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I want to show that a sequence of functions in $L^p$ converges $f_n\to f \in L^p$ if every subsequence has a further subsequence that converges to $f\in L^p$.

We are given that every subsequence has a further subsequence that converges to $f\in L^p$. My attempt is to suppose $f_n$ does not converge to $f$ in $L^p$. This means there is some $\varepsilon,\delta,$ and $\{f_{n_k}\}$ such that $$m(x\vert \|f_{n_k}-f\|_p>\delta)>\varepsilon$$ for all $k$. But we know that there is some $g_k\subseteq f_{n_k}$ such that $g_k\to f$ in $L^p$ since this is given, but this is a contradiction to the line above. Is this correct?

mathim1881
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1 Answers1

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Yes, that's correct. Actually what you propose works for any topological space with the same reasoning.

Moreover, you can see this question for more information on nets.

MikeG
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  • Is the converse reasoning true as well? i.e. if $f_n \to f$ in $L^p$, does this imply that every subsequence has a further subsequence that converges? I have seen something similar to this regarding the convergence of probability measures. – moonknight Oct 04 '21 at 20:02
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    @moonknight Yes, that's true; actually it's true for general topological space. See this question. – MikeG Oct 06 '21 at 00:29