I'm following a course on harmonic analysis, and I have a doubt on a proof.
Let $\mathcal{M}_p := \{ m\in S' : T_m: L^p\rightarrow L^p \text{ is bounded}\} $, where $T_m(f)=\mathcal{F}^{-1}(m\hat{f})$
I want to prove that $\ T_m\in \mathcal{M}_2$ if and only if $m\in L^\infty$.
One direction is easy: if $m\in L^\infty$ you can just apply Plancherel theorem and conclude that $\ T_m\in \mathcal{M}_2$. I don't know how to prove the other implication($\ T_m\in \mathcal{M}_2 \Longrightarrow m\in L^\infty$). My problem is that $m$ a priori is only a tempered distribution and not a function. If I could prove that $\ T_m\in \mathcal{M}_2 \Longrightarrow \text{m is a measurable function}$, it would be easy to conclude the proof(for example like in this question).
Thank you very much
Since the Fourier transform is unitary on $L^2$, it suffices to prove that if $m\in S'$ such that $f\in L^2$ implies $mf\in L^2$, then $m$ is a regular distribution.
If $A$ has finite measure, define $h_A=m1_A$, which is a measurable function by assumption. The resulting family is consistent in the following sense: If $A$, $B$ have finite measure, then $$ h_A 1_{A\cap B}=m 1_{A\cap B}=h_B 1_{A\cap B}. $$ In other words, $h_A$ and $h_B$ coincide a.e. on $A\cap B$. Thus you can glue the functions $h_A$ to get a measurable function $h$ that coincides with $h_A$ a.e. on $A$ (for example you can take a partition $(A_n)$ of $\mathbb R^n$ into sets of finite measure and define $h(x)=h_{A_n}(x)$ if $x\in A_n$).
In particular, since $h1_K=m1_K\in L^2(K)\subset L^1(K)$ for every compact set $K$, the function $h$ is in $L^1_{\mathrm{loc}}$.
Now if $\varphi\in C_c^\infty$ and $K=\operatorname{supp}\varphi$, then $$ m(\varphi)=m(1_K \varphi)=m 1_K(\varphi)=\int_K h_K\varphi=\int h\varphi. $$ Therefore $m$ is the regular distribution induced by $h$.