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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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    $\begingroup$ Sion, what have you tried so far and what are your thoughts on the problem? $\endgroup$
    – Rasmus
    Commented Jun 25, 2011 at 6:48
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    $\begingroup$ 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... $\endgroup$ Commented Dec 6, 2011 at 22:15
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    $\begingroup$ SION, please check Diestel's "Sequences and series in Banach spaces" . $\endgroup$
    – niyazi
    Commented Dec 7, 2011 at 5:27
  • $\begingroup$ See math.stackexchange.com/a/74877/442 $\endgroup$
    – GEdgar
    Commented Feb 5, 2014 at 22:15

3 Answers 3

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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 the continuum 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 the 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 Math., 46 (1973), 197–200.

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I recently saw it asserted that the second dual of $C[0,1]$ was $\mathcal L^\infty[0,1]$, the space of bounded Borel functions. I was surprised at this, tried to prove it, couldn't quite make the details work. Finally saw a simple proof it was not so. (See the Amusing Note below regarding what $\mathcal L^\infty[0,1]$ actually is.) Started to post something about that, this thread popped up as possibly related.

Here, here and here we see much more complete discussions of that second dual; it seems perhaps nonetheless worthwhile to post the counterexample here, just for the sake of giving a very elementary proof that it's not as simple as one might possibly think.

Motivation: A few approaches to proving the false result would have worked if only I could show that the function $x\mapsto\Lambda\delta_x$ was Borel. Finding an example where that was not so turned out to give a counterexample to the result itself:

Say $\delta_x$ is a point mass at $x$, and let $X$ be the span of $\delta_x$ for $x\in[0,1]$. Suppose $f:[0,1]\to\Bbb C$, $|f(t)|\le1$ for all $t$, and $f$ is not Borel. Define $\Lambda:X\to\Bbb C$ by $$\Lambda\left(\sum_{j=1}^nc_j\delta_{x_j}\right)=\sum_{j=1}^nc_jf(x_j)$$(or $\Lambda\mu=\int f\,d\mu$). Since $||\sum c_j\delta_{x_j}||=\sum|c_j|$ it follows that $||\Lambda||\le 1$; now Hahn-Banach extends $\Lambda$ to $\overline\Lambda\in\mathcal M[0,1]^*$. And $\overline\Lambda$ is not given by integration against $g\in\mathcal L^\infty$. Because if it were then $$g(t)=\int g\,d\delta_t=\overline\Lambda \delta_t=\Lambda\delta_t=f(t),$$but $f$ is not Borel.

In fact in retrospect it's clear that the norm-closed span of the $\delta_x$ is just $\ell^1[0,1]$, with dual $\ell^\infty[0,1]$. Already much larger than $\mathcal L^\infty[0,1]$, and of course an element of the second dual of $C[0,1]$ is not determined by its action on the $\delta_x$.

Amusing Note Of course $C[0,1]$ is weak* dense in the second dual. In fact $\mathcal L^\infty[0,1]$ is precisely the closure of $C[0,1]$ in the second dual under sequential weak* convergence. Hint: This is the same as showing that $\mathcal L^\infty[0,1]$ is the closure of $C[0,1]$ under "uniformly bounded sequential pointwise" convergence. (And note that of course you don't get the closure of a set under a sequential limit operation by adding all the sequential limits; in fact you have to repeat "add all the sequential limits" $\omega_1$ times.)

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  • $\begingroup$ Nice example. I guess one should observe that $\left\|\sum c_j \delta_{x_j}\right\| = \sum |c_j|$ to verify that $\Lambda$ is bounded on $X$. $\endgroup$ Commented Feb 10, 2016 at 3:10
  • $\begingroup$ Please see also math.stackexchange.com/a/1578784/17929 and math.stackexchange.com/questions/1532956/… $\endgroup$ Commented Feb 10, 2016 at 21:32
  • $\begingroup$ What do you mean by $\ell^\infty [0,1]$? The set of bounded $f:[0,1] \to \mathbb{C}$ with only countably many non-zero values? $\endgroup$
    – Andrei Kh
    Commented Nov 13, 2019 at 16:37
  • $\begingroup$ The first sentence says $\mathcal L^\infty([0,1])$ is the space of bounded Borel functions. I don't see the notation $\ell^\infty[0,1]$ anywhere... $\endgroup$ Commented Nov 13, 2019 at 16:48
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    $\begingroup$ @DavidC.Ullrich: three years late, but you did write and use $\ell^\infty[0,1]$ in your second to last paragraph. $\endgroup$ Commented Nov 14, 2022 at 12:41
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The following just regurgitates the links in David Ullrich's answer, but Ullrich doesn't give them the emphasis they deserve.

Let $S$ be a maximal set of mutually singular measures on $[0,1]$; such a set always exists by Zorn's Lemma. Then, for any $\mu\in C([0,1])^*$, we have $$\mu=\sum_{s\in S; s\ll\mu}{\frac{d\mu}{ds}s}$$ Moreover, by singularity, $$\|\mu\|_{C([0,1])^*}=\sum_{s\in S;s\ll\mu}{\left\|\frac{d\mu}{ds}\right\|_{L^1(s)}\|s\|_{C([0,1])^*}}$$ which shows that $C([0,1])^*$ is isometric to the $l^1$ direct sum $$C([0,1])^*\cong{\bigoplus_{s\in S}}^{l^1}{L^1(s)}\cong L^1\left({\oplus S}\right)$$ where $\oplus S$ denotes the measure on $(S;\text{discrete})\times[0,1]$ given by $$(\oplus S)(A)=\sum_{s\in S}{s(\{x:(s,x)\in A\})}$$ (That is, take $S$-many copies of $[0,1]$, each with their own natural measure. Conjoin them together into an incredibly long line.)

Since $(L^1)^*=L^{\infty}$, $$C([0,1])^{**}\cong L^{\infty}(\oplus S)$$

(Of course, whether you can find a useful description of $S$ is a separate question.)

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