In this notes we show that bounded translation invariant operators from to are essentially convolution singular integral operators for certain . More precisely, let be a bounded linear operator from to , we say that is a translation invariant operator, if
where denotes the translation operator on , and is a Schwartz function. The object of this notes is only find a tempered distribution such that
if is translation invariant operator and is bounded on certain Lebesgue spaces. Firstly, let us recall some well know results.
 Let , . Then there exist a positive constant such that
In fact, let be an unity sphere and let which is given by . Then the function is a continuous function in and for
we get , for all and . Since
we obtain (3).
 Let , we have
for all .

Indeed, by inequality (6) we have
Lemma 1 Let be a tempered distribution. Suppose that for all . Then is a function and there exists a positive constant such that
Proof: Let be the space of all bump functions with support on euclidean sphere of radius . Since , it follows from Hölder inequality that . Moreover, we have that lives in too. This follows by means of the inequality (8) and some properties of Fourier transform on euclidean spaces. Indeed,
Integrating the last inequality, we obtain
It follows that Fourier inverse of agrees, almost everywhere, with an uniformly continuous function (see Folland p. 239). Since on , so the tempered distribution can be redefined as an uniformly continuous function, namely , which is defined on for all . Since the choice of is not relevant, we get
Lemma 2 Let and let . If is a bounded linear operator which is translation invariant, then commutes with derivatives, that is,
for all multiindex .
Proof: For readers convenience, let . The general case, follows by interactions. Firstly, let and set
Let and note that . It follows that,
which converges to in as . The continuity of in join with give us
Therefore, by uniqueness of limit in one has
as we desired.
Let us to introduce the reflection operator in which is given by,
for all . Notice that, for the integral
for all . This motivates to define convolution operator between a Schwartz function and a tempered distribution as follows
Theorem 3 Let be a bounded linear operator. If is translation invariant, there exists such that (2) holds.
Proof: Let be a Schwartz function such that . In view of Lemma 1 and Lemma 2, is a continuous function which satisfy the estimate
for . Here stands for Schwartz norm
In other words, there exists a tempered distribution such that . As is translation invariance, in view of (17) we have
because of .
In the very last part of the post there is a typo: you say “Hence $(A\phi)(0)$ is a bounded linear functional … and so $(A\phi)(0)=T(R\phi)$ for **any** tempered distribution $T$”. The correct wording is something like “and so *there exists* a tempered distribution $T$ such that …”.
thanks Giuseppe, I will follow these suggestions