Show that convolution product $f*g$ is $\mathcal{C}^{\infty}$ if $f,g\in L^{1}$ with $g\in \mathcal{C}^{\infty}.$
My attempt: First observe that, since $f*g=g*f$ we have that, $$(f*g)(x) = \int f(x-y)g(y)dy$$ and so, $$(f*g)'= (f*g').$$ Now we have to justify why $F(x) = (f*g')(x)$ is continuous. For this we look at the difference, \begin{align*} |F(x)-F(y)|\leq \int |f(t)|\cdot |g'(x-t) - g'(y-t)|dt\leq \epsilon ||f||_1 \end{align*} where given $\epsilon>0$ we found a $\delta>0$ such that, $$|x-y|<\delta \implies |g'(x-t)-g'(y-t)|<\epsilon.$$ So we use the the same $\delta$ for showing the continuity of $F.$ Now we just apply this argument repeatedly to get that $f*g$ is $\mathcal{C}^{\infty}.$ I don't know if this argument is correct so any feedback is much appreciated.