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Let $D$ be a bounded domain and let $f$ be analytic function from $D$ into $D$. Show that if $z_{0}$ $\in D$ is a fixed point for $f$ , then $|f' (z_{0} )| \leq 1$.

WHAT I WAS THINKING:

To have a conformal mapping from $D$ to the unit disc & after composing it with $f$ ; I was thinking to use Pick's Lemma. So, can anyone tell me any conformal map from any bounded domain to the unit disc?? OR Is there any other way out to solve this problem??

user92360
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    If the domain isn't simply connected, you're never going to get a conformal mapping to the unit disk. But is there a way to localize this to a disk that is a neighborhood of $z_0$? – Ted Shifrin Oct 02 '14 at 16:39
  • $D$ is bounded, so the family of iterates ${ f^n : n \in \mathbb{N}}$ is a normal family [Montel]. The family of the derivatives ${(f^n)'}$ is then also normal. What is $(f^n)'(z_0)$? – Daniel Fischer Oct 04 '14 at 14:32
  • @DanielFischer Hi.. this is a really old comment of yours, but your remark seems to be something that will help me in a similar problem. Can you elaborate? So we know that $f^{n_{k}}(z_{0})$ converges to $F'(z_{0})$, where F is the limit function of some convergent subsequence.. and by computation it is 1.. – Elliot Apr 14 '15 at 02:55
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    @Elliot If $z_0$ is a fixed point of $f$, then $(f^n)'(z_0) = \bigl(f'(z_0)\bigr)^n$ (proof by induction). By normality, it follows that that is a bounded sequence, hence $\lvert f'(z_0)\rvert \leqslant 1$. If $z_0$ is an attractive fixed point ($\lvert f'(z_0)\rvert < 1$), then $f^n$ converges locally uniformly to the constant $z_0$. If $\lvert f'(z_0)\rvert = 1$, all limit functions of subsequences are non-constant and have $\lvert F'(z_0)\rvert = 1$, but not necessarily $F'(z_0) = 1$. – Daniel Fischer Apr 14 '15 at 08:40

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If $D$ is simply connected, the Riemann mapping theorem and the Schwarz lemma give a nice proof of this (think chain rule). If not, the easiest answer in my mind is a bit dynamical...

If $D$ is bounded then it's hyperbolic as a Riemann Surface. From Montel's Theorem, we know $D$ must be in the Fatou set of $f$. Because it contains a fixed point, the component that $z_0$ is in needs to be fixed. The classification theorem for hyperbolic complex dynamics (see Milnor's Dynamics in One Complex Variable) tells you it needs to be (after dismissing some cases) either attracting ($|f(z_0)|<1$), identity ($|f(z_0)|=1$), or irrational rotation ($|f(z_0)|=1$).

Edit: Without assuming some of the hammers from complex dynamics, this can be done (with some effort) using the fact that $D$ is hyperbolic as a Riemann surface. First, show (or know from Pick's theorem) that a holomorphic function between hyperbolic Riemann surfaces doesn't increase Poincare distance. Then write down the definition of the derivative at $z_0$. Finally, show that Poincare distance and the standard metric give the same answer for derivatives at fixed points.

(Essentially this is a method to reprove classification in the slightly simpler case where there is a fixed point. Also it would probably make your life easier to assume that $z_0 =0$.)

Joe Manlove
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