I have seen the following statement:

Let $G\subset \mathbb{R}^n$ be open, bounded and $f:\overline{G}\rightarrow \mathbb{R}^n$ a continuous and open map. Then $\|f\|$ gets its maximum on the boundary of $G$.

Why does this holds? Could you explain this to me?


I suppose that you mean that $f|_G$ is open.

If the maximum was attained at a point $p$ ouside the boundary, then $p$ would belong to $G$. But, since $f|_G$ is open, the range of $f$ must contain an open ball centered at $p$, contradicting the assumption about the maximum.

  • $\begingroup$ Do you mean that it should be $f|_G:\overline{G}\rightarrow \mathbb{R}^n$ ? Or do you mean $f|_G:G\rightarrow \mathbb{R}^n$ or something else?If the maximum is attained at $p$ why is it not possible that there exists an open ball centered at $p$ ? I got stuck right now. $\endgroup$ – Mary Star Jun 17 '18 at 18:52
  • $\begingroup$ @MaryStar I mean that $f$ is continuous and that $f|_G$ is open. And if the range of $f$ contains an open ball of radius $r$ centered at $f(p)$, then there will be points $q$ such that $\|f(q)\|\in(\|f(p)\|,\|f(p)\|+r)$. In particular, $\|f(q)\|>\|f(p)\|$. $\endgroup$ – José Carlos Santos Jun 17 '18 at 19:05
  • $\begingroup$ I understand!! Thank you so much!! :-) $\endgroup$ – Mary Star Jun 17 '18 at 19:40
  • $\begingroup$ Do you maybe have also an idea for my other question: math.stackexchange.com/questions/2822587/… $\endgroup$ – Mary Star Jun 17 '18 at 19:40

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