Is $C_0(\mathbb{R})$ a Banach space? Let $C(\mathbb{R})$ be a Banach space of continuous real-valued functions defined on $\mathbb{R}$, with supremum norm, and let  $C_0(\mathbb R)$ be the subspace of functions vanishing at infinity. Is $C_0(\mathbb{R})$ a Banach space?
I try to see it using:
$f\in C_0(\mathbb{R})$ iff for any $\epsilon>0$ there exists $K>0$ such that $|f|<\epsilon$ whenever $|x|>K$. But I think it is not Banach.
Please I need a counter-example or a proof.

$C_0(\mathbb R)$ is a Banach space because it may be identified with a closed subspace of some $C(K)$, the real vector space of continuous functions on the compact Hausdorff space $K$, equipped with the $\|\cdot\|_\infty$ norm. Since uniform convergence on compact sets is so well-behaved, $C(K)$ is a Banach space (belonging to the "classical" ones).
The continuous embedding
$$\mathbb R\hookrightarrow S^1,\: x\mapsto\frac{x+i}{x-i}\in\mathbb C\;\text{ or }
\left(\frac{x^2-1}{x^2+1},\frac{2x}{x^2+1}\right)\in\mathbb R^2$$
of $\,\mathbb R\,$ into its 1-point compactification $S^1$ lets one identify $C_0(\mathbb R)$ with the subspace $\big\{f\in C(S^1)\mid f(1)=0\big\}$. It is  closed being the kernel of the continuous linear map
$\operatorname{eval}_{x=1}:C(S^1)\to\mathbb R, f\mapsto f(1),\,$ and has codimension $1$.
