20

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?

  • 1
    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
  • @richard: yes, I thought about that, but why is $h'(0)=1$ ? – Georges Elencwajg Jan 10 '13 at 18:27
  • 2
    @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
  • 1
    @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
  • @GeorgesElencwajg: Note that $z_0$ is a fixed point of $\varphi$. The derivative at a fixed point is invariant under conjugation. – 23rd Jan 11 '13 at 09:08
  • Very true and very interesting, richard : I'm learning a lot with you! – Georges Elencwajg Jan 11 '13 at 09:24
  • 1
    @Jack why is $z_{0} = 0$? – u_any_45 Sep 22 '20 at 01:57
  • @u_any_45 consider $\varphi(z) - z_0$ and $\Omega - z_0$ to shift the setup to $0$ WLOG. – al2000 Sep 22 '22 at 03:26
  • @23rd I think the assumption that $\Omega$ is simply connected is redundant, WLOG assume $z_0 = 0$ and consider a disc $D_{\epsilon}$ centered at origin s.t $\varphi$ maps $D_{\epsilon}$ to itself. The rest is only a matter of using Schwarz lemma. But there is something suspicious to me, I don't use the fact that $\Omega$ is bounded! – Khaled Alekasir Mar 16 '24 at 07:13

1 Answers1

26

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.

23rd
  • 16,234
  • 44
  • 70
  • 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. –  Jan 10 '13 at 21:42
  • 1
    @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 09:06
  • A different proof: this is even better than what I encouraged you to do! +1, of course. – Georges Elencwajg Jan 11 '13 at 09:28
  • I don't think this proof works, since the radius of convergence of $\varphi^k$'s power series expansion in (1) may (a priori) be smaller than $r$ (i.e. that of $\varphi^1$'s expansion)... (For example, for $k=2$, we can only expand $\varphi(\varphi(z))$ if $|z-z_0|\le r$ [to expand $\varphi(z)$] AND $|\varphi(z)-z_0| \le r$ [to expand $\varphi(\text{expansion of }\varphi(z))$].)

    (BTW, this is Ch. 2, Exercise 9 in Stein & Shakarchi, and they hint at a solution along these lines... but again, I don't think it works?)

     – Victor Wang Aug 07 '14 at 01:44
  • 1
    @VictorWang: Sorry, I just read your comment because it is very long time since the last time I visited this website. Please note that since $\varphi(\Omega)\subset \Omega$, $\varphi^k$ is a well-defined holomorphic function on $\Omega$ for every $k$. Then from the choice of $r$ we know the radius of convergence of the Taylor expansion of $\varphi^k$ around $z_0$ is no less than $r$. – 23rd Dec 25 '14 at 11:31
  • @23rd: thanks; that should've been obvious. I misinterpreted your line "by induction, it follows that for every $k\ge1$..." (for (1)) as saying we should actually compose the power series inductively, rather than (more indirectly) showing the Taylor expansion exists and then only inductively computing the Taylor coefficients. – Victor Wang Dec 25 '14 at 12:06
  • @VictorWang: You are welcome. – 23rd Dec 25 '14 at 15:37
  • 1
    @23rd I have a basic question - How is $\phi^{k}(z) = z+ka_n(z-z_0)^n+O(z-z_0)^{n+1}$.? Even if I took k=2, I won't have a linear term in z. – u_any_45 Sep 22 '20 at 04:32