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Let $X$ be a locally compact Hausdorff space, and $(C_0(X),d)$ be the metric space of (complex) continuous functions on $X$ which vanish at infinity (definition below). The metric $d$ is such that for $f,g\in C_0(X)$, $$d(f,g) = \sup_{x\in X}|f(x) -g(x)|$$ Let $\{f_n\}$ be a Cauchy sequence in $C_0(X)$. Show that $\{f_n\}$ converges uniformly.

(Rudin): A complex function $f$ on a locally compact Hausdorff space is said to vanish at infinity if for every $\epsilon > 0$ there exists a compact set $K\subset X$, such that $|f(x)| < \epsilon$ for every $x\notin K$.


My work: Let $\{f_n\}$ be Cauchy in $C_0(X)$. So, $$\forall \epsilon > 0\ \exists N\in \mathbb N\ \forall m,n\ge N\ d(f_m,f_n) < \epsilon$$ We already know that uniform convergence and convergence are equivalent notions in the sup metric as defined above. To show uniform convergence, it suffices to show convergence. To that end, let us define a function $f$ as follows. For every $x\in X$, $\lim_{n\to\infty} f_n(x)$ exists, since $\mathbb C$ is a complete metric space. Define $f(x) = \lim_{n\to\infty} f_n(x)$. It suffices to show that $f \in C_0(X)$ and $f_n \xrightarrow{n\to\infty} f$ in the sup metric.

  1. $f\in C_0(X)$: Let $\epsilon > 0$. Since all $f_n\in C_0(X)$, there exists a compact set $K_n$ such that $|f_n(x)| < \epsilon$ for every $x\in K_n^c$. Now how do I find a compact set $K$ such that $|f(x)| < \epsilon$ for every $x\in K^c$? We must also show that $f$ is infact continuous.
  2. To show $f_n \xrightarrow{n\to\infty} f$ in the sup metric, I need $$\forall \epsilon > 0\ \exists N\in\mathbb N\ \forall n\ge N\ d(f_n,f) < \epsilon$$ Pick $x\in X$ and observe that there exists $N_x \in\mathbb N$ such that for every $n\ge N_x$, we have $|f_n(x) - f(x)| < \epsilon/2$. Define $N:= \sup_{x\in X} N_x$, and note that for every $n\ge N$ and $x\in X$, $|f_n(x) - f(x)| < \epsilon/2$. Taking supremum over all $x$, we have $d(f_n,f) = \sup_{x\in X} |f_n(x) - f(x)| \le \epsilon/2 < \epsilon$. Does this work? I'm a little unsure because $N = \sup_{x\in X} N_x$ may also be $\infty$.

I would appreciate your help in filling gaps in the above proof. Thanks a lot!

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  • $\begingroup$ $C_0(X) \subset C_b(X)$, so if $(f_n)$ is Cauchy in $C_0$, it is Cauchy in $C_b$, hence convergent in the sup norm. $\endgroup$ May 28 at 16:40
  • $\begingroup$ What is $C_b(X)$? Also, it may be helpful to proceed from scratch, the way I have done. Could you help me complete the attempt? @JoseAvilez $\endgroup$ May 28 at 17:18
  • $\begingroup$ $C_b(X)$ is the space of bounded continuous functions on $X$; it's a Banach space. Have you encountered it before? $\endgroup$ May 28 at 17:21
  • $\begingroup$ Haven't encountered it yet. I'm just starting off with Functional Analysis. That might be evident from my solution! $\endgroup$ May 28 at 17:24
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Ad 1:

Fix $\epsilon>0$. By the Cauchyness we find $N \in \mathbb{N}$ such that $\forall m,n\ge N\ d(f_m,f_n) < \epsilon/2$. Note that this implies $|f_n(x) - f_N(x)| < \epsilon/2$ for all $n>N$ and all $x \in X$. Hence, by passing to the limit, we also have $|f(x) - f_N(x)| \leq \epsilon/2$ for all $x \in X$. Now, using that $f_N$ vanishes at $\infty$, we can find $K \subseteq X$ such that $|f_N(x)| < \epsilon/2$ on $X \setminus K$. Combining these two inequalities, we get $$|f(x)| \leq |f(x) - f_N(x)| + |f_N(x)| < \epsilon, \hspace{2cm} x \notin K.$$ So we have proved that $f$ is indeed in $C_0(X)$. It is a standard fact that a uniform limit of continuous functions is continuous, so the continuity of $f$ will follow from 2.

Ad 2:

Your $N$ can indeed be $\infty$, we need to somehow use the fact that the sequence is Cauchy in the uniform metric, not just pointwise:

Fix $\epsilon > 0$. We can find $N \in \mathbb{N}$ such that for all $n \geq N$ and all $x \in X$ we have $|f(x) - f_n(x)| \leq \epsilon/2$ by the same argument as above (using Cauchyness to get $|f_m(x) - f_n(x)| < \epsilon/2$ for all $m,n \geq N$ and all $x \in X$, and then passing to the limit $m \rightarrow \infty$). But this works for all $x \in X$, therefore $$d(f,f_n) = \sup_{x \in X} |f(x) - f_n(x)| \leq \epsilon/2 < \epsilon$$ for $n > N$, which is exactly what we wanted.

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  • $\begingroup$ Thank you, that's amazing! Do similar statements also hold for other spaces of continuous functions, such as $C_c(X)$ and $C_b(X)$? $\endgroup$ May 29 at 5:50
  • $\begingroup$ $C_c$ is not complete in the sup norm. $C_b$ is complete. $\endgroup$
    – user932138
    May 29 at 6:06
  • $\begingroup$ @epsilon-emperor As Peters said, $C_c$ is not complete. It actually holds that the closure of $C_c(X)$ is $C_0(X)$. The fact that $C_b(X)$ is complete can be shown using proof very similar to the answer above :) . $\endgroup$ May 29 at 7:11

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