Linear operator commutes with translation. Observation in $L^1\cap L^2$. Stein 
Hello again. Continuing with the book Stein, there is an observation that I would like to try.
I have the following:
For any $f\in L^1\cap L^2$, by proposition 1 in Bounded linear operators that commute with translation
$T(f)=f\ast \mu$ and by prop 2. in this post, $T(f)=\mathcal{F}^{-1}(m(x)\hat{f})$ then $\mathcal{F}^{-1}(m(x)\hat{f})=f\ast \mu$ implies $m(x)\hat{f}=F(f\ast\mu)=\hat{f}\hat{\mu}$.
Therefore  $\hat{f}(m-\hat{\mu})=0$ for any $f\in L^1\cap L^2$.
Therefore $m-\hat{\mu}=0$?
 A: Yes what are you not sure about. As usual look at $g_k = k^n e^{-\pi k^2 |x|^2}$ and $f\in C^\infty_c$, approximating $f_k=f\ast g_k$ with a sequence of finite sums of translates of $g_k$, since $T$ is continuous you get $$T(f_k) = T(g_k)\ast f,\qquad \widehat{T(f_k)} = \widehat{T(g_k)}\widehat{f}$$ by density this stays true for $f\in L^2$. 
Taking $\widehat{f}= 1$ on $|x|<r$ you obtain your function $$m=\lim_{k\to \infty}\widehat{T(g_k)}\qquad (\text{convergence in } L^2_{loc})$$
It is (represented by) a measurable function because it is in $L^2_{loc}$. 
This function is bounded because if the measure of $\{ x,|\widehat{T(g_k)}(x)| \ge A\}$ is non-zero then $\|T\|\ge A$.
A: A bounded linear operator $ T: L^2 \to L^2$ that commutes with translation
I will prove it for a Schwartz function $f$ ,you can easily extend the result for an $L^2$ function.
For  a Schwartz function I will show :$f$ ,$T= f* \mu$ where $\mu $ is a tempered distribution.
A known result from Riesz Representation theorem see here is :$\int Tf(s)u(s) ds=\int f(s)T^*u(s)ds$ written as $<Tf,u>=<f,T^*u>$, where $ T^*$ is the linear adjoint operator of $ T $ , $ u \in L^2$
now for $f(s)$ translated  by $x$ ,$f(s+x)$
we have $\int Tf(s+x)u(s)ds=\int f(s+x)T^*u(s)ds$
Now take Let $u_{\epsilon}=\phi_\epsilon(s)=\phi(s/\epsilon)\epsilon^{-1}$ where
$\phi(s)$ is normalized gaussian function with zero mean.
$\int Tf(s+x)u_{\epsilon}(-s) ds=\int Tf(s+x)u_{\epsilon}(s)ds=\int f(s+x)T^*u_{\epsilon}(s)ds$
$H(s)=Tf(s)$ , by translation commutation $H(s+x)=Tf(s+x)$
$R(-s)=T^*u_{\epsilon}$
we have $\lim_{\epsilon \to 0} H* u_{\epsilon}=\lim_{\epsilon \to 0} f*R$
$\lim_{\epsilon \to 0} H*u_{\epsilon}=H(x)$ ae
for $u \in L^1$ function, we have the following theorem regarding fourier transform:
$v(x)=g*s=\int g(s)u(x-s) ds$
$L^2$ : for $g \in L^2$ define $\hat{g}=lim_{n \to \infty}\int_{n}^{-n}g(s)e^{2i\pi x z} dx$ the limit exists in the sense of $L^2$. If $u \in L^1$ and $\int |g(s)||u(x-s)| ds \le P$ where $P$ is a non negative integrable function in $L^2$ then $\hat{v}=\hat{g}\hat{u}$
Proof : $g_n=g1_{[-n,n]},v_n=g_n*u$
it's known that $\hat{v_n}=\hat{g_n}\hat{u}$
By dominated convergence theorem $lim_{n \to \infty} ||v-v_n||^2=0$
and by Plancherel theorem $lim_{n \to \infty} ||v-v_m||^2= lim_{n \to \infty} ||\hat{v}-\hat{v_n}||^2=0$
This implies $lim_{n \to \infty} \hat{v_n}=\hat{v}$
Now from our equation when $u$ is a normalized  Gaussian we have $|H|* |u_{\epsilon}| \le MH(x)$ where $MH(x)$ is the Hardy Littlewood Maximal function (also $ ||MH(x)||^2 \le ||H(x)||^2$)
using Fourier transform $\hat{H}(z)\hat{u_{\epsilon}}(z)=\hat{f}(z)G(z)$ where $G(z)=\hat{R}$
$\lim_{\epsilon \to 0}\hat{ u_{\epsilon}}=1$
$\lim_{\epsilon \to 0}G=m(z)$
consequently we have $\hat{Hf}(z)=m(z)\hat{f}(z)$
