Timeline for The Schwartz space is not normable
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2022 at 18:57 | comment | added | user1110 | @priel could you please elaborate a little more? I mean each of the functions are bounded, but why is the whole family bounded? And to conclude rel. comapctness don't we need equicontinuity as well? | |
| Aug 8, 2019 at 17:23 | comment | added | Adrián González Pérez | Complete overkill: $S(\mathrm{R}^n)$ is a nuclear space (in the sense of Grothedieck) and therefore it can not be infinite dimensional Banach space. | |
| Sep 11, 2015 at 11:24 | answer | added | Simon Henry | timeline score: 10 | |
| Sep 11, 2015 at 9:39 | comment | added | priel | Well, bounded sets are relatively compact (follows from Ascoli), something which cannot happen in infinite dimensional normed spaces. Not sure if you would call that simple. | |
| Sep 11, 2015 at 9:01 | history | asked | Bazin | CC BY-SA 3.0 |