Is a Lipschitz function differentiable?

I have been wondering whether or not this property applies to all functions.

I do not need a formal proof, just the concept behind it.
Let $f: [a,b] \to [c,d]$ be a continuous function (What is more - it is uniformly continuous!) And let's assusme that it's also Lipschitz continuous on this interval.

Does this set of assumptions imply that $f$ is differentiable on $(a,b)$?

  • 6
    $\begingroup$ It only implies it is differentiable almost everywhere. $\endgroup$
    – Ian
    Jan 17, 2018 at 22:13
  • 6
    $\begingroup$ "Let $f: [a,b] \to [c,d]$ be a continuous function (What is more - it is uniformly continuous!) And let's assusme that it's also Lipschitz continuous on this integral." You should be aware that these assumptions are highly redundant. Lipschitz implies all preceding assumptions $\endgroup$
    – zhw.
    Jan 17, 2018 at 22:42
  • 2
    $\begingroup$ Lipschitz continuous does not imply differentiability. In fact, we can think of a function being Lipschitz continuous as being in between continuous and differentiable, since of course Lipschitz continuous implies continuous. If a function is differentiable then it will satisfy the mean value theorem, which is very similar to the condition to be Lipschitz continuous. $\endgroup$ Jan 18, 2018 at 9:43

3 Answers 3


It is not always true indeed, good counterexample could be $x\mapsto |x-a|$. But rather, we have

Theorem: Radamacher theorem says every Lipschitz function is almost everywhere differentiable

Fine a nice proof of this theorem here: An Elementary Proof of Rademacher's Theorem - James Murphy or Here using distribution theory

  • 1
    $\begingroup$ those do not include a proof of the Rademacher theorem in dimension 1. Terrence Tao has a reasonably self-contained proof of this and a couple other related here: terrytao.wordpress.com/2010/10/16/… $\endgroup$ Feb 2, 2018 at 23:08
  • $\begingroup$ @CalvinKhor I don't think those proof exclude the one dimensional case. Or am I wrong . one the proof use distributional derivative $\endgroup$
    – Guy Fsone
    Feb 2, 2018 at 23:49
  • $\begingroup$ Well I may have missed it but both links start with "in dimension one its because Lipschitz implies absolutely continuous/bounded variation so the result is true", which isn't wrong but I doubt someone who would ask this question on MSE would know this result for AC or BV functions $\endgroup$ Feb 3, 2018 at 0:00
  • $\begingroup$ yes many proofs of Rademacher in R^n rely on the fact having been established on R. To prove it on R, one of the the underlying truths is that a continuous and increasing function is differentiable almost everywhere. In fact an increasing function is differentiable almost everywhere, but this isn't needed for Rademacher's theorem. $\endgroup$
    – T_M
    May 15, 2020 at 15:13

The function $$x \mapsto \left|x\right|$$ is Lipschitz-continuous (with $k=1$) but not differentiable at $0$.


No, it does not imply $f$ is differentiable.

Try $f(x) = |x|$ as an example!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.