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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – reuns
    Jul 21 '16 at 19:46
  • $\begingroup$ @user1952009 Oh I think that it is enough that $X$ is compact. Does that argument need Hausdorff space? $\endgroup$
    – John S
    Jul 21 '16 at 19:47
  • $\begingroup$ If anyone wants to see a short proof that $ f $ is continuous regardless of the properties of $ X , $ let me know. $\endgroup$ Jul 22 '16 at 0:25
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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.

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