# 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.

• 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 $m\ge N$. – David C. Ullrich Feb 3 '16 at 15:33
• $C(\mathbb R)$ 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 $D(f,g)=\sup\lbrace \min\lbrace |f(x)-g(x)|,1\rbrace: x\in\mathbb R\rbrace$. – Jochen Feb 4 '16 at 8:45

## 1 Answer

$$C_0(\mathbb R)$$ 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 $$1$$.