# Differentiability in a neighborhood of a strictly convex continuous function

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly convex continuous function, and let $f$ be differentiable at the point $x_0\in \mathbb{R}$. Can we say that $f$ is differentiable on some neighborhood $U$ of $x_0$?

I know that strictly conxex functions have a countable number of non-differentiable points.

No, a counter-example is the function $f$ such that $f(0):=0$, and $f(\frac{1}{2^n}):=\frac{1}{4^n}$ and $f(\frac{-1}{2^n}):=\frac{1}{4^n}$ for all $n \in \mathbb{Z}$. And such that for all $x\in]1/2^n,1/2^{n-1}[ \cup ]-1/2^{n-1},-1/2^n[$, $$f(x):=1/4^n+(|x|-1/2^n)\times3\times\frac{1}{2^n}+\frac{1}{2}(|x|-1/2^n)\times(|x|-1/2^{n-1})$$
$f$ is strictly convex and differentiable at the point $x_0=0$ (because $f'(0)=0$ ), but not differentiable in a neighborhood of $0$, because $f'(\frac{1}{2^n}^-)=7/ 2^{n+2}$ and $f'(\frac{1}{2^n}^+)=5/2^{n+1}$, for all $n \in \mathbb{Z}$.
$f$ is strictly convex on $\mathbb{R}^+$ because, for all $n \in \mathbb{Z}$, $f'$ is increasing on $]1/2^n,1/2^{n-1}[$ and $f'(\frac{1}{2^n}^-)<f'(\frac{1}{2^n}^+)$.
$f$ is strictly convex on $\mathbb{R}^-$ because, for all $n \in \mathbb{Z}$, $f'$ is increasing on $]-1/2^{n-1},-1/2^n[$ and $f'(-\frac{1}{2^n}^-)<f'(-\frac{1}{2^n}^+)$.