If function $f$ is continuously differentiable at some point, say $x=0$, and is Lipschitz in some neighborhood of $x=0$, is that true there is an open neighborhood of $x=0$ in which $f$ is continuously differentiable?
I know there is a function which is differentiable at just one point and continuous everywhere else. I also know the set of continuity of a derivative of a function is dense. I also familiar with a differentiable function which is not $C^1$ On the Cantor set.
No it is not true.
Let $$ f(x)=\max\{1-\lvert x\rvert,0\}, $$ and define $$ g(x)=\sum_{n=1}^\infty 2^{-n}f\Big(nx-\frac{1}{n}\Big). $$ Then $f$ is Lipschitz everywhere, differentiable a $x=0$, and not differentiable at $x=1/n^2$, for all $n\in\mathbb N$.
Construct any everywhere differentiable function $f$ such that $f'$ has a dense set of discontinuities. The function $f'$ has to be continuous at some point $x_0$ (since $f'$ is Baire 1) and so at that point you won't have an open neighborhood where $f'$ is continuous. For the construction see
Note that in some neigborhood of $x_0$ the function $f'$ is bounded and so the function $f$ is Lipschitz in that interval. (So, in fact, asking for $f$ to be Lipschitz in a neigborhood was redundant since continuity of the derivative at a point supplies that anyway.)
