Does Lipschitz continuous function only has a countable non-differentiable points?

Question

If $f$ is Lipschitz, we know by Rademacher's theorem that it is differentiable a.e. That is, the set of non-differentiable points is of measure zero. But this does not necessarily imply that it is a countable set.

Can we find a Lipschitz continuous function that has an uncountable differentiable points?

This seems helpful. – MathematicsStudent1122 Sep 25 '17 at 05:23 — MathematicsStudent1122, Sep 25 '17 at 05:23

score 5 · Answer 1 · edited Dec 10 '24 at 16:00

5

No; as one can read in the paper

Differentiability of Lipschitz functions, structure of null sets, and other problems. http://pagine.dm.unipi.it/alberti/ricerca/2010-12/acp-icm2010.pdf

we know that for a given set $E$, there exists a Lipschitz function such that $f$ is not differentiable at $E$ if and only if $E$ is of Lebesgue null. It is easy to find a uncountable set of Lebesgue null, say Cantor set.

edited Dec 10 '24 at 16:00

rod

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answered Sep 25 '17 at 05:56

Ice sea

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Worth mentioning is that not every Lebesgue null set can be the set of non-differentiability of a Lipschitz function, since the set of non-differentiability for any function is a Borel set (more precisely, a $G_{\delta \sigma}$ set). Thus for some Lebesgue null sets $E,$ any Lipschitz function not differentiable on $E$ will also not be differentiable at some points outside of $E.$ See also this mathoverflow answer and its comments. – Dave L. Renfro Dec 10 '24 at 16:43

Does Lipschitz continuous function only has a countable non-differentiable points?

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