# Almost everywhere convergence and convergence of $L^{p}$ norms implies weak convergence

Let $(f_n)$ be functions in $L^p(\Omega), 1<p<\infty$ 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

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!

-
when I said "the $f_n$'s are bounded," I of course meant that the $\Vert f_n \Vert$'s are bounded. – jake Nov 2 '11 at 23:44
The 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". – Nate Eldredge Nov 3 '11 at 0:10
@jake: You can edit your question using the "edit" link just above the comments, you know. – Henning Makholm Nov 3 '11 at 0:11