For the problem $latex Lu:=div(A\nabla u)=div(g)$, we will show $latex L: W_{0}^{1,p}\to W^{-1,p} $ (cf. Meyers “An Lp-estimate for the gradient of solutions of second order elliptic divergence equations “). Assume that we have coercivity: $latex \inf_{\phi\in L^{1/q’}}\sup_{\psi\in L^{1/q}}|B_{A}(\psi,\phi) |\geq \frac{1}{K}> 0$.
- Suppose that $latex g\in L^{q’}$ and consider sequence $latex g_{k}\in L^{2}$ such that $latex |g_{k}-g|_{L^{q’}}\to 0$.
- For $latex div(A\nabla u_{k})=div(g_{k})$ the solutions $latex u_{k}\in W^{1,2}_{0}$. From the coercivity assumption we get: $latex | \nabla u_k |_{q’}\leq K| g_k |_{q’}$.
- Therefore, there exists $latex u\in W^{1,q’}$ s.t. $latex \nabla u_{k}\to \nabla u$ weakly in $latex L^{q’}$.
- Therefore, u solves the original problem $latex B_{A}(\psi, u)-\int_{\Omega} (\nabla \psi, g)dx=B_{A}(\psi, u-u_{k})+\int_{\Omega} (\nabla \psi, g_{k}-g)dx\to 0$.
- and satisfies the estimate $latex |\nabla u |_{q’}\leq K |g |_{q’}$.
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