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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.

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up vote 6 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


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.

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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

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