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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 $fin C_{0}rightarrow |f|$ 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 $h = fphi in C_{00}(X)$ 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.

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 $Usubseteq X$ 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 $Vsubseteq overline{U_1} subseteq F subseteq U$, $V$ 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 $Usubseteq X$ 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 $Vsubseteq X$ 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 $psi|_{Xsetminus overline{V}} = 0$. Define a continuous function $phi : X to [0,1]$ as $phi = psi|_{X}$. Then $phi|_K = 1$ and $phi(x) = 0, forall x in Xsetminus V$ 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 $overline{V}$.