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
$$
D\_x\left[ \int\_{\mathbb{R}}f(y)\psi(x-y)dy\right] = \int\_{\mathbb{R}}f(y)D\_x\psi(x-y)dy.
$$
Use that $$ \widehat{d/dx f(x)} = 2 \pi i \xi \hat{f}(\xi)$$ where $\hat{}$ denotes the Fourier transform and that $\widehat{f \ast g} = \hat{f} \cdot \hat{g}$.

Then 
$$ \widehat{(d/dx (f \ast g)(x))} = 2 \pi i \xi \hat{f} \cdot \hat{g}$$

Applying the inverse Fourier transform to both sides with either $(2 \pi i \hat{f}) \cdot \hat{g}$ or $ \hat{f} \cdot (2 \pi i \hat{g})$ gives you 
$$ d/dx (f \ast g) = (d/dx f) \ast g = f \ast (d/dx g)$$

From which you get that $f \ast \psi$ is smooth since $\psi$ is.