If $f:U\rightarrow\mathbb{R}$, $U\subset\mathbb{R}^m$ open and bounded, is continuous and has partial derivatives, when does $f$ have global extrema?

Question

My actual question was a bit too long to state on the title, but I'm trying to prove the following statement, which is a generalization of Rolle's theorem:

Let $f:U \rightarrow \mathbb{R}$ be continuous on the open bounded set $U \subset \mathbb{R}^m$, having partial derivatives at every point of $U$. If, for every $a \in \partial U$, we have $\lim_{x \rightarrow a} f(x)=0$, then there exists some $c \in U$ such that $\frac{\partial f}{\partial x_i} (c)=0$ for $i=1, \dots, m$.

I'm not yet looking for direct hints to the above statement as I feel that I need to work more on it before asking for help. However, I was thinking that, if I proved that $f$ attains some global extrema at some point $c \in U$, then I could prove that the partial derivatives vanish there. However, unlike the case for closed bounded sets, I know that continuity on an open bounded set doesn't guarantee the existence of global extrema, but I was wondering whether the additional condition $\lim_{x \rightarrow a} f(x)=0$ for every $a \in \partial U$ was enough to give us that result. This definitely seems to be the case when $m=1$, and perhaps when $m=2$ as well, though I'm having some trouble visualizing the latter. Regardless, I was also wondering if we could at least say that the function $f$ can be extended to a continuous function $g$ that has partial derivatives and is defined on a closed bounded set (perhaps the closure of $U$?).

Answer 1

... the function can be extended to a continuous function $g$ ...

Yes, you can extend $f$ to a continous function $g$ on $\overline U$ by defining $g(x) = f(x)$ for $x \in U$ and $g(x) = 0$ for $x \in \partial U$.

... that has partial derivatives and is defined on a closed bounded set (perhaps the closure of $U$?).

No, $g$ does not necessarily have partial derivatives at points on the boundary (example below). However, that is not needed for the desired conclusion.

If $g$ is identically zero then so is $f$ and you are done. Otherwise $g$ attains a strictly positive maximum or a strictly negative minimum at a point $c \in U$. Then $g$ (and consequently, $f$) has a global extremum at $c$ and all partial derivatives of $f$ at $c$ are zero.

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Example: $f(x) = \Vert x \Vert (1-\Vert x \Vert)$ on $U = \{ x \in \Bbb R^m \mid 0 < \Vert x \Vert < 1 \}$. $f$ is continuously differentiable in $U$ and has zero limits at the boundary. However, the extended function $g$ does not have partial derivatives at the origin.

Answer 2

Suppose $\sup f > 0$, then let $C = \{ x \in U | f(x) \ge {1 \over 2} \sup f \}$. Note that $C$ is closed (why?) hence compact and so $f$ attains a $\max$ at some $c \in C$, and we must also have that $f$ attains a $\max$ on $U$ at $c$ (again, why?). Since $c \in U^\circ = U$, we must have ${\partial f(c) \over \partial x_k} = 0$ for all $k$.

If $\sup f < 0$, repeat the above in a similar manner.

Otherwise pick any point in $U$.