Cyl(E) = Borel(E) for E non-reflexive Grothendieck Banach space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:47:11Z http://mathoverflow.net/feeds/question/32785 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32785/cyle-borele-for-e-non-reflexive-grothendieck-banach-space Cyl(E) = Borel(E) for E non-reflexive Grothendieck Banach space santker heboln 2010-07-21T10:51:49Z 2010-12-26T05:00:48Z <p>This is sort of a follow-up to <a href="http://mathoverflow.net/questions/24432/borelx-sigmax-for-x-non-separable" rel="nofollow">http://mathoverflow.net/questions/24432/borelx-sigmax-for-x-non-separable</a></p> <p>PROBLEM: Given a Banach space $E$ over <code>$\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$</code> 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). </p> <hr> <p>Some definitions:</p> <ul> <li><p>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.</p></li> <li><p>The $\sigma$-algebra $\hat{C}(E)$ is the $\sigma$-algebra generated by sets of the form <code>$\mathcal{C}_{u_1, \cdots, u_n; C} := \{x \in E : (u_1(x), \cdots, u_n(x)) \in C\}$</code> where $u_1, \cdots, u_n \in E'$, $C \in \mathcal{B}(\mathbb{K}^n)$ and $n \in \mathbb{N}$. </p></li> </ul> <p>Notes:</p> <ul> <li><p>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).</p></li> <li><p>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.]</p></li> <li><p>A reflexive space is automatically Grothendieck.</p></li> <li><p>For a separable Grothendieck space $E$ we have that the identity is weakly compact so $E$ becomes reflexive</p></li> <li><p>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.</p></li> <li><p>The example $E = \ell^2(\mathbb{R})$ shows that there is a reflexive and non-separable space with $\mathcal{B}(E) \not= \hat{C}(E)$</p></li> <li><p>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?</p></li> <li><p>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.</p></li> </ul> <hr> <p>REFERENCES:</p> <p>[1] N. N. VAKHANIA, V. I. TARIELADZE, S. A. CHOBANYAN, Probability Distributions on Banach Spaces, Mathematics and its applications (D. Reidel Publishing Company), 1987</p> <p>[2] <a href="http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1977/26/26053" rel="nofollow">http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1977/26/26053</a></p> <p>[3] R. HAYDON, A non-reflexive Grothendieck Space that does not contain $\ell^{\infty}$, Israel Journal of Mathematics, Vol 40, No. 1, 1981</p> <hr> <p>EDIT: I rephrased the question and added some information. </p> http://mathoverflow.net/questions/32785/cyle-borele-for-e-non-reflexive-grothendieck-banach-space/32830#32830 Answer by Gerald Edgar for Cyl(E) = Borel(E) for E non-reflexive Grothendieck Banach space Gerald Edgar 2010-07-21T18:31:32Z 2010-07-21T18:31:32Z <p>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.</p>