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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?

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  • $\begingroup$ Not for $n=1$, of course... $\endgroup$ – Nate Eldredge Sep 6 at 19:02
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    $\begingroup$ "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! $\endgroup$ – Dirk Sep 6 at 19:38
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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?.

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

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