Do Sobolev spaces contain nowhere differentiable functions?

Does the Sobolev space $$H^1(R^n)$$ of weakly differentiable functions on a bounded domain in $$R^n$$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function?

• Not for $n=1$, of course... – Nate Eldredge Sep 6 at 19:02
• "The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question! – Dirk Sep 6 at 19:38

1 Answer

As Nate Eldredge pointed out, $$W^{1,2}(\mathbb{R})$$ functions are absolutely continuous on $$\mathbb{R}$$, and therefore differentiable a.e., and so the answer is no.

For $$n\geq 2$$, the answer is yes.

When n=2, this is a classical result of L. Cesari

Cesari, Lamberto, Sulle funzioni assolutamente continue in due variabili, Ann. Sc. Norm. Super. Pisa, II. Ser. 10, 91-101 (1941). ZBL0025.31301.

see also an explicit construction applicable to $$W^{1,n}(\mathbb{R}^n)$$ in J. Serrin,

Serrin, J., On the differentiability of functions of several variables, Arch. Ration. Mech. Anal. 7, 359-372 (1961). ZBL0109.03904.

The paper of Cesari also contains an a.e. differentiability result for $$W^{1,p}$$, $$p>2$$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder\'on in Riv. Mat. Univ. Parma, 1951.

For $$n>2$$, it looks like a suitable construction is given at the question Are functions of bounded variation a.e. differentiable?.

• I'll be honest, my first thought after reading this answer was "what about $1<n<2$?" – Wojowu Sep 6 at 22:12