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
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!
Best Answer
The boundedness of $\|f_n\|_p$, together with the reflexiveness of $L^p(\Omega)$ for $p\in (1,\infty)$, allows us to conclude that $f_{n_k}\rightharpoonup g$ weakly in $L^p(\Omega)$ for some $g\in L^p(\Omega)$. Recall that $f_n\rightarrow f$ a.e. in $\Omega$. Using these facts and Egorov's theorem we can show $g=f$ a.e. in $\Omega$. Then via arguing by contradiction (a standard way) one can show the original sequence $f_n\rightharpoonup f$ weakly in $L^p(\Omega)$.