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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.

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    $\begingroup$ Certainly it's a Banach space. This is a very standard thing. Hint: Say $f_n$ is a Cauchy sequence. Say $||f_n-f_m||<\epsilon$ for all $n,m\ge N$. Choose $K$ compact so $|f_N|<\epsilon$ on $\Bbb R\setminus K$. It follows that $|f_m|<\epsilon+\epsilon$ on $\Bbb R\setminus K$, for every $m\ge N$. $\endgroup$ – David C. Ullrich Feb 3 '16 at 15:33
  • $\begingroup$ $C(\mathbb R)$ is not a Banach space: The supremum "norm" is not a normed because it is not real valued. You can make $C(\mathbb R)$ a complete metric space under uniform convergence by $D(f,g)=\sup\lbrace \min\lbrace |f(x)-g(x)|,1\rbrace: x\in\mathbb R\rbrace$. $\endgroup$ – Jochen Feb 4 '16 at 8:45
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$C_0(\mathbb R)$ 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 $1$.

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