Problem of convolution.

Question

If we are given with a polynomial $\mathcal P$ and a compactly supported distribution $g$. Can we prove that their convolution will be a polynomial again?

Answer 1

It should be somewhat easy to show the following:

Lemma

If $f:\Bbb{R}^n\rightarrow\Bbb{R}$ is $m$ times continuously differentiable and $\partial^m_{x_j}f\equiv 0$ for all $1\leq j\leq n$, then $f$ is a polynomial of degree at most $m-1$.

Then, you should be able to differentiate under the integral sign to obtain your result.

Answer 2

If you take $p(x) = x^k$ (where $x^k = x_1^{k_1}\cdots x_n^{k_n}$, and $\binom{k}{j} = \binom{k_1}{j_1} \cdots \binom{k_n}{j_n}$), then $(p \ast g)(x) = \int (x-t)^k dg(t) = \sum_{j \le k } \binom{k}{j} x^j \int (-t)^{k-j} dg(t)$, which is a polynomial.

It follows by linearity that polynomials will be mapped to polynomials.

Compact support of $g$ ensures that the coefficients $\binom{k}{j} \int (-t)^{k-j} dg(t)$ are finite.