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?

$\begingroup$ Not for $n=1$, of course... $\endgroup$– Nate EldredgeSep 6, 2019 at 19:02

4$\begingroup$ "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! $\endgroup$– DirkSep 6, 2019 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, 91101 (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, 359372 (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?.

7$\begingroup$ I'll be honest, my first thought after reading this answer was "what about $1<n<2$?" $\endgroup$– WojowuSep 6, 2019 at 22:12

$\begingroup$ There's a statement in the literature that Sobolev functions are absolutely continuous on almost all lines and differentiable almost everywhere. How does that square with the results that you cite? There seems something missing $\endgroup$– shuhaloDec 19, 2022 at 12:59