Let $ \mathbb{D} = \{ z : |z|<1 \} $ and $ f $ an holomorphic function on $ \mathbb{D} $ and continuous on $ \overline{\mathbb{D}} $ such that $ f(\overline{\mathbb{D}}) \subset \mathbb{D} $.

Prove the following:

  1. There exists single point $ z^* \in \mathbb{D} $ such that $ f(z^*)=z^* $ (obvious by Rouche theorem).
  2. Let $ f_1=f,...,f_{n+1}=f(f_n) $ show that $ f_n(z) \longrightarrow z^* $ uniformly.
  • $\begingroup$ Since this is homework, you should explain your own effort to solve this problem. That alone may give you the last hint needed to answer it yourself. $\endgroup$ – hardmath Aug 2 '12 at 15:38
  • $\begingroup$ How is Rouche's theorem used for 1? I can see a contraction, but not Rouche... $\endgroup$ – copper.hat Aug 2 '12 at 15:46
  • $\begingroup$ @copper.hat: I guess the idea is that $\lvert f(z) \rvert \lt \lvert -z\rvert$, so $z \mapsto -z$ and $z \mapsto f(z) - z$ have the same numbers of roots in $\mathbb{D}$ (but this would need $f$ to be holomorphic in a neighborhood of the closed disk, I believe). $\endgroup$ – t.b. Aug 2 '12 at 15:53
  • 2
    $\begingroup$ Or at least a further word than "obvious" should be said... $\endgroup$ – t.b. Aug 2 '12 at 15:59
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    $\begingroup$ @GunnarMagnusson, the map $f(z)=z/2$ is constant? $\endgroup$ – JSchlather Aug 2 '12 at 17:54

The key to all this is that $f(\bar{D}) \subset D$:

  1. Since $f(\bar{D})$ is compact, there exists $r_0>0$ such that $f(\bar{D})\subset D_{r_0}$. So for any $r_0<r<1$ we have $|f(z)|=|(f(z)-z)+z|<|-z|$ on $D_r$ so by Rouché's theorem $-z$ and $f(z)-z$ have the same zeroes, which is one. Since this is valid for any $r>r_0$ the uniqueness result follows.

  2. Since $|f(z)/z|=|f(z)|$ is continuous on $\partial D$ it has a maximum $M$, and by hypothesis $M<1$. So, assuming for the moment that $f(0)=0$, by the maximum principle we get $|f(z)|\leq M|z|$ for $z\in D$. This gives that $|f_n(z)|\leq M|f_{n-1}(z)|$ in $D$, and so $|f_n(z)|\leq M^n|z|$. Taking supremums over $\bar{D}$ and then the limit as $n\to \infty$ the result follows.

Assume now that $f(0)\neq 0$ then everything just said applies to $g=h\circ f\circ h^{-1}$ (with $h$ an appropiate automorphism of the disk), and the result follows in general.


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