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I'm working through a section of Ambrosio's Functions of Bounded Variation and Free Discontinuity Problems and have gotten stuck at a step in the proof of the closure theorem for $SBV(\Omega)$ (details of this aren't very important for my question, I think).

We have a sequence $(u_k)_k$ converging in $L^1$ to some $u$, a uniformly integrable sequence $(\nabla u_k)_k$ converging weakly in $L^1$ to some $a$ and a Lipschitz function $\psi \in C^1(\mathbb{R})$. The author claims that we can then use Vitali's dominated convergence theorem to conclude $$ (\psi'(u_k) - \psi'(u)) \nabla u_k\xrightarrow[] {L^1} 0 \,. $$

I've been thinking about this for a couple of days now, but I really can't see how to get there. In particular, I don't know how to conclude that this sequence goes to 0 almost everywhere without having bounds on $\Vert \nabla u_k \Vert$. What am I missing?

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    $\begingroup$ $\|\nabla u_k\|$ is bounded, since it converges weakly in $L^1$. $\endgroup$
    – Giorgio Metafune
    Commented Apr 19, 2022 at 14:33
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    $\begingroup$ Take any book of functional analysis: weak convergence of sequences implies boundedness; it is a consequence of the Banach-Steinhaus theorem. $\endgroup$
    – Giorgio Metafune
    Commented Apr 19, 2022 at 17:41
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    $\begingroup$ Given $\epsilon>0$ choose $\delta>0$ such that $\int_F |\nabla u_k| \leq \epsilon$ for every $k$, whenever $|F| <\delta$. Next use Egorov theorem to get $G \subset \Omega$ such that $u_k \to u$ (up to a subsequence) uniformly in $G$ and $|\Omega \setminus G| <\delta$. Next, split the integral over $\Omega$ as over $\Omega \setminus G$ and $G$. In the first use $\epsilon$ and the bound for $\phi'$, in the second the uniform convergence of $\psi'(u_k)$ to $\psi'(u)$ and the $L^1$ bound on the gradients. $\endgroup$
    – Giorgio Metafune
    Commented Apr 21, 2022 at 8:38
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    $\begingroup$ Yes, it follows from the uniform continuity of $\psi'$. If this last is only supposed to be continuous you can use Egorov from the beginning for both $(u_k)$ and $\psi'(u_k)$ $\endgroup$
    – Giorgio Metafune
    Commented Apr 22, 2022 at 13:05
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    $\begingroup$ Yes true, however you need that $u_k \to u$ in some sense. $\endgroup$
    – Giorgio Metafune
    Commented Apr 25, 2022 at 13:04

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Using the helpful advice of @GiorgiMetafune in the comments, this is the solution I came up with:

Fix $\epsilon > 0$, and denote the Lipschitz constant of $\psi$ by $K$. By uniform integrability of $(\nabla u_k)_k$, there exists $\delta > 0$ such that $$ \sup_{k \geq 1} \int_E \Vert \nabla u_k \Vert \, dx \leq \epsilon / 4K $$ whenever $\lambda(E) \leq \delta$. Using Egorov's theorem, we obtain a measurable subset $G \subseteq \Omega$ such that $\lambda(\Omega \setminus G) \leq \delta$ and $\psi'(u_k) \xrightarrow[] {L^\infty} \psi'(u)$ on $G$. Now we can split the integral $$ \int_\Omega \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx $$ over $G$ and its complement. For the first component, we use the uniform convergence and the boundedness of the $L^1$-norms of $(\nabla u_k)_k$ to obtain $$ \int_G \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx \leq \vert \psi'(u_k) - \psi'(u) \vert_\infty \sup_{k \geq 1} \int_G \Vert \nabla u_k \Vert \, dx \,, $$ and thus we have $\int_G \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx \leq \epsilon / 2$ for $k$ sufficiently large. The second component can be estimated by using the boundedness of $\psi'$ and the fact that $\lambda(\Omega \setminus G) \leq \delta$: $$ \int_G \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx \leq 2K \sup_{k \geq 1} \int_G \Vert \nabla u_k \Vert \, dx \leq \epsilon / 2 $$

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