TITLE: Locally Lipschitz does not imply $C^1$
QUESTION [6 upvotes]: Let $A$ be open in $\mathbb{R}^m$; let $g:A\rightarrow\mathbb{R}^n$. If $S\subseteq A$, we say that $S$ satisfies the Lipschitz condition on $S$ if the function $\lambda(x,y)=|g(x)-g(y)|/|x-y|$ is bounded for $x\neq y\in S$. We say that $g$ is locally Lipschitz if each point of $A$ has a neighborhood on which $g$ satisfies the Lipschitz condition.

Show that if $g$ is locally Lipschitz, then $g$ is not necessarily of class $C^1$.

I've thought about functions $f:\mathbb{R}\rightarrow\mathbb{R}$. Functions like $f(x)=x^a$ for $a\geq 1$ are locally Lipschitz, but they're also continuously differentiable, so don't quite work.

REPLY [3 votes]: By Rademacher's theorem   any Lipschitz continuous function is differentiable a.e. 
The derivative
of a lipschitz function is a bounded measurable function, but this derivative need not be continuous.
Hence the locally lipschitz function   need not be $C^1$. 
For example $f(x) = \frac {x^i x^j}{ |x|}  $ on the unit $n$-ball.