TITLE: Do Sobolev spaces contain nowhere differentiable functions?
QUESTION [26 upvotes]: 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?

REPLY [18 votes]: 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?.