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This is sort of a follow-up to Borel(X) = \sigma(X') for X non-separable

PROBLEM: Given a Banach space $E$ over $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$ that has the Grothendieck property. Does $\hat{C}(E) = \mathcal{B}(E)$ imply $E$ is reflexive? (This would in turn imply that $E$ is separable).

Some definitions:

  • A Banach space is a Grothendieck space if a sequence in $E'$ which is $\sigma(E', E)$-convergent is automatically $\sigma(E', E'')$-convergent. Or equivalently: every $\sigma(E', E)$ zero sequence has subsequence which is $\sigma(E', E'')$-convergent, or equivalently: every linear, bounded operator from $E$ to $c_0$ (or any separable Banach space) is automatically weakly compact.

  • The $\sigma$-algebra $\hat{C}(E)$ is the $\sigma$-algebra generated by sets of the form $\mathcal{C}_{u_1, \cdots, u_n; C} := \{x \in E : (u_1(x), \cdots, u_n(x)) \in C\}$ where $u_1, \cdots, u_n \in E'$, $C \in \mathcal{B}(\mathbb{K}^n)$ and $n \in \mathbb{N}$.


  • The $\sigma$-algebra $\hat{C}(E)$ equals the $\sigma$-algebra of weak Baire sets $\mathcal{B}_0(E, \sigma(E, E'))$ for every locally convex space $E$ (see [2], Theorem 2.3).

  • The inclusion $\hat{C}(E) \subset \mathcal{B}(E)$ is trivially true. If $E$ is separable then $\hat{C}(E) = \mathcal{B}(E)$. [To see this use the Hahn-Banach theorem to show that $\mathcal{B}_E \in \hat{C}(E)$. As translations and scalar multplications are measurable with regard to the cylindrical $\sigma$-algebra the other inclusion follows.]

  • A reflexive space is automatically Grothendieck.

  • For a separable Grothendieck space $E$ we have that the identity is weakly compact so $E$ becomes reflexive

  • A reflexive space $E$ with $\hat{C}(E) = \mathcal{B}(E)$ is automatically separable ([1], Prop. 2.6, p.19). Without reflexivity the equality $\hat{C}(E) = \mathcal{B}(E)$ does not imply $E$ is separable in general.

  • The example $E = \ell^2(\mathbb{R})$ shows that there is a reflexive and non-separable space with $\mathcal{B}(E) \not= \hat{C}(E)$

  • Edgar's example below or $E = \ell^{\infty} = C(\beta \mathbb{N})$ gives a non-reflexive Grothendieck space with $\mathcal{B}(E) \not= \hat{C}(E)$. The question therefore: does every non-reflexive Grothendieck space have that property?

  • There are non-reflexive Grothendieck spaces which do not contain $\ell^{\infty}$ (cf. [3]). So we can't simply reduce to this case. I don't know much more about Grothendieck spaces though or characterizations of them that might be helpful.


[1] N. N. VAKHANIA, V. I. TARIELADZE, S. A. CHOBANYAN, Probability Distributions on Banach Spaces, Mathematics and its applications (D. Reidel Publishing Company), 1987


[3] R. HAYDON, A non-reflexive Grothendieck Space that does not contain $\ell^{\infty}$, Israel Journal of Mathematics, Vol 40, No. 1, 1981

EDIT: I rephrased the question and added some information.

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Clarification needed. According to , every reflexive space is Grothendieck. – Gerald Edgar Jul 21 '10 at 11:56
Yes. I better edit the question title to "non-reflexive" Grothendieck space if that makes it clearer. – santker heboln Jul 21 '10 at 13:40

Space $C(K)$ of continuous functions on a Stone space $K$ is Grothendieck, right? So take $K$ so large that countably many continuous functions do not separate points in $K$. Then (as in the $l^2(I)$ answer to Question 24432 cited) the weak Baire sets (= the cylindrical sigma-algebra) is not equal to the weak Borel sets, and certainly not equal to the norm Borel sets. Since the closed unit ball is not a weak Baire set.

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I'm not sure I understand you correctly, but I wanted an example where the cylindrical algebra is equal to the Borel sigma algebra. In the $\ell^2(I)$ case and the $C(K)$ case for e.g. $K = \beta \mathbb{N}$ it is not, as we know from the other question. Is there some characterisation of Grothendieck spaces I'm not aware of you are using here? As in all non-reflexive Grothendieck spaces are of $C(K)$ type (where $K$ is Stonean)? – santker heboln Jul 22 '10 at 6:08

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