Question: 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.

Answer: 
With a change of variable you obtain
$$
\int\_{\mathbb{R}}f(x-y)\psi(y)dy = \int\_{\mathbb{R}}f(y)\psi(x-y)dy.
$$
Now, using the fact that $\psi\in\mathscr{C}^{\infty}$ ($\psi$ and all its derivatives are bounded because $\mathrm{supp}(\psi)\subset\mathbb{R}$) and $f\in L^1$ with compact support, the Dominated Convergence Theorem gives the differentiability you need. In particular
$$
D\_x\left[ \int\_{\mathbb{R}}f(y)\psi(x-y)dy\right] = \int\_{\mathbb{R}}f(y)D\_x\psi(x-y)dy.
$$
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 $f\*(\psi^{(n)})$.