Crawling ink spots lemma and its application

By | June 13, 2018

The crawling ink spots lemma is useful to exploit a level set argument. The argument is orginally due to Safonov and Krylov (1980).

Lemma 1. Let 0<\delta<1 and E\subset F\subset B_{1} be two open sets satisfying

  1. for all x\in F, there is a ball B\subset B_{1} such that x\in B and

        \[ \left|E\cap B\right|\le\left(1-\delta\right)\left|B\right|. \]

  2. for all B\subset B_{1} such that

        \[ \left|E\cap B\right|>\left(1-\delta\right)\left|B\right|, \]

    we have B\subset F.

Then \left|E\right|\le\left(1-c\delta\right)\left|F\right| for some constant c depending only on the dimension.

Proof. Let x\in F. Then by (i), we have a ball B^{0}\subset B_{1} such that x\in B^{0} and

    \[ \left|E\cap B^{0}\right|\le\left(1-\delta\right)\left|B^{0}\right|. \]

Since F is open, choose a maximal ball B^{x} such that x\in B^{x}, B^{x}\subset B^{0} and B^{x}\subset F. If B^{x}=B^{0}, then \left|B^{x}\cap E\right|\le\left(1-\delta\right)\left|B^{x}\right| by (i).

If B^{x}\subsetneq B^{0}, there is a ball \tilde{B} such that B^{x}\subset\tilde{B}\subset B^{0}. By maximality of B^{x}, \tilde{B}\not\subset F. Then

    \[ \left|E\cap\tilde{B}\right|\le\left(1-\delta\right)\left|\tilde{B}\right| \]

from the contrapositive of second condition. Now choose a decreasing sequence of balls converging to B^{x}. Then by continuity of measure, we have

    \[ \left|E\cap B^{x}\right|\le\left(1-\delta\right)\left|B^{x}\right|. \]

So we construct a cover \left\{ B^{x}\right\} of the set F so that for all x\in F,

  • x\in B^{x};
  • B^{x}\subset F;
  • \left|B^{x}\cap E\right|\le\left(1-\delta\right)\left|B^{x}\right|.

Now we apply the Vitali covering lemma to get a countable subcollection of balls B_{j} such that F\subset\bigcup_{j=1}^{\infty}5B_{j}. Here 5B_{j} denotes the cocentric ball with radious 5 times of the radius of ball B_{j}. Since B_{j}\subset F and \left|E\cap B_{j}\right|\le\left(1-\delta\right)\left|B_{j}\right|, we have

    \begin{align*} \left|B_{j}\cap\left(F\setminus E\right)\right| & =\left|B_{j}\cap F\right|-\left|B_{j}\cap E\right|\\ & \ge\left|B_{j}\right|-\left(1-\delta\right)\left|B_{j}\right|\\ & =\delta\left|B_{j}\right|. \end{align*}


    \begin{align*} \left|F\setminus E\right| & \ge\sum_{j=1}^{\infty}\left|B_{j}\cap\left(F\setminus E\right)\right|\\ & \ge5^{-n}\delta\sum_{j=1}^{\infty}\left|5B_{j}\right|\\ & \ge5^{-n}\delta\left|F\right|. \end{align*}


    \begin{align*} \left|F\right| & =\left|F\setminus E\right|+\left|E\right|\\ & \ge5^{-n}\delta\left|F\right|+\left|E\right| \end{align*}

and so

    \[ \left|E\right|\le\left(1-5^{-n}\delta\right)\left|F\right|. \]

This completes the proof of Lemma 1.

Application of crawling ink spots lemma

The crawling ink spots lemma helps us to get \Leb{p}-estimates. The application of crawling ink spots lemma can be found at e.g. Dong and Kim (arXiv:1806.02635v1).

Fix 1<p<\infty. Suppose that D^2 u ,f \in \Leb{p}(\mathbb{R}^n). For s>0, define

    \[ \mathcal{A}(s) = \{ x \in \mathbb{R}^n : |D^2 u(x)|>s \} \]


    \[ \mathcal{B}(s) = \{ x \in \mathbb{R}^n : Mf(x)>s \}. \]

Let 0<R<\infty and 0<\gamma<1. Suppose that there exists a constant \kappa>1 such that the following hold: x_0 \in \mathbb{R}^n and s>0, if

    \[ |B_{R/4}(x_0) \cap \mathcal{A}(\kappa s)| \ge \gamma |B_{R/4}(x_0)|, \]

then we have

    \[ B_{R/4} (x_0) \subset \mathcal{B}(s). \]

Then by Lemma 1, we see that

    \[ |\mathcal{A}(ks)| \le (1-c\gamma)|\mathcal{B}(s)|. \]


    \[ \norm{f}{p}^p = \int_0^\infty p \lambda^{p-1} | \{ x \in \mathbb{R}^n : |f(x)|>\lambda \} | d\lambda, \]

we have

    \begin{align*} \norm{D^2 u}{p}^p &= \int_0^\infty p s^{p-1} |A(s) | ds \\ &=p\kappa^p \int_0^\infty s^{p-1} |A(\kappa s) | ds \\ &\leq C \int_0^\infty s^{p-1} |\mathcal{B}(s)| ds\\ &=C \norm{Mf}{p}^p \\ &\leq C \norm{f}{p}^p. \end{align*}

So we obtain the following estimates:

    \[ \norm{D^2 u}{p} \leq C \norm{f}{p} \]

for some constant C>0.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.