Timeline for If $u_n\rightharpoonup u$ in $L^2(0,T;L^2)$ then there is a subsequence such that $u_n(t)\rightharpoonup u(t)$ almost everywhere?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2022 at 6:26 | comment | added | username | What about the Aubin-Lions Lemma? Would that be suifficient for your purpose? | |
| Jan 13, 2022 at 5:19 | history | edited | demlevi33 | CC BY-SA 4.0 |
added 533 characters in body
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| Jan 13, 2022 at 5:09 | comment | added | Fedor Petrov | Ah, this symbol means weak convergence! | |
| Jan 12, 2022 at 23:30 | comment | added | Bill Johnson | Consider the Rademachers or $\sin (nt)$ sequence in $L^2(0,1)$. | |
| Jan 12, 2022 at 22:42 | comment | added | Ben Deitmar | @FedorPetrov I don't think that's enough. The question is about weak convergence, where we do not have the convergence of $||u_n-u||$ to zero in measure. | |
| Jan 12, 2022 at 21:16 | comment | added | Fedor Petrov | a sequence of scalar functions $\|u_n-u\|$ converges to 0 in measure, thus a subsequence converges to 0 a.e. | |
| Jan 12, 2022 at 20:59 | history | edited | demlevi33 | CC BY-SA 4.0 |
edited title
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| Jan 12, 2022 at 20:46 | history | asked | demlevi33 | CC BY-SA 4.0 |