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Let $\Omega$ be a compact metric space and $C(\Omega,\mathbb{R})$ the space of borel continuous real valued functions. I would like to know if there is any real Banach space $V$ such that its dual space (topological)is exactly $C(\Omega,\mathbb{R})$.

My main interest is when $\Omega$ is an infinite cartesian product as for example, $\Omega=E^{\mathbb{Z}^d}$, where $E$ is a compact metric space. If the answer for this case is also negative are we in a better situation if $E$ is a finite set ?

Thanks for any comment or reference.

Edition. Remove the superfluous hypothesis pointed out by Philip.

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    @Yemon: yes, that's the canonical example. You can obtain the corresponding $X$ from $[0,1]$ by taking the projective limit over reverse inclusion of the Stone-Cech compactifications of the open dense subsets of $[0,1]$.2012-02-22

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Firstly, let me just point out that every compact metric space is automatically separable.

Secondly, note that $c_0$ does not embed into any separable dual space, hence neither does any Banach space containing a subspace isomorphic to $c_0$. Since every $C(K)$ space ($K$ compact Hausdorff) contains a subspace isomorphic to $c_0$, no $C(K)$ space embeds isomorphically into a separable dual space. In particular, since metrizability of $K$ is equivalent to $C(K)$ being norm separable, the answer to your question is always no.

For a reference for all of the above claims, look up $C(K)$ and $c_0$ in the index of Albiac and Kalton's book Topics in Banach Space Theory.

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    Hi Philip thanks a lot for your help. I will have a look at the reference you provided.2012-02-10
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The result is false in general. For example, consider $\Omega=\{\frac 1 n: n\in \mathbb N\}\cup \{0\}$, then $C(\Omega,\mathbb R)=c_0$ which is not a dual space. Edit: I'm adding some references: My claim follows from Proposition 4.4.1 of Albion and Kalton's book.

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    @Yemon: Thank you for pointing that out, that comment didn't make sense. In fact, I will delete it. (You can alert people of comments directed to them using `@username`; I only happened to look back at this just now.)2012-02-12