Corollary 7.11.2925E0E9371996

Proof. By the moment estimate (7.11.4) with $p=2$, the integrals $\int _ B\varphi _ j^2\, \mathrm d\nu $ are bounded uniformly in $j$. Since the $\varphi _ j$ are nonpositive, so is $u$, by the submean inequality. Fix $T>0$. By lemma 7.11.9, the truncations $\max \{ \varphi _ j,-T\} $ converge to $\max \{ u,-T\} $ in $L^1(\nu )$, and so do their squares, since squaring is Lipschitz on $[-T,0]$. Hence $\int _ B\max \{ u,-T\} ^2\, \mathrm d\nu \leq \sup _ j\int _ B\varphi _ j^2\, \mathrm d\nu $, and letting $T\to \infty $ shows that $u\in L^2(\nu )$ with the same bound. For a function $v\leq 0$ we have $0\leq \max \{ v,-T\} -v\leq v^2/T$, so the $L^1(\nu )$ distances between $\varphi _ j,u$ and their truncations are at most $T^{-1}$ times a bound independent of $j$. Letting first $j\to \infty $ and then $T\to \infty $ proves that $\varphi _ j\to u$ in $L^1(\nu )$. $\square$