I have some question on global boundedness of weak solution to Neumann problems.
Let $u\in W^{1,2}(\Omega)$ is a weak solution for Neumannn problem $$ \mathrm{div} (A \nabla u )= \mathrm{div}\, F\quad \text{in } \Omega,\quad A\nabla u\cdot \nu = F \cdot \nu$$ i.e., $u$ satisfies $$ \int_{\Omega} A\nabla u\cdot \nabla \phi\, dx =\int_{\Omega} F \cdot \nabla \phi\, dx $$ for all $\phi \in W^{1,2}(\Omega)$. Here $A(x)=(a^{ij}(x))$ is uniformly elliptic, i.e., $$ \Vert A \Vert_{L^\infty (\Omega)} \leq B, \quad a^{ij}(x) \xi_i \xi_j \geq \delta |\xi|^2$$ for all $x\in \Omega$ and $\xi \in \mathbb{R}^n$.
I want to show that if $F \in L^q(\Omega)$, $q>n$, then $u\in L^\infty(\Omega)$ and $$ \Vert u \Vert_{L^\infty(\Omega)} \leq C \Vert F \Vert_{L^q(\Omega)} $$ for some constant $C>0$.
In the case of Dirichlet problem, we can prove similar conclusion via De Giorgi type argument due to Stampaccia and Moser type argument. It seems that the result is quite classical, but I cannot find exact reference on this result. I tried to prove it via Moser type argument, but I fail to conclude the desired result.
Thanks in advance.
Add : Here is my trial (not rigorous for the check whether I can use Moser's approach) Set $w=u^+$. Note that $$ w^\beta \nabla w = \frac{1}{\beta+1} \nabla(w^{\beta+1})$$ Testing $\phi=w^{\beta}$, we obtain via Young's inequality that $$ \frac{4}{(\beta+1)^2} \int_\Omega |\nabla w^{(\beta+1)/2}|^2 dx =\int_\Omega w^{\beta-1} |\nabla w|^2 dx \leq \int_\Omega w^{\beta-1} |F|^2 dx.$$
Using Sobolev's embedding theorem and Holder's inequality we get \begin{align*} \Vert w^{(\beta+1)/2} \Vert_{L^{2^*}(\Omega)} &\leq C\Vert w^{(\beta+1)/2} \Vert_{L^2(\Omega)}+C\Vert \nabla(w^{(\beta+1)/2}) \Vert_{L^2(\Omega)}\\ &\leq C\Vert w\Vert_{L^{(\beta+1)}}^{(\beta+1)/2} + C(\beta+1) \Vert{w^{\beta-1}}\Vert_{L^{p/(p-2)}(\Omega)}^{1/2} \Vert F \Vert_{L^p(\Omega)}^{1/2} \end{align*} I'm not sure to conclude the iteration due to the term $\Vert w\Vert_{L^{(\beta+1)}}^{(\beta+1)/2}$.
