# Completion of $C_c(X)$ with respect to $\|\cdot\|_\infty$

Notation:

All functions here are from $X$ to $\mathbb R$.

$C_c(X)$ = compactly supported continuous functions.

$C_b(X)$ = bounded continuous functions.

$B(X)$ = bounded functions.

$C_0(X)$ = continuous functions that tend to zero (so $X$ has to be locally compact and Hausdorff)

Today I proved that $C_c(\mathbb R)$ is not complete with respect to $\|\cdot\|_\infty$. One can do this by taking $g_n$ to be the function that is zero on $(-\infty,-n]$, linear on $[-n, -n+1]$, $1$ on $[-n+1, n-1]$ and symmetric with respect to the $y$-axis. For $f(x) = e^{-x^2}$ one can show that $\|fg_n - f\|_\infty \to 0$ but $f \notin C_c(\mathbb R)$.

Then I read about completions and wanted to work out the completion of $C_c$(X). (I did this all for $X = \mathbb R$). In any case, my thoughts were as follows: $C_c(X)$ is contained in $B(X)$. But it is a proper subset because the uniform limit of continuous functions is continuous but there are discontinuous bounded functions. The next candidate then seems to be $C_b(X)$ but $f(x) = 1$ is in there and not the uniform limit of $f_n$ in $C_c$. (cannot be because if $f_n$ is zero outside a compact set then $\|f_n - 1\|_\infty = 1$ for all $n$ so this doesn't converge in norm).

The next candidate then is $C_0(X)$ and I'm quite sure that that's the completion of $C_c$ with respect to $\|\cdot\|_\infty$ in $B(X)$. But now I need to show this by showing that $C_0(X)$ is isomorphic to the space of Cauchy sequences in $C_c(X)$ quotient Cauchy sequences that tend to the zero function and I don't really know how to think about this. Can someone please show me how to prove this? Thank you. I want to see this quotient construction and an isomorphism but if there are other ways to show that $C_0$ is the completion of $C_c$ then go ahead and post it, I will upvote it.

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If $f \in C_0(X)$ find a compact set $K$ such that $|f(x)| \lt \varepsilon$ whenever $x \notin K$. Consider the restriction $f|_K and remember Tietze (again :)). – t.b. Jul 11 '12 at 18:48 Yes, that's right. Note that the completion is unique up to unique isometry, so you only have to find a complete space in which$C_c(X)$is dense with respect to$\|\cdot\|$and I suggested how to show that$C_c$is dense in$C_0$. And yes, we've convinced ourselves at various points over the last half-a-year that$B(X) = \ell^\infty(X)$is complete... – t.b. Jul 11 '12 at 18:54 Of course you can use Tietze. Take an open set$U$with compact closure containing$K$(you can do that by local compactness of$X$). Define a function$g$on$K \cup (X \smallsetminus U)$by$g = f|_K$on$K$and$g = 0$on$X \smallsetminus U$. Observe that$g$is continuous. Now extend. – t.b. Jul 11 '12 at 19:24 My suggestion works on any locally compact Hausdorff space. But on$\mathbb{R}$, you can use those$g_n$'s, yes. – t.b. Jul 11 '12 at 19:27 Is this link good enough? ProofWiki is usually not very good but this entry looks reasonable (I haven't read it). – t.b. Jul 11 '12 at 19:57 ## 1 Answer In a locally compact space, the$C_0$functions are complete in the$\|\cdot\|_\infty$norm; they constitute a Banach Space in this norm. The space$C_c(X)$is dense in$C_0$if$X$is locally compact and Hausdorff. This is sufficient to tell you that$C_0(X)$is the completion of$C_c(X)$in a locally compact Hausdorff space. - Yes. +1 I still would like to see the construction with the quotient space and an isomorphism. : ) – Rudy the Reindeer Jul 11 '12 at 19:06 Mainly because I have not proved that completions are unique up to unique isometry so I don't understand properly why it's enough to show that it's dense in a complete space. But I have a good idea of what the quotient construction looks like (like the construction of the reals from$\mathbb Q$). – Rudy the Reindeer Jul 11 '12 at 19:09 Look at pp. 17 ff in this: google.com/… – ncmathsadist Jul 11 '12 at 19:18 Thank you.${}{}\$ – Rudy the Reindeer Jul 11 '12 at 19:23
They seem to skip the proof of isometry of any two completions. But they do it in the proof linked to in the comments to the question. – Rudy the Reindeer Jul 12 '12 at 5:37