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Due to a previous question I wonder if one knows the dual space of $C_b(X)$. Here $C_b(X)$ is the space of all continuous bouded functions with values in $\mathbb{R}$. Of course this depends on the space $X$ itself. One can assume that this space is nice, for example a complete separable metric space. It would be nice if we can exclude compactness of $X$. Is then anything known about the dual? In particular, we have for $P(X)$, the space of Borel probability measure on $X$, that $P(X)\subset C_b(X)'$, where the latter denotes the dual. The pairing is given by

$$\phi_\mu(f):=\langle \mu,f\rangle=\int fd\mu$$

which defines for every $f\in C_b(X), \mu\in P(X)$ bounded linear functional. But do we have "equality"? Are there example for structures of $X$ such that $C_b(X)$ is reflexive?

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    $\begingroup$ Have you heard, $C_b(X)\cong C(\beta X)$? en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification. That might help. $\endgroup$ – Jonas Meyer Oct 5 '13 at 8:58
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    $\begingroup$ @JonasMeyer That could be an answer, with a very little fleshing out to pass the character limit. $\endgroup$ – Daniel Fischer Oct 5 '13 at 9:03
  • $\begingroup$ I would try to get this book: books.google.lt/books/about/…, $\endgroup$ – mpiktas Oct 5 '13 at 9:10
  • $\begingroup$ @JonasMeyer Using $C(\beta X)$ you would then use Riesz? The problem about applying Riezs is we get for a positive functional $\Lambda\in C(\beta X)'$ a finite positive Borel measure. In general I get signed measures which does not fit with $P(X)$. Moreover, $\beta X$ seems rather complicated to work with in application. $\endgroup$ – math Oct 5 '13 at 9:41
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See theorem 6 in section IV of Linear Operators: General theory volume 1 by Nelson Dunford and Jacob T. Schwartz.

The theorem says that for a normal topological space $X$ the continuous dual of $C_b(X)$ is isometrically isomorphic to the space $rba(X)$ of finititely additive regular bounded Borel measures on $X$.

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  • $\begingroup$ Thanks for your answer. Do you know if one can strengthen the assumptions, such that we actually get $P(X)$ instead of $rba(X)$? $\endgroup$ – math Oct 5 '13 at 11:42
  • $\begingroup$ You can't get $P(X)$ as a dual of some space of functions, because $P(X)$ not even a linear space $\endgroup$ – Norbert Oct 5 '13 at 11:52
  • $\begingroup$ Ah, I see. Thanks a lot. $\endgroup$ – math Oct 5 '13 at 11:55

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