Help in a proof for a lemma 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.
 A: I answer questions 1-4 below. It's possible that after reading this you will be able to sort out the page 6 part. If not, leave a comment and I'll be happy to continue. 


*

*I think $S$ should be $Q$. The probably worked with "square $S$" in an earlier version of the paper, and then changed to "cube $Q$". 

*Please don't introduce extra confusion by using $A$ in two ways in the same sentence. :) The assumption $Q_k\subset \overline{Q}_k\subset \frac{1}{4} Q$ says that: we have some cube $\overline{Q}_k$ contained in $\frac{1}{4} Q$, and $Q_k$ is one of its dyadic children. The lemma is about the values of the maximal function in two cubes: a parent and a child. The subscript $k$ is really unnecessary, they use it just to implicitly tell the reader that these are dyadic cubes. (I disapprove of this notation.)

*Yes, they sometimes drop the (unnecessary) subscript $k$. Read $\overline{Q}$ as $\overline{Q}_k$. Note that $\tilde Q=4\overline{Q}_k$ is contained in $Q$ by assumption of the lemma. 

*Recall that the proof is by contradiction; it begins by assuming that the maximal function is $\le\lambda$ somewhere in $\overline{Q}_k$. Divide the cubes containing $x$ into three groups: 


*

*"large" cubes: those that contains $\overline{Q}_k$. On them the 
the average of $|\nabla u|^p$ is  $\le\lambda$ by the above assumption.

*"small" cubes: those that are contained in $2\overline{Q}_k$. On them the average is at most $M^*(|\nabla u|^p)$.

*other cubes. If $x\in Q_k$ is contained in some cube $Q'$ which is not "small", then $2Q'$ is "large": draw a picture of this. Since the average over $2Q'$ is  $\le\lambda$, it follows that  the average over $Q'$ is  $\le 2^N\lambda$. (The authors write $Q$ in the definition of $M^*$, forgetting that $Q$ already has a meaning: it was introduced in the statement of lemma. I'm writing $Q'$ instead.)



(Page 6 questions)


*

*We want to prove that $M^*(|\nabla u_h|^p)\le \frac{A \lambda}{2^{p+1}}$ on $Q_k$. Since $M^*$ takes averages only over the cubes contained in $2\overline{Q}_k$, it suffices to show that $\|\nabla u_h\|^p\le \frac{A \lambda}{2^{p+1}}$ on $2\overline{Q}_k$. Applying (1.1) with $2\overline{Q}_k$ in place of $Q$, we get an upper bound in terms of the average over $4\overline{Q}_k$. Recall that $4\overline{Q}_k$ is also denoted by $\tilde Q$, and the average over it was bounded by $\lambda$ on the previous page. This yields $\|\nabla u_h\|^p\le \frac{B \lambda}{2}\le \frac{A \lambda}{2^{p+1}}$ on $2\overline{Q}_k$, as required. We do not need $A\ge 2^N$ here; it will be used to answer your next question.

*From the last line on page 5 we get the following: if $\mu$  is a number such that $\mu>2^N\lambda$, then $\{x \in Q_k : M(| \nabla u|^{p}) \ge \mu \} \subset \{x \in Q_k : M^{*}(| \nabla u|^{p}) \ge \mu\}$. (In fact, the sets are the same but we don't need this.) Now we want to apply this with $\mu = A\lambda$. Apparently, the authors missed the fact that they need strict inequality $A\lambda>2^N\lambda$ for the logic to work. So, you should  fix this by replacing the definition of $A$ with, for example, $A=\max(A^N+1, 2^{p+1}B)$.
