— 1. A fractional Hörmander type multiplier theorem —
In a few months ago we deal with translation invariant operators
in
, see previous post. More precisely, was shown that there exists a tempered distribution
such that
. However, what does happens with
to get
-boundedness of
? Such questions belongs to a fruitiful area of Fourier analysis which many personages work in that, such as Marcinkiewicz, Minhklin, Hörmander, Lizorkin and more recentily Fefferman’s work about ball multiplier conjecture. In this post will be initiated the saga related to
spaces so far. Before, we recall some important definitions. Let
, we set

for any
. Also, recall that

Then via Theorem 3 the operator
may be written in
as
 \ \ \ \ \ (3) \displaystyle Tu(x)=\mathcal{F}^{-1}[\sigma(\xi)\mathcal{F}u(\xi)](x) \ \ \ \ \ (3)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+Tu%28x%29%3D%5Cmathcal%7BF%7D%5E%7B-1%7D%5B%5Csigma%28%5Cxi%29%5Cmathcal%7BF%7Du%28%5Cxi%29%5D%28x%29+%5C+%5C+%5C+%5C+%5C+%283%29&bg=ffffff&fg=000000&s=0)
where
. In fact, by observing that

we have from (2) that

being
the inverse Fourier transform of
. We call the operator
in (3) of multiplier operator associated to symbol
and we will written it as
. Also, we say that
is a multiplier on
, if for all
the map
is bounded on
.
Example 1 The Fourier transform in the classical sense of

is not possible, because
not belong to any
space. However, Fourier transform
in distribution sense is given by

for more details, see notes of Iannis Parissis.
Let
, the Bessel space
is the space of
functions such that
which endowed with norm

is a Banach space. The following proposition give a condition on symbol
for which
is a multiplier on
.
Proposition 1 Let
and
. Then
is a multiplier on
, for
.
Proof: In view of
, where
. From Young’s inequality, just we need to show
. To do this, let
. As
, from Cauchy-Schwartz inequality one gets


Example 2 Let
be bounded, that is,
. We claim that

It’s quite easy to get
via Plancherel’s theorem. It remains to get the other inequality. Let
be a ball in
with radius
. Let
such that
and
on
. Note that

The Lebesgue differentiation’s theorem yields

in other words,
.
Let
be the set of all tempered distributions
away from origin of
satisfying the estimate

for all multi-index
with
. In the next sections we give a much weaker hypotheses on symbol
, via Littlewood-Paley theory, which imply such as so-called symbol-H\”{o}rmader condition and
. In all that follows, the class
denotes the set of multipliers operators with symbol
belonging to
.
— 1.1. Littlewood-Paley theory —
Let
be radial Schwartz function supported in
which is identically
on
. To get that, recall of the topological Urysohn’s lemma. For all
, set

Notice that
, because
and
for, respectively,
and
. Moreover, the family of functions
forms a partition of unity

Next, for
we define the cut-off operator or Littlewood-Palay operator as

where
. Notice that
is supported in
. Indeed, by Fourier property

it follows that
. Therefore, if
is supported in the annulus
then the inverse Fourier transform is supported in the annulus

In other words, the operator
isolate the part of a function being each one concetrated near annulus
.
Proposition 2 The cut-off operator is a self-adjoint operator,

for any
and
.
Proof: The prove is a consequence of the radial property of
. For that, firstly, shall be observed that

when we have in mind (1). Now, radial property of
and (4) implies that


The decomposition of a tempered distribution
into a sum of “long pieces”
supported in the annulus
is known as Littlewood-Paley decomposition, because of the famous works de Littlewood and Paley related to Fourier and Power series, see xx. More precisely, Littlewood-Paley decomposition of a tempered distribution
is defined as follows

However, there is tempered distributions for which Littlewood-Paley decomposition fail on
. For instance take the distribution
and note by from Proposition 2 and
that

that is,
,
. To overcome this discomfort, we introduce in
the equivalence class
given by

where
denotes the set of polynomials
,

with complex coefficients
and
a non-negative integer. The space
endowed with
will be denoted by
.
Proposition 3 Let
be the space of Schwartz functions such that

Then
is a subspace of
which has

in sense of isomophism.
Proof: The prove is standard, consider the identification map
which goes an element
into the equivalence class
that contains it. The kernel of the map
is
. 
The aim is that, for any
we have

This follows by observing that

where
is that radial Schwartz function suported into a compact set and equal to
in
. To show (7), recall that a distribution
is a tempered distribution if

Easily, via identification (6), this indue in nature’s way the convergence in
. Thus, it is quite easy to show
. It remains to show

Let us fixe
such that
and write

because for
we have
and
on ball
. Thus, given
one has
and the its Littlewood-Paley decomposition is null in
, because of

and

The last equality is obtained via

where
is the Dirac mass and don’t forget that
, which is consequence of properties
- (i)
for every multi-index 
- (ii)
for every multi-index 
for all
.
Proposition 4 If
, then
agrees with
function away from the origin and satisfies the following pointwise estimate

for all multi-index
, where
.
Proof: Let
, then for all
this give us

where each peace
is supported on annulus
. So makes sense to define,

If
, for
one has

It follows that

If we put
into the last equality above, for all multi-indices
and non-negative
we have

We split
into two sums as

For
and
into (10), respectively, we get

and

This inequalities says us that
converges absolutely and uniforms in
. It follows that
converges in
to a function
which also satisfies the estimate

As
converges to
, then
and we obtain the desired estimate. 
In what follows,
denotes the space of
Bochner integrable functions
from
to a Banach space
and we write

Theorem 5 (Littlewood-Paley Theorem) Let
the Littlewood-Paley operator. If
, there exists a positive constant
such that

Moreover,

Proof: The inequality (13) follows from Theorem 5.4 in [1]. In fact,

in other words, the map
is bounded from
to
. It remains to show that the kernel
of the map
satisfies the
H\”{o}rmander condition, which can be obtained from estimate

Indeed, by mean value theorem one has

The inequalities above show us that the kernel
is a singular kernel from
to
and the inequality (13) follows from Theorem 5.4 in [1].
From now on, will be showed the estimate (14). To this end, we remember of
which give us

Recall that
, where
is a Schwartz function. Therefore,
for all multi-index
, that is,
. And by Proposition 4 follows that there exists a positive constant
for which

Setting

and recalling that
, we have via inequalities (15) and (16), respectively, that

Let
be in
, where
. For each
, let
and
. From self-adjointness of
, follows that

By duality of
and Cauchy-Schwartz inequality, respectively, one has

where we use the Hölder inequality and first Littlewood-Paley inequality (13) above. If we replace
by
in (16) such that
and we put
, it follows that

as desired. It remains to get
. To do this, let

and set

Then
on annulus
, that is,
. Indeed,
since
and
since
. This finish our prove. 
Lemma 6 Let
and
. Set
. If
, then there exists a positive constant
(independent of
,
and
) such that

where
denotes the Hardy-Littlewood maximal function,

Proof: Let
, then the kernel of
satisfy
. Therefore, by Hölder inequality and Plancharel theorem we obtain

where
and
. It follows by Fubini theorem that

The last inequality is obtaind from splitting
into lattices
of length
, see figure below.

Indeed,

where
is a integer such that
. Setting
it leads us to
. By definition of
and (21), it follows that

the last
series converges, because of
. 
The next theorem is a fractional variant of Hörmander multiplier theorem.
Theorem 7 Let
be such that

for
and let
. Then the operator
is a multiplier on
.
Proof: From Littlewood-Paley inequality (14),

Thus,

where
and
is a Littlewood-Paley operator such that
given in paragraphs above. Now, we write

If
, by Riesz representation theorem there exists
satisfying
and
for which

As the symbol of
is
and
, for all
. The Lemma 6 gives

As
, by Hardy-Littlewood theorem and Hölder’s inequality we get

because the Littlewood-Paley inequality (13) still hold with
in place of
. Inserting the last inequality into (23) we have

for all
. The case
follows by standard duality argument and shall be remarked only that the symbol of form adjuint operator
is
and hence is bounded by arguments above. The case
it’s a small exercise to reader. 
In the next post shall be showed that fractional condition (22) implies so-called symbol Hörmander condition in Theorem 2.5 and also Minhklin condition for
. Some small applications to PDEs equations will be mentioned finally.
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