Let $n\ge 2$ and $f\in C^1(\mathbb{R}^n, \mathbb{R})$ with $f(0)=0$ being a strict local maximum for the function $f$. Denote by $$F=\lbrace x\in\mathbb{R}^n : f(x)<0\rbrace \cup \{0\}$$ and let $F_1$ be the component of $F$ containing $0$.
First question: What is meant by "component of $F$"? I would say that $F_1\subset F$ where, if for example $f$ is negative for $x\in A\subset\mathbb{R}^n$, it is $$F_1 = A\cup\{0\}.$$ I am not sure about that.
Hence it is said that, if $F_1$ is bounded, we can conclude that $$ f(x) < f(0)=0 \quad \forall x\in F_1$$ and $$f(x) = f(0)=0 \quad\text{ for } x\in\partial(F_1).$$
I think that $f(x) < f(0)=0 \ \forall x\in F_1$ comes from the definition of $F$ and the fact that $F_1\subset F$ is a component. The condition $f(x) = f(0)=0$ for $x\in\partial(F_1)$ it is not clear (to me) instead.
Could anyone please help to understand that?
${\bf EDIT:}$ The question concerns the paper https://link.springer.com/article/10.1007/BF02571356
I am trying to understand why the assumptions of Theorem 2.22 imply what the authors state in (1.3) and the 2 subsequents in the introduction at page $1$.
In this context component means connected component. Since $0$ is the strict local maximum of the function $f^{-1}(0)$ is necessarily the boundary. To see this note that for a continuous function the inverse image of an open set is open so the preimage $f^{-1}((-\infty , 0))$ is open and all of its points are interior. Since the boundary points are not interior we have $\partial(F_1) \subset f^{-1}(0)$. This shows for $x \in \partial F_1$ we have $f(x)=f(0)=0$.