When is the convolution between a Schwartz function and a smooth function again a Schwartz function? Consider a distribution $u\in\mathscr{S}'(\mathbb{R}^n)$ which is given by a smooth, polynomially bounded function $u$, i.e.
\begin{align}
u(f)=\int_{\mathbb{R}^n}u(x)f(x)~d^nx.
\end{align}
for all $f\in\mathscr{S}(\mathbb{R}^n)$.
Furthermore, $u$ is such that $u\ast f\in\mathscr{S}(\mathbb{R}^n)$ for all $f\in\mathscr{S}(\mathbb{R}^n)$.
Does this imply $u\in\mathscr{S}(\mathbb{R}^n)$?
My incomplete Idea:
Choose $f$ in a way to probe the approximate value of $u$ in a given point and show that this is rapidly decreasing. For the derivatives choose $\partial^\alpha f$ for $\alpha\in\mathbb{N}^n$.
 A: No, it's not necessarily the case that $u \in \mathscr{S}$. Here's one possible counterexample:
Let $u \in \mathscr{S}'$ be the distribution given by the smooth function
$e^{-i |x|^2}$. It's not too hard to see that $u$ and its derivatives are all polynomially bounded. Note that $\hat{u}$ is given by the very similar smooth function $C_n e^{-i |\xi|^2}$, modulo a real constant factor in the exponential depending on your preferred definition of the Fourier transform.
We can also make the following argument to demonstrate that $u * f \in \mathscr{S}$ for all $f \in \mathscr{S}$. First, we of course have $(u * f) \hat{} = \hat{u} \hat{f}$,
again possibly modulo a constant factor. Since $\hat{f} \in \mathscr{S}$, and $\hat{u}$ is smooth and polynomially bounded along with all its derivatives, it follows that $(u * f) \hat{} \in \mathscr{S}$. Thus $u * f \in \mathscr{S}$ as well.
So $u$ and its derivatives are polynomially bounded, and $u * f \in \mathscr{S}$ for all $f \in \mathscr{S}$. But certainly $u \notin \mathscr{S}$ since $u$ does not decay at infinity.
