TITLE: Convolution of an integrable function of compact support with a bump function.
QUESTION [5 upvotes]: Let $f\in L^1(\mathbb{R})$ be of compact support and $\psi(x)=C \exp(-(1-x^2)^{-1})$ where $C$ is chosen so that $\int_{\mathbb{R}} \psi =1$. Show that the convolution $f*\psi(x)=\int_{\mathbb{R}} f(x-y)\psi(y) dy$ is infinitely differentiable.

REPLY [3 votes]: First prove the following statement.

If $g$ is any compactly supported smooth function on $\mathbb{R}$, then $f*g$ is differentiable, and $(f*g)' = f*(g')$.

To do this, you appeal directly to the limit definition of the derivative.$$
(f*g)'(x) = \lim_{h\to 0}\frac{(f*g)(x + h) - (f*g)(x)}{h} = \lim_{h\to 0}\int f(y)\left\{\frac{g(x+h-y) - g(x-y)}{h}\right\}\,dy.$$ The quantity in braces converges uniformly (in $y$) to $g'(x-y)$ as $h\to 0$, so you can pass the limit inside the integral to get that $(f*g)'(x)$ exists and is equal to $$\int f(y)g'(x - y)\,dy = f*(g')(x).$$
This proves the statement. Repeated application of it proves that the derivatives $(f*\psi)^{(n)}$ exist and are equal to $f*(\psi^{(n)})$.