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.
Best Answer
The Euclidean norm $f(x)=\|x\|$ is probably the simplest example: the Lipschitz condition comes straight from the triangle inequality, and the non-differentiability is seen from the fact that the restriction of $f$ to the first coordinate axis is $|x_1|$.