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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.
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    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.2012-08-02
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    How is Rouche's theorem used for 1? I can see a contraction, but not Rouche...2012-08-02
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    @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).2012-08-02
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    Or at least a further word than "obvious" should be said...2012-08-02
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    @t.b.: Thanks! Just slow on my part, maybe...2012-08-02
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    I was using Cauchy's estimate.2012-08-02
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    unfortunatly i have no idea how to solve this i tried using hurwitz theorem but i cant use that because i cant prove the sequence is uniformly convenging2012-08-02
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    i think the main problem here is proving the sequence is uniformly converging2012-08-02
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    I think Brouwer's fixed point theorem applies here directly to show there's a fixed point, and using Lefschetz fixed point theorem (with homology...I can't see any other way) we get unicity...For a moment I thought Schwarz Lemma could be of some help here but I can't see how we could possibly deduce that $\,f(0)=0\,$, so perhaps this lemma isn't the thing here.2012-08-02
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    we could assume $ f(0)=0 $ and use mobius transform otherwise to convert our function to $ g = L^{-1}\circ f \circ L $ and then $ g(0) = 0 $ i think schwarz lemma is correct here but i dont know how to use it in this case2012-08-02
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    This is a very minor nitpick: the condition on the image of $f$ should read $f(\overline{\mathbb D}) \subset \overline{\mathbb D}$. If the image is only contained in $\mathbb D$, then $f$ is constant.2012-08-02
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    @GunnarMagnusson, the map $f(z)=z/2$ is constant?2012-08-02
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    Oh dear. *filler*2012-08-02

1 Answers 1

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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 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.