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.
Analysis – Convolution of an Integrable Function with a Bump Function
analysisconvolutionmeasure-theory
Best 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 $$ D_x\left[ \int_{\mathbb{R}}f(y)\psi(x-y)dy\right] = \int_{\mathbb{R}}f(y)D_x\psi(x-y)dy. $$