Timeline for DF-algebras and DF-modules
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 11, 2017 at 12:09 | vote | accept | Krzysztof | ||
| Jul 6, 2017 at 11:32 | answer | added | Jochen Wengenroth | timeline score: 2 | |
| Jul 3, 2017 at 12:35 | comment | added | Krzysztof | In fact I am assuming completeness. As for reflexivity, I would like to omit this assumption, if possible. I guess barrelledness can help but this is again something I would like to assume only if there is a non-barrelled counterexample. | |
| Jun 30, 2017 at 12:47 | comment | added | Jochen Wengenroth | If you don't require completeness, a counterexample would be any non-complete DF space $X$ embedded into its completion $Y$. | |
| Jun 30, 2017 at 6:59 | comment | added | Jochen Wengenroth | The closed range theorem is perhaps overkill: $\phi^*$ is injective since $\phi$ has dense range and if you assume that it is surjective then it is an isomorphism for the strong duals. This means that for every (closed absolutely convex) bounded set $B\subseteq Y$ there is $C\subseteq X$ with the same properties such that $C^\circ\subseteq \phi^*(B^\circ)$ which yields $\phi(C)^\circ = (\phi^*)^{-1}(C^\circ) \subseteq B^\circ$ and hence $B\subseteq \phi(C)^{\circ\circ}=\overline{\phi(C)}$. I don't think that, in general, this implies surjectivity of $\phi$ but I don't have an example. | |
| Jun 30, 2017 at 6:24 | comment | added | Jochen Wengenroth | The closed range theorem implies that the bitransposed map $\phi^{\ast\ast}$ has closed range, if you knew that $X$ and $Y$ were reflexive this would be enough. | |
| Jun 30, 2017 at 6:20 | comment | added | Jochen Wengenroth | I don't see anything ''more specific'': $X$ is a quotient of itself and $Y$ is a closed subspace of itself. | |
| Jun 29, 2017 at 10:51 | history | asked | Krzysztof | CC BY-SA 3.0 |