In a paper I saw this expression: "The function $F:R^*\rightarrow R$ with $F(s)=s|s|^{p-2}\log(|s|), p>1$ is locally Lipschitz." To show this we have to prove that for any $\rho>0$ there exists a constant $C(\rho)>0$ such that $$ |F(s_{1})-F(s_{2})|\leq C(\rho)|s_{1}-s_{2}|,\qquad\forall s_{1}, s_{2}: 0<|s_{1}|,|s_{2}|<\rho.\qquad\qquad (E) $$ However, the above function isn't differentiable at $s=0$. I think the term $F:R^*\rightarrow R$ means the continuous extension of $F$ at $s=0$ (the paper dosn't say any thing about this notation.) I mean the function $$ F_1(s)=F(s),\quad for\quad s\neq0;\qquad F_1(s)=0,\quad for\quad s=0. $$ Easily, we can see $F_{1}$ is differentiable at $s=0$ and $F_1'(0)=0$.
My Question:
If my understanding of "locally Lipschitz" is true, then how we can show (E) for $F_{1}$? Is it enough to say that every continuously differentiable function is locally Lipschitz? (This is a theorem.)
Thank you very much in advance
'Locally Lipschitz' means Lipschitz in some neighborhood of each point of the domain. If $s \in \mathbb R^{*}=\mathbb R \setminus \{0\}$ then $s>0$ throughout some neighborhood of $s$ or $s<0$ throughout some neighborhood of $s$. Hence $F$ is continuously differentiable in the neighborhood. By MVT any continuously differentiable function on a closed interval is Lipschitz. Hence $F$ is locally Lipschtz.
Note that differentiablity at $0$ and inequality (E) are not involved at all in proving this result.