The page 20 of the book "Topological Fixed Point Theory for Singlevalued and Multivalued Mappings and Applications" (written by Ben Amar and O'Regan) has the following lemma.

Lemma: Let $(X,\Sigma,\mu )$ be a finite measure space and $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$. Suppose that

  1. $\sup_{n\in\mathbb{N}}\Vert f_n\Vert _{L^1}<\infty$
  2. $(\forall \varepsilon >0)(\exists \delta >0)(\forall E\in \Sigma )\left(\mu (E)<\delta\Rightarrow \sup_{n\in\mathbb{N}}\int _E|f_n|d\mu <\varepsilon \right)$.

Then there's a subsequence $(f_{k_n})_{n\in\mathbb{N}}$ of $(f_n)_{n\in\mathbb{N}}$ such that $\left(\int _Ef_{k_n}d\mu \right)_{n\in\mathbb{N}}$ is a Cauchy sequence for all $E\in \Sigma$.

Unfortunately, the book doesn't give any tips on how to prove this lemma.

My question is: how can I prove that lemma?

I couldn't do anything worth mentioning, but I know that the conclusion of that lemma is true if and only if there's a subsequence $(f_{k_n})_{n\in\mathbb{N}}$ of $(f_n)_{n\in\mathbb{N}}$ that converges weakly.

Thank for your attention!

EDIT: Please don't use the Dunford-Pettis Theorem because I want to use that lemma to prove this theorem.

Please don’t use either the Eberlein-Smulian Theorem, because I want to avoid advanced theorems of functional analysis.


1 Answer 1


For each fixed $k$, the sequence $\left(f_n\mathbf{1}_{\{\lvert f_n\rvert\leqslant k\}}\right)$ is bounded in $L^2$ hence using the diagonal extraction, we can find an increasing sequence of integers $(n_j)$ such that for each $k$, $\left(f_{n_j}\mathbf{1}_{\{\lvert f_{n_j}\rvert\leqslant k\}}\right)$ converges weakly to some $g_k$ in $L^2$.

For each $E\in \Sigma$, the sequence $(\int_E f_{n_j})$ is Cauchy. Indeed, for a fixed $\varepsilon$, pick $k$ such that $\sup_n \int \lvert f_n\rvert\mathbf{1}_{\{\lvert f_n\rvert> k\}}<\varepsilon$. Then use the fact that $(\int_E f_{n_j}\mathbf{1}_{\{\lvert f_{n_j}\rvert\leqslant k\}})$ converges to reach the conclusion.

  • $\begingroup$ Could please explain how to perform this diagonal extraction? Because I don’t know what that means. $\endgroup$
    – rfloc
    Dec 4, 2023 at 21:30
  • $\begingroup$ See math.stackexchange.com/questions/2621135/… $\endgroup$ Dec 4, 2023 at 22:20
  • $\begingroup$ Thanks to the answer in this link, I understood better what a diagonal extraction is. However I still have some questions. Why did you choose to perform the diagonal extraction in $L^2$ instead of $L^1$? Are you using the fact that $L^2$ is reflexive? If yes, how to use this fact to obtain a list of subsequences with which the diagonal extraction procedure is applied? Thanks for your attention! $\endgroup$
    – rfloc
    Dec 4, 2023 at 23:28
  • $\begingroup$ "Why did you choose to perform the diagonal extraction in 𝐿2 instead of 𝐿1? " Because we want weak convergence in $L^2$. "Are you using the fact that 𝐿2 is reflexive? " Yes. Extract a sub sequence which works for $k=1$. Then extract a further subsequence that works for $k=2$, and so one. A subsequence that works for each $k$ is given by taking the $j$-th element of the $j$th subsequence. $\endgroup$ Dec 5, 2023 at 17:27

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