# Prove a sequence of Weak Solution is Cauchy

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.