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I am interested in proving the following:

Given a measure space $(\Omega, \mu, \mathcal{F})$, then the following set is compact: $$\{ X \in L^2: \int |X|^2 d \mu \leq K \}$$

but I have no idea where to start given my lack of familiarity with dual spaces and my tremendously weak understanding of weak topologies.

Any references to an "elementary" proof of the Dunford-Pettis theorem would be massively appreciated, since I am trying to understand the claim here, in particular the part where it states "by compactness in $L^2$", which is the only part I do not understand.

qp212223
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  • You need to clarify the topology before mentioning something is "compact". For example, in general, that set is not compact with respect to the norm topology. – Danny Pak-Keung Chan Oct 25 '21 at 22:41
  • Ok fair enough. I suppose I can rephrase my question as follows: Why is it that for any sequence of random variables in the set I specify in the question, there exists a subsequence which converges weakly in $L^2$? In particular, is there any reference you know of which has an elementary proof of this? – qp212223 Oct 25 '21 at 23:02
  • Another comment: Sequential compactness and compactness, in general, are different. Which one do you want to prove? You should specify it in you question. – Danny Pak-Keung Chan Oct 26 '21 at 14:33
  • In this case they are the same since sequential compactness is equivalent to compactness on a metric space and weak convergence is metrizable. But I wish to prove sequential compactness (via compactness or otherwise) in this case. – qp212223 Oct 27 '21 at 16:08
  • In general, the weak topology is not metrizable. To make the discussion precise. Let $X=L^{2}(\Omega,\mathcal{F},\mu)$ be the usual Hilbert space (with real scalar field $\mathbb{R})$. By Riesz Representation, the dual space of $X$ is itself, i.e., $X^{\ast}=X$. The topology we are talking about is $\mathcal{T}=\sigma(X^{\ast},X)$. – Danny Pak-Keung Chan Oct 27 '21 at 18:16
  • Let $B^{\ast}$ be the unit ball of $X^{\ast}$. Let $\mathcal{T}{1}$ be the relative $\mathcal{T}$-topology on $B^{\ast}$. There is a result that if $X$ is separable (i.e., containing a countable dense subset. In Hilbert space setting, it is equivalent to having a countable orthonormal base), then $\mathcal{T}{1}$ is metrizable. However, in general Hilbert space $L^{2}(\Omega,\mathcal{F},\mu)$ is not separable, even if $\mu(\Omega)<\infty$. – Danny Pak-Keung Chan Oct 27 '21 at 18:16
  • Also see https://math.stackexchange.com/questions/914570/regarding-metrizability-of-weak-weak-topology-and-separability-of-banach-spaces and https://mathoverflow.net/questions/42310/when-is-l2x-separable#:~:text=L2%20on%20any%20Lebesgue,space%20is%20not%20separable%20either. – Danny Pak-Keung Chan Oct 27 '21 at 18:17
  • Also this one: https://math.stackexchange.com/questions/3380759/when-is-l2-mu-separable – Danny Pak-Keung Chan Oct 27 '21 at 18:27

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