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Feb 9, 2023 at 19:26 history edited Adi CC BY-SA 4.0
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Feb 9, 2023 at 19:24 comment added Adi Thanks for the suggestion, I'll do that.
Feb 9, 2023 at 17:07 comment added Willie Wong If you are changing the goals, I would suggest asking this as a new question, and not just editing the current one. When asking the new question you can link back to this one.
Feb 9, 2023 at 16:56 history edited Adi CC BY-SA 4.0
The question has been updated.
S Feb 8, 2023 at 15:48 vote accept Adi
Feb 7, 2023 at 9:23 history edited YCor
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Feb 7, 2023 at 0:18 history became hot network question
Feb 6, 2023 at 22:17 answer added Iosif Pinelis timeline score: 4
S Feb 6, 2023 at 21:22 vote accept Adi
S Feb 8, 2023 at 15:48
S Feb 6, 2023 at 21:22 vote accept Adi
S Feb 6, 2023 at 21:22
Feb 6, 2023 at 21:22 vote accept Adi
S Feb 6, 2023 at 21:22
Feb 6, 2023 at 21:11 answer added Willie Wong timeline score: 7
Feb 6, 2023 at 20:22 comment added leo monsaingeon There you go! ;-)
Feb 6, 2023 at 20:16 vote accept Adi
Feb 6, 2023 at 20:36
Feb 6, 2023 at 19:46 answer added Christian Remling timeline score: 5
Feb 6, 2023 at 19:35 history edited Adi CC BY-SA 4.0
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Feb 6, 2023 at 19:33 comment added Adi So if there's a counterexample, i would be very much interested to know what features it might have.
Feb 6, 2023 at 19:32 comment added Adi You are right in the sense that it does not see the gradient and that the gradient replaced with any other generic function, the question remains the same.
Feb 6, 2023 at 18:36 comment added leo monsaingeon This does not see at all the fact that the integrand is a gradient, does it? So you could ask if a function $v\in L^2$ whose square average cannot blow up too fact is in fact better than $L^2$. In which case the answer is probably no, just fiddling around with the usual logarithms and borderline integrability should give a counterexample, I guess?
Feb 6, 2023 at 16:11 history asked Adi CC BY-SA 4.0