Relation between Lipschitz and differentiablity

Question

Let $f:\Bbb R \to \Bbb R$ be a Lipschitz function. Show that the set of all reals at which $f$ is differentiable in non-empty.

I know that if $f$ is differentiable and derivative is bounded then it is Lipschitz. I know that the converse is FALSE. For example $f(x)=|x|$.

But I've no idea how to prove this statement. Can anyone give some hint ?

Answer 1

The set would be better than being merely non-empty.

If $f:\mathbb{R}\to\mathbb{R}$ is Lipschitz, then it differentiable almost everywhere in $\mathbb{R}$ by Rademacher's theorem.

Answer 2

Without Rademacher: $f$ Lipschitz implies $f$ is of bounded variation on $[0,1]$ , and so it can be written as the difference of two increasing functions, hence $f$ is differentiable there, a.e.