# Lipschitz function and continuously differentiable function

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.

• What does it mean for a function to be continuously differentiable at a point?
– user258700
Commented Dec 25, 2015 at 23:44
• It means $\lim_{x\to x_0}f'(x)=f'(x_0)$. Commented Dec 26, 2015 at 10:45
• This doesn't make sense unless $f'$ exists in a neighbourhood of $x_0$.
– user258700
Commented Dec 26, 2015 at 10:54
• Yes, of course, we suppose $f'$ exists in a neighborhood of $x_0$ Commented Dec 26, 2015 at 11:08

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 $n\in\mathbb N$.

• Thank you for your answer, but I need the function be continuously differentiable at x=0, not just differentiable. Commented Dec 26, 2015 at 10:39

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

Discontinuous derivative.

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.)