Let $(f_n)$ be functions in $L^p(\Omega), 1 such that $(f_n) \rightarrow f$ almost everywhere and $\Vert f_n\Vert_p \rightarrow \Vert f\Vert_p$. How does one show that $f_n \rightarrow f$ weakly in $L^p(\Omega)$ without first showing that $f_n \rightarrow f$ strongly in $L^p(\Omega)$? (I know that under these hypotheses we in fact get that $f_n \rightarrow f$ strongly, as explained here If $f_k \to f$ a.e. and the $L^p$ norms converge, then $f_k \to f$ in $L^p$ , but there must be an easier, direct argument that $f_n \rightarrow f$ weakly.) I know that some subsequence $(f_{n_k})$ converges weakly to some $g$ in $L^p(\Omega)$ since the $f_n's$ are bounded, but then (1) how do we pass from the subsequence to the original sequence, and (2) how do we show that $g=f$? Thanks!
Almost everywhere convergence and convergence of $L^{p}$ norms implies weak convergence
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functional-analysis
banach-spaces
weak-convergence
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0when I said "the $f_n$'s are bounded," I of course meant that the $\Vert f_n \Vert$'s are bounded. – 2011-11-02
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2The proof of strong convergence is about three lines, and your proposed argument invokes Alaoglu's theorem, which in turn invokes Tychonoff's. I'm not sure I would call that "easier". – 2011-11-03
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0@jake: You can edit your question using the "edit" link just above the comments, you know. – 2011-11-03