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Let $\Omega\subset \mathbb{R}^n$ be bounded and Lebesgue measurable, $p \in [1,\infty]$, and $a,b \in L^p(\Omega)$. Consider the set $$ K = \big\{ u \in L^p(\Omega):\, a(x) \leq u(x) \leq b(x) \, \text{ a.e. in } \Omega \big\}. $$ Is $K$ weakly sequentially closed?

In [1] the case $p=2$ was partially studied. In Fredi Tröltzsch book [2 - p263] the proof of the set to be nonempty, closed, bounded, and convex for any $p$ is presented as an exercise. In Manzoni et al [3 - p285] a simple argument is taken for proving that indeed $K$ is convex and closed.

Moreover, it can be shown [4 - Th 3.7] that if a subset $K$ of a Banach space $X$ is convex and closed, then it is weakly sequentially closed.

Up to this point, it would seem that yeah, $K$ is w.s.c., but in Manzoni's example it is argued that only the cases $p \in (1,\infty)$ are actually w.s.c. due to reflexivity, and there is a treatment for the $p=\infty$ case via weakly–$\star$ sequences. However, I don't see how reflexivity is a limitation?

Andymt
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  • I don't understand your question. You already seem to know that convex and closed implies weakly sequentially closed. Where is this concern about reflexivity coming from? – Severin Schraven Jun 12 '24 at 22:01
  • In fact, in your setting, one does not even need the completeness (see here https://math.stackexchange.com/questions/1151148/a-convex-subset-of-a-banach-space-is-closed-if-and-only-if-it-is-weakly-closed). – Severin Schraven Jun 12 '24 at 22:04
  • Thanks for the link @SeverinSchraven, I didn't know that fact about general normed spaces. Regarding my question, I guess I am just puzzled that [3] stopped from concluding the w.s.c. for any $L^p$, making me believe there could be some reasoning for $L^1$ and $L^\infty$ to not satisfy the property, so I am asking if anyone can see what fails or confirmation that the set is indeed w.s.c. – Andymt Jun 12 '24 at 22:14
  • I don't know how Manzoni proofs his stuff. Which part of his argument relies on reflexivity? Is he using for example the sequential version of Banach-Alaoglu? – Severin Schraven Jun 13 '24 at 07:17
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    I think the author got confused as they previously discussed weak* (sequential) compactness. In that setting you need reflexivity if you want to use Banach-Alaoglu in the proper form. – Severin Schraven Jun 13 '24 at 09:09
  • That sounds feasible. Indeed, they mention Banach-Alaoglu a little bit before.

    Just for completeness, let me include the passage I am referring to after they conclude closedness:

    Since $L^p(\Omega)$, for $p\in (1,\infty)$, is reflexive, then $K$ is weakly sequentially closed. As $L^1(\Omega)$ and $L^\infty (\Omega)$ are not reflexive, in principle we cannot draw any conclusion. However if $p=\infty$ and $\Omega$ is bounded, we can deduce that $K$ is weakly* sequentially closed.

     – Andymt Jun 13 '24 at 20:41

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There is no difference in proving this result for $p \in (1,\infty)$ or $p=1,\infty$.

To prove closedness of the set, take a converging sequence. This has a pointwise a.e. converging subsequence. (this is true for all $p$). Hence, the limit satisfies the constraints. Then use that closed plus convex implies weakly (sequentially) closed. Also this result has nothing to do with reflexivity.

daw
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  • Thanks for agreeing on this, that was exactly why I got so puzzled about the mention of reflexivity! – Andymt Jun 13 '24 at 20:45