Using convolution to impose differentiablilty.

Question

If I had a function $g$ that was not differentiable at a known point, is it possible to convolute it with say a $C^{\infty}$ function $f$, resulting in a differentiable function?

Thanks in advance!

Answer 1

As explained in comments: this works quite well. We only need $g$ to be locally integrable. Let $f$ be $C^\infty$ smooth and compactly supported; normalize by $\int f=1$. Then

• $f*g$ is $C^\infty $ smooth, with derivatives given by $D^k(f*g) = (D^kf)*g$;
• if $f$ is rescaled as $\lambda^{-d}f(x/\lambda)$, where $d$ is the dimension of ambient space, then the convolutions converge to $g$ as $\lambda\to 0$. The convergence holds a.e., and also in $L^p$ norm provided $g\in L^p$.

The above facts are found in any decent book on real analysis; e.g., Real Analysis by Folland.