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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.

Leandro
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  • In the infinite Cartesian product example you mention, I don't think you have a separable metric space. Or are you taking some kind of restricted direct product? –  Feb 10 '12 at 03:00
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    What do you mean by "Borel continuous"? Are they continuous, or only Borel? If you mean continuous, then e.g. if $\Omega=[0,1]$ or another connected space, then the only extreme points of the unit ball of $C(\Omega,\mathbb R)$ will be the constant functions $f=1$ and $f=-1$, and by the Krein-Milman theorem this implies that $C(\Omega,\mathbb R)$ is not a dual space. If $\Omega$ is the one point compactification of the integers then you get a space isomorphic to the space of convergent sequences, which I believe is not a dual space but don't know how to show it. – Jonas Meyer Feb 10 '12 at 03:02
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    In fact: in the complex-valued case, C(X) for X compact Hausdorff is a dual Banach space precisely when it is an abelian von Neumann algebra, which happens if and only if X is extremely disconnected. I have a feeling that the only Von Neumann algebras that are separable in norm topology are the finite dimensional ones. Now all this should carry over to the real-valued case, implying that $\Omega$ is finite... –  Feb 10 '12 at 03:03
  • @Yemon: Your feeling about separability of von Neumann algebras is correct. There's maybe an easier way to show it than this, but every infinite dimensional C*-algebra contains an element with infinite spectrum, and with Borel functional calculus of a self-adjoint element with infinite spectrum it isn't hard to show that the von Neumann algebra generated by such an element is nonseparable. – Jonas Meyer Feb 10 '12 at 03:09
  • @YemonChoi I think that this cartesian space is always separable in this case, because $E^{\mathbb{Z}^d}$ is compact in the product topology, which is in this case generated by the metric $d_1(\omega,\eta)=\sum_{i\in\mathbb{Z}^d}\frac{1}{2^{|i|}} d(\omega_i,\eta_i)$, where $\omega=(\omega_i)_{i\in\mathbb{Z}^d}$ and $d$ is the metric on $E$. – Leandro Feb 10 '12 at 10:27
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    @Yemon: your "if and only if" is not exactly right. There are extremely disconnected compact spaces $X$ such that $C(X)$ is not a von Neumann algebra. The right iff is given by hyperstonean spaces; these are extremely disconnected but they have the additional property that $C(X)$ has normal states. – Martin Argerami Feb 22 '12 at 07:41
  • @Martin You are of course correct - the embarrassing thing is that I even pointed this out on a MO answer (the Dixmier example which is AW but not von Neumann, right?) –  Feb 22 '12 at 10:41
  • @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 $[0,1]$. – Martin Argerami Feb 22 '12 at 14:51

2 Answers2

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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.

Philip Brooker
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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.

azarel
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  • hi azarel, you mean that $c_0$ has no a predual space ? could you elaborate more or give me a reference ? – Leandro Feb 10 '12 at 03:00
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    $c_0$ is a pre-dual space (it is the predual of its dual), however it is not a dual space (which is what you mean to say). – Martin Wanvik Feb 10 '12 at 03:01
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    In your example, $C(\Omega,\mathbb R)\cong c$, the space of real convergent sequences, not $c_0$. It is easier to see that $c_0$ is not a dual space, because its unit ball has no extreme points. I don't know how to show that $c$ is not a dual space. – Jonas Meyer Feb 10 '12 at 03:04
  • Note that $c_0=c_0\oplus \mathbb R=c$ – azarel Feb 10 '12 at 03:45
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    @azarel: No, $c_0\neq c$. – Jonas Meyer Feb 10 '12 at 04:16
  • @JonasMeyer. Are you sure? The last paragraph on page 87 on Albiac and Kalton's book says otherwise. – azarel Feb 10 '12 at 06:41
  • @JonasMeyer Fair enough – azarel Feb 10 '12 at 06:52
  • Jonas, if you now move to the isomorphic rather than the isometric case, what do you mean by "is a dual space"? It's not clear to me why knowing that c_0 has no isometric predual immediately implies that all spaces isomorphic to it also fail to have isometric preduals... –  Feb 11 '12 at 09:19
  • @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.) – Jonas Meyer Feb 12 '12 at 19:19