**— 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

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 classical sense of

is not possible, because not belong to any space. However, Fourier transform of 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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