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If $f$ is a polynomial and $z\in\mathbb{C}$, show that either $f^n(z)\rightarrow\infty$ or $\{f^n(z) : n\geq 1\}$ is a bounded set.

Here, $f^2(z)=f(f(z))$ and $f^n(z)=f(f^{n-1}(z))$ for $n\geq 2$

I had a proof structured as followed:

1) Suppose $\{f^n(z) : n\geq 1\}$ is unbounded. Then there exists a subsequence $(n_k)$ such that $f^{n_k}(z)\rightarrow\infty$.

2) We are done if we can show that $|f^{k}(z)|$ is monotone everntually.

But the trouble is 2) is really tedious to verify. I am just wondering whether there is a more pretty way to do this.

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up vote 5 down vote accepted

I will not treat the case where $f$ is of degree at most 1. So assume that $f$ is a polynomial of degree at least 2.

Then $|f(z)|$ grows faster than $|z|$ as $|z|\to +\infty$. This means that there is some $M>0$ such that for all $z\in\mathbb{C}$ with $|z|>M$ we have $$|f(z)|\geq |z|.$$ (Try to make this precise if you have doubts.) We can conclude that as soon as the sequence gets further than $M$ from the origin, it will only get even further from the origin. This proves the result.

Remark. We have indeed shown that if $\{f^n(z)\}$ is unbounded, then $|f^k(z)|$ is monotone eventually. Note that you could strengthen the argument above to show that if $\{f^n(z)\}$ is unbounded, then $|f^k(z)|$ will eventually grow at least exponentially (for example, take $M$ big enough to make sure that $|f(z)|\geq 2|z|$).

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For the linear case $f(z)=az+b$ with $a\neq 1$ you could say that $f$ fixes $\frac{b}{1-a}$ and each iteration multiplies distances by $|a|$, proving that the sequence is bounded if $|a|\leq 1$ and wanders off to infinity otherwise. Then only the case $f(z)=z+b$ remains (and that is not a very hard case). – Daan Michiels May 9 '12 at 11:49

Here's an alternative proof. I'm certain that it's not the proof expected in an elementary class, as the tools it uses are a bit heavier. It has the advantage, though, of introducing the idea of conjugation and how it can be used to treat the point at $\infty$ - an important step towards extending polynomial dynamics to the dynamics of rational functions on the Riemann sphere.

The idea is to replace the polynomial $f(z)$ with the rational function $F(z)=1/f(1/z)$ and study the orbit of $F$. It turns out that $F$ has a super-attractive fixed point at zero and the statement about orbits of $f$ being attracted to $\infty$ can be recast in terms of orbits of $F$ being attracted to zero.

More precisely, let $f$ be a complex polynomial of degree at least two and define

$$F(z) = \left\{\begin{array}{ll} 1/f(1/z) & z \neq 0 \\ 0 & z=0. \end{array}\right. $$

Essentially, we've simply conjugated by the reciprocal function and filled the resulting removable discontinuity at the origin. $F$ is, in fact, differentiable at the origin with $F'(0)=0$, since

$$F'(0)=\lim_{z\rightarrow 0} \frac{F(z)-F(0)}{z-0} = \lim_{z\rightarrow 0} \frac{1}{zf(1/z)}=0.$$

That last equality is true because $f$ has degree at least two and, thus, $f(1/z)\rightarrow \infty$ faster than $z\rightarrow 0$.

Next, note that

$$F(F(z)) = 1/f (1/(1/f(1/z))) = 1/f(f(1/z))$$

and, more generally,


Substituting $z$ for $1/z$, we get


As a result, an orbit of $f$ generates an orbit of $F$ by taking the reciprocals of the terms in the orbit and vice-versa. Thus, an orbit in the neighborhood of $\infty$ for $f$ can be treated as an orbit in the neighborhood of $0$ for $F$.

We can now establish the result using the fact that zero is an attractive fixed point for $F$, for then there is an $r>0$ such that $|z|<r$ implies that $z$ converges to zero under iteration of $F$. Thus, if $|z|>1/r$, then $z$ converges to $\infty$ under iteration of $f$.

Note that the number $1/r$ becomes the crucial bound in the original question.

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