TITLE: Convolution with a polynomial is a polynomial. Why?
QUESTION [5 upvotes]: Let $P:\mathbb{R}\to\mathbb{R}$ such that $\deg P=N$. Let $f$, an integrable-$2\pi$-periodic function. Show that $f\star P$ is also a polynomial.

So we can prove it for an arbitrary $x^n$ (Since a linear combination of polynomial is obviously a polynomial).
$$f\star P = \frac{1}{2\pi}\int_0^{2\pi} f(x-t)t^n \ dt$$
It looks like I don't have any other information in order to proceed.
What am I missing?

REPLY [4 votes]: First notice that the all $C^{\infty}$ solutions of the differential equation $ y^{(N+1)}=0$ are polynomials.
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We know that convolution of an $C^{\infty}$ function is again $C^{\infty}$.
From commutativity and dominated convergence we see that $$(f\ast P)^{(N+1)} =(P\ast f )^{(N+1)}=\frac{1}{2\pi} \int _{0}^{2\pi} (P(x-t))^{(N+1)}f(t)\rm{d}t=0,$$
hence we conclude the result.