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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