I was reading Theorem 8.15 about Moser Iteration in Gilbarg and Trudinger's monograph. I understand all the steps of the given proof, but I have the following doubts which could not be cleared by a careful reading.
The Authors, as hypotheses for the theorem, require that $f^i\in L^q(\Omega)$, $i=1,\ldots,n$ and $g\in L^{q/2}(\Omega)$ for some $q>n$ but it seems they haven't used these facts anywhere in the proof: is this so and, if not, in which steps are these facts used?
Does the theorem fail for $q\le n$?
Please help me to fully understand this proof.
Here I have uploaded a snapshot of the theorem.
Equation 8.3
\begin{equation} Lu=D_i(a^{ij}(x)D_ju+b^i(x)u)+c^i(x)D_iu+d(x)u \end{equation}.
Equation 8.30
\begin{equation} \int_{\Omega}\left(D_ivA^i-vB\right)dx=(\le,\ge)0 \end{equation}
Equation 8.32
\begin{equation} \bar z=|z|+k,\qquad \bar b=\lambda^{-2}(|b|^2+|c|^2+k^{-2}|f|^2)+\lambda^{-1}(|d|+k^{-1}|g|) \end{equation}
Equation 8.33
\begin{align} p_iA^i(x,z,p) & \ge \frac{\lambda}{2}(|p|^2-2\bar b\bar z^2) \\ | \bar zB(x,z,p) | &\le \frac{\lambda}{2}\left( \epsilon|p|^2+\frac{\bar b}{\epsilon}\bar z^2\right) \end{align}
Any Help Hint will be greatly appreciated
it definitely needs the condition $f^i\in L^q(\Omega)$ and $g\in L^{q/2}(\Omega)$.
During the proof, one needs to choose $\chi=\hat{n}(q-2) / q(\hat{n}-2)>1$ (above equation (8.37)). This is possible if and only if $q>\hat n$.
The theorem in general fails for $q\leq n$. One can get some clue from the $W^{2,p}$ estimates of elliptic equations. Conside a special case, $f=0$ and $Lu=g$ with $u=0$ on the boundary. The $W^{2,p}$ roughly says $$||u||_{W^{2,q/2}}\leq C||g||_{L^{q/2}}$$ Recall the Sobolev embedding theorem, $W^{2,q/2}\in L^\infty$ if $q>n$, while this is not true when $q\leq n$.
For a counterexample, one can just take one element $g\in W^{2,n/2}$ but not in $g\not\in L^\infty(\Omega)$. Then $$\Delta u=\Delta g$$ has a solution $u$ while (8.34) can not be true.