Weak convergence and domination implies strong convergence in $L^1$

Question

Take a sequence $\{u_n\} \subset L^1(\Omega)$, where $\Omega$ is an open subset in $\mathbb{R}^d$. Assume that $u_n$ converges weakly to $u \in L^1$ and that $|u_n|(x) \leq |u|(x)$ for almost every $x \in \Omega$. I want to prove that $u_n$ strongly converges to $u$. This shouldn't be that hard, but I can't solve it. Observe that you can easily obtain the convergence of the norms, but that's not sufficient in $L^1$. I'm pretty sure some clever application of Fatou is involved... any idea?

Answer 1

On the set $\Omega^+ := \{x: \ u(x) \ge0\}$, we have $u-u_n \ge0$, and by weak convergence $$ \int_{\Omega^+} |u-u_n| \ dx =\int_{\Omega^+} u-u_n \ dx \to 0. $$ Similarly, on $\Omega^- := \{x: \ u(x) <0\}$, we have $u_n-u \ge0$, so that $$ \int_{\Omega^-} |u-u_n| \ dx = \int_{\Omega^-} u_n-u \ dx \to 0. $$ This implies $\|u-u_n\|_{L^1(\Omega)} \to 0$. A more general result can be found in

A. Visintin (1984) Strong convergence results related to strict convexity, Communications in Partial Differential Equations, 9:5, 439-466, http://dx.doi.org/10.1080/03605308408820337