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The second dual or double dual of the space of all continuous functions on $[0,1]$, $C[0,1]$ is von Neumann algebra. Can anyone help me identifying this space?

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Sion, what have you tried so far and what are your thoughts on the problem? –  Rasmus Jun 25 '11 at 6:48
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I think Rasmus asks a good question: it's possible to give rather abstract answers: e.g. it's $C(K)$ where $K$ is the hyperstonian spectrum of $C[0,1]^{**}$. I suspect this is not the sort of answer you want; so some hints as to what you mean by "identify" would really help... –  Matthew Daws Dec 6 '11 at 22:15
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SION, please check Diestel's "Sequences and series in Banach spaces" . –  niyazi Dec 7 '11 at 5:27
    
See math.stackexchange.com/a/74877/442 –  GEdgar Feb 5 at 22:15
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1 Answer

The Banach space dual of $C[0, 1]$ is $M[0, 1]$, the space of complex Borel measures on $[0, 1]$. The dual of $M[0, 1]$ is the “enveloping von Neumann algebra” of $C[0, 1]$ and is fairly intractable. However, assuming continuu hypothesis, R. D.Mauldin [1] proved the following representation theorem for bounded linear functionals on $M[0, 1]$: for every such functional $T$ there is a bounded function from the set of Borel subsets of $[0, 1]$ to $\mathbb{C}$ such that $$ T(\mu)=\int\psi d\mu $$ for all $\mu\in M[0,1]$. Here the integral notation signifies the limit over the directed set of Borel partitions $\{B_1,\ldots,B_n\}$ of $[0,1]$ of the quantity $\sum\psi(B_i)\mu(B_i)$, and part of the assertion is that $\psi$ can be chosen so that this limit always exists. The proof begins by choosing a maximal family of mutually singular Borel measures on $[0,1]$; using continuum hypothesis, this family is then indexed by countable ordinals and $\psi$ is then defined by transfinite recursion.


[1] R. D. Mauldin, A representation theorem for the second dual of $C[0, 1]$, Studia Mathematica, vol. 46 (1973), pp. 197–200.

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