Suppose I have a distribution $E$ such that $\phi \ast E$ is square-integrable for all $\phi \in C\_c^\infty \left( \mathbb{R}^d \right)$. Is it possible to prove that $E$ is tempered? It seems plausible to me, but I only get so far:


For brevity, define


\begin{equation}
G\_\phi = \phi \ast E
\end{equation}


for any $\phi \in C\_c^\infty$.
Now, because convolution is commutative, for all $\phi, \psi \in C\_c^\infty$ we have


\begin{equation}
\mathcal{F} \left( \phi \ast \psi \ast E \right)
= \tilde{G}\_{\phi \ast \psi}
= \left( 2 \pi \right)^{d/2} \tilde{\phi} \cdot \tilde{G\_\psi}
= \left( 2 \pi \right)^{d/2} \tilde{\psi} \cdot \tilde{G\_\phi}
\end{equation}


where the tildes denote Fourier transformed quantities.
Then, we get


\begin{equation}
\frac{\tilde{G}\_\phi}{\tilde{\phi}} = \frac{\tilde{G}\_\psi}{\tilde{\psi}} =: F
\end{equation}


which is of course what we expect, since we would like to interpret $F$ as the Fourier transform of our distribution $E$.
We can now deduce that $F \in L^1\_{\mathrm{loc}} \left( \mathbb{R}^d \right) \cap L^2\_{\mathrm{loc}} \left( \mathbb{R}^d \right)$ by


\begin{equation}
\left \Vert F \right \Vert\_{L^1 \left( K \right)}
\le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert\_{L^1 \left( K \right)}
\le \left \Vert \frac{1}{\tilde{\phi}} \right \Vert\_{L^2 \left( K \right)} \left \Vert \tilde{\phi} F \right \Vert\_{L^2 \left( K \right)}
< \infty
\end{equation}


and


\begin{equation}
\left \Vert F \right \Vert\_{L^2 \left( K \right)}
\le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert\_{L^1 \left( K \right)}
\le \sqrt{\left \Vert \frac{1}{\left \vert \tilde{\phi} \right \vert^2} \right \Vert\_{L^\infty \left( K \right)}} \left \Vert \tilde{\phi} F \right \Vert\_{L^2 \left( K \right)}
< \infty
\end{equation}


for any compact $K$ by using some $\phi \in C\_c^\infty \left( \mathbb{R}^d \right)$ with strictly positive Fourier transform (These do [exist](https://math.stackexchange.com/questions/1653134/an-example-of-smooth-compactly-supported-function-with-everywhere-non-vanishing)).


I seem to be unable to get any further though.
It is now evident, that $F$ is in fact a distribution (as it is locally integrable) and it remains to show that we can approximate any Schwartz function under the integral in a (Schwartz)-continuous way.



Prove that a given distribution is tempered
In Laurent Schwartz's *Théorie des Distributions* (page 245, chap. VII, §5) you can find something similar: A distribution $T\in \mathscr D'(\mathbb R^d)$ is tempered if and only if all regularizations $T \ast \varphi\in \mathscr O\_M$ for
$\varphi\in\mathscr D(\mathbb R^d)$, where $\mathscr O\_M$ is the space of slowly increasing $C^\infty$-functions $f$, i.e., for every $\alpha\in\mathbb N\_0^d$ there is $k\in\mathbb N$ such that $\partial^\alpha f(x)/(1+|x|)^k$ is bounded.


