We put, $C_{0}(\mathbb R)=$ The space of continuous functions on $\mathbb R$ vanishing at $\infty$; $C^{k}(\mathbb R)=$ The space of all functions $\mathbb R$ whose derivative of order $\leq k$ exist and are continuous; $C_{c}^{\infty}(\mathbb R)=$ The space of $C^{\infty}$ functions on $\mathbb R$ whose support is compact.
It is well-known that, If $f\in L^{1}(\mathbb R), g\in C^{k}(\mathbb R),$ and $\partial^{\alpha}g$ is bounded for $|\alpha| \leq k,$ then $f\ast g\in C^{k}(\mathbb R)$ and $\partial^{\alpha} (f\ast g) = f\ast (\partial^{\alpha} g)$ for $|\alpha| \leq k.$ (Bit roughly speaking, convolution of smooth functions with $L^{1}-$ integrable functions gives us smooth functions)
My Question is: Suppose $f\in L^{p}(\mathbb R)\cap C_{0}(\mathbb R); (1<p<\infty), g\in C^{\infty}_{c}(\mathbb R).$ Then (1) Can we expect $f\ast g \in C^{k}(\mathbb R),$ and $\partial^{\alpha} (f\ast g)= f\ast (\partial^{\alpha}g)$ for $|\alpha| \leq k$ for every $k\in \mathbb N.$ (2) Also, assume, $f$ has a compact support. Then can expect $f\ast g$ also has a compact support ?
Thanks,
Assume $g$ has support in $[-M,M]$. Then, for $x\in[-N,N]$, $$ h(x)=\int f(x-y)g(y)dy=\int_{[-M,M]} f(x-y)g(y)dy=\int \tilde f(x-y)g(y)dy $$ where $\tilde f=1_{[-N-M,N+M]}f$, $1_K$ denoting the indicator function of the set $K$. Now show that $\tilde f\in L^1$ and conclude that $h\in C^k$.