Complex dynamics: iterates of rational function : Julia and fatou set

Question

Prove that attracting fixed points of rational map lies in fatou set and repelling fixed point of rational map lies in the julia set

Answer 1

Could you state the definitions of these sets that you are using? I ask because there are equivalent definitions which say different things.

Assuming the Fatou set $F$ of a rational function $R:\mathbb C \longrightarrow \mathbb C$ is the maximal open set on which the family of iterates $\{R^n : n \in \mathbb N\}$ is equicontinuous we may show that an attracting fixed point $z$ is in $F$ as follows:

Suppose that $|R'(z)|<1$ (definition of an attracting fixed point). Let $\varepsilon>0$ be given. By definition of the derivative, there is $\delta'>0$ so that $$\left|\frac{R(y)-R(z)}{y-z}-R'(z)\right|<\frac{1-|R'(z)|}2 \,\text{ whenever }\, |y-z|<\delta'.$$ For $y\neq z$ we may multiply across the inequality by $|y-z|$ and simplify to obtain $$|R(y)-R(z)-R'(z)(y-z)|<\frac{1-|R'(z)|}2\cdot|y-z| \,\text{ whenever }\, |y-z|<\delta'.$$ So by the triangle equality we have that if $|y-z|<\delta'$, then $$|R(y)-R(z)|<\frac{1+|R'(z)|}2\cdot|y-z|.$$ Now select $\delta=\min\{\varepsilon,\delta'\}$, and write $k=\frac{1+|R'(z)|}2<1$. So if $|y-z|<\delta$ and $n \in \mathbb N$, then $$|R^n(y)-R^n(z)|=|R\circ R^{n-1}(y)-R(z)|<k|R^{n-1}(y)-z|=k|R^{n-1}(y)-R^{n-1}(z)|<k^n\delta<\varepsilon.$$ Therefore the family of functions $\{R^n : n \in \mathbb N\}$ is equicontinuous at $z$, and so $z \in F$.

Hint to show that a repelling fixed point $z$ is in the Julia set $J$ of $R$:

There is an open neighborhood $N$ of $z$ such that for $y \in N$ we have $$|R(y)-R(z)|>\frac{1+|R'(z)|}2\cdot|y-z|.$$