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Aug 24, 2023 at 16:46 history edited Ben Francis CC BY-SA 4.0
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Aug 22, 2023 at 22:53 history edited Ben Francis CC BY-SA 4.0
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Aug 22, 2023 at 22:36 history edited Ben Francis CC BY-SA 4.0
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Aug 22, 2023 at 22:13 history edited Ben Francis CC BY-SA 4.0
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Aug 22, 2023 at 22:12 comment added Ben Francis I am unable to delineate under what circumstances this definition would (or rather would not) be appropriate, but it is the natural generalization of every definition of linear operator norm I have ever seen.
Aug 22, 2023 at 22:08 comment added Ben Francis There is no reason why the supremum need be finite, even for linear $A$; that's precisely the concept of an unbounded operator: an operator that does not have a finite norm.
Aug 17, 2023 at 22:02 comment added Alex M. @MaxHorn: And, if that quantity is finite, why is it a norm?
Aug 17, 2023 at 21:38 comment added Max Horn If $A$ is nonlinear, why should this supremum be finite in general? You hint it might be useful to do this "in some cases". Could you elaborate which cases you have in mind?
Aug 17, 2023 at 21:31 comment added Ben Francis Corrected to not be a question.
Aug 17, 2023 at 21:31 history edited Ben Francis CC BY-SA 4.0
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Aug 17, 2023 at 21:23 comment added Max Horn This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
S Aug 17, 2023 at 20:42 review Low quality posts
Aug 18, 2023 at 1:25
S Aug 17, 2023 at 20:42 review Late answers
Aug 17, 2023 at 22:02
Aug 17, 2023 at 20:24 history edited Ben Francis CC BY-SA 4.0
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S Aug 17, 2023 at 20:22 review First answers
Aug 17, 2023 at 20:41
S Aug 17, 2023 at 20:22 history answered Ben Francis CC BY-SA 4.0