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This is an exercise in complex analysis:

Let $\Omega\subset{\Bbb C}$ be open and bounded, and $\varphi:\Omega\to\Omega$ a holomorphic function. Prove that if there exists a point $z_0\in\Omega$ such that $$ \varphi(z_0)=z_0\qquad\text{and }\qquad \varphi'(z_0)=1 $$ then $\varphi$ is linear.

I'm trying work out the case $z_0=0$ first, in which $$ \varphi(z)=z+\sum_{n=2}^{\infty}a_nz^2. $$ It suffices to show that $a_n=0$ for all $n\geq 2$. If let $$ \varphi(z)=z+a_2z^2+O(z^3) $$ then $$ \varphi^k(0)=z+ka_2z^2+O(z^3), $$ and $$ \varphi^k(0)=0,\quad (\varphi^k)'(0)=1. $$ If one can show that $\{ka_2\}_{k=1}^{\infty}$ is uniformly bounded, then one at least has $a_2=0$. But I don't know how to go on. Any idea?

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Are you familiar with universal covering and uniformization theorem? By the way, $\Omega$ should be connected. –  23rd Jan 10 '13 at 17:33
Dear @richard, if you can solve this problem I would be grateful to you for posting an answer. I am not unacquainted with universal coverings and the uniformization theorem but I don't see how your hint suffices to answer the question. –  Georges Elencwajg Jan 10 '13 at 18:10
@GeorgesElencwajg: $\varphi$ can be lifted to a holomorphic map $h$ from the unit disk to itself, which satisfies $h(0)=0$ and $h'(0)=1$. By Schwarz lemma, $h$ is the identity map. Then when $\Omega$ is connected, $\varphi$ is also the identity map. –  23rd Jan 10 '13 at 18:22
@GeorgesElencwajg: You may choose the covering map $p$ such that $p(0)=z_0$ and apply chain rule to $p\circ h=\varphi\circ p$ at $0$. –  23rd Jan 10 '13 at 18:38
@richard: ah yes, that's correct ( although surprizing to me, because étale maps don't preserve derivatives in general). I still believe that a full-fledged answer would be a welcome addition to your first comment, given that the various steps of the proof are not so evident . –  Georges Elencwajg Jan 10 '13 at 19:07

1 Answer 1

up vote 6 down vote accepted

Following your thoughts, when $\Omega$ is connected, if $\varphi$ is not linear, then there exists $n\ge 2$ and $a_n\ne 0$, such that
$$\varphi(z)=z+a_n(z-z_0)^n+O((z-z_0)^{n+1}).$$ As you have noticed, by induction, it follows that for every $k\ge 1$, $$\varphi^k(z)=z+ka_n(z-z_0)^n+O((z-z_0)^{n+1}). \tag{1}$$

Let $r>0$ be such that when $|z-z_0|\le r$, then $z\in\Omega$. Then by $(1)$,

$$ka_n=\frac{1}{2\pi i}\int_{|z-z_0|=r}\frac{\varphi^k(z)}{(z-z_0)^{n+1}}dz.\tag{2}$$ Since $\varphi^k(\Omega)\subset\Omega$ and since $\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $|\varphi^k|\le M$ on $\Omega$. Then by $(2)$, $$k|a_n|\le Mr^{-n}.$$ Since $k$ is arbitrary, $a_n=0$, a contradiction.

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Thank you very much! I didn't the key point that $\varphi^k$ is bounded since $\Omega$ is bounded and $\varphi^k(\Omega)\subset\Omega$. I haven't seen "universal coverings and the uniformization theorem" before. Is it an alternative quick approach for the proof? What's more, could you explain why $\Omega$ should be connected? Indeed, $\Omega$ is used as a region (connected open set) on the complex plane in some chapters of the textbook. But I don't see the explicit assumption of connectedness of $\Omega$ in this exercise. –  Jack Jan 10 '13 at 21:42
@Jack: You are welcome! With the help of uniformization theorem, we may reduce $\Omega$ to the unit disk and apply Schwarz lemma. If $\Omega$ is disconnected, denote the connected component of $\Omega$ containing $z_0$ by $\Omega_0$. Then on $\Omega_0$, $\varphi$ is the identity map, but we know nothing about the behavior of $\varphi$ on $\Omega\setminus\Omega_0$. –  23rd Jan 11 '13 at 9:06
A different proof: this is even better than what I encouraged you to do! +1, of course. –  Georges Elencwajg Jan 11 '13 at 9:28

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