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I have two doubts in the proof of the theorem below. If you want the detaIls can be found here.

Theorem A Let $q$ a given real number $q>p$. Let $u \in W^{1,p}$ be a solution to (E2). Assume that (H1) holds. Then there exists $\varepsilon_0>0, \varepsilon_0(q)$, such that if (H2) holds for some $0 < \varepsilon < \varepsilon_0$, then $u \in W^{1,q}$.

Given $q>p$ we study when $g\equiv M(| \nabla u|) \in L^{q/p}$ by standards arguments of measure theory, $g \in L^{q/p}$ if

\begin{equation} \sum_{k=1}^{\infty} A^{kq/p}w_g(A^{k}\lambda_0)< \infty, \tag{1} \end{equation} where $w_g$ is the distribution function of g.

Now take $A,\delta$ and the corresponding $\varepsilon>0$ given by lemma 1.3; by lemma 1.2 we obtain that $w_g(A\lambda_0) \le \delta w_g(\lambda_0) $ and by recurrence $w_g(A^k\lambda_0) \le \delta^k w_g( \lambda_0)$.

  1. (My Ask)Then the candidate to $\varepsilon_0(q)$ is $\varepsilon=\varepsilon(\delta)$ given by lemma 1.3. But what fixed $0<\delta<1$? To apply the lemma 1.2 we need that i) $w_g(A \lambda_0)< \delta|Q´|$ where $w_g(t) = |\{x \in Q':|g(x)|>t\}|$ and $Q'$ is such that we hope obtain $g \in L^{q/p}$ in $Q'$ and ii) below, but by lemma 1.3 ii)holds only to dyadic cubes $Q_k \subset \overline{Q_k} \subset \dfrac{1}{4}Q$ from $Q$. To answer this, I think that is because we want to obtain local results, then we can consider $Q'$ such that the dyadic cubes $Q_k \subset \overline{Q_k} \subset \dfrac{1}{4}Q$ be also dyadic cubes from $Q'$. Am I right here?.

    Then by (1) implies that $$ \sum_{k=1}^{\infty} A^{kq/p}w_g(A^{k}\lambda_0) \le w_g(\lambda_0)\sum_{k=1}^{\infty} A^{kq/p} \delta^k.$$ We need that $\delta A^{q/p}<1$. If $M(|\nabla u|^p)\in L^{q/p}$ a fortiori $\nabla u \in L^q$.

    2.(Ask) We need $\delta <<1$ to obtain $\delta A^{q/p}<1$, but $\delta$ satisfies $ w_g(A\lambda_0) < \delta |Q´|$. Is not clear for me that we can choose $Q',\lambda_0$ that we can find $\delta$ such that $\delta A^{q/p}<1$ and $w_g(A\lambda_0) < \delta |Q´|$.

Lemma 1.2: Let $Q$ be a bounded cube in $\mathbb{R}^{N}$. Assume that $A$ and $B$ are measurable sets, $A \subset B \subset Q$, and that there exists a $\delta>0$ such that

i) $|A| < \delta|Q|$ and

ii) for each $Q_k$ dyadic cube obtained from $Q$ such that $|A \cap Q_k|> \delta_k$, its predecessor $\tilde{Q}_k \subset B$.

Then $|A|< \delta|B|$.

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 an $\varepsilon= \varepsilon(\delta)>0$ 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 $Q$.

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1 Answer 1

Once $Q'$ is chosen in order to satisfy the condition $Q_k\subset \widetilde Q_k\subset \frac 14Q$, we can choose $\lambda_0$ in order to make $\omega_g(A\lambda_0)$ small enough, say than $\frac 12|Q'|A^{-q/p}$. Indeed, we have $\lim_{n\to +\infty}w_g(An)=0$, using the fact that the sequence $Q'\cap \{x,|g(x)|\geq n\}=:S_n$ is decreasing and consists of set of bounded measure. Then we pick $\delta \in (\frac 12A^{-q/p}, A^{-q/p})$.

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I thought the following.I introduced $Q$ in the problem because, I dont know how to apply the lemma 1.2 which has not restriction on dyadic cubes( like $Q_k \subset 1/4 Q$ ). Then I think $Q$ as subset of $ Q$ such that the dyadic cubes from $Q$ are also dyadic cubes from $Q$.For example $Q´= \dfrac{1}{4}Q$. I am not sure if can I to do this. Take $\delta$ such that $\delta A^{q/p}<1$ and choose $\lambda_0 >> 1$ such that $w_g(A\lambda_0) < \delta |Q´|$. We can to do this because $w_g(t)\le \dfrac{C}{t}\|f\|_{L^{1}(Q`)}$. Finally, $\varepsilon(q)=\varepsilon(\delta)$. Do you agree with me? –  user29999 Aug 14 '12 at 19:27
    
I did not understand very well your answer. Maybe you can write more details. In any way, I thank you. If you agree with me, Ok. –  user29999 Aug 14 '12 at 19:28
    
What would we have the last inequality in your comment? We have $\lim_{n\to +\infty}w_g(An)=0$, using the fact that the sequence $Q'\cap \{x,|g(x)|\geq n\}=:S_n$ is decreasing and consists of set of bounded measure. –  Davide Giraudo Aug 14 '12 at 19:33
    
What's the problem? What do you mean? –  user29999 Aug 14 '12 at 19:54
    
How do you know the inequality $w_g(t)\leq \frac Ct\lVert f\rVert_{L^1(Q)}$? –  Davide Giraudo Aug 14 '12 at 20:05

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