Let $\Omega$ be a bounded domain with smooth boundary in $\mathbb{R}^n$. Fix $u_0 \in H^1(\Omega) - \{ 0\}$.
Define
$$ J(u):= \int_{\Omega} \langle A(x) \nabla u(x), \nabla u(x)\rangle dx,$$ $$u \in K:=\{ v \in H^1(\Omega); v-u_0 \in H^{1}_{0}(\Omega)\}$$
where $A(x) = a_{ij}(x), i,j=1,...n , x \in \Omega$ is a matrix with smooth coeficients and there are constants $0<\lambda_1 <\lambda_2 < +\infty$ such that
$$\lambda_1 |\xi|^2 \leq \langle A(x) \xi,\xi\rangle \leq \lambda_2 |\xi|^2 .$$
Let $(u_n) \in K$ and $u \in K $ a sequence such that $u_n \rightarrow u$ weakly in $H^1(\Omega)$, $u_n \rightarrow u $ in $L^{2}(\Omega)$ and $u_n \rightarrow u$ a.e in $\Omega.$
I am reading a paper and the author says that
$$ \int_{\Omega} \langle A(x) \nabla u(x) , \nabla u(x) \rangle dx \leq \ \ \underline{\lim} \int_{\Omega} \langle A(x) \nabla u_n(x) , \nabla u_n(x) \rangle dx.$$
I have no idea of how to prove this. Someone could help me to prove this?
Thanks in advance
It is well known that for functionals of the form $J(u) = \int_{\Omega}f(x,\nabla u)\,dx$ a sufficient condition for sequential lower semicontinuity is that $\xi \mapsto f(x,\xi)$ is convex. Proofs of this result are a little bit involved, for a reference look at Direct Methods in the Calculus of Variation by B. Dacorogna or Modern Methods in the Calculus of Variation: $L^p spaces$ by I. Fonseca and G. Leoni.
You can apply the result I mentioned above since $f(\xi) = \langle A\xi,\xi\rangle$ is convex if and only if $f(\xi) \ge 0$. This is obviously satisfied thanks to the ellipticity of the matrix $A$.