Timeline for Sequential Continuity in dual spaces of separable Banach Spaces
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2019 at 18:21 | comment | added | DJA | That is extremely helpful, thank you very much! | |
| Nov 13, 2019 at 18:01 | comment | added | Robert Furber | @DJA An example that works in ordinary set theory is to take $X = \aleph_1$ with the order topology. Then by the Riesz representation theorem and the fact that $X$ is a scattered space, $C_0(X)^* \cong \ell^1(X)$, which defines a weak-* topology on $\ell^1(X)$. Then the mapping defined by summation $\ell^1(X) \rightarrow \mathbb{R}$ is sequentially continuous but not continuous. | |
| Nov 13, 2019 at 17:56 | comment | added | Robert Furber | @DJA It is not true in general that sequential continuity implies continuity. There is a nice counterexample in Solovay's model where all sets are Lebesgue measurable: the linear map $\int : \ell^\infty([0,1]) \rightarrow \mathbb{R}$ defined by Lebesgue integration is sequentially continuous with $\ell^\infty([0,1])$ given the weak-* topology as the dual of $\ell^1([0,1])$, essentially by the dominated convergence theorem, but not continuous. | |
| Nov 13, 2019 at 17:46 | comment | added | Robert Furber | @DJA I use the separability to go from sequentially closed to weak-* closed. What I'm using is the fact that the weak-* topology on bounded subsets of the dual of a separable Banach space is metrizable, so the intersection of the kernel of $f(y)$ with each closed bounded set is closed. Krein-Šmulian/Banach-Dieudonné can then be applied to deduce that the kernel of $f(y)$ is weak-* closed. | |
| Nov 12, 2019 at 22:45 | comment | added | DJA | did you use the separability anywhere? it seems this holds without separability? | |
| Oct 24, 2017 at 16:09 | vote | accept | Manish Kumar | ||
| Oct 24, 2017 at 12:12 | history | answered | Robert Furber | CC BY-SA 3.0 |