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!
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.