Let $f:[0,1]\to\mathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0\in [0,1]$. Is it true that there exists $L>0$ such that $\lvert f(x)-f(x_0)\lvert\leq L\lvert x-x_0\lvert$?
I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?
Best Answer
From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |\leq L_1| x-x_0|$ for $|x-x_0| < \delta$. Since $f$ is continuous on a compact set, $|f(x)|_{\infty} < \infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |\leq L_2| x-x_0|$ for $|x-x_0| \geq \delta$. Take $L = \max \{L_1, L_2\}$.