Given a locally compact Hausdorff space $\Omega$ and a Banach space $E$.
Denote functions with compact support by: $$\mathcal{C}_0(\Omega,E):=\{F\in\mathcal{C}(\Omega,E):\operatorname{supp}F\subseteq K\}$$ and those vanishing at infinity by: $$\mathcal{C}_\infty(\Omega,E):=\{F\in\mathcal{C}(\Omega,E):F(\omega)\stackrel{\omega\to\infty}{\to}0\}$$
Clearly, the former is a dense subspace $$F\in\mathcal{C}_\infty(\Omega,E):\quad\|F-Fh\|<\varepsilon\quad(\chi_{K(\varepsilon)}\leq h\leq\chi_{W(\varepsilon)})$$ Moreover, is the latter a Banach space?
Yes, it is.
It is obvious that $C_\infty (\Omega , E)$ is a normed linear space with norm $$||F||_\infty := \sup_{x\in \Omega} ||F(x)||_E . $$
To show that it is complete, let $F_n$ be a sequence in $C_\infty(\Omega , E)$ so that $\{F_n\}$ is Cauchy with respect to the norm.
The completeness of $E$ imply that $F_n(x)$ converges to some $F(x) \in E$, for all $x$. $F$ is continuous as $F$ is uniform limit of continuous function.
To show that $F$ vanish at infinity, let $\epsilon >0$. Then there is $n$ so that $||F - F_n|| <\epsilon/2$. As $F_n \in C_\infty (\Omega, E)$, there is a compact set $K \subset \Omega$ so that $||F_n(x)||_E < \epsilon /2$ for all $x \notin K$. So
$$||F(x)||_E \leq ||F(x) - F_n(x)||_E + ||F_n(x)||_E < \epsilon$$
whenever $x\notin K$.