Timeline for Nuclear vs Banach spaces: compactness properties
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12 at 9:10 | vote | accept | user267839 | ||
| Oct 7 at 10:50 | comment | added | Robert Furber | Wikipedia has a bit of a problem in certain places with people copying things from textbooks and papers without understanding. Unfortunately there is also the requirement of "no original research" that means you can't necessarily use your own expertise to fix it. | |
| Oct 7 at 10:43 | comment | added | Robert Furber | @JochenWengenroth A necessary and sufficient condition is quasicompleteness, that every closed bounded set is complete. | |
| Oct 1 at 16:41 | comment | added | user267839 | let me also add this script adressing Jochen Wengenroth's point, see 7 ff | |
| Oct 1 at 16:32 | comment | added | user267839 | hmm, it seems that the term "compactness properties" from above is phrased a bit misleading, even if replace it by "precompact". It seems that the seeked feature is that these spaces have a weakened version of Heine Borel property. But, is there maybe also some connection to concept of compact operators with resp. induced canonical seminorm maps? | |
| Oct 1 at 8:15 | comment | added | Jochen Wengenroth | This is not true without an additional completeness property: Take any non-reflexive Banach space $X$. Then $(X,\sigma(X,X^*)$ is nuclear (the local Banach spaces corresponding to weak neighbourhoods are finite-dimesnional), the unit ball $B$ of $X$ is $\sigma(X,X^*)$-bounded but not weakly compact. It is true that bounded subsets of nuclear spaces are precompact. | |
| Sep 30 at 17:12 | history | answered | Gerald Edgar | CC BY-SA 4.0 |