Let $\, f:\mathbb{R}^n \to \mathbb{R}$. Assume that for every path $\alpha:[0,1] \to \mathbb{R}^n$ such that $\alpha(0)=\bar 0$, there exists an $\epsilon >0$ such that $$ (*) \, \,f(\bar 0) \le f(\alpha(t)) $$ for all $t \in (-\epsilon,\epsilon)$ .
Is it true that $\bar 0$ is a local minima of $f$? (i.e is there a ball around the origin where the value of $f$ is not lower than $f(\bar0)$?)
When assuming $f$ is twice differentiable, one gets (via the Taylor expansion of $f$) that $\operatorname{Hess}f$ is positive semi-definite**. However, this does not imply $\bar 0$ is a local minimum. I would like to know if there are smooth counter-examples.
**Indeed:
First note that condition (*) implies $df_{\bar 0}=0$ ($\bar 0$ is a critical point of $f$). This can be seen by choosing $\alpha(t)=te_i$.
Then, by Taylor's theorem:
$$ f(\alpha(t))=f(\bar 0) +\frac{1}{2}\alpha(t)^T \operatorname{Hess}f(\bar 0) \, \alpha(t),$$ so
$$ \alpha(t)^T \operatorname{Hess}f(\bar 0) \, \alpha(t) \ge 0 $$ for every path $\alpha$ and $t$ small enough.
Let's prove the contraposition: If $f({\bf 0})=0$, and ${\bf 0}$ is not a local minimum of $f$, then there is a path such that $\ldots$
Proof. If $f({\bf 0})=0$, and ${\bf 0}$ is not a local minimum of $f$ then for each $n\geq1$ there is a point ${\bf x}_n$ with $|{\bf x}_n|<2^{-n}$ and $f({\bf x}_n)<0$. Now put $$ \alpha(2^{-n}):={\bf x}_n\quad(n\geq1)\ ,\qquad \alpha(0):={\bf 0}\ ,$$ and interpolate linearly. For this path (of finite length!) there is no $\epsilon>0$ such that $f\bigl(\alpha(t)\bigr)\geq0$ for $0\leq t\leq\epsilon$.