4

I have a question that asks

If $S \subset \mathbb{C}$ is a bounded domain and $f : S \to S$ is a holomorphic map such that $f(p) = p$ and $|f'(p)| = 1$ for some $p \in S$, then $f$ is bijective.

I am aware of this question: Bijective holomorphic map with a fixed point ,which seems to suggest that the problem is wrong. However, in that question the suggested counterexample is $g(z) = z + z^2$ and I can't think of a bounded domain such that $g(S) \subseteq S$.

This question is the second part of a larger question, the first part of which states that (under the same notation) $f(z) = z$ for all $z \in S$ if $f(p) = p$ and $f'(p) = 1$, so I'm assuming this result is going to be useful somehow. Also, since there is a fixed point involved, the answer is surely going to use some complex dynamics ideas somewhere.

I know that the multiplier of a fixed point is conjugation invariant, so could I conjugate $f$ by some conformal map $\varphi$ to get $(\varphi \circ f \circ \varphi^{-1})'(p) = 1$? If so then $f(z) = z$ for all $z \in S$ by the previous result. This is pretty optimistic, and I have my doubts, but it is all I have at this point. Any help would be seriously appreciated!!

Community
  • 1
user131576
  • 305

2 Answers2

7

Since $S$ is bounded, the family

$$\mathscr{F} = \left\{f^n : n \in \mathbb{N}\right\},$$

where $f^n$ denotes the $n$-fold iterate, $f^0 = \operatorname{id};\; f^{n+1} = f\circ f^n$, is normal.

Let $c = f'(p)$. Pick a strictly increasing sequence $(n_k)$ of natural numbers such that $c^{n_k} \to 1$. Since $\mathcal{F}$ is normal, by passing to a subsequence, we may assume that the sequence $\left(f^{n_k}\right)$ is locally uniformly convergent. Let $g$ be the limit function. Then $g(p) = p$ and $g'(p) = \lim\limits_{k\to\infty} c^{n_k} = 1$, so $g$ is not constant. By pointwise convergence $g(S) \subset \overline{S}$ and by the open mapping theorem, since $g$ isn't constant, we have $g\colon S \to S$. The family of iterates $\mathscr{G} = \{g^n : n \in \mathbb{N}\}$ is also normal, and that implies $g = \operatorname{id}$, since if

$$g(z) = p + (z-p) + a_k(z-p)^k + \dotsc,$$

with $k > 1$, we have

$$g^m(z) = p + (z-p) + m\cdot a_k(z-p)^k + \dotsc,$$

so $\frac{d}{dz}^k (g^m)(p) = m\cdot g^{(k)}(p)$. The normality of $\mathscr{G}$ implies the normality of the $k$-th derivatives of the iterates of $g$, and the sequence $\left(m\cdot g^{(k)}(p)\right)_{m\in\mathbb{N}}$ only has convergent subsequences when $g^{(k)}(p) = 0$. (This is the first part of the exercise.)

So $f^{n_k} \to \operatorname{id}$ locally uniformly. And that implies that $f$ is bijective by Hurwitz's theorem.

Daniel Fischer
  • 206,697
  • Thank you for your answer. I understand the last bit, but where are you using the assumption that $|f'(p)| = 1$? Is this used to create the normal family $\mathcal{F}$? – user131576 Apr 29 '14 at 21:14
  • The condition $\lvert f'(p)\rvert = 1$ ensures that the limit function is not constant. – Daniel Fischer Apr 29 '14 at 21:17
  • @user131576 Sorry, the argument I had in mind works only for some special cases, the actual argument is more involved, I've edited. – Daniel Fischer Apr 29 '14 at 22:21
  • I don't see why the following is true "So $f^{n_k}\rightarrow id$ locally uniformly. And that implies that $f$ is bijective by Hurwitz's theorem." – M.G Mar 10 '16 at 19:25
  • @M.G Which part don't you see? If $f(z_1) = f(z_2)$ then $z_1 = \lim f^{n_k}(z_1) = \lim f^{n_k}(z_2) = z_2$, so $f$ is injective. If $w\in S$, then $w = \operatorname{id}(w)$, and Hurwitz's theorem says $f^{n_k}$ attains the value $w$ for all large enough $k$, so $w \in f^{n_k}(S) = f(f^{n_k-1}(S)) \subset f(S)$, hence $f$ is surjective. – Daniel Fischer Mar 10 '16 at 19:36
  • @Daniel Fischer, I'm not familiar with Hurwitz's theorem. I'm reading the theorem statement on wikipedia trying to figure out why $f^{n_k}$ attains the value $w$ for all large enough $k$. – M.G Mar 10 '16 at 19:45
  • 1
    @M.G Let $g(z) = z-w$, and $g_k(z) = f^{n_k}(z) - w$. Then $g_k \to g$ locally uniformly on $S$, and $g$ has a simple zero at $w$. The statement of the theorem on wikipedia (apart from the naming of the functions) says that for small enough $\rho > 0$, $g_k$ has precisely one zero in ${ z : \lvert z-w\rvert < \rho}$ for all large $k$. But a zero of $g_k$ is a point where $f^{n_k}$ attains the value $w$. – Daniel Fischer Mar 10 '16 at 19:51
  • Right, very many thanks. This is an interesting theorem indeed. – M.G Mar 10 '16 at 19:53
1

Here are two hints if we assume $S$ is simply connected: (1) By the Riemann mapping theorem, you may assume $S$ is the unit disk. (2) Then what does the Schwarz Lemma tell you?

I'll have to ponder the non-simply connected case.

Ted Shifrin
  • 115,160