Sliding possibility of the Julia sets

Hiromichi Nakayama and Takuya Takahashi

A. Sannami constructed an example of the $C^{\infty}$ -Cantor set embedded in the real line $\mathbf{R}$ whose difference set has a positive measure in [3], which was an answer of the question given by J. Palis ([2]). In this paper, we generalize the definition of the difference sets for sets of the two dimensional Euclidean space $\mathbf{R}^{2}$ and estimate the measure of the difference sets of the Julia sets.

Here the difference set for two subsets $X$ and $Y$ of $\mathbf{R}^{2}$ is defined by

X-Y=\{x-y\in\mathbf{R}^{2}\,;\,x\in X,y\in Y\}.

For a vector $\mu$ of $X-X$ , there are two points $x$ and $y$ of $X$ satisfying $x=y+\mu$ . Thus the slided set $X+\mu$ intersects $X$ itself. In this point of view, the measure of $X-X$ is closely related to the problem whether the set $X$ can slide easily.

Let ${\cal C}$ be a Cantor set in $\mathbf{R}$ whose difference set has a positive measure ([3]). Then we have $m(({\cal C}\times{\cal C})-({\cal C}\times{\cal C}))=m(({\cal C}-{\cal C})\times({\cal C}-{\cal C}))=m({\cal C}-{\cal C})^{2}>0$ for the Lebesgue measure $m$ . Thus the measure of the $2$ -dimensional difference set is not always zero. In this paper, we consider the condition when the measures of the difference sets of the Julia sets vanish.

Let $c$ be an element of $\mathbf{C}$ . We define the quadratic map $Q_{c}:\mathbf{C}\to\mathbf{C}$ by $Q_{c}(z)=z^{2}+c$ . Let $D=\{z\,;\,|z|\leq|c|\}$ . The filled Julia set is the set of the bounded orbits of $Q_{c}$ , denoted by $K_{c}$ , and the Julia set $J_{c}$ is the boundary of $K_{c}$ . In the following, we assume that $|c|>2$ . Then it was shown that $K_{c}$ is a Cantor set and $J_{c}$ coincides with $K_{c}$ ([1]). Furthermore, $J_{c}=\bigcap_{n\geq 0}{Q_{c}}^{-n}(D)$ .

Theorem 1.

If $|c|>3+\sqrt{3}$ , then the Lebesgue measure $m$ of the difference set of the Julia set $J_{c}$ vanishes $($ i. e. $m(J_{c}-J_{c})=0)$ .

The authors thank to S. Matsumoto who gave the attention on the difference sets to the author.

1 Preparation for the Julia sets

Let us define $F_{+}:\mathbf{C}\to\mathbf{C}$ by $F_{+}(re^{i\theta})=\sqrt{r}e^{\frac{i\theta}{2}}$ for $0\leq\theta<2\pi$ and $F_{-}:\mathbf{C}\to\mathbf{C}$ by $F_{-}(re^{i\theta})=-\sqrt{r}e^{\frac{i\theta}{2}}$ for $0\leq\theta<2\pi$ . For the map $L(z)=z-c$ , the composition $F_{+}\cdot L$ is denoted by $G_{0}$ and the composition $F_{-}\cdot L$ is denoted by $G_{1}$ . Then $Q_{c}\cdot G_{0}=\operatorname{id}$ and $Q_{c}\cdot G_{1}=\operatorname{id}$ . Thus $G_{0}$ and $G_{1}$ are the partial inverse maps of $Q_{c}$ .

When $|c|>2$ , $G_{0}(D)\subset D$ and $G_{1}(D)\subset D$ . Let $I_{0}=G_{0}(D)$ and $I_{1}=G_{1}(D)$ . Then $I_{0}\cup I_{1}$ is the set contained in the well-known $8$ -figured set. Let $\{s_{0},s_{1},\cdots,s_{n}\}$ be a sequence consisting of the numbers $0$ and $1$ . Let $I{s_{0}s_{1}\cdots s_{n}}=\{z\in D\,;\,z\in I{s_{0}},Q_{c}(z)\in I{s_{1}},\cdots,{Q_{c}}^{n}(z)\in I{s_{n}}\}$ . Then $I{s_{0}s_{1}\cdots s_{n}}=G_{s_{0}}\cdot G_{s_{1}}\cdot\cdots\cdot G_{s_{n}}(D)$ and ${Q_{c}}^{-n}(D)=\bigcup_{\{s_{0},s_{1},\cdots,s_{n}\}}I{s_{0}s_{1}\cdots s_{n}}$ . The set $\bigcap_{n\geq 0}I{s_{0}\dots s_{n}}$ is called the component of the Julia set $J_{c}$ . Then $J_{c}$ is the union of the components. The diameter of $I{s_{0}s_{1}\cdots s_{n}}$ is defined by

\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}=\max\{|z-w|\,;\,z,w\in I{s_{0}s_{1}\cdots s_{n}}\}.

We define the sequences $\{R_{n}\}_{n=1,2,\cdots}$ and $\{r_{n}\}_{n=1,2,\cdots}$ by $R_{n+1}=\sqrt{|c|+R_{n}}$ , $R_{1}=\sqrt{2|c|}$ and $r_{n+1}=\sqrt{|c|-R_{n}}$ and $r_{1}=0$ . Then $r_{n}\leq|z|\leq R_{n}$ for $z\in{Q_{c}}^{-n}(D)$ . When $|c|>2$ , $\{R_{n}\}$ is monotone decreasing and $\{r_{n}\}$ is monotone increasing by induction.

Lemma 1.

$\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}<\frac{1}{\sqrt{2}r_{n+1}}\operatorname{diam}I{s_{1}\cdots s_{n}}$ for $n\geq 1$ . In particular,

\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}<2^{-\frac{n}{2}}\frac{1}{r_{n+1}r_{n}\cdots r_{2}}\operatorname{diam}I_{0}.

Proof.

Let $z^{\prime}$ and $w^{\prime}$ be points of $I{s_{0}s_{1}\cdots s_{n}}$ such that $|z^{\prime}-w^{\prime}|=\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}$ . By definition, $I{s_{0}s_{1}\cdots s_{n}}=G_{s_{0}}(I{s_{1}\cdots s_{n}})$ , and thus $Q_{c}(I{s_{0}s_{1}\cdots s_{n}})=I{s_{1}\cdots s_{n}}$ . Therefore,

			$\displaystyle\operatorname{diam}I{s_{1}\cdots s_{n}}$
		$\displaystyle\geq$	$\displaystyle\|Q_{c}(z^{\prime})-Q_{c}(w^{\prime})\|$
		$\displaystyle=$	$\displaystyle\|({z^{\prime}}^{2}+c)-({w^{\prime}}^{2}+c)\|$
		$\displaystyle=$	$\displaystyle\|{z^{\prime}}^{2}-{w^{\prime}}^{2}\|$
		$\displaystyle=$	$\displaystyle\|z^{\prime}-w^{\prime}\|\ \|z^{\prime}+w^{\prime}\|.$

Let $z=Q_{c}(z^{\prime})$ and $w=Q_{c}(w^{\prime})$ . Then $z^{\prime}=G_{s_{0}}(z)$ and $w^{\prime}=G_{s_{0}}(w)$ are contained in $I{s_{0}}$ . Thus the angle between $z^{\prime}$ and $w^{\prime}$ is less than $\pi/2$ . Since $\angle(z^{\prime},w^{\prime})<\pi/2$ , $|z^{\prime}|\geq r_{n}$ and $|w^{\prime}|\geq r_{n}$ , we have

|z^{\prime}+w^{\prime}|=\sqrt{|z^{\prime}|^{2}+|w^{\prime}|^{2}+2|z^{\prime}|\,|w^{\prime}|\cos\angle(z^{\prime},w^{\prime})}>\sqrt{|z^{\prime}|^{2}+|w^{\prime}|^{2}}\geq\sqrt{2}r_{n+1}.

Thus

			$\displaystyle\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}$
		$\displaystyle<$	$\displaystyle\frac{1}{\sqrt{2}r_{n+1}}\operatorname{diam}I{s_{1}\dots s_{n}}$
		$\displaystyle<$	$\displaystyle\left(\frac{1}{\sqrt{2}r_{n+1}}\right)\cdots\left(\frac{1}{\sqrt{2}r_{2}}\right)\operatorname{diam}I{s_{n}}$
		$\displaystyle=$	$\displaystyle\left(\frac{1}{\sqrt{2}}\right)^{n}\frac{1}{r_{n+1}r_{n}\cdots r_{2}}\operatorname{diam}I_{0}.$

∎

Let $K_{n}$ denote $2^{-\frac{n}{2}}\frac{1}{r_{n+1}r_{n}\cdots r_{2}}\operatorname{diam}I_{0}$ in Lemma 1. Thus $\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}<K_{n}$ for $n\geq 1$ .

2 Estimate of the measure of the difference sets

Lemma 2.

For any $I{s_{0}s_{1}\cdots,s_{n}}$ , there is a closed disk $D{s_{0}s_{1}\cdots s_{n}}$ containing $I{s_{0}s_{1}\cdots s_{n}}$ such that the radius of $D{s_{0}s_{1}\cdots s_{n}}$ is equal to $\frac{\sqrt{3}}{2}\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}$ .

Proof.

Let $x$ and $y$ be points of $I{s_{0}s_{1}\cdots s_{n}}$ whose distance is equal to $\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}$ . For any point $z$ of $I{s_{0}s_{1}\cdots s_{n}}$ , we have $|z-x|\leq|x-y|$ and $|z-y|\leq|x-y|$ . Thus $z$ is contained in the closed disk with the center $x$ whose radius is $|x-y|$ , and furthermore $z$ is contained in the closed disk with the center $y$ whose radius is $|x-y|$ . The intersection of these closed disks is contained in the closed disk whose center is the middle point of $x$ and $y$ and radius is $\frac{\sqrt{3}}{2}|x-y|$ . Then this disk satisfies the conditions of $D{s_{0}s_{1}\cdots s_{n}}$ . ∎

The above estimate of the radius of $D{s_{0}s_{1}\cdots s_{n}}$ can be made more strict, but it does not improve the condition of the main theorem in the following proof.

Lemma 3.

Let $D_{1}$ and $D_{2}$ be closed disks whose radii are equal to $R$ . Let $O_{1}$ and $O_{2}$ denote the centers of $D_{1}$ and $D_{2}$ respectively. Then $D_{2}-D_{1}$ is the closed disk $\{v\,;\,|v-\overrightarrow{O_{1}O_{2}}|\leq 2R\}$ , and thus the center is $\overrightarrow{O_{1}O_{2}}$ and the radius is $2R$ .

Proof.

Let $A$ denote the middle point of $O_{1}$ and $O_{2}$ . Denote by $r$ the length between $O_{1}$ and $O_{2}$ . Let $\ell_{1}$ denote the straight line passing through $O_{1}$ and $O_{2}$ . Let $O_{3}$ denote the point in $\ell_{1}$ satisfying that the distance between $O_{3}$ and $A$ is equal to $r$ and $O_{3}$ is closer to $O_{2}$ than $O_{1}$ . Denote by $D_{3}$ the closed disk with the center $O_{3}$ whose radius is $2R$ . Then $D_{3}$ and $D_{2}$ are similar, and the ratio is $2$ .

Let $\ell_{2}$ be a straight line passing through $A$ and intersecting $D_{2}$ . Let $P$ denote the point among the intersection points of $\ell_{2}$ and $\partial D_{1}$ which is further from $D_{2}$ (Fig 1). The other intersection point of $\ell_{2}$ and $\partial D_{1}$ is denoted by $S$ if $\ell_{2}\cap\partial D_{1}$ is not a single point. If $\ell_{2}\cap\partial D_{1}$ is a single point, then the point $S$ is given by $P$ . Let $U$ denote the point of $\ell_{2}\cap\partial D_{2}$ which is further from $D_{1}$ . The other point of $\ell_{2}\cap\partial D_{2}$ is denoted by $T$ ( $T=U$ if $\ell_{2}\cap\partial D_{2}$ is a single point). Furthermore, let $W$ denote the intersection point of $\ell_{2}$ and $\partial D_{3}$ further from $D_{1}$ . The other intersection point is denoted by $V$ ( $V=W$ when $\ell_{2}\cap\partial D_{3}$ is a single point).

We take the orientation of $\ell_{2}$ so that the direction from $A$ to $U$ is positive. Then the maximum vector of $D_{2}-D_{1}$ in $\ell_{2}$ with respect to this orientation is $\overrightarrow{PU}$ . By symmetry, $\overrightarrow{PA}=\overrightarrow{AU}$ . Since the triangle $AUO_{2}$ is similar to the triangle $AWO_{3}$ , we have $2\overrightarrow{AU}=\overrightarrow{AW}$ . Thus we obtain $\overrightarrow{PU}=2\overrightarrow{AU}=\overrightarrow{AW}$ . On the other hand, the minimum vector of $D_{2}-D_{1}$ in $\ell_{2}$ with respect to the positive orientation of $\ell_{2}$ is $\overrightarrow{ST}$ , where $\overrightarrow{ST}$ can be in the negative direction with respect to $\ell_{2}$ when $D_{1}\cap D_{2}\neq\emptyset$ . By symmetry, $\overrightarrow{AT}=\overrightarrow{SA}$ , and the triangle $ATO_{2}$ is similar to the triangle $AVO_{3}$ . Thus $\overrightarrow{ST}=2\overrightarrow{AT}=\overrightarrow{AV}$ . As a consequence, we obtain $\ell_{2}\cap(D_{2}-D_{1})=\ell_{2}\cap\{v\,;\,|v-\overrightarrow{O_{1}O_{2}}|\leq 2R\}$ . This is true for any straight line $\ell_{2}$ passing through $A$ and intersecting $D_{2}$ . Thus we conclude that $D_{2}-D_{1}=\{v\,;\,|v-\overrightarrow{O_{1}O_{2}}|\leq 2R\}$ . ∎

Lemma 4.

$m({Q_{c}}^{-n}(D)-{Q_{c}}^{-n}(D))<12\pi 4^{n}{K_{n}}^{2}$

Proof.

We have

			$\displaystyle m({Q_{c}}^{-n}(D)-{Q_{c}}^{-n}(D))$
		$\displaystyle=$	$\displaystyle m(\bigcup_{\{s_{0},s_{1},\cdots,s_{n},{s_{0}}^{\prime},{s_{1}}^{\prime},\cdots,{s_{n}}^{\prime}\}}(I{s_{0}s_{1}\cdots s_{n}}-I{{s_{0}}^{\prime}{s_{1}}^{\prime}\cdots{s_{n}}^{\prime}}))$
		$\displaystyle\leq$	$\displaystyle\sum_{\{s_{0},s_{1},\cdots,s_{n},{s_{0}}^{\prime},{s_{1}}^{\prime},\cdots,{s_{n}}^{\prime}\}}m(I{s_{0}s_{1}\cdots s_{n}}-I{{s_{0}}^{\prime}{s_{1}}^{\prime}\cdots{s_{n}}^{\prime}})$
		$\displaystyle\leq$	$\displaystyle\sum_{\{s_{0},s_{1},\cdots,s_{n},{s_{0}}^{\prime},{s_{1}}^{\prime},\cdots,{s_{n}}^{\prime}\}}m(D{s_{0}s_{1}\cdots s_{n}}-D{{s_{0}}^{\prime}{s_{1}}^{\prime}\cdots{s_{n}}^{\prime}})$

(see Figure 1).

Here $\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}<K_{n}$ by Lemma 1. The radius of $D{s_{0}s_{1}\cdots s_{n}}$ is equal to $\frac{\sqrt{3}}{2}\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}$ by Lemma 2, which is smaller than $\frac{\sqrt{3}}{2}K_{n}$ . Furthermore, the radius of $D{{s_{0}}^{\prime}{s_{1}}^{\prime}\cdots{s_{n}}^{\prime}}$ is equal to $\frac{\sqrt{3}}{2}\operatorname{diam}I{{s_{0}}^{\prime}{s_{1}}^{\prime}\cdots{s_{n}}^{\prime}}$ , which is also smaller than $\frac{\sqrt{3}}{2}K_{n}$ . By Lemma 3,

			$\displaystyle m({Q_{c}}^{-n}(D)-{Q_{c}}^{-n}(D))$
		$\displaystyle<$	$\displaystyle\sum_{\{s_{0},s_{1},\cdots,s_{n},{s_{0}}^{\prime},{s_{1}}^{\prime},\cdots,{s_{n}}^{\prime}\}}\pi(\sqrt{3}K_{n})^{2}$
		$\displaystyle=$	$\displaystyle 2^{2n+2}3\pi{K_{n}}^{2}$
		$\displaystyle=$	$\displaystyle 12\pi 4^{n}{K_{n}}^{2}$

∎

Proof of Theorem 1

When $|c|>3+\sqrt{3}$ , we have

$\displaystyle\|c\|^{2}-6\|c\|+6$	$\displaystyle>$	$\displaystyle 0$
$\displaystyle 1+4\|c\|$	$\displaystyle<$	$\displaystyle(2\|c\|-5)^{2}$
$\displaystyle\sqrt{\frac{2\|c\|-1-\sqrt{1+4\|c\|}}{2}}$	$\displaystyle>$	$\displaystyle\sqrt{2}.$

Therefore, there is a small $\varepsilon>0$ such that $\sqrt{\frac{2|c|-1-\varepsilon-\sqrt{1+4|c|}}{2}}>\sqrt{2}$ . Let $\delta$ $(>0)$ denote $\sqrt{\frac{2|c|-1-\varepsilon-\sqrt{1+4|c|}}{2}}-\sqrt{2}$ .

On the other hand, $\lim_{n\rightarrow\infty}R_{n}=\frac{1+\sqrt{1+4|c|}}{2}$ and $\lim_{n\to\infty}r_{n}=\sqrt{\frac{2|c|-1-\sqrt{1+4|c|}}{2}}$ . Since $r_{n}$ is monotone increasing, there exists a positive integer $N$ such that, if $n>N$ , then $r_{n}\geq\sqrt{\frac{2|c|-1-\varepsilon-\sqrt{1+4|c|}}{2}}(=\sqrt{2}+\delta)$ .

By Lemma 1, $\operatorname{diam}I{s_{0}s_{1}\cdots s_{n}}<K_{n}$ , where $K_{n}=2^{-\frac{n}{2}}\frac{1}{r_{n+1}r_{n}\cdots r_{2}}\operatorname{diam}I_{0}$ . For $n>N$ , we obtain

			$\displaystyle m({Q_{c}}^{-n}(D)-{Q_{c}}^{-n}(D))$
		$\displaystyle<$	$\displaystyle 12\pi 4^{n}{K_{n}}^{2}$
		$\displaystyle=$	$\displaystyle 12\pi 4^{n}\frac{1}{2^{n}}\left(\frac{1}{r_{n+1}r_{n}\cdots r_{N+1}}\right)^{2}\left(\frac{1}{r_{N}r_{N-1}\cdots r_{2}}\right)^{2}\operatorname{diam}^{2}I_{0}$
		$\displaystyle\leq$	$\displaystyle 12\pi 4^{n}\frac{1}{2^{n}}\frac{1}{(\sqrt{2}+\delta)^{2(n+1-N)}}\left(\frac{1}{r_{N}r_{N-1}\cdots r_{2}}\right)^{2}\operatorname{diam}^{2}I_{0}.$

Thus there is a constant $K>0$ such that $m({Q_{c}}^{-n}(D)-{Q_{c}}^{-n}(D))\leq K\left(\frac{\sqrt{2}}{\sqrt{2}+\delta}\right)^{2n}$ for $n>N$ . As a consequence, $m(J_{c}-J_{c})=\lim_{n\to\infty}m({Q_{c}}^{-n}(D)-{Q_{c}}^{-n}(D))=0$ . $\blacksquare$

References

[1] R. Devaney, A first course in chaotic dynamical systems, Addison-Wesley Studies in Nonlinearity. Addison-Wesley Publishing Company, 1992.
[2] J. Palis, Fractional dimension and homoclinic bifurcations, Colloquium-Hokkaido University (October, 1988).
[3] A. Sannami, An example of a regular Cantor set whose difference set is a Cantor set with positive measure, Hokkaido Mathematical Journal 21 (1992) 7–24.

Hiromichi Nakayama
E-mail: [email protected]
Takuya Takahashi
Department of Physics and Mathematics,
College of Science and Engineering,
Aoyama Gakuin University,
5-10-1 Fuchinobe, Sagamihara, Kanagawa, 252-5258, Japan