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Based on Iteration of Quadratic Polynomials, Julia Sets:

Q1. It seems like that the behaviour of $f(z) = z^2$ on any 2 of the following 3 sets $\{z||z|=1\}$, $\{z||z|<1\}$ and $\{z||z|>1\}$ is already sufficient to conclude that $\{z||z|=1\}$ is the Julia set of $f$, denoted $J(f)$. Am I wrong?

If $f = z^2$, then $f^{\circ n}(z) = z^{2^n}$. Thus

$$|f^{\circ n}(z)| = |z^{2^n}| = |z|^{2^n}$$

For $|z| < 1$, $$\lim_{n \to \infty} |f^{\circ n}(z)| = 0 \tag{A1}$$

For $|z| > 1$, $$\lim_{n \to \infty} |f^{\circ n}(z)| = \infty \tag{A2}$$

For $|z| = 1$, $$\lim_{n \to \infty} |f^{\circ n}(z)| = 1 \tag{A3}$$

So for example, if we have $A1$ and $A3$, can we skip $A2$ and then already conclude $\{z||z|=1\} = J(f)$?

Q2-3. Based on slides 5 and 6, it seems like author deduces $A1, A2, A3$ and then concludes $J(f)$ based on $A2$ and $A1$ and doesn't use $A3$. Is that right?

If not: Where is $A3$ used?

If so: Can we use instead $A1$ and $A3$? $A3$ and $A2$?

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    I agree with you. It's interesting, nonetheless, to examine the iterates right on the unit circle. I guess that's all the author is doing. – 2017-01-01
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    Technically speaking, there's nothing indicating that some of the numbers you have not investigated can not replicate some of the behaviour observed in the other sets, until you try. For $z^2$ this is rather trivial, but generally there could be another convergent basin with a disconnected boundary. (Did you really mean skipping the investigation of $A_2$ or was $A_3$ what was intended?) – 2017-01-01
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    @TheVee 1. 'Technically speaking, there's nothing indicating that some of the numbers you have not investigated can not replicate some of the behaviour observed in the other sets, until you try.' --> Thus we must investigate for the 3 sets? 2. 'Did you really mean skipping the investigation of A2 or was A3 what was intended?' --> What do you mean? I don't think it really matters. The choice was arbitrary. Author picked $A_1$ and $A_2$. I was wondering if $A_3$ can replace either. – 2017-01-01
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    @MarkMcClure Thanks! ^-^ Post as answer? Merry Christmas, Happy Holidays, and Happy New Year! – 2017-01-01
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    @TheVee Edited. – 2017-01-06

1 Answers 1

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We just need (A1) and (A2) to conclude about the Fatou set and Julia set of the function $f(z)=z^2$. In detail, let derive from (A1) and (A2) that the Fatou set is the $\mathbb{C} \setminus \{ |z|=1\}$

By (A1), we deduce that $\{ |z| <1\}$ is included in the Fatou set $F_f$ of $f$.

By (A2), we deduce that $\{ |z| >1\}$ is included in the Fatou set.

Now suppose that $A=\{|z|<1\} \cup \{|z|>1\} \neq F_f$. Then there exist a point $z_0$ such that $z_0 \in F_f$ and $|z_0|=1$. $z_0 \in F_f$ means that there exist an open neighborhood $U$ of $z_0$ in $A$ such that $\mathcal{F}=\{f^{\circ n}\}$ is normal on $U$. It means that there is a subsequence $\{f^{\circ n_m} \}$ converges locally uniformly to a holomorphic map or diverges from every compact subset of $U$.

The family $\mathcal{F}$ is normal on $U$ so $\mathcal{F}$ is normal on $V=U \cap \{|z|<1\}$. But $\{f^{\circ n_m}\}$ converges uniformly to $0$ in $V$ so $\{f^{\circ n_m}\}$ can not diverges from every compact subset of $U$. Hence $\{f^{\circ n_m}\}$ converges locally uniformly to a holomorphic map $g$ on $U$.

Then it means that $\{f^{\circ n_m}\}|_{V}$ converges locally uniformly to $g|_V$, hence $g|_V$ is a constant function $0$. It implies that $g$ is constant function on $U$.

But it is contradiction because on $U \cap \{|z|>1\}$, $\{f^{\circ n_m}\}$ converges to $\infty$. Then $A=\{|z|<1\} \cup \{|z|>1\}$ is indeed the Fatou set $F_f$.

Actually, this is the general situation for polynomial $P(z)$, when we have the phenomenon of escaping radius. It means there exist an constant $R> 0$ such that $$ |P(z)|>|z|$$ forvery $|z| >R$. Then $$ I_P= \bigcup\limits_{i=0}^{\infty} P^{-i} (\{|z|>R\})$$ is called basin of attraction of $\infty$. Equivalently, we can take $$ K_P=\{z \in \mathbb{C}| P^{\circ n}(z) \, \mbox{is bounded}\}$$ which has name as Filled-in Julia set. Then by the same argument, we can prove that $$J_f =\partial K_P = \partial I_P$$

Another solution: By observing that $$A \subset F_f$$ then $$J_f = \mathbb{C} \setminus F_f \subset \{|z|=1\} = \mathbb{C} \setminus A$$ We prove that $\{|z|=1\} \subset J_f$ and then done. Here (A3) can be used.

By Marty's criterion of normality:A family $\{f_{\alpha}: U \rightarrow \mathbb{P}^1\}_{\alpha}$ of holomorphic map on an open subset $ U \subset \mathbb{P}^1$ is normal if and only if the spherical derivative $$\{ ||f_{\alpha}'(z)||_{\mathbb{P}^1} = |f_{\alpha}'(z)|\frac{1+|z|^2}{1+|f_{\alpha}(z)|^2} \}_{\alpha}$$ is uniformly bounded on every compact subset of $U$

Take a point $z_0$ in the unit circle, then for every open neighborhood $U$ of $z_0$, we compute the spherical derivative of $\mathcal{F}=\{f^{\circ n}(z)\}=\{z^{2^n}\}$ $$\{ ||(f^{n})'(z_0)||_{\mathbb{P}^1} = |(f^{n}) '(z_0)|\frac{1+|z|^2}{1+|f^{n} (z_0)|^2} \}_{n}=\{|2^n z_0^{2^n-1}|\frac{1+|z_0|^2}{1+|z_0^{2^n}|^2}\}_n =\{2^n\}_n$$ Which is tend to infinity as $n \rightarrow \infty$