Is the convolution of a function $f(x)$ and a polynomial $p(x)$ always a polynomial?

Question

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 $p(x)$.

Answer 1

Depends. If you take a convolution per the Fourier transform, $$\begin{align}(f*p)(x)&= \int_{-\infty}^\infty f(y) p(x-y) dy\\&= \int_{-\infty}^\infty f(y) \sum_{k=0}^n a_k (x-y)^k dy\\&= \int_{-\infty}^\infty f(y) \sum_{k=0}^n \sum_{j=0}^k a_k {k \choose j} x^{j} y^{k-j} dy\end{align}$$ or, to make it more pronounced, $$ \sum_{j=0}^n\left(\sum_{k=j}^n a_k {k \choose j}\int_{-\infty}^\infty f(y)y^{k-j} dy\right) x^{j}$$ which is clearly a polynomial in $x$. All this assumes $(f*p)(x)$ is well defined to begin with.

Alternatively, in this case, noting that derivatives can be passed through when convolving with a smooth function,

$$\frac{d^n}{dx^n} (f*p)(x)=(f*p^{(n)})(x)$$ and choose $n$ so that $p^{(n)}=0$.

However, if you take the convolution Laplace transform style, take $p(x)=1$ to get

$$(f*p)(t) = \int_0^t f(\tau) p(t-\tau) d\tau = \int_0^t f(\tau)d\tau.$$

If you pick $f(\tau) = e^\tau$, for instance, you do not get a polynomial.

Answer 2

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 $n$.