|
12
|
|
edited Dec 26 2010 at 5:00 |
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
[2] http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1977/26/26053
[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
|
|
edited Dec 26 2010 at 4:29 |
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
[2] http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1977/26/26053
[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
|
|
edited Dec 26 2010 at 4:23 |
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
[2] http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1977/26/26053
EDIT: I rephrased the question and added some information.
|
|
|
|
9
|
|
edited Dec 25 2010 at 2:54 |
|
|
|
|
|
|
8
|
|
edited Nov 5 2010 at 17:28 |
|
|
|
|
|
|
7
|
|
edited Aug 3 2010 at 4:57 |
Cyl(E) = Borel(E) for E non-reflexive Grothendieck Banach space
|
|
|
|
|
6
|
|
edited Aug 3 2010 at 3:14 |
|
|
|
|
|
|
5
|
|
edited Jul 23 2010 at 16:15 |
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).
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). 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 $\mathcal{B}(E) \not= \hat{C}(E)$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.
EDIT: I rephrased the question and added some information.
|
|
|
|
|
4
|
|
edited Jul 23 2010 at 16:04 |
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).
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). 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 $\mathcal{B}(E) \not= \hat{C}(E)$
EDIT: I rephrased the question and added some information.
|
|
|
|
|
3
|
|
edited Jul 23 2010 at 15:58 |
PROBLEM: Give an example of Given a Banach space $E$ over $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$ that has the Grothendieck property, is not reflexive and is such that . Does $\hat{C}(E) = \mathcal{B}(E)$ holds. There's probably a simple constructionimply $E$ is reflexive?(This would in turn imply that $E$ is separable, but I can't find it at the moment.see Vakhania). 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). 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). 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 $\mathcal{B}(E) \not= \hat{C}(E)$ EDIT: I rephrased the question and added some information.
|
|
|
|
|
2
|
|
edited Jul 21 2010 at 13:41 |
Cyl(E) = Borel(E) for E non-separable non-reflexive Grothendieck
This is sort of a follow-up to http://mathoverflow.net/questions/24432/borelx-sigmax-for-x-non-separable
PROBLEM: Give an example of a Banach space $E$ over $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$ that has the Grothendieck property, is not separable (equivalently not reflexive ) and is such that $\hat{C}(E) = \mathcal{B}(E)$ holds.
There's probably a simple construction, but I can't find it at the moment..
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 $n \in \mathbb{N}$.
|
|
|
|
|
1
|
|
asked Jul 21 2010 at 10:51 |
Cyl(E) = Borel(E) for E non-separable Grothendieck
This is sort of a follow-up to http://mathoverflow.net/questions/24432/borelx-sigmax-for-x-non-separable
PROBLEM: Give an example of a Banach space $E$ over $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$ that has the Grothendieck property, is not separable (equivalently not reflexive) and is such that $\hat{C}(E) = \mathcal{B}(E)$ holds.
There's probably a simple construction, but I can't find it at the moment..
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 $n \in \mathbb{N}$.
|
|
|
