Timeline for Compactness of the unit ball of a Banach space for topologies finer than the weak* topology
Current License: CC BY-SA 4.0
20 events
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| Dec 14, 2022 at 20:45 | comment | added | Robert Furber | @NikWeaver The statement of your answer omits the necessary assumption in the question and Jochen's comment that $\tau$ be coarser than $\sigma(X',X'')$, which has led the OP of this question astray. If $X$ is infinite-dimensional, then the bounded weak-* topology $\tau$ on $X'$ is locally convex, strictly finer than $\sigma(X',X)$, and agrees on the unit ball. Of course $\tau$ is not coarser than $\sigma(X',X'')$ (it's finer if $X$ is reflexive, and incomparable otherwise). | |
| Jul 31, 2022 at 11:01 | comment | added | Goulifet | The Krein-Smulian theorem is indeed decisive here, thanks! Thanks also to @JochenWengenroth for completing the proof after knowing that the continuous linear forms coincide. | |
| Jul 31, 2022 at 9:55 | vote | accept | Goulifet | ||
| Jul 30, 2022 at 20:38 | comment | added | Nik Weaver | Thank you for the kind words. | |
| Jul 30, 2022 at 17:12 | comment | added | Jochen Wengenroth | Don't worry, Nik, you have spotted the crucial argument Krein-Smulian=Banach-Dieudonné. You intuition thus still works very well. | |
| Jul 30, 2022 at 13:01 | comment | added | Nik Weaver | You're right again. Guess I'm getting sloppy ... (or maybe I always was) | |
| Jul 30, 2022 at 13:00 | history | edited | Nik Weaver | CC BY-SA 4.0 |
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| Jul 30, 2022 at 9:34 | comment | added | Jochen Wengenroth | ... $U^\circ \subseteq E^{\circ\circ}$ which by the bipolar theorem is the closed absolutely convex hull of $E$ and thus contained in the finite dimensional linear span of $E$. On the other hand, $U^\circ$ is contained in $X$ (since $(X',\tau)'=X$). It is $\sigma(X,X')$ compact by Alaoglu and finite dimensional and hence contained in the absolutely convex hull of a finite set $F\subseteq X$. This shows that $U=U^{\circ\circ}$ contains $F^\circ$ which is a $\sigma(X',X')$-neighbourhood of $0$. | |
| Jul 30, 2022 at 9:27 | comment | added | Jochen Wengenroth | What you and wikipedia call Krein-Smulian theorem is also know as the Banach-Dieudonné theorem and it implies indeed that the new topology $\tau$ yields to the dual $(X',\tau)'=X$ (or, more precisely, every $\tau$-continuous linear functional on $(X',\tau)$ is an evaluation in a point of $X$. The additional assumption $\tau\subseteq \sigma(X',X'')$ then indeed implies $\tau=\sigma(X',X)$: Every $\tau$-closed absolutely convex $\tau$-neighbourhood $U$ contains a $\sigma(X',X'')$-neighbourhood of $0$ so that there is a finite set $E\subseteq X''$ with $E^\circ\subseteq U$ and hence ... | |
| Jul 30, 2022 at 9:01 | comment | added | Jochen Wengenroth | Your argument does not use that the new topology is coarser than the weak topology $\sigma(X',X'')$. | |
| Jul 30, 2022 at 8:48 | comment | added | Jochen Wengenroth | I don't understand the last sentence: A locally convex space is not determined by its continuous linear functionals! The weak topology of a locally convex space always has the same dual. | |
| Jul 29, 2022 at 20:55 | history | edited | Nik Weaver | CC BY-SA 4.0 |
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| Jul 29, 2022 at 20:55 | comment | added | Nik Weaver | Oh gosh, you are absolutely right! I'll have to correct my answer. | |
| Jul 29, 2022 at 18:52 | comment | added | Goulifet | About the last point, we see that the sequence $u \in \ell_1$ should satisfy that $\langle u , e_n \rangle = 0$ for any $n$ and $\langle u, 1 \rangle = 1$, the two requirements being impossible. This seems to prove that it is not possible to extract from $(u_n)$ a subsequence (of sequences) that converges in $\ell_1$. Do you agree? | |
| Jul 29, 2022 at 18:51 | comment | added | Goulifet | I have some problem with your construction. Take the following example where $\mathcal{X} = c_0$ is the set of vanishing sequences. Then, $\mathcal{X}' = \ell_1$ and $\mathcal{X}'' = \ell_\infty$. Consider the sequence of sequences $u_n = e_n = (0,\ldots , 0, 1 , 0 ,\ldots) \in \mathcal{B}$ and $f = 1 \in \ell_\infty \backslash c_0$. Clearly, $u_n \rightarrow 0$ for the weak* topology and $\langle u_n , f \rangle = 1$ for any $n$. However, I cannot say that there exists $u \in \ell_1$ such that $u_n \rightarrow u$. | |
| Jul 28, 2022 at 14:10 | history | edited | Nik Weaver | CC BY-SA 4.0 |
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| Jul 28, 2022 at 14:09 | comment | added | Nik Weaver | Thank you! I was having doubts about that point and trying to figure it out. I'll correct my answer. | |
| Jul 28, 2022 at 14:03 | comment | added | Jochen Wengenroth | For the James space $J$ there is essentially only one continuous linear functional on $J'$ which is not weak$^*$-continuous (all others are multiples since $J$ is co-one-dimensional in $J''$). | |
| Jul 28, 2022 at 13:41 | history | edited | Nik Weaver | CC BY-SA 4.0 |
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| Jul 28, 2022 at 13:35 | history | answered | Nik Weaver | CC BY-SA 4.0 |