an analytic function from unit disk to unit disk with two fixed point

Question

prove that if $f:\mathbb{D}\rightarrow\mathbb{D}$ is analytic with two distinct fixed point then $f$ is identity.

I thought if one of the fixed points was zero by schwarz lemma this statement is easily proved.

but what can I do if fixed points were nonzero?

score 21 · Accepted Answer · answered May 19 '14 at 14:01

21

I thought if one of the fixed points were zero by the Schwarz lemma this statement is easily proved.

Good. So it remains to reduce the general case to the known special case.

Let $T$ be an automorphism of $\mathbb{D}$. What do you know about $T^{-1}\circ f \circ T$? Can you find a condition on $T$ that reduces the problem to the known case?

answered May 19 '14 at 14:01

Daniel Fischer

206,697

it is a function from $\mathbb{D}\rightarrow\mathbb{D}$ that fixes zero! am i right? – user115608 May 19 '14 at 17:56
1

Whether it fixes zero depends on $T$. What condition on $T$ makes it fix zero? – Daniel Fischer May 19 '14 at 18:01
no sorry!it is a function from $\mathbb{D}\rightarrow\mathbb{D}$ that fixes $T^{-1}$ (fixed point of f)! am i right? – user115608 May 19 '14 at 18:04
Yes. That's right. So what condition on $T$ do you want? – Daniel Fischer May 19 '14 at 18:05
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i want $T(0)=$ fixed point of f. – user115608 May 19 '14 at 18:08
Yup. I guess you know that that is always possible, and you can probably write down such a $T$ explicitly. – Daniel Fischer May 19 '14 at 18:09
I don't understand. If $h:=T^{-1} \circ f \circ T$, we have $h(0)=0$. But if we use Schwarz lemma, we get $h(z)=cz$ for all $z \in \mathbb D$ and some $c \in \mathbb C$ and that implies $f(z)=T(cT^{-1}(z))$ for all $z \in \mathbb D$, but why does that imply $f(z)=z$? – Twnk Nov 03 '14 at 08:23
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@Twink $f$ has (at least) two fixed points, hence so has $h$. Let $\zeta\neq 0$ be another fixed point. Then you can deduce $c =:?$ – Daniel Fischer Nov 03 '14 at 11:02

an analytic function from unit disk to unit disk with two fixed point

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