Journal of Mathematical Analysis
ISSN: 2217-3412, URL: http://www.ilirias.com
Volume 1 Issue 2(2010), Pages 18-27
WEIGHTED INEQUALITIES FOR ONE-SIDED OPERATORS ON
EVEN AND NON-INCREASING FUNCTIONS
(COMMUNICATED BY LARS-ERIK PERSSON)
ÁLVARO CORVALÁN AND LILIANA DE ROSA
Abstract. The purpose of this paper is to find necessary conditions and sufficient conditions on the pair of weights (u, v) for the boundedness of the
+
one-sided operators Nα,β
from Lpdec (v) into Lq (u) in the case 1 < p ≤ q < ∞.
1. Introduction and Result
Let 0 ≤ β ≤ α ≤ 1. For each function f locally integrable on R are defined the
operators,
Z x+h
1
|f (s)|
+
Nα,β
(f )(x) = sup β
ds.
(s − x)1−α
h>0 h
x
These operators were considered by F. J. Martı́n–Reyes, in [9] (see also [11] and
[8]).
+
In the case β = 1 = α the operator N1,1
is equal to the one-sided maximal Hardy+
+
Littlewood operator M and, Nα,β is equivalent to M + whenever 0 < β = α ≤ 1.
+
+
If 0 < β < α = 1, N1,β
= M1−β
is the one-sided fractional maximal operator of
+
order 1 − β, and Nα,β is equivalent to a one-sided fractional maximal operator in
the cases 0 < β < α ≤ 1.
+
When 0 = β < α < 1 we have that, Nα,0
is the one-sided fractional integral
+
operator Iα .
F. J. Martı́n–Reyes and A. de la Torre gave, in [10], a condition on the pair of
weights (u, v) equivalent to the boundedness of the one-sided fractional maximal
operator from Lp (v) into Lq (u), in the case 1 < p ≤ q < ∞, (see also [7] and [1]).
A. Meskhi obtained, in [12], a necessary and sufficient condition on the weight
+
v for the boundedness of the Weyl transform Wα (f )(x) = Nα,0
(f )(x), x > 0, from
1
p
q
L (v, R+ ) into L (R+ ), in the cases q0 < α < 1 or α > 1.
V. Kokilashvili and A. Meskhi gave, in [5], necessary and sufficient conditions
on the weight u for the boundedness of the one-sided fractional integral operator
from Lp to Lp (u), 1 < p < ∞, 0 < α < 1/p.
2000 Mathematics Subject Classification. 26A33, 42B25.
Key words and phrases. One-sided operators, weighted inequalities.
c
2010
Ilirias Publications, Prishtinë, Kosovë.
Submitted December 14, 2010. Published February 6, 2011.
18
A. Corvalán and L. de Rosa
19
M. Lorente characterized, in [6], the good weights for the boundedness of the
one-sided fractional integral operator from Lp (v) into Lq (u), 1 < p ≤ q < ∞. For
1 < p < q < ∞, this problem was studied by D. E. Edmunds, V. Kokilashvili and
A. Meskhi in [2].
V. Kokilashvili considered, in [4], the boundedness of the Weyl transform from
Lp (v) into Lq (u), (see also [3]).
The boundedness of the operators,
Z
1
|f (y)| dy
Mα (f )(x) = sup n−α
t>0 t
kx−yk<t
and
Z
f (y)
dy , (x ∈ Rn )
n−α
Rn kx − yk
from the space Lpdec (v) of all radial and non-increasing functions into Lq (u), was
studied by Y. Rakotondratsimba in [13], [14] and [15].
The conditions of our theorem imply the ones given by Y. Rakotondratsimba in
[13] and [14] for the maximal Hardy-Littlewood operator and the fractional maximal
operator, respectively. For the one-sided fractional integral operator, we give a new
1
sufficient condition which do not require the locally integrability of v − p−1 .
p
In the sequel, given a weight v, we denote Ldec (v) the set of all non-negative
functions in Lp (v), even and non-increasing on (0, ∞).
For any measurable set E, χE will denote the characteristic function of E.
If p > 1, its conjugate exponent will be denoted by p0 .
Throughout this paper, the letters A and C will denote positive constants, not
necessarily the same at each occurrence.
With these notations we state our result.
Iα (f )(x) =
Theorem 1.1. Let β = α = 1, 0 < β < α = 1 or 0 = β < α < 1, and 1 < p ≤ q <
∞. If there exists a finite constant A > 0 such that for every R > 0 the inequalities,
!1/q
!1/p
Z
Z
1
α
u(x)
+
−β
R q q0
dx
≤A
v(x)dx
(1.1)
1−α
R
|x|≤R |x|
2 ≤|x|≤R
and,
Z
−R
−∞
u(x)
|x|(1−α+β)q
!1/q Z
dx

R
Z
|x|
v(s)ds
−R
−|x|
1/p0
!−p0
0
|x|p v(x)dx
≤A
(1.2)
hold then, for every non-negative function f, even and non-increasing on (0, ∞) the
inequality,
Z ∞
1/q
Z ∞
1/p
+
Nα,β
(f )(x)q u(x)dx
≤C
f (x)p v(x)dx
−∞
−∞
holds. Reciprocally, if the operator
is bounded from Lpdec (v) into Lq (u) then,
there exists a finite constant A > 0 such that for every R > 0 the inequalities (1.2)
and,
!1/q
!1/p
Z R/2
Z
α
1
u(x)
+
−β
0
dx
≤A
v(x)dx
Rq q
1−α
−R (R + |x|)
|x|≤R
+
Nα,β
hold.
20
One-sided operators
Remark 1.2. If β = α = 1 the condition (1.1) can be substituted by,
!1/q
!1/p
Z
Z
u(x)dx
≤A
.
v(x)dx
|x|≤R
|x|≤R
Remark 1.3. Let be the weights u(x) = |x|δ−1 and v(x) = |x|γ−1 where 1 − α <
δ < (1 − α + β)q and 0 < γ < p. In the cases β = α = 1, 0 < β < α = 1 or
0 = β < α < 1 the pair of weights (u, v) satisfies the conditions (1.1) and (1.2) of
our theorem if, and only if, α − β + qδ = γp .
2. Proof of the Theorem
In order to prove Theorem 1.1, we need the following definitions.
Definition 2.1. Let 0 ≤ β ≤ α ≤ 1. For each non-negative and locally integrable
function f on R we define the operator,
Z |x|
1
+
Hα,β (f )(x) =
f (s)ds.
|x|1−α+β 0
Definition 2.2. Let 0 ≤ β ≤ α ≤ 1. For each non-negative function f we define
for each x > 0 the operator,
x α−β x
+
Tα,β
(f )(x) = sup j
f x+ j
2
j≥0 2
whenever β > 0 and,
+
Tα,β
(f )(x)
∞ X
x
x α f
x
+
=
2j
2j
j=0
in the case β = 0.
Definition 2.3. Let 0 ≤ β ≤ α ≤ 1. For each non-negative and locally integrable
function f on R we define the operator,
Z 2j+1 |x|
1
+
Nα,β (f )(x) = sup j
f (s)ds
1−α+β
j≥1 (2 |x|)
2j |x|
whenever 0 < β < α = 1,
+
Nα,β
(f )(x) =
∞
X
j=1
1
j
(2 |x|)1−α
Z
2j+1 |x|
f (s)ds
2j |x|
+
in the case 0 = β < α < 1 and, if β = α = 1 then, Nα,β
(f )(x) = 0.
With these notationes we state the following proposition.
Proposition 2.4. Let β = α = 1, 0 < β < α = 1 or 0 = β < α < 1. There
exist constants ci , 1 ≤ i ≤ 6, which only depend on α and β, such that for every
non-negative function f, even and non-increasing on (0, ∞) the inequalities,
+
+
+
Nα,β
(f )(x) ≤ c1 Hα,β
(f )(x)χ(−∞,0) (x) + c2 Tα,β
(f )(x)χ(0,∞) (x)
(2.1)
+
+c3 Nα,β
(f )(x)
and,
+
+
+
Nα,β
(f )(x) ≥ c4 Hα,β
(f )(x)χ(−∞,0) (x) + c5 Tα,β
(f )(x)χ(0,∞) (x)
(2.2)
A. Corvalán and L. de Rosa
21
+
+c6 Nα,β
(f )(x)
hold.
Proof. The case β = α = 1 is trivial. We assume that 0 < β < α = 1 or
0 = β < α < 1.
First of all we will prove (2.1). Suppose that x > 0. If 0 < h < 2x then, there
x
. In the case 0 < β < α = 1 we have that,
exists j ≥ 0 such that 2xj ≤ h < 2j−1
1
hβ
Z
x+h
2j
x
f (s)ds ≤
x
≤
β X
∞ Z
i=j−1
∞ X
x+
x+
x
2i
β 1
21+i−j
i=j−1
f (s)ds
x
2i+1
x 1−β
2i+1
x +
f x + i+1 ≤ CT1,β
(f )(x).
2
For 0 = β < α < 1, we obtain that,
Z x+h
∞ Z x+ xi
X
2
f (s)
f (s)
ds
≤
ds
1−α
x
(s
−
x)
(s
−
x)1−α
x
i=j−1 x+ i+1
2
≤
∞ X
x
x α +
f
x
+
≤ Tα,0
(f )(x).
2k
2k
k=j
If 2x ≤ h we have that,
Z x+h
1
f (s)
ds ≤
hβ x
(s − x)1−α
=
Z 2x
Z x+h
1
f (s)
1
f (s)
ds + β
ds
(2x)β x (s − x)1−α
h 2x (s − x)1−α
I + II.
In the case 0 < β < α = 1, we obtain the estimates,
x
∞ Z
∞
1 X x+ 2i
1 X x
x I ≤
f
(s)ds
≤
f
x
+
x
(2x)β i=0 x+ i+1
(2x)β i=0 2i+1
2i+1
2
∞
β
1
1 X
+
+
≤
T1,β
(f )(x) = CT1,β
(f )(x).
2β i=0 2i+1
When 0 = β < α < 1, we have,
∞ ∞ Z x+ xi
X
X
2
f (s)
x α x
+
I≤
ds
≤
f
x
+
≤ Tα,0
(f )(x).
1−α
k
k
x
(s
−
x)
2
2
x+
i+1
i=0
k=1
2
In order to estimate B we observe that if 2x ≤ h then, there exists j ≥ 1 such that
2j x ≤ h < 2j+1 x. For 0 < β < α = 1, we have that,
II
≤
≤
1
j
(2 x)β
j+1
X
k=1
Z
2j+2 x
f (s)ds ≤
2x
j+1
X
k=1
1
1
j−k
β
k
(2 ) (2 x)β
1
+
N + (f )(x) ≤ CN1,β
(f )(x).
(2j−k )β 1,β
Z
2k+1 x
f (s)ds
2k x
22
One-sided operators
In the case 0 = β < α < 1, we obtain the bounds,
2j+2 x
Z
II
≤
2x
21−α
≤
j+1 Z 2k+1 x
X
f (s)
ds ≤ 21−α
1−α
(s − x)
k=1
j+1
X
k=1
Z
1
(2k x)1−α
2k x
f (s)
ds
s1−α
2k+1 x
2k x
+
f (s)ds ≤ CNα,0
(f )(x).
In summary, for each x > 0 we have the estimate,
+
+
+
Nα,β
(f )(x) ≤ c2 Tα,β
(f )(x) + c3 Nα,β
(f )(x).
Now, let us consider x < 0. If 0 < h ≤ 3|x| it is easy to check that,
Z x+h
f (s)
1
+
ds ≤ CHα,β
(f )(x).
β
h x
(s − x)1−α
If 3|x| < h we can decompose,
=
1
hβ
Z
0
x
x+h
1
hβ
Z
f (s)
1
ds + β
(s − x)1−α
h
Z
x
2|x|
0
f (s)
ds
(s − x)1−α
f (s)
1
ds + β
(s − x)1−α
h
Z
x+h
2|x|
f (s)
ds
(s − x)1−α
= III + IV + V.
We have the estimates,
III ≤ C
|x|/2
Z
1
|x|1−α+β
+
f (s)ds ≤ CHα,β
(f )(x),
0
and,
IV ≤
1
1
(3|x|)β |x|1−α
2|x|
Z
0
+
f (s)ds ≤ CHα,β
(f )(x).
If 3|x| < h then, there exist j ≥ 1 such that 2j |x| < h ≤ 2j+1 |x|. In the case
0 < β < α = 1, we have the bounds,
j−k
j Z 2k+1 |x|
j X
X
1
1
+
+
V ≤ j
f (s)ds ≤
N1,β
(f )(x) ≤ CN1,β
(f )(x).
β
(2 |x|)β
2
k
2 |x|
k=1
k=1
If 0 = β < α < 1 then,
Z
2j+1 |x|
V ≤
2|x|
j
X
f (s)
1
ds ≤
1−α
k
(s − x)
(2 |x|)1−α
k=1
Z
2k+1 |x|
2k |x|
+
f (s)ds ≤ Nα,0
(f )(x).
In consequence, for every x < 0 we have that,
+
+
+
Nα,β
(f )(x) ≤ c1 Hα,β
(f )(x) + c3 Nα,β
(f )(x).
This concludes the proof of (2.1).
The proof of (2.2) is very simple and it will be omitted.
A. Corvalán and L. de Rosa
23
Lemma 2.5. Let 0 ≤ β ≤ α ≤ 1 and 1 < p ≤ q < ∞. The inequality,
1/p
Z 0
1/q
Z ∞
+
p
q
Hα,β (f )(x) u(x)dx
f (x) v(x)dx
≤C
−∞
(2.3)
−∞
holds, for every non-negative function f, even and non-increasing on (0, ∞) if,
and only if, there exists a finite constant A > 0, such that for every R > 0 the
inequalities,
!1/p
1/q
Z 0
Z R
(α−β)q
v(x)dx
|x|
u(x)dx
≤A
−R
−R
and
Z
−R
!1/q Z
u(x)
|x|(1−α+β)q
−∞
dx

R
Z
0
|x|p v(x)dx
v(s)ds
−|x|
−R
1/p0
!−p0
|x|
≤A
hold.
Proof. The inequality (2.3) is equivalent to,
!q
!1/q
Z ∞
1/p
Z |x|
Z ∞
1
(α−β)q
p
f (s)ds |x|
u(x)dx
≤C
f (x) v(x)dx
|x| 0
0
0
(2.4)
where u(x) = u(−x) and v(x) = v(x) + v(−x). By Theorem 2 in [16], the inequality
(2.4) holds if, and only if,
!1/q Z
!−1/p
Z R
R
(α−β)q
sup
u(x)dx
v(x)dx
<∞
|x|
R>0
0
0
and
Z
∞
sup
R>0
u(x)
|x|(1−α+β)q
R
1/q Z
R
Z
dx
0
x
!1/p0
−p0
p0
v(s)ds
x v(x)dx
<∞
0
hold, which are the conditions given in our lemma.
Lemma 2.6. Let 0 ≤ β ≤ α ≤ 1 and 0 < p ≤ q < ∞. If there exists a finite
constant A > 0 such that for every R > 0 the inequality,
!1/q
!1/p
Z R
Z R
(α−β)q
x
u(x)dx
≤A
v(x)dx
(2.5)
0
−R
holds, then for every non-negative function f, even and non-increasing on (0, ∞)
we have,
Z ∞
1/q
Z ∞
1/p
+
q
p
Tα,β (f )(x) u(x)dx
≤C
f (x) v(x)dx
.
(2.6)
−∞
0
Reciprocally, if (2.6) holds then, for every R > 0 the inequality,
!1/q
!1/p
Z R/2
Z R
x(α−β)q u(x)dx
≤A
v(x)dx
0
holds.
−R
(2.7)
24
One-sided operators
Proof. It is easy to prove that there exists a constant C > 0 such that for every
x > 0 the inequality,
+
Tα,β
(f )(x) ≤ Cxα−β f (x)
holds. Then, by Proposition 1 part (a), in [16], the condition (2.5) on the pair of
weights implies that the inequality (2.6) holds.
Reciprocally, taking f = χ(−R,R) it can be seen that there exists a constant
C > 0 such that for every 0 < x < R/2 the inequality,
+
Tα,β
(f )(x) ≥ Cxα−β
holds. Then, (2.6) implies (2.7).
Proposition 2.7. Let 0 = β < α < 1 or 0 < β < α = 1, and 1 < p ≤ q < ∞. If
there exists a finite constant A > 0 such that for every R > 0 the inequality,
R
α
1
q + q0
Z
−β
|x|≤R
u(x)
dx
|x|1−α
!1/q
!1/p
Z
≤A
v(x)dx
(2.8)
R
2 ≤|x|≤R
holds then, for every non-negative function f, even and non-increasing on (0, ∞)
the inequality,
Z
1/q
∞
+
Nα,β
(f )(x)q u(x)dx
−∞
Z
1/p
f (x) v(x)dx
∞
p
≤C
−∞
+
holds. Reciprocally, if the operator Nα,β
is bounded from Lpdec (v) into Lq (u) then,
there exists a finite constant A > 0 such that for every R > 0 the inequality,
R
1
α
q + q0
−β
Z
|x|≤R/4
u(x)
dx
(R + |x|)1−α
!1/q
!1/p
Z
≤A
v(x)dx
(2.9)
|x|≤R
holds.
Proof. The case 0 < β < α = 1 is similar to Theorem in [14] and we include it
for completeness. We have that,
Z ∞
+
N1,β
(f )(x)q u(x)dx
(2.10)
−∞
Z
∞
=
sup
−∞ j∈N
≤
XZ
i∈Z
≤
Z
1
j
(2 |x|)β
f (s)ds
sup
i∈Z j∈N
1
(2j+i )β
u(x)dx
2j |x|
2i <|x|≤2i+1 j∈N
XX
!q
2j+1 |x|
Z
1
(2j+i )β
2j+i+2
Z
2j+i+2
!q
f (s)ds
!q Z
f (s)ds
2j+i
u(x)dx
2j+i
u(x)dx.
2i <|x|≤2i+1
A. Corvalán and L. de Rosa
25
Changing indexes and the order of the sums we obtain the bounds,
X
(2.10) ≤
−mβ
Z
C
m−1
X
f (s)ds
2
2m
m∈Z
≤
!q
2m+2
X
Z
u(x)dx
i=−∞
f (2m )q 2m(1−β)q
2i <|x|≤2i+1
Z
u(x)dx.
|x|≤2m
m∈Z
Keeping in mind the condition (2.8), since 1 < p ≤ q < ∞ and f is a non-increasing
function on (0, ∞) we have,
q
≤ CA
(2.10)
X
m q
v(x)dx
f (2 )
2m−1 ≤|x|≤2m
m∈Z
≤ CAq
!q/p
Z
X
f (x)p v(x)dx
2m−1 ≤|x|≤2m
m∈Z
!q/p
XZ
q
≤ CA
m∈Z
q
!q/p
Z
Z
p
f (x) v(x)dx
2m−1 ≤|x|≤2m
q/p
∞
p
≤ CA
f (x) v(x)dx
,
−∞
which proves this case.
Now, suppose that 0 = β < α < 1. We have that,
Z
∞
−∞
∞
∞
X
−∞
k=1
Z
=
1
k
(2 |x|)1−α
∞
X
XZ
≤
i∈Z
+
Nα,0
(f )(x)q u(x)dx
2i <|x|≤2i+1
k=1
Z
(2.11)
!q
2k+1 |x|
f (s)ds
u(x)dx
2k |x|
1
(2k+i )1−α
Z
!q
2k+i+2
f (s)ds
u(x)dx.
2k+i
Applying Hölder’s inequality and recalling that f is a non-increasing function on
(0, ∞), it follows that,
(2.11)
≤
C
XZ
i∈Z
≤
C
X
i∈Z
1
2i <|x|≤2i+1
1
2i(1−α)
Z
2i(1−α)
∞
X
(k+i)
2
k=1
∞
X
2i <|x|≤2i+1 k=1
Z
1
(1−α)
q0
(k+i)q(1−
2
2k+i+2
!q
f (s)ds
2k+i
(1−α)
)
q0
f (2i+k )q u(x)dx.
u(x)dx
26
One-sided operators
Changing indexes and the order of the sums we obtain that,
Z
m−1
X mq( 1 + α )
X
1
q
q 0 f (2m )q
(2.11) ≤ C
2
u(x)dx
2i(1−α) 2i <|x|≤2i+1
i=−∞
m∈Z
≤ C
X
f (2m )q 2
mq( q1 + qα0 )
Z
|x|≤2m
m∈Z
u(x)
dx.
|x|1−α
Taking into account the condition (2.8) on the pair of weights (u, v) and the hypothesis 1 < p ≤ q < ∞ we have,
!q/p
Z
X
q
m q
f (2 )
v(x)dx
(2.11) ≤ CA
2m−1 ≤|x|≤2m
m∈Z
q
≤ CA
Z
X
f (x) v(x)dx
2m−1 ≤|x|≤2m
m∈Z
≤ CAq
XZ
m∈Z
≤ CAq
Z
!q/p
p
∞
!q/p
f (x)p v(x)dx
2m−1 ≤|x|≤2m
q/p
f (x)p v(x)dx
,
−∞
as we wanted to prove.
+
Reciprocally, if Nα,β
is bounded from Lpdec (v) into Lq (u), taking f = χ(−R,R) it
is obtained (2.9).
Proof of Theorem 1.1. It is a consequence of Proposition 2.4, Lemmas 2.5
and 2.6, and Proposition 2.7.
Acknowledgments. The authors would like to thank the anonymous referee for
his/her comments that helped us improve this article.
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Álvaro Corvalán
Instituto del Desarrollo Humano, Universidad Nacional de General Sarmiento, Juan
Marı́a Gutiérrez 1150, Los Polvorines, Malvinas Argentinas, 1613 Provincia de Buenos
Aires, Argentina.
E-mail address: [email protected]
Liliana de Rosa
Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad
de Buenos Aires, Ciudad Universitaria, Pabellón I 1428 Ciudad de Buenos Aires. Argentina.
E-mail address: liliana de [email protected]