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Prove that if a sequence $\{x_n\}_{n \in \mathbb{N}} \subset L^p$, $1<p<\infty$ converges in the weak topology to an element $x \in L^p$, then there is a subsequence $\{x_{n_k}\}_{k \in \mathbb{N}}$ of it that converges almost everywhere to $x$. From Mazur's lemma, we can find convex combinations that converge strongly to $x$, but I couldnt finish the problem from there. Any hint is appreciated.

M. Rubick
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  • Strong convergence implies convergence almost everywhere (at least for a subsequence). Convex combinations must then converge almost everywhere to...? – Evangelopoulos Foivos Sep 25 '23 at 18:30
  • @EvangelopoulosPhoevos thanks for your comment. The problem is that the number of elements in the convex combination increases. I cant see how to conclude. – M. Rubick Sep 25 '23 at 18:48

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This is not true. The sequence $$ x_n := sign(\sin(n \pi t) $$ converges weakly in $L^p(0,1)$ to $x=0$. But $x_n\ne0$ almost everywhere.

daw
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  • Great answer. Could you please elaborate on the weak convergence? I wasnt able to follow that step. Thank you! – M. Rubick Sep 26 '23 at 05:28
  • see eg https://math.stackexchange.com/questions/3536412/how-to-show-weak-convergence-of-mathrmsign-sinnx-as-n-to-infty?noredirect=1&lq=1 – daw Sep 26 '23 at 21:20