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Hi,

is the space $C_0(\mathbb{R}^m)$, $m \in \mathbb{N}$ of continuous functions with compact support separable? If yes: where can I find a proof for that?

Please note: this is not a duplicate of this question (click), since I asked for compact support.

The reason I'm asking is: here (click) the author (L.C. Evans) seems to use the fact on page 18 (page 20 of the pdf) first line without mentioning the difficulty, that $\mathbb{R}$ isn't compact. The entire proof is senseless without this fact.

Your help is very much appreciated! Thanks!

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    $\begingroup$ Isn't $C_0(\mathbb{R}^m)$ usually continuous functions that vanish at infinity? $\endgroup$ Jun 14, 2011 at 14:15
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    $\begingroup$ This is not a research-level question. $\endgroup$ Jun 14, 2011 at 14:29
  • $\begingroup$ Oh... he doesn't clarify his notation. But that could very well be. But doesn't that make it even worse? $\endgroup$
    – fjodor_d
    Jun 14, 2011 at 14:30
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    $\begingroup$ The notation for $C_0(U)$ is indeed explained at page 7, line 18 (functions with compact support, as you said). It is not stated if a topology is chosen. In any case, in the proof at page 18 just uniform convergence is used. So the separability you need comes e.g. from the separability of the Banach space $C(S)$ where $S$ is the one-point compactification of $\mathbb{R}^m$ (you can see $C_0(\mathbb{R}^m)$ as a subset of $C(S)$). $\endgroup$ Jun 14, 2011 at 15:12

1 Answer 1

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Each $C([-N,N]^m)$ is separable, and so is the subspace $X_N\subseteq C([-N,N]^m)$ consisting of functions which vanish on the boundary $\partial[-N,N]^m$. Take a countable dense subset of each $X_N$, and take the union over all integers $N$. This is then a countable dense subset of $C_0(\mathbb R^m)$.

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