Let $U$ be an open bounded in $\mathbb{R}^n$. For any function $w:U\rightarrow\mathbb{R}$ set $$ \operatorname{Argmax}(w):=\{x\in U:w(x)=\max_Uw\}. $$ I have to prove that, if $w\in C(U)$ (i.e. $w$ is continuous in $U$), $w<0$ near $\partial U$, while $w(x)>0$ for some $x\in U$, then $\operatorname{Argmax}(w)$ is nonempty. Some hints?
Thank Your
Since $ w < 0 $ near $ \partial U $ and $ w(x) > 0 $ for some $ x \epsilon U $, we know by the intermediate value theorem that $ 0 \epsilon w (U) $. Then, $ w^{-1} [0, \infty) $ is nonempty. But $ [0, \infty) $ is a closed set, then, because $ w $ is continuous, $ w^{-1} [0, \infty) $ is closed. Let's call this set A. Since $ U $ is bounded and $ A \subset U $, $ A $ is closed and bounded, and therefore compact. Then, $ w | A $ (w restricted to $ A $) being a continuous function in a compact set, it reaches it's maximum by Weierstrass' theorem, that is,
$$ \exists \: u \: \epsilon \: U \: st. w(u) = max_{x \epsilon A} w(x) $$
Since $ w < 0 $ in $ U-A $, this maximum in $ A $ should be a maximum in $ U $ too, that is,
$$ w(u) = max_{x \epsilon U} w(x) $$
Therefore
$$ u \: \epsilon \: argmax_{x \epsilon U} w(x) $$
showing this set is nonempty.
Edit: I think the nontrivial part is that the only thing I proved about $ A $ is that it is closed relative to $ U $. But to prove that $ A $ is compact I need to show that it is bounded and closed with respect to $ \mathbb R $. There you use that $ w < 0 $ close to the boundary of $ U $, because this means that the set $ w >= 0 $ is contained in a closed set inside $ U $. That is, $ A $ is a closed set with respect to a closed set and hence closed in $ \mathbb R $.