If $ f*g$ is a polynomial of degree at most $m$ for all $g \in C_{0}^{\infty} $. Show $f$ is a polynomial of degree at most $m$ almost everywhere

Question

$C_{0}^{\infty}$ denotes the set of smooth functions with compact support. In attempt of this, I've evaluated the convolution at the (m+1)th derivative to obtain $$ \frac{d}{d^{m+1}}(f*g) =\int f(x-y)g^{(m+1)}(y)dy = 0 $$ But am lost in how to show that this implies f is a polynomial of degree at most m.

Answer 1

In general $f$ is a distribution. If you assume $f$ is a $L^1_{loc}$ function it doesn't change the argument

• If its distributional derivative $f^{(m+1)} $ is not the $0$ distribution then $f \ast g^{(m+1)}=f^{(m+1)} \ast g \ne 0$ for some $g \in C^\infty_c$.

• If $f^{(m)}$ is not constant then $f^{(m+1)}$ is not the zero distribution.

• So the $m$-th derivative of $f-c_m x^m$ vanishes and by induction $f$ is a degree $\le m$ polynomial.

If you don't like distributions and their derivatives, pick some $\phi \in C^\infty_c, \int \phi = 1$ and replace $f$ by the smooth functions $f_n = f \ast n\phi(n.)$. You'll find that each $f_n$ is a degree $\le m$ polynomial which implies since $\lim_{n \to \infty} f_n$ converges (to $f$) in the sense of distribution that the same holds for the limit.