12 edited body

This is sort of a follow-up to http://mathoverflow.net/questions/24432/borelx-sigmax-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 $n \in \mathbb{N}$.

Notes:

• 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.

REFERENCES:

[1] N. N. VAKHANIA, V. I. TARIELADZE, S. A. CHOBANYAN, Probability Distrubitions 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.

11 deleted 74 characters in body; added 110 characters in body

This is sort of a follow-up to http://mathoverflow.net/questions/24432/borelx-sigmax-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, see [1] p.19)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 $n \in \mathbb{N}$.

Notes:

• 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 (see Vakhania et. al. Probability Distributions on Banach Spaces, [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?

• Haydon / Talagrand and others have shown that there

• There are non-reflexive Grothendieck spaces which do not contain $\ell^{\infty}$. \ell^{\infty}$(cf. [3]). So we can't simply reduce to this case. I don't know much more about Grothendieck spaces though or characterizations of them that might be helpful. REFERENCES: [1] N. N. VAKHANIA, V. I. TARIELADZE, S. A. CHOBANYAN, Probability Distrubitions on Banach Spaces, Mathematics and its applications (D. Reidel Publishing Company), 1987 [3] R. HAYDON, A non-reflexive Grothendieck Space that does not contain$\ell^{\infty}$, Israel Journal of Mathematics, Vol 40, No. 1, 1981 EDIT: I rephrased the question and added some information. 10 added 196 characters in body; added 2 characters in body; added 6 characters in body This is sort of a follow-up to http://mathoverflow.net/questions/24432/borelx-sigmax-for-x-non-separable PROBLEM: Given a Banach space$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, see Vakhania)[1] p.19). 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$n \in \mathbb{N}$. Notes: • 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$E$(as proven in Edgar G. A., Measurability in Banach spaces, see last post for reference)[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 (see Vakhania et. al. Probability Distributions on Banach Spaces, 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? • Haydon / Talagrand and others have shown that there are non-reflexive Grothendieck spaces which do not contain$\ell^{\infty}\$. 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.

REFERENCES:

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

EDIT: I rephrased the question and added some information.

9 edited tags
8 edited tags
7 title
6 edited tags
5 added 371 characters in body