Let $f:(-1,1) \to \mathbb R$ be a continuous function, which is not differentiable at zero. Suppose that $f$ is strictly decreasing on $(-1,0]$, and is strictly increasing on $[0,1)$.
Is $f$ convex in some neighbourhood of zero?
I know that if $f$ is smooth at zero, then it does not need to convex near zero. (One can construct examples using bump functions, see here and here.)
Does this functions fit to your conditions? $$ f(x) = \sqrt{|x|} $$
Just pointing out that the example in the first linked post is easily modified to produce a function that is non-differentiable at $0$, not convex at $0$, and $f(0)=0$ is a strict minimum (i.e. the smoothness was supposed to make it "harder" to prove):
$$f(x)=\begin{cases}x^2 + \max(0,x)^2\sin^2 \frac1x & x\neq 0\\ 0 & x=0\end{cases}$$ The $x^2$ makes it so that the unique zero of $f$ is $x=0$, and $f>0$ otherwise. Its not convex to the right of $0$ because $f$ "jumps" between the $x^2$ curve and $2x^2$ curve infinitely often. That it is not differentiable at $0$ is more or less standard.
An example similar to daw's is:
$$f(x)=\begin{cases}\arcsin(-2x-1) + \frac{\pi}{2} & -1\leq x< 0\\ \arcsin(2x-1) + \frac{\pi}{2} & 0\leq x\leq 1 \end{cases}$$
