TITLE: Is the convolution of a function $f(x)$ and a polynomial $p(x)$ always a polynomial?
QUESTION [6 upvotes]: After reading the following question:
How do I prove a convolution is a polynomial?
I want to ask if that is always the expected result, that is to say, does the following holds?

A convolution of a function $f(x)$ and a polynomial $p(x)$ will always
  result in a polynomial

Is that true?
If so, how to prove that? if not, can you give a counterexample?
If the answer depends on the properties of $f(x)$, continuity, differentiability, or anything else, please describe the required properties. 
For example, I've easily proved that if $f(x)$ is like:
$$f(x)=u(x)\,\mathrm{e}^{-x}\,q(x)$$
where $u(x)$ is the unit step function and $q(x)$ is any polynomial, then the convolution of $f(x)$ with $p(x)$ will be a polynomial with the same degree of $p(x)$.

REPLY [5 votes]: Let $p(x)=\sum_{k=0}^n a_kx^k$, with $a_n\ne 0$, and $\varphi$ a function for which $p*\varphi$ is definable - for example this is possible if $\varphi$ has compact support or dies sufficiently fast as $|x|\to \infty$.
Then
$$
(p*\varphi)(x)=\int_{-\infty}^\infty \varphi(t)\,p(x-t)\,dt.
$$
Now, if  $\varphi$ possesses the "nice" decay properties described earlier, we can differentiate inside the integral (by virtue of Lebesgue Dominated Convergence Theorem), and obtain
$$
(p*\varphi)^{(\ell)}(x)=\int_{-\infty}^\infty \varphi(t)\,p^{(\ell)}(x-t)\,dt,
$$
for every $\ell\in\mathbb N$. But $p^{(n+1)}(x-t)\equiv 0$, and hence $(p*\varphi)^{(n+1)}(x)\equiv 0$, which of course means that $p*\varphi$ is a polynomial of degree at most $n$.