I am running into some troubles with Lipschitz continuous functions.

Suppose I have some one-dimensional Lipschitz continuous function $f : \mathbb{R} \to \mathbb{R}$. How do I prove that its derivative exists almost everywhere, with respect to the Lebesgue measure?

I found on other places on the internet that any Lipschitz continuous function is absolutely continuous, and that this directly implies that the functions is differentiable almost everywhere. I don't quite see how this argument goes, though.

Any help with giving such a proof, or redirecting me to a source where I can find one, would be greatly appreciated.

  • $\begingroup$ Which part of the reasoning using absolute continuity do you not get? The fact that Lipshitz implies AC, or the fact that AC implies differentiability ae? (Or both?) $\endgroup$ – Thomas Jan 17 '17 at 19:33
  • $\begingroup$ The part that AC implies differentiability ae. :) $\endgroup$ – arriopolis Jan 17 '17 at 20:53

An excellent reference for the properties of Lipschitz functions is Lectures on Lipschitz Analysis by Juha Heinonen. The differentiability a.e. is Theorem 3.2, page 19. I state the steps of the proof here:

  1. Lipschitz continuity implies having bounded variation.
  2. A function of bounded variation can be written as the difference of two increasing functions
  3. An increasing function is differentiable a.e.: this is the main step of the proof, which uses the Vitali covering theorem. Concerning this step, see also $f$ continuous, monotone, what do we know about differentiability?
  • $\begingroup$ Wow, this turned out to be somewhat harder than I expected. Thanks for the references! $\endgroup$ – arriopolis Jan 18 '17 at 2:55
  • $\begingroup$ What if $f:R^n \to R$? $\endgroup$ – Red shoes Jun 10 '17 at 22:59
  • $\begingroup$ @Ashkan Theorem 3.1 in Lectures on Lipschitz Analysis that are linked in the post. $\endgroup$ – user357151 Jun 11 '17 at 0:57

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