Timeline for Why define Schwartz by supremum rather than limit?
Current License: CC BY-SA 4.0
10 events
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Nov 23 at 23:43 | comment | added | Ponder Stibbons | @LSpice Thanks. Understood. Which means it is still a mystery. My comment was an honest attempt to understand the reaction to the question and I asked you as my best bet for someone who might be involved. | |
Nov 21 at 22:37 | comment | added | LSpice | Re, I do not answer questions about specific votes. However, to be clear, my initial comment was not only mathematically foolish but based on careless reading, and I apologise for it. (I'd just delete it, except that it already has one explicit, and your implicit, response.) | |
Nov 21 at 22:30 | comment | added | Ponder Stibbons | @LSpice Did you downvote the question? Just wondering because while it is not a deep question leading to a long answer it was something that was bugging me and I got a good answer. Did you see anything wrong with the question that it would deserve a downvote? | |
Nov 20 at 18:12 | comment | added | LSpice | @FedorPetrov, re, right, of course, as is observed in the question itself. Sorry! | |
Nov 20 at 10:32 | comment | added | Ponder Stibbons | @Dirk That makes a lot of sense. As you say - the seminorms (based on the supremum) are good for making a topology, and so it is nice to have them there already. Thanks. It gives a motive for the approach. | |
Nov 20 at 10:02 | comment | added | Fedor Petrov | @LSpice if a function remains bounded after you multiply it by $x$, it need | |
Nov 20 at 9:38 | comment | added | Dirk | I think you could use the limit for the definition. But one wants to put a topology on the Schwartz space and this is achieved by the countable family of seminorms so these suprema are useful to have anyways… | |
Nov 20 at 4:37 | comment | added | LSpice | A bounded function need not go to $0$ at $\infty$. | |
S Nov 20 at 4:23 | review | First questions | |||
Nov 20 at 5:31 | |||||
S Nov 20 at 4:23 | history | asked | Ponder Stibbons | CC BY-SA 4.0 |