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Fall 2016 - Problem 10

entire functions, maximum principle

Consider the quadratic polynomial $f\p{z} = z^2 - 1$ on $\C$ . We are interested in the iterates $f^n$ of $f$ defined to be the identity on $\C$ for $n = 0$ and as

f^n = \underbrace{f \circ \cdots \circ f}_{n \text{ factors}}

for $n \in \N$ .

Find an explicit constant $M > 0$ such that the following dichotomy holds for each $z \in \C$ : either (i) $\abs{f^n\p{z}} \to \infty$ as $n \to \infty$ or (ii) $\abs{f^n\p{z}} \leq M$ for all $N \in \N_0$ .
Let $U$ be the set of all $z \in \C$ for which the first alternative (i) holds and $K$ be the set of all $z \in \C$ for which the second alternative (ii) holds.

Show that $U$ is an open set and $K$ is a compact set without "holes", i.e., $\C \setminus K$ has no bounded connected components.

Solution.

Notice that if $\abs{z} \geq 2$ , then
$\abs{f\p{z}} = \abs{z - \frac{1}{z}} \abs{z} \geq \abs{\abs{z} - \frac{1}{\abs{z}}} \abs{z} \geq \frac{3}{2}\abs{z}.$
Hence,
$\abs{f^n\p{z}} \geq \p{\frac{3}{2}}^n \abs{z} \geq 2\p{\frac{3}{2}}^n$
whenever $\abs{z} \geq 2$ . It follows that if $\abs{f^k\p{z}} \geq 2$ for some $k$ , then $\abs{f^n\p{z}} \to \infty$ . By contrapositive, if $\set{f^n\p{z}}_n$ is bounded, then $\abs{f^n\p{z}} \leq 2$ . Hence, $M = 2$ works.
Let $U_n = \p{f^n}^{-1}\p{M, \infty}$ , which is open as the continuous preimage of an open set. By (1), we know that $U = \bigcup_n U_n$ , so $U$ is open.

Since $K = U^\comp$ , we see that $K$ is closed. If $z \in K$ , then $\abs{f^0\p{z}} = \abs{z} \leq M$ , i.e., $K$ is bounded, hence compact.

Now suppose that $\C \setminus K$ has a bounded component $C$ . Since $\partial C \subseteq U$ , the maximum principle tells us that $\abs{f^n\p{z}} \leq M$ for all $z \in C$ and $n \in \N$ , but this is a contradiction as $\abs{f^n\p{z}} \to \infty$ in $C$ .