Let $f \in L^1_{loc}(\mathbb R^n)$ and $\varphi \in C^\infty_0(\mathbb R^n)$. I know that it is a very common result that $\varphi\star f$ is in $C^\infty(\mathbb R^n)$, but I am not able to prove it and I haven't found any answers on Stackexchange that proves the statement the way I want to. Perhaps one of you can redirect me to another question.
The function $x \mapsto \varphi(x - y) f(y) \in C^\infty (\mathbb R^n)$ almost everywhere. I want to use the Leibniz rule to prove that, for $\alpha \in \mathbb N^k$, $$\partial^\alpha(\varphi \star f) = (\partial^\alpha\varphi) \star f.$$ To this end, I am looking for an inequality that looks like the following $$|\partial_x^\alpha(\varphi(x - y)f(y))| = |(\partial^\alpha \varphi)(x - y)f(y)| \le \|\partial^\alpha\varphi\|_\infty |f(y)|.$$ Even if the last inequality is true, it does not help to prove the statement since $f$ is not integrable in the all $\mathbb R^n$. Is it the right way to do that ?
As $\varphi$ is compactly supported locally we can assume that $f\in L^1$
With $D_v$ the directional derivative in the direction $v\in \Bbb{R}^n$ then $$D_v (f\ast \varphi)=f\ast D_v\varphi$$ follows from
$$\lim_{h\to 0} \|f\ast T_{v,h}\|_\infty \le \lim_{h\to 0}\|f\|_{L^1} \|T_{v,h}\varphi\|_\infty=0$$ where $$T_{v,h}\varphi=\frac{\varphi(x+hv)-\varphi(x)}{h}-D_v\varphi(x)=\int_0^1 (D_v \varphi(x+thv)-D_v \varphi(x))dt$$ As $D_v\varphi$ is $C^\infty_c$ you can repeat to get $D_{v_1}\ldots D_{v_k}(f\ast \varphi)=f\ast D_{v_1}\ldots D_{v_k}\varphi$.