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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 $S$

  1. What is the $S$ above? They have not told about $S$

  2. 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?

  3. 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?

  4. 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

  1. 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 $A\ge 2^N$?

  2. 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.

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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.

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

  2. 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.)

  3. 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.

  4. 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)

  1. 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.

  2. 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, $A=\max(A^N+1, 2^{p+1}B)$.

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I still can not get the last part. Finally at the end we have $\cdots$. –  user29999 Aug 4 '12 at 1:45
    
When I did I draw for anwer 4. I saw a possible counterexemplo. Using the notation $Q_r(x_0)=\{x \in \mathbb{R}^{2}:|x-x_0|_{\infty}<\dfrac{r}{2}\}$ take $Q=Q_8((0,0)),Q_k=Q_1(1/2,1/2),\overline{Q}_k=Q_2(1,1)$ and $Q´=Q_1(-1/4,-1/4)$. Then $2\overline{Q}_k=Q_4(1,1)$ and $2Q´=Q_2(-1/4,-1/4)$.Then $Q´$ isn't a small cube, for example, $(1/8,-5/8) \in Q`\backslash 2\overline{Q}_k$ and $2Q´$ isn't a large cube, for example, $\{(0,0),(1,1),(1/2,1/2)\} \in \overline{Q}_k \backslash 2Q´$.Am I wrong? –  user29999 Aug 8 '12 at 14:34
    
By the way I left a bounty of 200 in math.stackexchange.com/questions/179156/… about a theorem $A$ of this article. –  user29999 Aug 8 '12 at 14:35
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