Derivability of convolution of a function with a finite measure We know that if $f\in C_c^k(\mathbb{R}^N)$ and $g\in L^1_{\text{loc}}(\mathbb{R}^N)$, then $f*g\in C^k(\mathbb{R}^N)$, where $*$ the operator convolution.
But I was wondering what happens if $\mu$ is a finite Radon measure and we define $f*\mu(x)=\int_{\mathbb{R}^N}f(x-y)d\mu(y)$. Is $f*\mu$ still in $C^k(\mathbb{R}^N)$?
 A: Well, the answer to your question is "yes"; and here is the reason:
To begin with, the function $f \ast \mu: \mathbb R^N \to \mathbb C$ is clearly well-defined as $f$ has compact support. To see that it is continuous, pick $R > 0$ such that $\operatorname{supp}(f) \subseteq \mathrm B(0, R)$. Further let $r > 0$ and $\delta > 0$. Then you have
\begin{align*}
\vert f \ast \mu(x) - f \ast \mu(z) \vert &= \left\vert \int_{\mathrm B(0, r + R)} f(x - y) - f(z - y) \, \mathrm d \mu(y)\right\vert \\
& \leq \sup \{\vert f(x - y) - f(z - y) \vert : \vert x - z \vert < \delta \} \cdot \mu(\mathrm B(0, r + R))
\end{align*}
for all $x, z \in \mathrm B(0, r)$ such that $\vert x - z \vert < \delta$. Now the second factor is finite, since $\mu$ is a finite measure and the first factor gets arbitrarily small since $f$ is uniformly continuous on $\mathbb R^N$ (since $f$ has compact support). Thus, we obtain $f \in C(\mathbb R^N)$.
Using very similiar arguments (or directly the theorem on the differentiability of parameter integrals that you prove this way), you can show in the same manner that the differentiability of $f$ implies the differentiability of $f \ast \mu$.
Notice that this is basicly the same proof as in the case of $g \in L^1_{\mathrm{loc}}(\mathbb R^N)$ you mentioned in the beginning; just that you do not use a locally integrable function anymore but a finite measure. Basicly you just need to assert that the second factor is finite and this is true in any of this two cases. I hope things got a little bit clearer :-)
