lipschitz function in a compact, is it differentiable?

Question

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 ?

Nope. A counterexample is $f(x)= |x|$. See https://en.wikipedia.org/wiki/Rademacher%27s_theorem for a similar result. — Crostul, Jan 18 '19 at 16:28
@Crostul minor correction: $|x|$ is differentiable on $[0,1].$ — zhw., Jan 18 '19 at 19:22

score 3 · Accepted Answer · answered Jan 18 '19 at 16:28

3

No. For example, try $f(x) = |x - 1/2|$.

answered Jan 18 '19 at 16:28

Robert Israel

score 2 · Answer 2 · answered Jan 18 '19 at 16:30

In general, no. The function $x \mapsto |x|$ gives a counterexample.

However, Lipschitz functions are differentiable almost everywhere:

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

2 Answers2