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We are working on a problem which is a variant of problem 16 from the Conformal Mappings chapter of Rudin's Real and Complex Analysis. We need assistance on part 3.

Suppose $\Omega$ is a bounded region, $a \in \Omega$, $f$ is an analytic function on $\Omega$, $f(\Omega) \subset \Omega$, and $f(a) = a$.

  1. Put $f_1:= f$ and $f_n := f \circ f_{n-1}$. Compute $f'_n(a)$ and prove that $|f'(a)| \leq 1$.

  2. If $f'(a)=1$, show that $f(z)=z$ for all $z \in \Omega$.

  3. If $|f'(a)| = 1$, prove that $f$ is one-to-one and that $f(\Omega) = \Omega$.

We proved part 1 and part 2. In both cases, we appeal to the fact that $\mathcal{F_m} = \big\{f_n^{(m)}: n\geq 1\big\}$ for $m \geq 0$ is locally bounded and thus a normal family. From part 1, we also have that $|f_n'(a)| = |f'(a)|^n = 1$. We aren't sure what to do from here.

Kroki
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TheSenate
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1 Answers1

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We follow the hint given in the book.

Let $\gamma:=f’(a)$. Since $f’_n(a)=\gamma^n$, there exists a sequence $n_1\lt n_2\lt\dots$ such that $\lim_{k\to \infty}f’_{n_k}(a)=1.$ By Arzelà-Ascoli, taking a subsequence of $\{n_k\}$, which we denote $\{m_k\}$, $f_{m_k}$ converge to a holomorphic function $g$ locally uniformly. In this case, differentiation and limit commute and $g’(a)=1$. By Ex 20, Chap 10 in the book, we also have $g(\Omega)\subset \Omega$. By 2., $g=\mathrm{id}$. By definition of $g$, $f$ must be one-to-one. To show $f(\Omega)=\Omega$, fix $\alpha\in\Omega$. The function $g-\alpha$ has a zero. Again by Ex20, Chap 10, there exists $k$ such that $f_{m_k}-\alpha$ has a zero, showing that $\alpha\in f(\Omega)$.

Acrobatic
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  • We weren’t given this hint unfortunately. Will work on it. Thank you! – TheSenate May 10 '23 at 08:47
  • Why must $f$ be one-to-one? Is it something relating to the limit and definition of $f$? Something about the functions being a normal family? – TheSenate May 10 '23 at 09:08
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    If $u, v\in \Omega$ and $ f(u)=f(v)$, $f_{m_k}(u)=f_{m_k}(v)$ so that $u=g(u)=g(v)=v$. – Acrobatic May 10 '23 at 09:17
  • Thank you so much Acrobatic. I think we understand now. That hint in the book was super helpful, and your explanation of the other details was clear. – TheSenate May 10 '23 at 09:33
  • Actually, could you explain why $f_{m_k} - \alpha$ has a zero for some $k$? I see that $(f_{m_k} - \alpha) \to (g - \alpha)$ and obviously $g(\alpha) - \alpha = 0$, but why does that give us that one of the functions $f_{m_k} - \alpha$ indeed has a zero instead of only that $(f_{m_k}(\alpha) - \alpha) \to 0$? – TheSenate May 10 '23 at 11:02
  • @TheSenate $f_{n}(a) = a$ for all $n$ – Kroki May 10 '23 at 11:18
  • I think $\alpha$ here is just a fixed arbitrary element, but $a$ is the specific fixed point. Notational confusion. – TheSenate May 10 '23 at 11:22
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    Suppose $f_{m_k}-\alpha$ has no zero for any $k$. By Ex 20, their locally uniform limit $g-\alpha$ either has no zero or is constant, leading to contradiction. – Acrobatic May 10 '23 at 11:34