Local minimum implies local convexity? Consider a real function $f$, and suppose it has a local minimum at $a\in \mathbb R$.
It typically looks like


What hypotheses can be added to $f$ so that there is some $\epsilon >0$ such that $f$ is convex over $(a-\epsilon,a+\epsilon)$ ?

The motivation for this question is intuition, but I can't find any valid criterion.

No in the continuous case, a part requiring in practice convexity. Yes in the smooth case.
In general local minima have nothing to do with convexity:
The function $\sqrt{|x|}$ has a local minimum in $0$ but it is not convex
The function $e^x$ is striclty convex everywhere but has no minimum.
On the other hand, as pointed out in comments, if $f$ is continuously differentiable at least twice, and if $f''(x)\neq 0$ at a local minimum $x$, then we have $f'(x)=0$ and $f''(x)>0$, which forces local convexity.
In fact, this issue is the core of the so called maximum principle which is very useful in the theory of differential equations:
Suppose $f$ is a smooth function and you are able to show that at every critical point $f''<0$. Then $f$ has no local minima. 
(Ex. $x''(t)=e^x\sin(x'(t)e^t) -1$. If $x'=0$ then $x''<0$)
