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I may be missing something very trivial,but cannot figure it out. To prove the fact that the space $C_0(X)$ is a Banach space under the usual sup norm when $X$ is locally compact and Hausdorff, where do we use the properties of $X$?

Thanks for any help.

egreg
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Ester
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  • $\mathbb{R}$ is locally compact and Hausdorff, but $C_0 (\mathbb{R})$ is not a Banach space, so you cannot prove such a result. – Crostul Jul 12 '14 at 12:19
  • Crostul: $C_0(\mathbb{R})$ probably refers to continuous functions with the max-norm, which is a Banach space. – Peter Franek Jul 12 '14 at 12:20
  • @PeterFranek: We are looking at function from $\mathbb Q$ to, say, $\mathbb C$ not function to $\mathbb Q$. – Rasmus Jul 12 '14 at 12:26
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    @Crostul: why would it not be a Banach space? – Rasmus Jul 12 '14 at 12:30
  • @Rasmus: Sorry, maybe this is just a matter of definitions. For me $C_0(\mathbb{R})$ is the space of all continuous functions $\mathbb{R} \rightarrow \mathbb{R}$ not necessarily bounded. – Crostul Jul 12 '14 at 12:39
  • @Ester: So what should be the norm of $e^x$? Do you admit functions with norm $+ \infty$? – Crostul Jul 12 '14 at 12:43
  • O sorry!I mistook C_0(R) according to my definition.I am deleting the comment. – Ester Jul 12 '14 at 12:45
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    @Crostul $C_0(X)$ usually denotes the space of continuous functions vanishing at infinity, functions for which $f^{-1}({x : \lvert x\rvert \geqslant \varepsilon})$ is compact for all $\varepsilon > 0$. – Daniel Fischer Jul 12 '14 at 12:56

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The point is that functions vanishing at infinity is a concept that is really only sensible on locally compact spaces.

See http://en.wikipedia.org/wiki/Vanish_at_infinity

and http://en.wikipedia.org/wiki/Locally_compact#The_point_at_infinity

In particular, it turns out that the isomorphism class of the $C^*$-algebra $C_0(X)$ remembers the homeomorphism class of the locally compact space $X$. In fact, every commutative $C^*$-algebra is of the form $C_0(X)$ for some locally compact space -- its spectrum. If $X$ is not locally compact, I guess you can consider the spectrum of $C_0(X)$ as a locally compact-ification of $X$.

See http://en.wikipedia.org/wiki/C*-algebra#Commutative_C.2A-algebras.

(Above, locally compact always implies Hausdorff.)

Rasmus
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  • But the definition makes sense for any topological space X,isn't it? – Ester Jul 12 '14 at 12:36
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    @Ester In some sense, the definition makes sense for an arbitrary topological space $X$, and the space $C_0(X)$ of continuous functions vanishing at infinity is a Banach space under the supremum norm regardless of $X$. But there are a lot of spaces where that is a rather boring Banach space, namely ${0}$. If $X$ is locally compact (locally quasicompact and Hausdorff), then $C_0(X)$ is sufficiently interesting. – Daniel Fischer Jul 12 '14 at 12:53
  • Thanks, @DanielFischer, I've added some remarks in a vein similar to your comment. – Rasmus Jul 12 '14 at 13:30
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@Ester, this isn't the best answer, but it will be enough for you to get the idea: consider $X$ to be an infinite-dimensional topological vector space. It is known from a theorem of Weil that it cannot be locally-compact. Can we have functions with compact support defined on it, then? Let $f \neq 0$ be such a function and $x$ such that $f(x) \neq 0$. Then there exist a whole neighbourhood $U$ of $x$ such that $f(y) \neq 0 \forall y \in U$. Then $U \subset supp(f)$ and, since $supp(f)$ was supposed compact, $U$ will be relatively compact. Thus, $x$ has a relatively compact neighbourhood and, by translating it, all points will have some relatively compact neighbourhood, and thus $X$ is locally-compact, which is a contradiction. So, you cannot have functions of compact support on a non-locally-compact topological vector space.

On a non-linear space the proof is more involved, but the idea stays the same.

Of course, this makes sense if by $C_0(X)$ you understand the space of functions with compact support. If not, then you'd better specify its meaning in your original question.

Alex M.
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