# I want to prove that $C_0(X)$ is Banach

Let $X$ be locally compact Hausdorff space. I'm trying to prove that $$C_0(X)=\{ f:X\to \mathbb{C} \; | \; f \text{ is continuous and }\forall \epsilon>0 \; \exists K(\text{compact}) \subset X \text{ s.t. } |f|<\epsilon \text{ on } X\setminus K\}$$ is complete space with $\sup$ norm. I tried as follows:

Let $f_n$ be a Cauchy sequence in $C_0(X)$. Then for all $x\in X$, $$|f_n(x) - f_m(x)| \le \|f_n - f_m\|_\infty \to 0$$ as $n,m\to \infty$. So $\{f_n\}$ is Cauchy in $\mathbb C$ and therefore the limit $f(x) := \lim_{n\to \infty} f_n(x)$ exsits. Now I'm trying to show that this $f(x)$ satisfies $f\in C_0(X)$ and $\|f - f_n\|_\infty \to 0$ as $n\to \infty$.

(1) Continuity:

Since $\forall \epsilon>0$, $\exists N$: $\forall n,m\ge N$, $\forall x \in X,$ $|f_n(x) - f_m(x)| \le \|f_n - f_m\|_\infty < \epsilon$, taking $m\to \infty$, we have $|f_n(x) - f(x)| \le \epsilon$ so that $\{f_n\}$ converges uniformly to $f$ on $\mathbb C$, therefore $f$ is continuous.

(2) $\forall \epsilon>0$ $\exists K(\text{compact}) \subset X \text{ s.t. } |f|<\epsilon \text{ on } X\setminus K$

Since $f_N \in C_0(X)$, $\exists K$(compact) $\subset X$ s.t. $|f_N|< \epsilon$ on $X\setminus K$. So $$|f| \le |f-f_N| + |f_N| \le \epsilon + \epsilon = 2\epsilon.$$

(3) From (1), we have $|f_n(x) - f(x)| \le \epsilon$ for $n \ge N$. So $\| f_n - f\|_\infty \le \epsilon$.

I'm wondering that my proof is correct or not. Would you please confirm my solution? In fact, I'm wondering that $2\epsilon$ argument works or not in (2). Also I think I did not use the property that $X$ is locally compact and Hausdorff in my solution.

• don't you have to use some properties of $X$ for saying that 'the $f_n$ are continuous and $f_n \to f$ uniformly' $\implies$ $f$ is continuous ? Jul 21 '16 at 19:46
• @user1952009 Oh I think that it is enough that $X$ is compact. Does that argument need Hausdorff space? Jul 21 '16 at 19:47
• If anyone wants to see a short proof that $f$ is continuous regardless of the properties of $X ,$ let me know. Jul 22 '16 at 0:25

You have the right ideas. To "fix" the $2\epsilon$ part you could just choose $N$ large enough so that $\|f-f_N\| < \epsilon/2$, and then derive a compact subset such that $\|f_N \| < \epsilon/2$ outside.
The "uniform limit of continuous functions functions is continuous" statement is true regardless of the topological properties of $X$. However when working with the space $C_0(X)$, the hypotheses that $X$ be locally compact and Hausdorff are usually included because then one can extend any function $f \in C_0(X)$ to the one-point compactification of $X$ by setting $f(\infty) := 0$, and this extension is continuous.