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Let $(\nu_{j} - \phi)$ be is a bounded sequence in $W^{1,p}_{0}(\Omega)$. By reflexivity, there is a function $u \in W^{1,p}(\Omega)$ such that, up to a subsequence $$ \nu_{j} \rightarrow u \ \mbox{weakly} \ \mbox{in} \ W^{1,p}(\Omega), \nu_{j} \rightarrow u \mbox{in} \ L^{p}(\Omega), \nu_{j} \rightarrow u \ \mbox{a.e in} \ \Omega.$$

I know that from lower semicontiuity of norms we obtain $$ \int_{\Omega} | \nabla u |^{p} dx \le \liminf_{j} \int_{\Omega} | \nabla \nu_{j} |^{p} dx. $$ But I don't undertand it. If I had $\nabla u(x) \rightarrow \nabla \nu_{j}(x)$ I could use Fatou's lemma. I would like to know this in the hope of showing \begin{equation} \int_{\Omega} \dfrac{1}{2} \langle A(x) \nabla u(x), \nabla u(x)\rangle \le \liminf_{j} \int_{\Omega} \dfrac{1}{2} \langle A(x) \nabla u_j(x), \nabla u_j(x)\rangle \end{equation} or \begin{equation} \int_{\{u>0\}} \dfrac{1}{2} \langle A(x) \nabla u(x), \nabla u(x)\rangle \le \liminf_{j} \int_{\{u_j >0\}} \dfrac{1}{2} \langle A(x) \nabla u_j(x), \nabla u_j(x)\rangle \end{equation} where $A$ is a continuous matrix.

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The first ynequality holds because the functional is strong continuous in $L^{p}$ and convex. –  user29999 Jun 6 '12 at 21:54

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