The lemma below has a small proof in the page 5. But I don't understand the details. I would like that someone help me. the details can be found here.
Lemma 1.3 Assume (H1) holds. Let $u \in W^{1,p}$ be a solutin to (E2) in $Q$. Denote by $A = \max(2^{N},2^{p+1}B\}$ with $B$ as in (H1). Then for $0<\delta<1$ fixed, there exist ad $\varepsilon= \varepsilon(\delta)>0$ such that if hjypothesis (H2) holds for such $\varepsilon$, and $Q_k \subset \overline{Q_k} \subset \dfrac{1}{4}Q$ satisfies \begin{equation} |Q_k \cap \{x :M(\nabla u|^{p}) < A \lambda| > \delta |Q_k|, \end{equation} the predecessor $\overline{Q}_k$ satisfies $\overline{Q}_k \subset \{x:M(|\nabla u|^{p}) > A \lambda\}.$ Remark: $A$ does not depend on $S$
What is the $S$ above? They have not told about $S$
The statement the predecessor of $\overline{Q}_k$ suggests that $Q_k$ is a Caldrón-Zygmund covering of a set $A \subset Q$ such that $|A| < \delta |Q|$. Maybe of $\{x:M(|\nabla u|^{p} > A \lambda\}$. Is this true?
What is $\overline{Q}$ in the solution of the corresponding problem (AP) $u_h$. Is it a typo? Was $\overline{Q} = \overline{Q}_k$ such that there exists $x \in \overline{Q}_k$ such that \begin{equation} \dfrac{1}{|Q|} \int_{Q} | \nabla u(y)|^{p}dy \le \lambda \quad \mbox{for all cubes} \quad Q \ni x. \end{equation} like above in the proof?
Consider the maximal operator \begin{equation} M^{*}(|\nabla u|^{p}) = \sup_{x \in Q, Q \subset 2 \overline{Q}_k} \dfrac{1}{|Q|} \int_{Q} | \nabla u(y)|^{p}dy ; \end{equation} then for $x \in Q_k, M(|\nabla u(x)|^{p}\le \max\{M^{*}(|\nabla u(x)|^{p}),2^N \lambda\}$. Why is the reason for this inequality?
In the page 5 we have
Since $A=\max\{2^N,2^{p+1}B\}$ a direct computation suggests that \begin{equation} |\{x \in Q_k : M^{*}| \nabla u|^{p})\}|>\dfrac{A \lambda}{2^{p+1}}. \end{equation} To explain this fact, I see the following. We have that $\|\nabla u\|^{p}_{L^{\infty}(\overline{Q}_k})\le \dfrac{\lambda B}{2}$. Then if $Q \subset \overline{Q}_k$ we have $\dfrac{1}{|Q|}\int_{Q} | \nabla u_h(y)|^{p}\dfrac{\lambda B}{2}$, and since $\dfrac{\lambda A}{2^{p+1}} \ge B \lambda$ this suggests that $M^{*}(| \nabla u|^{p}(x) \le \dfrac{A \lambda}{2^{p+1}}$. But if $Q \subset 2 \overline{Q}_k$ but $Q \not\subset \overline{Q}_k$ I don't what should I do. Where can I use that fact that $A\ge 2^N$?
Finally at the end we have \begin{equation} | \{x \in Q_k : M^{*}(| \nabla u|^{p}) \ge A \lambda\}| < \delta |Q_k| \end{equation} and we reach a contradiction. Can we prove \begin{equation} | \{x \in Q_k : M(| \nabla u|^{p}) \ge A \lambda\}| \le | \{x \in Q_k : M^{*}(| \nabla u|^{p}) \ge A \lambda\}| \end{equation} to obtain the contradiction?
Sorry if my English is bad.