**1. Mikhlin-Hormander type symbols**

Let be vector-spaces of measurable functions from to itself and let be a bounded linear operator from to . Recall that is called translation invariant if for all and . Let and with , we found that each such operator is determined by a certain tempered distribution such that for every (Schwartz space). So taking Fourier transform into we have . This motivate us to define a *Fourier multiplier* as a map given by

where is a tempered distribution and denotes the Fourier transform . We refer to as *symbol* of , sometimes one writes as to relate it with more general operators so-called pseudo-differential operators. No standard example of such *symbols* is, for ,

In the limit case, , the above symbol can be written as , where denotes the characteristic function of unit disk . It’s well-known that the condition and is necessary for be a *Fourier multiplier* on (see e.g., [1]). However, Fefferman (see [2],[3]) gave an intricate proof which show us that this condition is not sufficient, that is, he showed that the operator does not extend to a bounded operator on for any and . This result give us a negative answer to the famous disk conjecture which states that is bounded on for . In this post we will work with symbols more regular than (1) such as Minklin symbols .

In a few months ago, based on Littlewood-Paley theorem, we gave a proof that the operator is a *Fourier multiplier* from to itself (see Theorem 7) provided that and satisfies

for and . In this post will be showed that if , then its satisfies the inequality (2). Hence, by Theorem 7 one has the following classical Mikhlin-Hormander theorem.

Theorem 1 (Mikhlin-Hormander theorem)Let and away from the origin. If for we have

then is a Fourier multiplier on , . In particular, is a Fourier multiplier on if , that is,

Let us recall some important definitions. Let and , a function lies in *Sobolev spaces* if for every with there exists such that

Here, we use the standard notations

and we say that is the derivative of in distribution sense (more precisely in ) and we write in to mean (5). The space equipped with the norm

is a Banach space. Also, notice that *Sobolev spaces* coincide with inhomogeneous *fractional Sobolev spaces* because of the norm equivalence

It follows that

Using Leibniz’s formula we written the term as

By observing that on and zero otherwise, we get

Now making the change of variable one has . Hence, by (3) it follows that

Let . Therefore, inserting (7) into (6) easily gets

and Theorem 1 is a consequence of the Theorem 7 as we desired. Notice that if ,

Hence which implies (8).

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The name of the theorem is flawed: the guy’s name is Mikhlin, not Minklin.

Thanks ph.