Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.
share|cite|improve this question
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. – hardmath Aug 2 '12 at 15:38
How is Rouche's theorem used for 1? I can see a contraction, but not Rouche... – copper.hat Aug 2 '12 at 15:46
@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). – t.b. Aug 2 '12 at 15:53
Or at least a further word than "obvious" should be said... – t.b. Aug 2 '12 at 15:59
@GunnarMagnusson, the map $f(z)=z/2$ is constant? – JSchlather Aug 2 '12 at 17:54
up vote 6 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.