# 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.

• If $f \in C_0(X)$ find a compact set $K$ such that $|f(x)| \lt \varepsilon$ whenever $x \notin K$. Consider the restriction $f|_K and remember Tietze (again :)). – t.b. Jul 11, 2012 at 18:48 • Yes, that's right. Note that the completion is unique up to unique isometry, so you only have to find a complete space in which$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)$is complete... – t.b. Jul 11, 2012 at 18:54 • Of course you can use Tietze. Take an open set$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$is continuous. Now extend. – t.b. Jul 11, 2012 at 19:24 • My suggestion works on any locally compact Hausdorff space. But on$\mathbb{R}$, you can use those$g_n$'s, yes. – t.b. Jul 11, 2012 at 19:27 • Is this link good enough? ProofWiki is usually not very good but this entry looks reasonable (I haven't read it). – t.b. Jul 11, 2012 at 19:57 ## 1 Answer In a locally compact space, the$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)$in a locally compact Hausdorff space. • Yes. +1 I still would like to see the construction with the quotient space and an isomorphism. : ) Jul 11, 2012 at 19:06 • Mainly because I have not proved that completions are unique up to unique isometry so I don't understand properly why it's enough to show that it's dense in a complete space. But I have a good idea of what the quotient construction looks like (like the construction of the reals from$\mathbb Q\$). Jul 11, 2012 at 19:09
• Look at pp. 17 ff in this: google.com/… Jul 11, 2012 at 19:18
• 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. Jul 12, 2012 at 5:37