Let $\Omega$ be a bounded open set in $\mathbb{R}^N$ with smooth boundary conditions. Let $A_n \in C^\infty(\bar{\Omega})$ be a $N$ by $N$ symmetric matrix with $v^TA_nv \geq \alpha|v|^2$ for all $v \in \mathbb{R}^N$. Define $T_n u = -\operatorname{div}(A_n \nabla u)$ for $u \in H^1(\Omega)$ and let $\phi_n \in L^2(\Omega)$.
If $A_n \to A$ in $L^\infty$, $\phi_n \to \phi$ in $L^2$ and $u_n \in H^1_0$ are weak solutions to $T_n u_n = \phi_n$. Then $u_n$ is Cauchy in $H^1$.
I am not sure how to show this: I can not upper bound $\| u_n - u_m \|_{H^1}$ in a meaningful way so I can use $A_n \to A$ and $\phi_n \to \phi$. Any suggestions would be helpful.
Testing the weak formulation of $u_n$ with $u_n$ shows that $(u_n)$ is bounded in $H^1$.
Testing the weak formulations for $u_n$ and $u_m$ with $u_n-u_m$ und subtracting them yields $$ \int_\Omega (\phi_n-\phi_m)(u_n-u_m) = \int_\Omega \nabla(u_n-u_m) (A_n\nabla u_n - A_m\nabla u_m)\\ = \int_\Omega \nabla(u_n-u_m) (A_n\nabla (u_n-u_m) - (A_m-A_n)\nabla u_m) $$ which implies $$ \alpha \| \nabla(u_n-u_m)\|_{L^2}^2 \le \int_\Omega (\phi_n-\phi_m)(u_n-u_m) + \int_\Omega \nabla(u_n-u_m)(A_m-A_n)\nabla u_m. $$ Using Cauchy-Schwarz, Poincare inequality, and boundedness of $(u_n)$ proves the claim.