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It would be great if someone could name me a reference where I can find a proof for the following statement, thank you! :)

Given is a bounded sequence $(f_{n})_{n\in\mathbb{N}}\in L^{\infty}$. Then there exists a $f\in L^{\infty}$ and a subsequence $f_{n_{k}}$ of $f_{n}$ that convergeces weakly* to $f$ in $L^{\infty}$.

Lae
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You did not specify the basic measure space but I will assume that $L^{1}$ is separable and you are looking at $L^{\infty}$ as the dual of $L^{1}$. In this case Banach Alaoglu Theorem tells you that any closed ball in $L^{\infty}$ is compact in weak* topology. Further separability if $L^{1}$ makes this ball metrizable. Hence every sequence in it has a weak* convergent subsequence.

Kavi Rama Murthy
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  • Thank you for your answer! Yes, $L^1$ is separable and I am looking at $L^{\infty}$ the dual of $L^1$. But Banach Alaoglu Theorem just guaranties that the closed unit ball in $L^{\infty}$ is weak* sequentially compact. How do we get this property for $L^{\infty}$? – Lae Jul 20 '20 at 11:12
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    @Lae It is given that $(f_n)$ is bounded in $L^{\infty}$ So this sequence is contained in the closed ball of radius $R$ around $0$ for some $R$ and this closed ball is weak*ly sequentially compact. – Kavi Rama Murthy Jul 20 '20 at 11:36
  • But the theorem just says that the closed unit ball in $L^{\infty}$ is compact in weak* topology, why does it also hold for any closed ball in $L^{\infty}$? – Lae Jul 20 '20 at 14:07