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Let $\Omega$ is a bounded open set in $\mathbb R^n$ and $\{u_n\}$ be a bounded sequence with $||u_n||_2=1$ in the Sobolev space $H^1_0(\Omega)$. Suppose $A(x)$ be a non-constant $n\times n$ symmetric, uniformly bounded and uniformly positive definite matrix.
My question is that how to prove the following one $$\int_{\Omega}A\nabla u_n\cdot\nabla u_n\to\int_{\Omega}A\nabla u\cdot\nabla u $$ where $u$ is the weak limit of $u_n$ in $H^1_0(\Omega)$.
This should be false in general.
For example, take $A = \operatorname{Id}$. Then $[u,v] = \int_\Omega \nabla u \cdot \nabla v$ is an inner product on $H_0^1(\Omega)$ that induces a norm equivalent to the usual norm (by the Poincare inequality). Let $\|\cdot\|$ be the norm induced by $[\cdot,\cdot]$.
Since the norms are equivalent, without loss of generality, we may assume $u_n \to u$ weakly for the $\|\cdot\|$-topology. Then you want to show that $\|u_n\|^2 \to \|u\|^2$.
This can fail. In any infinite dimensional Hilbert space there is a sequence $u_n$ such that $\|u_n\| = 1$ and $u_n$ converges weakly to $0$ (See here). Such a sequence is then a counterexample to your claim.
Edit: I see that you've now edited to add extra conditions to try to exclude this example. However, since $A$ is symmetric, uniformly bounded and uniformly positive definite we still have that $[u,v] = \int_\Omega A\nabla u \cdot \nabla v$ is an inner product inducing a norm equivalent to the usual one and the exact same argument works. The point of taking $A = \operatorname{Id}$ is it simplifies notation for a simple counterexample. It is not in anyway crucial to the construction.
