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Suppose $X$ is an infinite-dimensional Banach algebra (hence not locally compact).

Does there exist any sort of Riesz representation theorem that says something about elements of $C(X)^*$?

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How do you make $C(X)$ a Banach space? –  András Bátkai Feb 3 '13 at 21:24
    
I didn't say $C(X)$ is a Banach space. –  Banach Feb 3 '13 at 21:27
    
Is $C(X)$ suppose to be a Banach space? –  Banach Feb 3 '13 at 22:10
    
Which norm do you take to get a Banach space? Or what do you mean by $C(X)$? If it is the continuous functions on $X$, then without compactness they are not bounded. –  András Bátkai Feb 3 '13 at 22:29
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No, it does not have to be a Banach space. Which topology do you take then? –  András Bátkai Feb 3 '13 at 22:30

1 Answer 1

I would guess "the" natural topology on this space should be that of uniform convergence on compact subsets. Since a function on $X$ is continuous iff it its restrictions on compact subsets are continuous, $C(X)$ is actually the projective limit of $C(K), K \subset X$, $K$ compact. Which means that its dual should be the inductive limit of spaces of measures on $K$, that is, the space of measures with compact support.

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What does it mean that $C(X)$ is the projective limit of $C(K)$? Does this require X to be a countable union of compact sets? –  Banach Feb 4 '13 at 9:53
    
@Banach: Of course not. A projective limit may be defined for any directed family, which may be neither countable nor linearly ordered. Intuitively, this means that a continuous function "is" a collection of functions on compact subsets that fit together nicely, and the topology on $C(X)$ is generated by taking these restrictions. For precise definitions, see, e.g., en.wikipedia.org/wiki/Inverse_limit and en.wikipedia.org/wiki/Initial_topology. –  Alexander Shamov Feb 4 '13 at 19:33
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According to my topology book, there is a restriction. The statement, "a function on X is continuous iff its restrictions on compact subsets are continuous" requires X to be compactly generated. However, in my case X is compactly generated. –  Banach Feb 4 '13 at 20:43
    
I looked up projective limit, but I don't see why $C(X)$ is the projective limit of $C(K)$. Would you please indicate what are the coordinate spaces and the bonding maps in the inverse limit? –  Banach Feb 4 '13 at 22:12
    
The set of compact subsets is naturally ordered by inclusion. To each compact $K$ we have to associate a space - namely, $C(K)$ - and for each pair $K_1 \subset K_2$ we need a map $C(K_2) \to C(K_1)$ - in this case it is the restriction map. –  Alexander Shamov Feb 4 '13 at 23:12

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