Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer
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… about a theorem $A$ of this article. – user29999 Aug 8 '12 at 14:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.