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 $A2$?
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 $n \rightarrow \infty$