I am confronted with 2 definitions Let $A \subseteq \mathbb R^n$ and $f:A\to \mathbb R^m$.
$f$ is said to be locally Lipschitz at a point $c \in A$ if $$ (\exists \delta>0)(\exists L \ge 0)(\forall x \in A)\\ \Vert x-c\Vert<\delta \implies \Vert f(x)-f(c)\Vert\le L\Vert x-c\Vert $$ and $f$ is said to be locally Lipscihtz around a point $c\in A$ if $$ (\exists \delta>0)(\exists L \ge 0)(\forall x,y \in K(c,\delta))\\ \Vert f(x)-f(y)\Vert\le L\Vert x-y\Vert $$ obviously the second definition asks for more and every locally Lipschitz function around a point is also locally Lipschitz in a point, but I am trying to prove that the opposite isn't true and have been failing for a while
the only thing the first definition gives is $$\Vert f(x)-f(y)\Vert \le L (\Vert x-c\Vert + ||y-c||)$$
edit: $K(c,\delta)$ is an open ball around $c$ with a radius $\delta$
Let $f(x)=x^{2}\sin(1/x^{2})$ for $x\ne 0$, $f(0)=0$, we have $f'(0)=0$, so it is locally Lipschitz at $0$, but $f'(x)$ is unbounded around $0$, so it is not locally Lipschitz around $0$.