Convolution with a polynomial is a polynomial. Why?

Question

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?

Answer 1

Use that convolution is commutative and linear, so it suffices to prove this for $P=x^n$. Then $$f\star P = \frac{1}{2\pi}\int_0^{2\pi} f(t)(x-t)^n dt$$

Now expand $(x-t)^n$ using the binomial theorem, and you'll get that $f\star P$ is a polynomial with coefficients of the form $$(-1)^k \binom nk\int_0^{2\pi} f(t)t^k dt$$

Answer 2

First notice that the all $C^{\infty}$ solutions of the differential equation $ y^{(N+1)}=0$ are polynomials. $\\ $ 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.