Weakly differentiable but classically nowhere differentiable

Question

Is there any example of a function which is weakly differentiable but none of its versions are classically differentiable anywhere (or differentiable only on a set of measure 0) ? Thanks

Answer 1

Surely, if you take a classically differentiable function and change its values on a dense (Lebesgue) null set, as the example with the characteristic function of the rationals in Wikipedia, that function would still have a weak derivative. But maybe you wanted less trivial examples?

Answer 2

This is rather a comment on the one-dimensional case and on $W^{1,\infty}_\mathrm{loc}(\Omega)$ than an answer.

In one dimension, this cannot happen. Take a function $f \in L^1_{\mathrm{loc}}(0,1)$, with weak derivative $f' \in L^1_{\mathrm{loc}}(0,1)$. Then, for any $[a,b] \subset (0,1)$, you have $f' \in L^1([a,b])$. Hence, $f$ is absolutely continuous on this interval and hence almost everywhere differentiable.

On the other hand, if $f \in W^{1,\infty}_{\mathrm{loc}}(\Omega)$ for some $\Omega \subset \mathbb{R}^n$, then $f$ is Lipschitz on compact subsets of $\Omega$ and therefore almost everywhere differentiable by Rademacher's theorem.

This shows, that one has to look for counterexamples in at least two dimensions, where the weak derivative is not in $L^\infty_\text{loc}(\Omega)$ (i.e. "nowhere bounded").