functional-analysisgeneral-topologylocally-compact-groups
How can I prove the following theorem?
Let $X$ locally compact, $K \subseteq X$ compact, $O \subseteq X$ open with $K \subseteq O$. Then there exists $f \in C_c(X)$ with $0 \leq f \leq 1$, $f|_K \equiv 0$ and $\operatorname{supp}(f) \subseteq O$.
(Where $C_c(X)=\{f ∈ C(X, \mathbb C) : \operatorname{supp}(f) \text{ compact}\}$).
I think that I would need the topological lemma for locally compact spaces, but I couldn't come up with a suitable proof. Thanks in advance!
Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.
Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \ \subset cl_X(V) \subset Y \ \text{ such that } cl_X(V)\text{ is $X-$compact.}$
$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$
As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.
By 2 and 4, $cl_Y(V) \subset cl_X(V)$.
As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:
Here is a proof of the Lemma in the OP based an a nice auxiliary result by Lynn H. Loomis.
Lemma (Loomis): Suppose $X,Y$ are locally compact Hausdorff (l.c.H) spaces and $Z$ a topological space, and let $f:X\times Y\rightarrow Z$ be a continuous function. If $K\subset X$ is compact and $U\subset Z$ open, then $$W=\{y\in Y: f(x,y)\in U\ \text{for all}\ x\in K\}$$ is open in $Y$.
Proof: Fix $y_0\in W$. We show that there is an open neighborhood $V_{y_0}$ of $y_0$ fully contained in $W$. The continuity of $f$ implies that for each $x\in K$, there are open neighborhoods $A_x\subset X$ containing $x$ and $B_x$ containing $y_0$ such that $f(A_x\times B_x)\subset U$. By compactness of $K$, there are finite $x_1,\ldots, x_m\in K$ such that $K\subset \bigcup^m_{j=1}A_{x_j}$. The set $V_{y_0}=\bigcap^m_{j=1}B_{x_j}$ is open, contains $y_0$. Suppose $y\in V_{y_0}$, and let $x\in K$. Choose $A_{x_j}$ containing $x$. It follows that $(x,y)\in A_{x_j}\times V_{y_0}\subset A_{x_j}\times B_{x_j}$ and so, $f(x,y)\in U$.
As in the OP, let $X$ be a l.c.H space and $G$ a l.c.H topological group that acts continuous on $X$, and let $f\in\mathcal{C}_c(X)$, ale let $K=\operatorname{supp}(f)$. Recall that an action can be seen as a continuous function $\Phi:G\times X\rightarrow X$ such that $\Phi(1,x)=x$ for all $x\in X$ and $\Phi(gh,x)=\Phi(g,\Phi(h, x))$ for all $g,h\in G$ and $x\in X$. It is usual to denote the group action as $g\cdot x:= \Phi(g,x)$.
Choose any symmetric neighborhood $U$ of $1\in G$ with compact closure. Define $F:X\times G\rightarrow \mathbb{R}$ by $$F(x,g)= f(g^{-1}\cdot x)-f(x)=f(\Phi(g^{-1},x))-f(x)$$ This is a continuous function. Notice that $\overline{U}\cdot K:=\Phi(\overline{U}\times K)\subset X$ is compact. The Lemma above shows that $$W=\{g\in G: |F(x,g)|<\varepsilon,\,\text{for all}\, x\in \overline{U}\cdot K\}$$ is an open subset of $G$ which contains $1$; hence, $V:=(W\cap W^{-1})\cap U$ is a symmetric open neighborhood of $1$.
Let $s\in V$, and let $x\in X$. If $x\in \overline{U}\cdot X$, then by definition of $W$, $$|f(s^{-1}\cdot x)-f(x)|=|F(x,s)|<\varepsilon$$ If $x\notin \overline{U}\cdot K$, then $g^{-1}\cdot x\notin K$ for all $g\in V$. Indeed, if $g^{-1}\cdot x\in K$ for some $g\in V$, then $x=ss^{-1}\cdot x=s\cdot(s^{-1}\cdot x)\in U\cdot K$ which is a contradiction. This means that $$|f(s^{-1}\cdot x)-f(x)|=|f(s^{-1}\cdot x)-f(1\cdot x)|=0<\varepsilon$$
This proves the Lemma in the OP.
Best Answer
You can see the proof of Urysohn's Lemma (a weaker version) for locally compact spaces on Walter Rudin's Real and Complex Analysis
The problem is that the Urysohn Lemma is valid for Normal Spaces and we are trying to prove something similar, in fact weaker, but for locally compact spaces. Recall that a locally compact space is completely regular, but there are locally compact spaces that are not normal (I am aware of the existence of such spaces, although a counterexample does not come to mind) and hence the full force of the Urysohn's Lemma does not apply.
Still, for what you are looking for, Rudin has it all covered up excellently and I couldn't do it in a more concise way here other than copy and pasting it, so you'd better look it upon pages 35 40 of the aforementioned book.