Lipchitz condition: $f:R→R$ is called Lipschitz continuous if there exists a $K>0$ such that, for all real $x_1$ and $x_2$, $|f(x_1 )-f(x_2 )|≤K|x_1-x_2 |$.
Is every function satisfying the Lipchitz condition differentiable?
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gbd
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1What about the absolute value function? – José Carlos Santos Jun 19 '21 at 15:40
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3No, but they are differentiable almost everywhere (Rademacher theorem). – copper.hat Jun 19 '21 at 15:41
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A counterexample is $$y=|x|$$. This function is not differentiable at $x=0$. However, it is Lipschitz continuous, since $|y(x_1)-y(x_2)|\le |x_1-x_2|$.
Mr.xue
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