# Functional analysis: locally compact hausdorff space

Let $$X$$ be a locally compact Hausdorff space. Show that every function in $$C_0(X)$$ (continuous functions that vanish at infinity) can be arbitrarily uniformly approximated by functions in $$C_{00}(X)$$ (continuous functions of compact support). In other words, show that the closure of $$C_{00}(x)=C_0(X)$$

My work so far: Let $$f\in C_{0}\rightarrow |f|$$ is continuous.

Let $$K_{n}=|f|^{-1}([1/n,+\infty[) n \in N$$

$$A_{n}=|f|^{-1}([0,1/2n])$$

Urysohn`s lemma states that $$\exists$$ a continuous function $$g_{n} : X \rightarrow [0,1]$$ such that $$g_n(K_n)=1$$ and $$g_n(A_n)=0$$

We now define $$f_n=g_n*f$$

I can't show that $$f_n$$ converges uniformly to $$f$$ and each $$f_n \in C_{00}.$$

Let $$\varepsilon > 0$$. Since $$f \in C_0(X)$$, set $$\{|f| \ge \varepsilon\} \subseteq X$$ is compact so by Urysohn's lemma there exists $$\phi \in C_{00}(X)$$ such that $$\phi(X) \subseteq [0,1]$$ and $$\phi|_{\{|f| \ge \varepsilon\}} = 1$$.

Define $$h = f\phi \in C_{00}(X)$$ and we claim that $$\|f-h\|_\infty \le \varepsilon$$. Indeed, on $$\{|f| \ge \varepsilon\}$$ we have $$h = f$$, and for $$x \in \{|f| < \varepsilon\}$$ we have $$|f(x) - h(x)| = |f(x) - f(x)\phi(x)| = |f(x)||1-\phi(x)| \le |f(x)| \le \varepsilon$$ so the claim follows.

Lemma:

Let $$X$$ be a locally compact Hausdorff space.

1. Let $$x \in X$$ and let $$U \subseteq X$$ be an open neighbourhood of $$x$$. Then there exists a precompact open neighbourhood $$V \subseteq X$$ of $$x$$ such that $$\overline{V} \subseteq U$$.
2. Assume $$K \subseteq X$$ is compact and $$U\subseteq X$$ open such that $$K \subseteq U$$. Then there exists a precompact open set $$V \subseteq X$$ such that $$K \subseteq V \subseteq \overline{V} \subseteq U$$.

Proof:

1. $$U$$ is an open neighbourhood of $$x$$ so by local compactness there exists a compact neighbourhood $$F \subseteq X$$ of $$x$$ such that $$F \subseteq U$$. Define $$U_1 = U \cap \operatorname{Int}(F)$$. Notice that $$\overline{U_1}$$ is a closed subspace of a compact set $$F$$ so it is also compact. Hence $$\overline{U_1}$$ is a compact Hausdorff space so in particular it is regular. $$\partial U_1$$ is closed in $$\overline{U_1}$$ so there exist $$V,W \subseteq \overline{U_1}$$ disjoint and open in $$\overline{U_1}$$ such that $$x \in V, \partial U_1 \subseteq W$$. Since $$V\subseteq \overline{U_1} \subseteq F \subseteq U$$, $$V$$ is open in $$X$$ and $$\overline{V}$$ is a closed subspace of $$F$$ and hence compact, which proves the lemma.
2. For every $$x \in K$$ there exists a precompact open set $$V_x$$ such that $$x \in V_x \subseteq U$$. $$(V_x)_{x \in K}$$ is an open cover of $$K$$ so there exist $$x_1, \ldots, x_n \in K$$ such that $$K \subseteq \bigcup_{i=1}^n V_{x_i} := V$$. Then $$V$$ is the desired set since $$\overline{V} = \bigcup_{i=1}^n \overline{V_{x_i}} \subseteq U$$ and is compact as a finite union of compact sets.

Urysohn's lemma for LCH spaces:

Let $$X$$ be a locally compact Hausdorff space, $$K \subseteq X$$ compact and $$U\subseteq X$$ open such that $$K \subseteq U$$. Then there exists a continuous function $$\phi : X \to [0,1]$$ such that $$\phi|_K = 1$$ and $$\operatorname{supp} \phi \subseteq U$$ is compact (i.e. $$\phi \in C_{00}(X)$$).

Proof: Let $$\tilde{X}$$ be the one-point compactification of $$X$$. Let $$V\subseteq X$$ be the open precompact set such that $$K \subseteq V \subseteq \overline{V} \subseteq U$$. Then $$K$$ and $$\tilde{X}\setminus V$$ are disjoint closed subspaces of $$\tilde{X}$$, which is a compact Hausdorff space so in particular it is normal. Hence by the classical Urysohn's lemma there exists a continuous function $$\psi : \tilde{X} \to [0,1]$$ such that $$\psi|_K = 1$$ and $$\psi|_{X\setminus \overline{V}} = 0$$. Define a continuous function $$\phi : X \to [0,1]$$ as $$\phi = \psi|_{X}$$. Then $$\phi|_K = 1$$ and $$\phi(x) = 0, \forall x \in X\setminus V$$ so $$\operatorname{supp}\phi = \overline{\phi \ne 0} \subseteq \overline{V} \subseteq U$$ and it is compact as a closed subspace of the compact set $$\overline{V}$$.

• Thank you for your answer. I think my problem is that I don't know what Urysohn's lemma says because i can't see why $\phi \in C_{00}(X)$ – Carlos Roger Jan 12 at 2:19
• @CarlosRoger We need a stronger version of Urysohn's lemma to ensure $\phi \in C_{00}(X)$. I have added the details, have a look. – mechanodroid Jan 12 at 15:34