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Let $X$ be some topological Hausdorff space and $C_b(X)$ the space of bounded complex continuous functions on $X$. Is there a Banach space $B$ such that $B^* \simeq C_b (X)$?

I know of a very similar result, namely that $L^\infty (\mu) \simeq L^1 (\mu) ^*$ whenever $\mu$ is $\sigma$-finite, but I work in a context where continuity is important, so this purely measure-theoretic result is good but not enough.

(Please note that I am not asking about $C_b (X) ^*$, that is a well-known result.)

Alex M.
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  • This question (particularly the comments) says that for compact $X$, that's hardly ever the case. Since for completely regular $X$ we have $C_b(X) = C(\beta X)$, the case of $X$ not completely regular remains. I doubt there's a nice theory about that. – Daniel Fischer Jul 25 '15 at 09:12
  • @DanielFischer: Yes, bounded, thank you. Concerning the question that you reference, how about this: "the space of continuous functions, say on a compact interval, is the dual of the space of measures thereon with the so-called bounded weak star topology"? That's what I'm trying to get a better view of. – Alex M. Jul 25 '15 at 09:23
  • Do you mean isometric isomorphism or arbitrary isomorphism of Banach spaces? – Norbert Jul 25 '15 at 09:32
  • I can't really parse that answer, but "bounded weak star topology" indicates a non-normable space. If you drop the normability requirement on the putative predual, then every Banach space $X$ is the dual of $(X',\sigma(X',X))$. – Daniel Fischer Jul 25 '15 at 09:33
  • @AlexM. As for your question in comments it is completely different. In fact the space of continuous functionals on $X^$ endowed with weak topology is exactly $X$. It is remains to note that for $X=C(K)$ we have $X^*=M(K)$. – Norbert Jul 25 '15 at 09:34
  • @DanielFischer: Oh, it may be my mistake, then. Since $C_b (X)$ is viewed as a Banach space, I assumed that it is natural to ask about a Banach predual. I was mistaken, it seems. Does the result that you cite have a well-known name? I'd like to take a look at its details, because in functional analysis these little details can ruin everything. – Alex M. Jul 25 '15 at 09:47
  • @Norbert: Is $M(K)$ the space called $rba(K)$ by Dunford and Schwartz, the space of regular bounded addditive set-functions? – Alex M. Jul 25 '15 at 09:52
  • @AlexM.exactly. – Norbert Jul 25 '15 at 10:19

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