TITLE: The local lipschitz condition implies differentiability?
QUESTION [7 upvotes]: I know differentiability implies the local lipschitz condition. however, I am not sure the converse. 
Actually, I think it might be. 
The definition of the local lipschitz condition is that for 
$$
f: A \subset \mathbb{R}^m \rightarrow \mathbb{R}^n
$$
and $ \forall x_{0} \in A$, there are a constant $M > 0$ and a $\delta_{0} > 0$ such that $\lVert x- x_{0}\rVert < \delta_{0}$ implies 
$$
\lVert f(x) - f(x_{0})\rVert \leq M\lVert x-x_{0}\rVert
$$
And the definition of differentialbility is that $\lVert x-x_{0}\rVert < \delta $ implies
$$
\lVert f(x) - f(x_{0}) - Df(x_{0})(x-x_{0})\rVert < \epsilon \lVert x-x_{0}\rVert
$$

REPLY [10 votes]: No. Consider $f(x))=|x|$, which is Lipschitz but not differentiable.
However, Rademacher's theorem states that a (locally) Lipschitz function is differentiable almost everywhere.

REPLY [5 votes]: Think about $f(x)=|x|$. Is it (globally) Lipschitz continuous? What about differentiable?