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?



Do Sobolev spaces contain nowhere differentiable functions?
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**](http://www.numdam.org/item?id=ASNSP_1941_2_10_2_91_0), Ann. Sc. Norm. Super. Pisa, II. Ser. 10, 91-101 (1941). [ZBL0025.31301](https://zbmath.org/?q=an:0025.31301).


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**](http://dx.doi.org/10.1007/BF00250769), Arch. Ration. Mech. Anal. 7, 359-372 (1961). [ZBL0109.03904](https://zbmath.org/?q=an:0109.03904).


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?](https://mathoverflow.net/questions/272282/are-functions-of-bounded-variation-a-e-differentiable).


