1. Introduction
For
, let
be the
d-dimensional Euclidean space and let
denote the unit sphere in
equipped with the normalized surface measure
. Let
be a homogeneous function of degree 0 on
which satisfies the cancellation property:
For
and
, we define the spherical averages:
and the corresponding maximal operator by:
An application of Minkowski’s inequality for integrals shows that the linear operators are bounded on with norms less than or equal to ; hence, is well-defined for almost every whenever .
Observe that when
,
is a constant multiple of the average value of
f on the sphere
. Spherical means are used to reformulate PDE’s in the form of integral equations whose kernels are generalized functions [
1]. Moreover, the spherical averages
play a crucial rule in solving certain types of partial differential equations that have several methods to solve, see [
2,
3]. For instance, in
, the function:
is a solution of the Darboux equation:
The corresponding maximal function is given by:
Many authors studied the boundedness issue of the maximal operator
in order to attain the point-wise convergence of the spherical averages. For instance, Stein [
4] proved that, for
, the maximal spherical operator is bounded on
if and only if
. After several years, the two-dimensional case was established independently by Bourgain [
5]. We mention here that different techniques were applied to conclude the theorems of Stein and Bourgain and others [
6]. For the former, we refer the reader to [
7,
8], and different proofs for the latter result can be found in [
9,
10,
11].
Chen and Lin [
12] introduced the maximal operator:
where
K is the closed unit ball of the space
and:
The authors of [
12] proved the boundedness of the maximal operator
for
, whenever
is a continuous function on the unit sphere of
. Motivated by the observation that:
we study operators of the form:
In particular, an inequality connecting this class of operators with some well-known Marcinkiewicz integral operators is proved by using the classical Hardy inequality. Then, the boundedness of the operator is determined for some whenever belongs to a certain class of Lebesgue spaces. In addition, for the purpose of outlining the -boundedness of , a novel proof of the optimality condition is proposed on .
In view of the previous results, it is natural to ask whether we can obtain the boundedness of the operator
for some
whenever
, the space of bounded complex-valued measurable functions on
. It is our goal in
Section 2 to attain a partial answer to this inquiry by establishing several initial results including the boundedness of the operator
some
. In addition, by establishing the boundedness of
, Al-Salman [
13] improved the result of Chen and Lin and obtained the boundededness of
for
whenever
belongs to certain Zygmund classes. The relations among Zygmund classes and other Lebesgue space are clearly stated in
Section 3. In addition, Al-Salman [
13] proved that the condition
is optimal in order to obtain the
-boundedness of the operator
. More precisely, he proved that there exists a function
that lies in
for all
such that
in not bounded on
. However, in
Section 3, we use the classical Hardy inequality to deduce the same result by proposing a new simple proof that uses the previously known results.
2. A First Look at Mapping Properties of
For a dense subspace in
, we first prove that:
for every
. More precisely, we have the following result.
Theorem 1. The operator maps into the space of locally integrable functions on .
Proof. It is enough to show that for
there exists constants
such that
for all
. Firstly, by the cancellation property (
1), we have:
By the mean value theorem and since
f is a Schwartz function, there exists a constant
such that:
On the other hand, the estimate:
gives:
Finally, by Equations (
2) and (
3) and the simple inequality
such that
, we obtain the required result. □
Remark 1. We notice that in case of Theorem 1, we have:where f is a Schwartz function and . So, by the properties of -norms, we have: Next, we wish to investigate for what finite values of
the inequality:
holds for all
. In order to do this, we split the operator
into dyadic sub-operators as follows. Let
for
. Pick a radial smooth function
such that
if
and
if
. Now, for
, define:
It can be readily seen that
is supported in the annulus
and:
for all
with the convention
. Let
for a suitable Schwartz function
, where:
For
, define:
where:
We shall need the following lemmas which can be found in [
2] and references therein.
Lemma 1. Let that satisfies: for some positive constants and for all . Then: for and all . Moreover, we have: for all .
Lemma 2. Let . Then, the following inequality holds: As a consequence of the previous results, we have the following lemma.
Lemma 3. Suppose that and satisfies the mean value zero condition (1). Then, satisfies the conclusions (7) and (8) in Lemma 1. Proof. It is clear that we need only to verify the assumptions of Lemma 1 for
to get the required result. The case
can be treated in a similar manner. Obviously,
is integrable on
. To prove (
4), notice that:
Since
is a Schwartz function, an application of Lemma 2 gives (5). More precisely, for
, we have:
Finally, by using the mean value theorem we get:
and thus we have (6) from which the result follows. □
Lemma 4. Let Ω
be as in Lemma 3. Then:for some and all . Proof. Firstly, by using Plancherel and Fubini theorems, we have:
Now, an application of Van der Corput’s Lemma gives:
for some
. Finally, by combining (
9) and (
10) we deduce our result. □
Now, we are ready to conclude the main result of this section. More precisely, in [
13] it was shown that the operator
is bounded for
. In the following Theorem, we attain the boundedness for
.
Theorem 2. Let Ω
be as in Lemma 3. Then: for .
Proof. Lemma 3 guarantees the boundedness of
for all
. So, we need only to take care of the sum on most right-hand side in (
11). Interpolating between the
and
estimates in Lemmas 3 and 4 yields:
for
and
. Finally, combining the fact that
is decreasing on
with (
11) and (
12) ends the proof. □
3. -Boundedness of
In this section, we introduce a new proof of the
-boundedness of
. The main feature of our proof is inequality (
24) which reveals the relationship between the spherical operators and the well-known parametric Marcinkiewicz integral operators which we derive by using the classical Hardy inequality. We start this section by deriving a multiplier formula for
and as a consequence of this derivation, we find a necessary and sufficient condition on
which guarantee the
-boundedness of
. We emphasis here that the proof of the next result follows the same lines of the one in [
13] but in great detail.
Theorem 3. Suppose that satisfies (1). Then, the operator is bounded on iff: Proof. In this proof, we follow the same lines as the one in [
14]. By Plancherel and Fubini theorems, we have:
On the other hand, consider:
By combining (
14) and (
15) and applying Lebesgue dominated convergence theorem, we obtain:
The sufficiency of (
13) is obvious. Assume that
is bounded on
and notice that:
The fact that Fourier transform is an isometry on
enables us to find a function
such that
and so an application of Lebesgue differentiation theorem proves the necessity of (
13). □
Remark 2. The following inclusions among Zygmund classes and Lebesgue spaces hold and are proper. For and , we have: The following lemma can be found in [
15].
Lemma 5. Let be satisfy (1). Then: Remark 3. Equation (14) and Lemma 5 show that is bounded on whenever . Therefore, by the relations among Zygmund classes it is natural to ask what is the optimal α such that is bounded on whenever . As a first step in answering the inquiry in Remark 3, we need to recall the decomposition.
Lemma 6. Suppose that for some , and satisfies the mean value zero condition. Then there exists a subset of , a sequence of nonnegative real numbers, and a sequence of functions in such that:
- (i)
for ,
- (ii)
,
- (iii)
for ,
- (iv)
,
- (v)
.
Theorem 4. Let be satisfy (1). Then is bounded on . Moreover, the condition is optimal for to be -bounded. Proof. We follow the same proof of Theorem 2 with a slight modification. By Lemma 6, we have:
In view of Lemma 3 and Remarks 3 and 6, we need only to show that:
Now, interpolating between the simple estimates:
and:
yields:
and:
Therefore, (
18) and (
23) complete the first part of the proof. To prove the optimality of the condition
, assume that
is a homogeneous function of degree 0 and recall that the parametric Marcinkiewicz integral operator [
16] is given by:
for
. It can be easily seen that:
for almost every
. To see this, we need the following version of Hardy inequality [
17]:
where
F is a measurable function,
and
. Switching to polar coordinates gives:
Thus, we get (
24) by applying (
25) with:
Finally, Al-Salman [
18] proved that there exists
such that
could not be bounded on
. Therefore, this fact and (
24) complete the proof. □