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Let

  • $X$ be a Hausdorff topological space,
  • $\mathcal C(X)$ the space of real-valued continuous functions,
  • $\mathcal C_b(X)$ the space of real-valued bounded continuous functions,
  • $\mathcal C_0(X)$ the space of real-valued continuous functions that vanish at infinity, and
  • $\mathcal C_c(X)$ the space of real-valued continuous functions with compact supports.

Then $\mathcal C_b(X)$ and $\mathcal C_0(X)$ are real Banach space with supremum norm $\|\cdot\|_\infty$. In Folland's Real Analysis: Modern Techniques and Their Applications, there is a theorem

4.35 Proposition. If $X$ is locally compact Hausdorff, then $\mathcal C_0(X)$ is the closure of $\mathcal C_c(X)$ in $\mathcal C_b(X)$.

Author's proof relies on a locally compact version of Urysohn's Lemma. However, in this proof, the local compactness is not needed, and I don't see anything unusual in the proof.

Could you confirm that above theorem indeed holds for arbitrary Hausdorff topological space $X$?

Update: I added related definitions from Folland's textbook.

If $X$ is a topological space and $f \in \mathcal C(X)$, the support of $f$, denoted by $\operatorname{supp}(f)$, is the smallest closed set outside of which $f$ vanishes, that is, the closure of $\{x: f(x) \neq 0\}$. If $\operatorname{supp}(f)$ is compact, we say that $f$ is compactly supported, and we define $$ \mathcal C_c(X)=\{f \in \mathcal C(X): \operatorname{supp}(f) \text { is compact}\}. $$

Moreover, if $f \in \mathcal C(X)$, we say that $f$ vanishes at infinity if for every $\epsilon>0$ the set $\{x:|f(x)| \geq \epsilon\}$ is compact, and we define $$ \mathcal C_0(X)=\{f \in \mathcal C(X): f \text { vanishes at infinity}\}. $$

Analyst
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  • What does "vanish at infinity" mean? It's a pretty "locally-compact" notion to consider – Giuseppe Negro Nov 01 '22 at 18:23
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    Vanishing at infinity means $\lim_U \lVert f\vert_U\rVert_\infty\to 0$, where $U$ ranges over all complements of compact sets, ordered by inclusion. It is not more "locally compact" than the notion of compactly supported, so that the statement also holds in spaces that are not locally compact, as the link shows. However, both $C_c$ and $C_0$ may be trivial in this case. – MaoWao Nov 01 '22 at 18:28
  • Consider a separable Banach space like $C[0,1]$, any compact in it has empty interior. If ${x : |f(x)| \geq \epsilon}$ is compact then the open set ${x : |f(x)| > \epsilon} \subset {x : |f(x)| \geq \epsilon}$ is necessarily the empty set. This holds for any $\epsilon > 0$ thus $\mathcal C_0(X)$ only contains the function that is identically zero everywhere. – Shiva Nov 01 '22 at 18:37
  • As the goal of the author is integration, locally compact Hausdorff (lcH) spaces is the main focus (the theory of integration developed by Bourbaki works well in lcH). Integration in more general topological spaces is riddle with complications. – Mittens Nov 01 '22 at 18:50
  • @OliverDíaz I may be wrong, but the Bochner integral works for arbitrary $\sigma$-finite measure spaces without any topological structure... – Analyst Nov 01 '22 at 18:58
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    @Analyst: I am referring to things such as Riesz representation theorem and Radon measures. Radon measures have stronger properties than abstract measures on $\sigma$-finite spaces (Radon measures are monotone continuous over increasing nets, not only sequences). – Mittens Nov 01 '22 at 19:01
  • @OliverDíaz Thank you so much! I got it. Riesz theorem is indeed very powerful. It helps us establish disintegration theorem. I'm excited that such an essential theorem in probability theory is due to a result from functional analysis. It seems working with pure measure space (without topological structure) is extremely daunting. – Analyst Nov 01 '22 at 19:05

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