$u\in W^{1,1}_0\cap L^\infty,\Delta u\in L^1$ imply $u\in W^{1,2}$ Assume that $u\in W^{1,1}(D^2)$ has zero trace on the boundary. Here $D^2$ denotes the unit disk in $\mathbb{R}^2$. Assume also that $u\in L^\infty(D^2)$ and $\Delta u\in L^1(D^2)$ in the distributional sense (i.e. there exists $v\in L^1(D^2)$ such that $\int_{D^2} u\Delta\phi=\int_{D^2} v\phi$ for any $\phi\in C^\infty_c(D^2)$). Show that $u\in W^{1,2}(D^2)$.
Heuristically, we would like to say that $\int_{D^2}|\nabla u|^2=-\int_{D^2}u\Delta u<+\infty$, but it is not easy to justify this equality.
Here is a failed attempt (which at least shows that the above equality holds a posteriori): by replacing $u$ with $u(r^{-1}\cdot)$ for an arbitrary $r<1$ (extending the rescaled function by zero outside $D^2_r$) and mollifying, we can build a sequence $u_n\in C^\infty_c(D^2)$ such that $\|u_n\|_\infty\le\|u\|_\infty$ and $u_n\to u$ in $W^{1,1}(D^2)$. Then
$$ -\int_{D^2}u\Delta u=-\lim_{n\to\infty}\int_{D^2}u_n\Delta u
=-\lim_{n\to\infty}\int_{D^2}(\Delta u_n)u=\lim_{n\to\infty}\int_{D^2}\nabla u_n\cdot\nabla u $$
but now I cannot conclude using Fatou's lemma since the integrand $\nabla u_n\cdot\nabla u$ could take negative values.
Any ideas for an alternative approach?
 A: You might want to check Lemma 3.3 in this paper: http://dx.doi.org/10.1051/cocv/2013084 (Preprint available somewhere else) and the references they cite.
Let me sketch the idea: We approximate $\Delta u$ be a sequence $(w_k)$ in $L^2(D)$ such that $w_k \to \Delta u$ in $L^1(D)$. Let $u_k\in H^1_0(\Omega)$ denote the weak solution of $-\Delta u_k = w_k$ in $D$ with homogeneous Dirichlet boundary conditions. These solutions are uniformly bounded in $W^{1,p}(D)$ for all $p<2$.
By compact embeddings, it follows $u_k \to u$ in $L^q(D)$ for all $1\le q<\infty$.
Set $M:=\|u\|_{L^\infty(\Omega)}$ and $u_k^M(x):=\max(-M,\min(u_k(x),M))$. Then $u_k^M \to u^M=u$ in $L^q(D)$ as well.
Testing the weak formulation with $u_k^M$ yields
$$
\int_D |\nabla u_k^M|^2=\int_D u_k^M w_k \le M \|w_k\|_{L^1(D)}.
$$
The right-hand side is uniformly bounded with respect to $k$, hence $(u_k^M)$ is bounded in $H^1_0(\Omega)$ and we can extract a weakly converging sequence. Since $u_k^M$ converges strongly to $u$ in $L^q(D)$, that weak limit is equal to $u$, and it follows $u\in H^1_0(D)$.
