I am wondering if a Lipschitz function $f:[a,b]\to\mathbb{R}$ is $C^1$, that is its derivative is also continuous? I have seen that in a text however I could not prove it and does not seem so obvious for me!
Any suggestion?
2026-03-26 06:18:29.1774505909
A Lipschitz function is $C^1$?
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The answer is negative. Take, for example, $f(x)=|x|$. Then $$ |f(x)-f(y)|=||x|-|y||\le|x-y| $$ and hence $f$ is Lipschitz. However, it is not differentiable at $x=0$.