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I am reading some paper, where they claim the following:

Let $G$ be a compact (Hausdorff) group and let $X$ be a locally, compact, Hausdorff space. Assume that $G$ acts on $X$ continuously. Denote by $C_0(X)$ the continuous functions on $X$ with complex values. Denote by $X/G$ the orbit space (it is Hausdorff as well), and let $\pi: X\to X/G$ be the canonical quotient map.

claim: Let $f\in C_0(X)$. The map $\pi(x)\mapsto sup_{g\in G}|f(g\cdot x)|$ is a continuous map $X/G\to \mathbb{R}$.

The proof starts like that: Let $f\in C_0(X)$, let $\epsilon>0$ and let $x\in X$. Use continuity of $f$ to find, for every $g\in G$, an open neighborhood $W_g$ of $g\cdot x$ such that $|f(g\cdot x)-f(y)|<\epsilon$ for all $y\in W_g$. By compactness of $G$, there exists an open neighborhood $W$ of $x$ such that $g\cdot W\subseteq W_g$ for all $g\in G$.

I can not see why the last claim is true or why it follows from compactness... it seems like they actually claim that $\bigcap\limits_{g\in G}g^{-1}\cdot W_g$ is open (of course, it contains $x$, but I don't see why should it be open...).

An alternative approach to the proof or counter example for the claim would be appreciated as well.

Thank for any help!

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  • $\begingroup$ Well, they only state that $x$ is an internal point of $\bigcap_{g\in G}g^{-1}W_g$, and.. that might somehow indeed follow from compactness of $G$. $\endgroup$ – Berci Feb 10 '18 at 14:54
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    $\begingroup$ @Berci Thanks for the comment. Yes, you are right. Do you know why? can you please give a hint for that? is it related to Baire Category theorem? $\endgroup$ – Kiko Feb 10 '18 at 15:14
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    $\begingroup$ It really sounds like an application of Arzelà-Ascoli Theorem: you have a compact set of continuous maps (namely $G\cdot f$) and you are looking for a uniform continuity neighborhood (i.e. equicontinuity). There might be easier way to deal with it though. But at least, you should not look for counter-examples. $\endgroup$ – Clément Guérin Feb 11 '18 at 4:33
  • $\begingroup$ @ClémentGuérin Thank you very much! Arzelà-Ascoli solves my problem. Would you like to write it as an answer? Then I can accept it. Otherwise, I can write the answer. $\endgroup$ – Kiko Feb 11 '18 at 10:24
  • $\begingroup$ You can post it as an answer. Once again, it might not be the most efficient way... Maybe wait for smarter answers than mine. $\endgroup$ – Clément Guérin Feb 11 '18 at 14:23
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As Clément Guérin suggested in the comments, a possible approach to prove this claim is by using Arzelà-Ascoli's Theorem in the generalised version for locally compact Hausdorff spaces, as it appears in Munker's book for topology. Can be found also here: Theorem of Arzelà-Ascoli.

With $f$, $x$ and $\epsilon>0$ as above, we have the family $\mathcal{F}=\{g\cdot f| g\in G\}$ of continuous functions (where I realise the action on $X$ as an action of $C_0(X)$, naturally). Since $G$ is compact and $\mathcal{F}$ is just the orbit of $f$ under the action, we get that $\mathcal{F}\subseteq C_0(X)$ is a compact subset. Therefore, we can apply Arzelà-Ascoli Theorem and obtain that this family is equicontinuous: There exists an open neighborhood $W\subseteq X$ of $x$ such that $|f(g\cdot x)-f(g\cdot y)|<\epsilon$ for all $g\in G$ and all $y\in W$.

Since the quotient map $\pi$ is open, $U:=\pi(W)$ is an open neighborhood of $\pi(x)$ in $X/G$. By the above, it follows that for all $y\in W$:

$sup_{h\in G}|f(h\cdot y)|-\epsilon\leq sup_{g\in G}|f(g\cdot x)|\leq sup_{h\in G}|f(h\cdot y)|+\epsilon$ , and therefore:

$sup_{\pi(y)\in U}|sup_{h\in G}|f(h\cdot y)|-sup_{g\in G}|f(g\cdot x)||<\epsilon$, as required.

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