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This thread is related to: Spectral Measure: Dominated Convergence

Given a measure space $\Omega$.

Consider a sequence of square integrables: $\int|f_n|^2\mathrm{d}\mu<\infty$

Suppose pointwise convergence: $f_n\to f$.

Does the following hold: $$\int|f_m-f_n|^2\mathrm{d}\mu\to0\implies\int|f-f_n|^2\mathrm{d}\mu\to0$$

The problem is that they may have no dominant at all: $$s_n:=\frac{1}{\sqrt{n}}\chi_{(n,n+1]}\to0:\quad\int|s_m-s_n|^2\mathrm{d}\mu\stackrel{m\neq n}{=}\frac{1}{m}+\frac{1}{n}\to0\quad\int\sup_n|s_n|\mathrm{d}\mu=\sum_{n=1}^\infty\frac{1}{n}=\infty$$

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  • Are the $f_n$ individually square integrable? What sort of convergence is your arrow indicating? What are your thoughts on the problem? – Jason Knapp Dec 10 '14 at 14:12
  • Yes, it is not clear what "an approximation $f_n\to f$" means. Does the sequence converge pointwise? Almost everywhere? Uniformly? In $L^2$? – Math1000 Dec 10 '14 at 14:13
  • @JasonKnapp: Pointwise convergence of square integrable functions. Will add these details. Thanks for pointing out! – C-star-W-star Dec 10 '14 at 14:16

2 Answers2

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See for a related question Does $L^p$ convergence imply pointwise convergence.

Since you have $(f_n)$ converging to some function $y$ in $L^2$, there exists a subsequence $(f_{n_k})$ converging to $y$ a.e., and therefore since this subsequence also converges pointwise to $f$ we must have $f = y$ a.e.. Since $y \in L^2$ we conclude that $f\in L^2$ as well.

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  • The point is don't have convergence a priori. I mean only that the sequence is cauchy; ok, the L^p spaces are complete, but are you sure that this applies to generic measures and not only to Lebesgue measure? – C-star-W-star Dec 10 '14 at 14:41
  • Sure, I am using critically that $L^2$ is complete. If this is a sort of in-time exercise where you are supposed to prove this before you know that fact, then you would have to somehow extract that information (or have a different argument!). – Jason Knapp Dec 10 '14 at 14:43
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Ok, I got it now: Fatou! :D

Extracting the step within the completeness proofs: $$\int\|F-F_n\|^2\mathrm{d}\mu\leq\liminf\int\|F_m-F_n\|^2\mathrm{d}\mu\to0$$ (Note that this works perfectly for Banach spaces!)

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