Let $A \subseteq \mathbb R^n $ and $f:A \to \mathbb R^m$. $f$ is locally Lipschitz in $c \in A$ if
$$(\exists \delta_c>0)(\exists L_c \le 0)(\forall x \in A)\\
(||x-c||<\delta_c \implies ||f(x)-f(c)||\le L_c||x-c||)$$
Lets call this property $(1)$
Also similarly $f$ is locally Lipschitz around $c \in A$ if
$$(\exists \delta_c>0)(\exists L_c \ge 0)(\forall x,y \in K(c,\delta_c))\\
(||f(x)-f(y)||\le L_c||x-y||)$$
Lets call this property $(2)$
These two concepts are similar but not the same which can be shown with $f:\mathbb R \to \mathbb R$
$$f(x)=
\begin{cases}
x^2\sin(\frac{1}{x^2}), & x\not = 0\\
0,& x= 0
\end{cases}
$$
which is locally Lipschitz in 0 but not around 0 since its derivative isn't bounded around 0. Aka it has property $(1)$ at 0 but not property $(2)$.
Now I am wondering if $f$ is locally Lipschitz in every point around some point $c\in A$ is that enough for a function to be locally Lipschitz around $c$. Or if a function has property $(1)$ in some Ball around point $c$, does it have property $(2)$ at $c$
What bothers me is that i can't just take infimum of all deltas since that could equal 0 since i am looking at uncountable set but i am unable to find a counterexample.
Best Answer
Your example $f$ is continuously differentiable in $\Bbb R \setminus\{0\}$, and as such it is locally Lipschitz in every $x\in \Bbb R \setminus\{0\}$.