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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 $(a,b)$?

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    $\begingroup$ It only implies it is differentiable almost everywhere. $\endgroup$ – Ian Jan 17 '18 at 22:13
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    $\begingroup$ "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 '18 at 22:42
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    $\begingroup$ 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. $\endgroup$ – TSF Jan 18 '18 at 9:43
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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

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    $\begingroup$ 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/… $\endgroup$ – Calvin Khor Feb 2 '18 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 '18 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 $\endgroup$ – Calvin Khor Feb 3 '18 at 0:00
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The function $$x \mapsto \left|x\right|$$ is Lipschitz-continuous (with $k=1$) but not differentiable at $0$.

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

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

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