≤ A.
A
Applying the Lospital test we note that
³R
´
t α+1 ¡ 1 ¢
µZ ∞
¶1−q/p
x
U
dx
x
0
p
ds
β−α U (t)
lim ³R
lim t
V (s) β+3
´q/p = t→∞
t→0
q
V (t)
s
t β+1 ¡ 1 ¢
t
x V x dx
0
and
³R
t
¡1¢
´
µZ ∞
¶1−q/p
dx
ds
p
β−α U (t)
V (s) β+3
lim ³
t
.
¡ 1 ¢ ´q/p = q lim
t→∞ R t
t→0
V (t)
s
t
β+1
x V x dx
0
0
x
α+1
U
x
e are
Since the functions involved are continuous we conclude that A and A
e < ∞ and the other way
comparable in the sense that if A < ∞ then A
round. The same is true for A and A2 . This does not give an answer to the
question if the two quantities are comparable or not in the general case.
4
Conclusions
By (8) and Theorem 3 one could prove results for power weights which
generalize some of the results proved in [1]. For further details see [2]. Also
for the case n = 2 the same results can be derived from Theorem 3 since
the power weights are also homogeneous functions. Observe that if p = q
the constant does not depend on the degree of homogeneity. Also we get
2
and v(x, y) = y2x+xy as
the same constant if the weights are u(x, y) = x2xy
+y 2
3
4
if they are u(x, y) = x2x+yy 2 and v(x, y) = y2x+xy For other related results as
well as for the results for increasing functions see also [6].
82
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References
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[2] S. Bârză, L.-E.Persson, J. Soria, Sharp weighted multidimensional integral inequalities for monotone functions, Math.Nachr. 210 (2000),
43-58.
[3] S. Bârză, L.-E.Persson, V. D. Stepanov, On weighted multidimensional
embeddings for monotone functions, Math. Scand. 88 (2001), 303-319.
[4] M. J. Carro, J. Soria, Weighted Lorentz spaces and the Hardy operator,
J.Func. Anal. 112 (1993), 1155-1166.
[5] C. P. Niculescu, Analiză matematică pe dreapta reală-O abordare contemporană , Edt.Universitaria, Craiova, Romania, 2002 (in romanian).
[6] J. Pečarić, I. Perić, L.-E.Persson, A multidimensional integral inequality for monotone functions of several variables, Acta. Sci. Math.
(Szeged) (62 (1996), 407-412.
[7] V. D. Stepanov, The weighted Hardy, s inequality for nonincreasing
functions, Trans. Amer. Math.Soc 338 (1993), 173-186.
Dept. of Eng. Sciences, Physics and Mathematics
Karlstad University
SE-65188 Karlstad, SWEDEN
E-mail: sorina.barza@kau.se