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.
On
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 $x''<0$)
As pointed out by the user above you certainly need more than just continuity. You also need more than differentiability since one could consider $f(x)=x^{2}(\sin(\frac{1}{x})+1)$ for $x\neq0$ and $f(0)=0$ which has a local minimum at $0$ but $f$ is not convex in any neighbourhood of $0$. By changing the exponent of the power of $x$ you can show further that to obtain local convexity it is not enough to be just $C^{2}$ unless you have positive derivative at the point of the minimum. Even worse, you can have a $C^{\infty}$ function with a local minimum at $x=0$ but still not be convex in any neighbourhood of $0$ provided you don't assume that $f''(0)>0$. This is shown by the function $f(x)=e^{\frac{-1}{x^{2}}}(\sin(\frac{1}{x})+1)$ for $x\neq0$ and $f(0)=0$.