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A complex number $w\in D$ is a fixed point for the map $f:D \to D$ if $f(w)=w$.
Prove that if $f:D\to D$ is analytic and has two distinct fixed points, then $f$ is the identity,that is,$f(z)=z$ for all $z\in D$.

If $f(0)=0$ , I can use Schwarz lemma to show that $f(z)=ze^{i\theta}$ and $\theta=0$. How could I deal with the condition when $f(0)=z_0 \neq 0$.

Jair Taylor
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J.Guo
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    What about $f(z):=z^2$ defined on $\Bbb C$? This is an entire (non-identity) function, yet it has two distinct fixed points at $z=0,1$. – Dave Sep 03 '18 at 05:41
  • What is $D$, exactly? (@Dave Not if $D$ is the open unit disc, hence my question.) – Arthur Sep 03 '18 at 05:44
  • @Arthur Yes that's true, but the OP doesn't define $D$ so I originally took it to just be an arbitrary domain, and so my comment was showing that the statement is not true in this case. – Dave Sep 03 '18 at 05:47
  • $\mathrm{Aut}(D)$ acts on $D$ transitively so without loss of generality, you may assume $f(0)=0$. in fact, this is exactly what Mr. Kavi Rama Murthy did in his answer. – QU Binggang Sep 03 '18 at 07:40

1 Answers1

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Proof assuming that $D$ is the open unit disk: Let $z_1,z_2$ be two distinct fixed points of $f$. Let $g(z)=\frac {z-z_1}{1-\overset {-}z_1 z}$. Let $h=g\circ f \circ g^{-1}$. Then $h$ maps $D$ into itself, vanishes at $0$ and it has a second fixed point, namely $g(z_2)$. By your argument based on Schwarz lemma you get $h=$ identity. This implies $f=$ identity. [I have used the fact that functions of the form $\frac {z-a}{1-\overset {-}a z}$ (where $|a| <1)$ are bijective, bi-holomorphic maps of the open unit disk].

EDIT: typo fixed

Lucius Aelius Seianus
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Kavi Rama Murthy
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  • Very good solution, not taking into account that I'm troubling to use the Schwarz lemma! Could you please extend a little bit more on this part, would be really appreciated. Thanks. – VIVID Jun 14 '20 at 10:33
  • Schwarz Lemma shows that $|h(z) | \leq |z|$ and equality holds for some $z\neq 0$ if $h(z)=ze^{ic}$ for some real number $c$. In our case equality holds at the second fixed point. Also $h(z)=ze^{ic}$ together with the existence of the second fixed point shows that $h(z)=z$. @VIVID – Kavi Rama Murthy Jun 14 '20 at 11:43
  • @KaviRamaMurthy why does $h(z)=z*e^{ic}$, show $h(z)=z$? thanks – james black Jan 25 '21 at 21:32
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    @jamesblack $h$ has a fixed point other than $0$. If $ae^{ic}=a$ for some $a \neq 0$ then we must have $e^{ic}=1$. – Kavi Rama Murthy Jan 25 '21 at 23:16
  • got it thank you – james black Jan 25 '21 at 23:21