Criterion for globality of extreme point of a scalar field? So, lets say we have $\DeclareMathOperator{\int}{int}$$\Omega \subset \mathbb{R}^n$, $ f:\Omega\rightarrow \mathbb{R}$ being continuous and partially differentiable on $\int\Omega$. Now there is $x_*\in \int\Omega$ such that $\forall x \in \int\Omega :\nabla f(x) = 0 \Leftrightarrow x = x_*$. Also, $x_*$ is a local maximum of $f$. Now I state that $x_*$ is also a global Maximum of $f$.
I think this could be true, at least for $n = 2$ (maybe with a little more regularities for f or $\Omega$), because I could not find an easy counterexample. If I try to get f bigger than $f(x_*)$ somewhere, it seems that there will always appear a point different from $x_*$ where the gradient is $0$. For example I thought of $f(x)=\exp ( \Vert x\Vert ^2)+\frac{1}{\Vert x - (2,\, 2)\Vert}$, $\Omega = \mathbb{R}^2\backslash\{(2,\,2)\}$. But here between the maximum and the singularity point there is a saddle point with vanishing gradient. Here the plot:
https://i.ibb.co/JFg1dbq/graphic.png
My question would be if my conjecture is correct and if it is, I'd like to know if this is a well known fact.
 A: I think we need to assume $-f$ is convex over $\Omega$.
Why?
Let us assume that there is a global maximizer $y\neq x^* : f(y) > f(x^*)$ on the boundary $\partial \Omega$. Then
$$\exists \lambda > 0,  c = \lambda x^*+(1-\lambda)y: f(c) = f(x^*)=q$$
by the Intermediate Value Theorem.
From differentiability of $f$ on $\text{int }\Omega$ we know that
$$\exists t >0, d = tx^*+(1-t)c: \nabla f(d)\cdot(c-x^*)=f(c)-f(x^*) = 0$$
by the multivariate Mean Value Theorem.
So we know that $d$ is a local extrema along $(y-x^*)$. However, by assumption $\nabla f(d) \neq 0$, so $d$ cannot be a critical point, which means there is some direction $u:\nabla f(d) \cdot u \neq 0$. Therefore either $-u$ or $u$ will result in an increase in $f$. Assuming no local critical point exists apart from $x^*$ we can trace out a path from $d$ to $y$ along the curve defined by the gradient (i.e., we perform gradient ascent starting at point $d$). Since there are no local critical points on the interior, the path will always end at a critical point on the boundary (assuming we switch to constrained gradient ascent along the boundary) because we will not get stuck any any local extrema. So your condition of a unique critical point is not sufficient to ensure $x^*$ is a global maximum.
However, if we assume that $-f$ is convex over $\Omega$, then we cannot have the above situation. Convex functions over a set $\chi$ have the property that if $r \in \chi$ is a local extremum, it is also the global extremum over $\chi$.
A: I found a paper with a concrete counterexample. The function has a "saddle point at infinity".
