Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is from an old qualifying examination question.

If f is holomorphic in the unit disk $D$ and $|f(z)|<1$ for all $z\in D$. Suppose also that $f$ has two distinct fixed points in $D$ then $f(z)=z$ for every $z\in D$

I know that I have to use the Schwarz's lemma and may be make use of Möbius transformations. I tried setting $g(z)=\phi_a\circ f\circ \phi_{-a}$. But that does not seem to work because I don't see why $|g(z)|=|z|$ for some non zero $z$.

Any helpful hints are greatly appreciated.

Edit: Of course the $a$ above is one of the fixed points.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Hint: Let $b$ be the other fixed point of $f$. What happens when you apply $g$ to $\phi_a(b)$?

Here's a full solution. Let $a$ and $b$ be the distinct fixed points of $f$ and let $\varphi$ be the conformal automorphism of the disk sending $a$ to $0$, recall

$$\varphi(z)=\frac{z-a}{1-\bar{a}z}.$$

In particular it's not hard to check that $g=\varphi\circ f \circ \varphi^{-1}$ fixes $0$. We also have

$$ g(\varphi(b))=\varphi[f(\varphi^{-1}[\varphi(b)])]=\varphi(f(b))=\varphi(b)$$

by Schwar'z lemma $g(z)=cz$. Now $\varphi(b)\neq 0$ since $\varphi(a)=0$ thereby $g(z)=z$, in particular this gives that $\phi^{-1}=f \circ \phi^{-1}$ so $f$ must be the identity on the disk.

share|improve this answer
    
Thanks for your answer. But how do we know that the constant in $g(z)=az$ is the same constant as the fixed point. In Schwarz's lemma one of the conclusions is $g(z)=cz$ for some $|c|=1$. But then $a\in D$ and therefore it's definitely not unimodular. Can you explain a little more. Thanks again. –  Jack Dawkins Jan 7 '13 at 0:47
    
We know that $g(\varphi(b))=\varphi(b)$ and we have that $g(\varphi(b))=c\varphi(b)=\varphi(b)$ since $\phi(b) \neq 0$ by division $c=1$. –  JSchlather Jan 7 '13 at 0:49
    
@Jacob: I think the OP's concern is in the line "by Schwar'z lemma $g(z)=az$". You probably wanna use a letter different from $a$ there. –  user641 Jan 7 '13 at 4:01
    
@Steve Ah, yeah I see what you mean. Fixed it. –  JSchlather Jan 7 '13 at 4:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.