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Let $f$ be a lipschitz function in $[0,1]$,

(it exists a $C>0$ that we have for all $x,y \in [0,1]$ $|f(x)-f(y)|<C.|x-y|$)

Can we prove that $f$ is differentiable ?

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    $\begingroup$ Nope. A counterexample is $f(x)= |x|$. See en.wikipedia.org/wiki/Rademacher%27s_theorem for a similar result. $\endgroup$ – Crostul Jan 18 at 16:28
  • $\begingroup$ @Crostul yeh you re right, thank you $\endgroup$ – Anas BOUALII Jan 18 at 16:29
  • $\begingroup$ @Crostul minor correction: $|x|$ is differentiable on $[0,1].$ $\endgroup$ – zhw. Jan 18 at 19:22
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No. For example, try $f(x) = |x - 1/2|$.

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  • $\begingroup$ Oh, yeh thats right, $\endgroup$ – Anas BOUALII Jan 18 at 16:29
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In general, no. The function $x \mapsto |x|$ gives a counterexample.

However, Lipschitz functions are differentiable almost everywhere:

Lipschitz continuity implies differentiability almost everywhere.

Also note that if the derivative is bounded, the converse is true.

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