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Let $u_n:[0,1]\to[0,\infty)$ are functions such that $u_n\rightharpoonup u$ in $L^2[0,1]$ and $u_n \log u_n\rightharpoonup u\log u$ in $L^1[0,1]$. Does this imply $u_n$ strongly converges to $u$ in $L^1[0,1]$?

I remember doing this problem a few years back in the affirmative direction but completely forgot how I got the conclusion. I probably used the de la vallée-poussin criterion (for uniform integrability) and Mazur lemma somewhere at that time. Currently I have a feeling that the conclusion derived then isn't correct. Can someone confirm if this result is true?

Here $\rightharpoonup$ signifies weak convergence in Banach spaces.

mudok
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  • Under the extra assumption of a.e. convergence this is true by Vitalli's convergence theorem. Note that boundedness in $L^2$ implies uniform integrability (Holder's inequality) so you don't really need de la vallée-poussin criterion. – Evangelopoulos Foivos Jan 02 '25 at 12:33
  • a.e convergence is very strong. That assumption is not there in the question. – mudok Jan 02 '25 at 12:34
  • I agree, it is. It was just a quick observation of mine - not sure if it helps – Evangelopoulos Foivos Jan 02 '25 at 12:35
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    Actually with a.e convergence you don't even require both the weak convergences. Only one would suffice. – mudok Jan 02 '25 at 12:40
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    This follows from Theorem 2 in this paper https://doi.org/10.1080/03605308408820337 by Visintin, not sure whether it is contained in any textbook. – daw Jan 03 '25 at 18:07
  • @daw absolutely fantastic. Much thanks for this neat reference. – mudok Jan 04 '25 at 07:41

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