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If $u_n\rightharpoonup u$ in $L^2(0,T;L^2(\Omega))$. Can we find a subsequence such that $u_{n_k}(t)\rightharpoonup u(t)$ almost everywhere on $[0,T]$?

I'm not sure if this question is trivial or not, and how to even start! Any hints would be great!

Edit 1: I would add that $\Omega$ is a bounded open subset of $\mathbb{R}^n$.

Edit 2: I want to apologize for the confusion caused by the symbol "$\rightharpoonup$", which means weak convergence. I used it under the assumption that it is standard.

Thank you for the counter-example and the answer. Bill Johnson mentioned in the comments that the sequence $\{t\mapsto \sin(nt)\}$ converges weakly in $L^2(0,T;L^2(0,1))$, but no subsequence has the sought after property.

Which urges the question: what is a non-trivial sufficient condition so that the question has a positive answer?

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  • $\begingroup$ a sequence of scalar functions $\|u_n-u\|$ converges to 0 in measure, thus a subsequence converges to 0 a.e. $\endgroup$
    – Fedor Petrov
    Commented Jan 12, 2022 at 21:16
  • $\begingroup$ @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. $\endgroup$
    – Ben Deitmar
    Commented Jan 12, 2022 at 22:42
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    $\begingroup$ Consider the Rademachers or $\sin (nt)$ sequence in $L^2(0,1)$. $\endgroup$
    – Bill Johnson
    Commented Jan 12, 2022 at 23:30
  • $\begingroup$ Ah, this symbol means weak convergence! $\endgroup$
    – Fedor Petrov
    Commented Jan 13, 2022 at 5:09
  • $\begingroup$ What about the Aubin-Lions Lemma? Would that be suifficient for your purpose? $\endgroup$
    – username
    Commented Sep 19, 2022 at 6:26

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