Let $(X,T)$ be a topological Hausdorff space. By $C_b(X)$ denote the continuous bounded function $f\colon X\to\mathbb{R}$, by $C_c(X)$ the continuous functions $f\colon X\to\mathbb{R}$ which have compact support $\text{supp}f=\overline{\left\{x\in X: f(x)\neq 0\right\}}$ and, finally, by $C_0(X)$ denote the continuous functions $f\colon X\to\mathbb{R}$ that vanish at infinity.
Show that the closure of $C_c(X)$ with respect to the Banachspace $(C_b(X),\lVert\cdot\rVert_{\infty})$ is $C_0(X)$, i.e.
$$
\overline{C_c(X)}=C_0(X).
$$
I think the inclusion "$\subseteq$" is the easier one.
Let $\varepsilon >0$ be arbitrary and let $(f_n)$ be a sequence in $C_c(X)$ that converges to $f$ w.r.t. to $\lVert\cdot\rVert_{\infty}$. This implies uniform convergence of $f_n$ to $f$. Hence, $f$ is continuous. Moreover, there exists some $N(\varepsilon)\in\mathbb{N}$ such that
$$
\lvert f_n(x)-f(x)\rvert < \varepsilon~\forall n\geq N(\varepsilon)
$$
for all $x\in X$. For all $x\notin\text{supp}f_{N(\varepsilon)}$, $f_{N(\varepsilon)}(x)=0$, hence
$$
\lvert f(x)\rvert=\lvert f(x)-f_{N(\varepsilon)}\rvert <\varepsilon
$$
for all $x\notin\text{supp}f_{N(\varepsilon)}$.
This means that $f\in C_0(X)$.
Now, I do not know exactly how to prove "$\supseteq$". I guess the general idea is the following. Let $f\in C_0(X)$. Now, construct a sequence $(\varphi_n)$ in $C_c(X)$ with $\varphi_n\to f$ w.r.t. the supremum norm $\lVert\cdot\rVert_{\infty}$.
Best Answer
I don't think you need to assume that $X$ is locally compact:
Let $f\in C_{0}\left(X\right)$ and $\varepsilon>0$. Note that $U:=\left\{ x\in X\,\mid\,\left|f\left(x\right)\right|>\frac{\varepsilon}{2}\right\} $ is open with closure $\overline{U}\subset\left\{ x\in X\,\mid\,\left|f\left(x\right)\right|\geq\frac{\varepsilon}{2}\right\} $, which is compact as a closed subset of a compact set. Furthermore, $K:=\left\{ x\in X\,|\,\left|f\left(x\right)\right|\geq\varepsilon\right\} $ is a compact subset of $U$, so that the Urysohn-Lemma (applied to the compact Hausdorff space $\overline{U}$, with $U\subset\overline{U}$ open and $K\subset U$ compact) yields $\varphi\in C(\overline{U})$ with $\varphi|_{K}\equiv1$, $0\leq\varphi\leq1$ and $\operatorname{supp}\varphi\subset U$, where the closure for the support is taken in $\overline{U}$.
Now, define $$ g:X\to\left[0,1\right],x\mapsto\begin{cases} 0, & x\in X\setminus U,\\ \varphi\left(x\right), & x\in\overline{U}. \end{cases} $$ Note that $X\setminus U$ and $\overline{U}$ are both closed, so that $g$ is continuous by the gluing lemma (note that $g$ is well-defined, since for $x\in\left(X\setminus U\right)\cap\overline{U}$, we have $\varphi\left(x\right)=0$ since $\operatorname{supp}\varphi\subset U$.
Thus, $g$ has compact support (in fact $\operatorname{supp}g=\operatorname{supp}\varphi$), so that the same is true of $\tilde{f}:=g\cdot f\in C_{c}\left(X\right)$. It is not hard to see $\left\Vert \tilde{f}-f\right\Vert _{\sup}\leq\varepsilon$.