WEIGHTED INEQUALITIES
FOR THE LAPLACE TRANSFORM
Yves Rakotondratsimba
(Received November 11, 1997)
Abstract. Necessary conditions and sucient conditions are given in order
that the Laplace Transform is bounded between two Lebesgue spaces with weights. Such a boundedness is characterized for a large class of weights. AMS1991 Mathematics Subject Classication. 26D15, 26A33.
Key words and phrases. Weighted inequalities, Laplace transform, Hardy op-erators.
x
1 Introduction and Results
The Laplace Transform is dened as (L
f
)(x
) =Z
1
0
f
(y
)exp;xy
]dy
0< x <
1:
Throughout this paper it is assumed that 1
< pq <
1p
0 =p
p
;1q
0 =q
q
;1r
=p
qp
;q
andu
(:
)v
(:
)v
1;p 0 (:
)are weight functions, i.e. nonnegative and locally integrable functions. Our purpose is to derive necessary conditions and sucient conditions on
u
(:
) andv
(:
) for which L is bounded from the Lebesgue spaceL
pv =L
p(]01
v
(x
)dx
) intoL
qu =L
q(]01u
(x
)dx
). That is for some constantC >
0 (1.1) Z 1 0 (Lf
)q(x
)u
(x
)dx
1 qC
Z 1 0f
p(x
)v
(x
)dx
1 p for allf
(:
) 0:
For convenience this boundedness is also denoted by L:
L
pv!L
qu.This problem has been investigated by many authors. Indeed a sucient condition ensuring (1
:
1) was given by K. Andersen and H. Heinig An-Hg],Hg] (see also Hz2], An]). And a sucient condition close to be necessary
was found by S. Bloom Bm].
Our present contribution is rst to provide variant sucient conditions for
L :
L
pv !L
qu improving the result of S. Bloom Bm] wheneverq >
4. Thetechnic used in Bm] is to reduce the problem (1
:
1) to weighted estimates involving Hardy operator and Hardy antidierentiation operator. As in An-Hg] and An-Hg], our approach is based on the weighted inequality(1.2) Z 1 0 (
Tf
)q(x
)u
(x
)dx
1 qC
Z 1 0f
p(x
)v
(x
)dx
1 p for allf
(:
) 0 where (Tf
)(x
) = Z x 0f
(y
)exp;xy
;1]dy
0< x <
1:
Although a sucient condition for (1
:
2) is available in An-Hg], Hg] and Hz2], here we give a new one which is not too far to be necessary forT
:L
pv!L
qu.The second contribution in this work is to provide another approach for L :
L
pv !L
qu which can be extended to treat boundednesses problems for manyintegral operators in higher dimension.
Our method for dealing with (1
:
2) is rst to break the operatorT
into small pieces and next to do summations just by using Holder inequalities and the fast increase of the exponential function.Our rst result reads as
Theorem 1.
Suppose that L:L
pv!L
qu. Then for some constantA >
0(1.3) Z R ;1 0
u
(x
)dx
1 q Z R 0v
1;p 0 (y
)dy
1 p0A
for allR >
0 and (1.4) Z 1 R;1u
(x
)exp;qRx
]dx
1 q Z R 0v
1;p 0 (y
)dy
1 p0A
for allR >
0:
Conversely let
p
q
. Then L :L
pv !L
qu whenever for someA >
0 and0
< "
1 (1.5) Z 1 R;1u
(x
)exp;4 ;1 (1;"
)qRx
]dx
1 q Z R 2 ;1Rv
1;p 0 (y
)dy
1 p0A
for allR >
0and the condition (1
:
3) is satised.A similar result was previously obtained by S. Bloom Bm] with the condi-tion (1
:
5) replaced by (1.6) Z 1 R;1u
(x
)exp;Rx
]dx
1 q Z R 0v
1;p 0 (y
)dy
1 p0A
for allR >
0:
Therefore our Theorem improves Bloom's one whenever
q >
4.In order to get a characterization result for certain weights, it would be useful to see connections between conditions (1
:
3) and (1:
5).Lemma 2.
Condition (1:
3) implies (1:
5) under one of the following assump-tions i)u
(:
) is a decreasing function ii)v
1;p 0 (:
)2D
iii)u
(:
)2D
iv)
v
(:
) is a increasing function.Here
w
(:
)2D
means that for someC >
0 Z 2R 0w
(y
)dy
C
Z R 0w
(y
)dy
for allR >
0:
Theorem 1 and Lemma 2 can be combined to get
Corollary 3.
Forp
q
, the boundedness L :L
pv !L
qu is equivalent tocondition (1
:
3) whenever one of the following assumptions is satised: i)u
(:
) is a decreasing functionii)
v
1;p 0(
:
)2D
iii)
v
(:
) is an increasing function iv)u
(:
)2D
.Our second main result concerning the case
q < p
isTheorem 4.
Letq < p
. Suppose that L:L
pv!L
qu. Then(1.7) Z 1 0 Z x ;1 0
u
(y
)dy
1 q Z x 0v
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1 and (1.8) Z 1 0 Z 1 x;1u
(y
)exp;qxy
]dy
1 q Z x 0v
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
ConverselyL:
L
pv!L
qu whenever for some 0< "
1(1.9) Z 1 0 Z 1 x;1
u
(y
)exp;4 ;1(1 ;"
)qxy
]dy
1 q Z x 2 ;1xv
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
and the condition (1
:
7) is satised.Following S. Blom Bm] for
q < p
, the boundedness L :L
pv !L
qu holdswhenever (1.10) Z 1 0 Z 1 x;1
u
(y
)exp;xy
]dy
1 q Z x 0v
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
and the condition (1
:
7) is satised. So this author's result is improved in Theorem 4 wheneverq >
4.As we have announced above, Theorems 1 and 4 are based on a sucient condition for the boundedness
T
:L
pv!L
qu, which is described byTheorem 5.
Forp
q
, the boundednessT
:L
pv !L
qu holds if for someconstants
A >
0 and 0< "
1 (1.11) Z R 2 ;1Ru
(x
)dx
1 q Z R 0v
1;p 0 (y
)exp;4 ;1(1 ;"
)p
0Ry
;1]dy
1 p0A
for allR >
0.And for
q < p
, thenT
:L
pv!L
qu whenever(1.12) Z 1 0 Z x 2 ;1x
u
(z
)dz
1 p Z 2x 0v
1;p 0 (y
)exp;4 ;1(1 ;"
)p
0xy
;1]dy
1 p0 ru
(x
)dx <
1:
Actually the above boundedness holds under the condition (1.13) 1 X k=;1 Z 2 ;k 2 ;(k+1)
u
(x
)dx
1 q Z 2 ;k 0v
1;p 0 (y
)exp;4 ;1(1 ;"
)p
02;ky
;1]dy
1 p0 r<
1 which is implied by (1:
12).A necessary condition for
T
:L
pv !L
qu to be true is that(1.14) Z R 2 ;1R
u
(x
)dx
1 q Z 2 ;1R 0v
1;p 0 (y
)exp;p
0Ry
;1 ]dy
1 p0A
for allR >
0.Condition (1
:
11) seems new and quite dierent from those given in An-Hg], An] and Hz1]. Moreover it appears not too far from the necessary condition (1:
14).While this paper was submitted, the author has been extended the present approach to treat boundednesses problems for the two-dimensional Laplace transform.
x
2 Proofs of Results
First we prove Theorem 1, Lemma 2 and Theorem 4. And next we give the proof of Theorem 5 on which Theorems 1 and 4 are based.
Proof of Theorem 1
The Necessary Part.
Suppose that L :
L
pv !L
qu. To get condition (1:
4), letR >
0 andf
(:
) anonnegative function whose support is included in ]0
R
. Since for 0< x <
1(L
f
)(x
) = Z R 0f
(y
)exp;yx
]dy
exp;Rx
] Z R 0f
(y
)dy
inequality (1:
1) yields Z R 0f
(y
)dy
q Z 1 R;1u
(x
)exp;qRx
]dx
C
q Z R 0f
p(x
)v
(x
)dx
qp:
Takingf
(:
) =v
1;p 0(
:
) on its support ]0R
and using the identity (1;p
0)
p
+1 =(1;
p
0) then condition (1
:
4) appears immediately.To get condition (1
:
3) the key is to observe that (Lf
)(x
)exp(;1)](Hf
)(x
;1) = exp( ;1)] Z x ;1 0f
(y
)dy
0< x <
1:
So inequality (1
:
1) yieldsH
:L
pv !L
qw withw
(x
) =u
(x
;1)x
;2:
As it is well-known in Oc-Kr], this last boundedness implies that
(2.1) Z 1 R
w
(x
)dx
1 q Z R 0v
1;p 0 (y
)dy
1 p0A
for allR >
0and for some xed constant
A >
0. By the denition ofw
(:
), (2:
1) is nothing else than condition (1:
3).The SucientPart.
As used in An-Hg] and Hg], the main point is to observe that (L
f
)(x
)
(
Hf
)(x
;1) + (T
f
)(x
;1)0
< x <
1where the dual
T
of the operatorT
is dened by(
T
f
)(y
) = Z 1 yf
(x
)exp;xy
;1]dx
0< y <
1and Z 1 0 (
T
f
)(y
)g
(y
)dy
= Z 1 0f
(x
)(Tg
)(x
)dx
. With the above observation to derive (1:
1) it remains to get(2.2.)
H
:L
pv !L
qw andT
:
L
pv !L
qwAs it is known in Oc-Kr], for
p
q
, the boundednessH
:L
pv!L
qw holdswhenever the test condition (2
:
1) (which is the same as (1:
3)) holds. By duality arguments,T
:L
pv !L
qw is equivalent toT
:L
q0 w1;q 0 !L
p0 v1;p 0.And by Theorem 5, this last boundedness holds whenever for some
A >
0 and 0< "
1 Z R 0w
(y
)exp;4 ;1(1 ;"
)qRy
;1]dy
1 q Z R 2 ;1Rv
1;p 0 (x
)dx
1 p0A
for all
R >
0. Using the denition ofw
(:
), clearly the validity of this last inequality is ensured by condition (1:
5).Proof of Lemma 2
Condition (1
:
3) implies (1:
5) wheneveru
(:
) is a decreasing function since forR >
0 Z 1 R;1u
(x
)exp;4 ;1(1 ;"
)qRx
]dx
u
(R
;1) Z 1 R;1 exp;4 ;1(1 ;"
)qRx
]dx
=u
(R
;1 )R
;1 Z 1 1 exp;4 ;1 (1;"
)qz
]dz
=c
1u
(R
;1 )R
;1c
1 Z R ;1 0u
(y
)dy:
The implication (1
:
3) =)(1:
5) forv
1;p0
(
:
)2D
can be seen as follows Z 1 R;1u
(x
)exp;4 ;1(1 ;"
)qRx
]dx
Z R 2 ;1Rv
1;p 0 (y
)dy
qp 0 = 1 X k=1 Z 2kR ;1 2k ;1R;1u
(x
)exp;4 ;1(1 ;"
)qRx
]dx
Z R 2 ;1Rv
1;p 0 (y
)dy
qp 0 1 X k=1 h exp;8 ;1(1 ;"
)q
2k] i Z 2kR ;1 2k ;1R;1u
(x
)dx
Z R 0v
1;p 0 (y
)dy
qp 0 1 X k=1 2kh exp;8 ;1(1 ;"
)q
2k] i Z 2kR ;1 0u
(x
)dx
Z 2 ;kR 0v
1;p 0 (y
)dy
qp 0A
q 1 X k=1 2kh exp;8 ;1(1 ;"
)q
2k] i =c
2A
q:
Here we have used the fact that
w
(:
)2D
implies that for some>
0 Z 2mR 0w
(y
)dy
2m Z R 0w
(y
)dy
for allR >
0 and all integersm
1:
The implication (1
:
3) =) (1:
5), foru
(:
) 2D
, can be easily seen as above.This implication is also true provided
v
(:
) is an increasing function, since in this casev
1;p0
(
:
) is a decreasing function and consequentlyv
1;p 0(
:
)2D
.Proof of Theorem 4
The Necessary part
As in the proof of Theorem 1, to get the condition (1
:
7), the key is to observe that the boundedness L :L
pv !L
qu implies the weighted Hardy inequalityH
:L
pv!L
qw withw
(:
) dened as in (2:
2). As it is well-known in Oc-Kr], forq < p
, this last boundedness implies that(2.3) Z 1 0 Z 1 x
w
(y
)dy
1 q Z x 0v
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
Using the denition of
w
(:
), then (2:
3) is nothing else than condition (1:
7). Condition (1:
8) can be derived immediatly from (Lu
)(z
)(T
u
)(z
;1) and the inequality Z 1 0 Lu
)(qx
) 1 q Z x 0v
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
The fact that this last is a necessary condition forL:
L
pv!L
qu was proved inBm].
The Sucientpart.
As we have explained in the proof of Theorem 1, it remains to get the boundednesses of
H
andT
as in (2:
2).For
q < p
, again by well-known results as written in Oc-Kr], thenH
:L
pv!L
qw whenever the test condition (2:
3) (which is the same as (1:
7)) holds.Theorem 5 applied to
q < p
andT
:L
q0w1;q 0
!
L
p0
v1;p
0, lead to state that
the boundedness
T
:L
pv !L
qw holds whenever Z 1 0 Z x 2 ;1xv
1;p 0 (y
)dy
1 q0 Z 2x 0w
(z
)exp;4 ;1(1 ;"
)qxz
;1]dz
1 qrv
1;p 0 (x
)dx <
1:
Clearly, after using the denition of
w
(:
), the validity of this inequality is equivalent to Z 1 0 Z 1 (2x) ;1u
(y
)exp;4 ;1 (1;"
)qxy
]dy
1 q Z x 2 ;1xv
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
This inequality will be a consequence of condition (1
:
9) and Z 1 0 Z x;1 (2x) ;1u
(y
)exp;4 ;1(1 ;"
)qxy
]dy
1 q Z x 2 ;1xv
1;p 0 (z
)dz
1 q0 rv
1;p 0 (x
)dx <
1:
This last inequality is true because of condition (1
:
7) and exp;4 ;1(1 ;"
)qxy
]<
exp;8 ;1(1 ;"
)q
] whenever (2x
) ;1< y < x
;1.Proof of Theorem 5
The Sucientpart
The main point to get
T
:L
pv!L
qu is to cut the operator as(2.4) Z 1 0 (
Tf
)q(x
)u
(x
)dx
1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
1 p k qfor all functions
f
(:
)0. Here j = exp;"
2 j] with 0< "
1 and k=k(pqvu
) = Z 2 ;k 2 ;(1+k)u
(y
)dy
1 q Z 2 ;k 0v
1;p 0 (x
)exp;4 ;1(1 ;"
)p
02;kx
;1]dx
1 p0:
It can be noted that 1 X
l=0
l =c
0<
1. We will postpone below the proof of
(2
:
4).Now consider the case
p
q
. By condition (1:
11) (withR
= 2;k) then
k
A:
Using this last fact and the cutting out (2
:
4), the boundednessT
:L
pv !L
quappears as follows Z 1 0 (
Tf
)q(x
)u
(x
)dx
1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
1 p k qA
q 1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
1 pqA
q 1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
qp h 1 X l=0 l iqp 0c
1A
q 1 X j=0 j 1 X k=;1 Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
qp sinceq
p
1c
2A
qh 1 X j=0 j iqp Z 1 0f
p(x
)v
(x
)dx
qp =c
3A
q Z 1 0f
p(x
)v
(x
)dx
qp:
For the case
q < p
, observe that by condition (1:
13)1
X
k=;1
rk
< A
r:
We will dier below the proof of the fact that (1
:
13) is implied by condition (1:
12).Using this observation, the boundedness
T
:L
pv!L
qu appears as follows Z 1 0 (Tf
)q(x
)u
(x
)dx
1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
1 pkq 1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
qp qkh 1 X l=0 l iqp 0c
1 1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
qp qkc
1 1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
qp 1 X m=;1 rm 1;qpc
1A
q 1 X j=0 j 1 X k=;1 Z 2 ;(j+k) 2 ;(2+j+k)f
p(x
)v
(x
)dx
qpc
2A
qh 1 X j=0 j iqp Z 1 0f
p(x
)v
(x
)dx
qpc
3A
q Z 1 0f
p(x
)v
(x
)dx
qp:
It is now time to prove (2
:
4). This inequality is true sinceZ 1 0 (
Tf
)q(x
)u
(x
)dx
= Z 1 0 h 1 X j=0 Z 2 ;jx 2 ;(j+1)xf
(y
)exp;xy
;1]dy
iqu
(x
)dx
= 1 X k=;1 Z 2 ;k 2 ;(k+1) h 1 X j=0 Z 2 ;jx 2 ;(j+1)x
f
(y
)exp;xy
;1]dy
iqu
(x
)dx
1 X k=;1 1 X j=0 Z 2 ;(j+k) 2 ;(j+2+k)f
(y
)dy
exp;2j] q Z 2 ;k 2 ;(k+1)u
(x
)dx
1 X k=;1 1 X j=0 Z 2 ;(j+k) 2 ;(j+2+k)f
p(y
)v
(y
)dy
1 p Z 2 ;(j+k) 2 ;(j+2+k)v
1;p 0 (z
)dz
1 p0 Z 2 ;k 2 ;(k+1)u
(x
)dx
1 q exp ;2j] q 1 X k=;1 1 X j=0 exp;"
2j] Z 2 ;(j+k) 2 ;(j+2+k)f
p(y
)v
(y
)dy
1 p Z 2 ;(j+k) 2 ;(j+2+k)v
1;p 0 (z
)exp;4 ;1 (1;"
)p
0 2;kz
;1 ]dz
1 p0 Z 2 ;k 2 ;(k+1)u
(x
)dx
1 qq 1 X k=;1 1 X j=0 exp;"
2j] Z 2 ;(j+k) 2 ;(j+2+k)f
p(y
)v
(y
)dy
1 p Z 2 ;k 0v
1;p 0 (z
)exp;4 ;1(1 ;"
)p
02;kz
;1]dz
1 p0 Z 2 ;k 2 ;(k+1)u
(x
)dx
1 qq = 1 X k=;1 1 X j=0 j Z 2 ;(j+k) 2 ;(j+2+k)f
p(y
)v
(y
)dy
1 p k q:
Now we can show how does the condition (1
:
12) imply (1:
13). Here the main point is the elementary equalityZ b a
u
(z
)dz
rq =r
q
Z b a Z x au
(z
)dz
rpu
(x
)dx
0< a < b <
1:
For shortness the term 4;1(1 ;
"
)p
0 is merely denoted by
C
. The implication(1
:
12) =)(1:
13) can be seen as follows 1 X k=;1 Z 2 ;k 0v
1;p 0 (y
)exp;C
2 ;ky
;1]dy
1 p0 Z 2 ;k 2 ;(k+1)u
(x
)dx
1 qr 1 X k=;1 Z 2 ;k 0v
1;p 0 (y
)exp;C
2 ;ky
;1]dy
rp 0 Z 2 ;k 2 ;(k+1) Z x 2 ;(k+1)u
(z
)dz
rpu
(x
)dx
= 1 X k=;1 Z 2 ;k 2 ;(k+1) Z x 2 ;(k+1)
u
(z
)dz
1 p Z 2 ;k 0v
1;p 0 (y
)exp;C
2 ;ky
;1]dy
1 p0 ru
(x
)dx
1 X k=;1 Z 2 ;k 2 ;(k+1) Z x 2 ;1xu
(z
)dz
1 p Z 2x 0v
1;p 0 (y
)exp;Cxy
;1]dy
1 p0 ru
(x
)dx
= Z 1 0 Z x 2 ;1xu
(z
)dz
1 p Z 2x 0v
1;p 0 (y
)exp;Cxy
;1 ]dy
1 p0 ru
(x
)dx
A
r by using (1:
12):
The Necessary part
To get the condition (1
:
14) from the boundednessT
:L
pv !L
qu, by a dualityargument, it can be assumed that
T
:L
q 0 u1;q 0 !L
p0 v1;p 0 or equivalently for some constantC >
0 Z 1 0 h Z 1 yf
(x
)u
(x
)exp;xy
;1]dx
ip 0v
1;p 0 (y
)dy
1 p0C
Z 1 0f
q0 (z
)u
(z
)dz
1 q0 for allf
(:
)0:
Take
R >
0 andf
(:
) a nonnegative function whose support is included ]2;1RR
. Since for 0< y <
2;1R
Z 1 yf
(x
)u
(x
)exp;xy
;1]dx
= Z R 2 ;1Rf
(x
)u
(x
)exp;xy
;1]dx
Z R 2 ;1Rf
(x
)u
(x
)dx
exp;Ry
;1]the above inequality yields
Z R 2 ;1R
f
(x
)u
(x
)dx
p 0 Z 2 ;1R 0v
1;p 0 (y
)exp;p
0Ry
;1]dy
C
p 0 Z R 2 ;1Rf
q0 (z
)u
(z
)dz
p 0 q0:
Taking
f
(:
) = 1 on its support ]2;1RR
then condition (1:
14) appearsimmedi-ately.
Acknowledgement
The author would like to thank the referee for his suggestions on improve-ments of some results in the preliminary version of this work.
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Yves Rakotondratsimba
Institut polytechnique St Louis, EPMI
13 bd de l'Hautil 95 092 Cergy Pontoise France E-mail: y.rakoto@ipsl.tethys-software.fr