Let $(f_k)_{k\in \mathbb{N}}$ a sequence in $L^2(\Omega)$ which converges weakly to $\overline{f}\in L^2(\Omega)$, where $\Omega\subset \mathbb{R}^n$ with $n=2,3$ a Lipschitz bounded domain. I know that the weak convergence of $(f_k)$ does not implies that $f_k(x)\to f(x)$ a.e. in $\Omega$. Howevere, I was wondering if there exists a subsequence of $(f_k)$ which converges a.e. to $f$. Please any hint will be aprecciated.
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No, such a subsequence need not exist. You can consider the sequence $f_n :[0,1] \to \Bbb R$ defined by $$ f_n(x) := g(nx) $$ where $g:\Bbb R\to\Bbb R$ is the function $$ g(x):=\begin{cases} 1 &; x\in[2n-1,2n) \text{ for some }n\in\Bbb Z \\ -1 &; x\in [2n,2n+1) \text{ for some }n\in\Bbb Z . \end{cases} $$
It is not difficult to prove that $(f_n)_{n\in\Bbb Z}$ converges weakly to the constant function $0$ on $[0,1]$ in the $L^p$ topology for any $p\in[1,\infty)$. However, it is clear that no subsequence of it can converge to $0$ pointwise at any $x\in[0,1]$ since $0$ is not even in the range of any $f_n$.
BigbearZzz
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1+1. On the other hand, by the Banach-Saks theorem, a subsequence of averages of terms in the sequence does converge pointwise. – Jose27 Oct 23 '23 at 20:23