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I am confronted with the following problem:

Let $M$ be the set of all holomorphic maps $f$ from the unit disk to itself with $f(0)=\frac{2}{3}$. Find $\sup_M|f'(0)|$.

I feel that Schwarz Lemma might help, but I am not getting anywhere. Can anyone help?

-best regards.


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yes, sorry, $f(0)=\frac{2}{3}$. My bad. – Marie. P. Nov 11 '11 at 4:42

2 Answers 2

up vote 7 down vote accepted

It helps to be familiar with the Möbius transformations that map the disk onto itself. One of these is $g(z)=\frac{\frac{2}{3}-z}{1-\frac{2}{3}z}$. Then $g$ is holomorphic on the disk and maps the disk bijectively to itself, and it swaps $\frac{2}{3}$ and $0$. Applying Schwarz's lemma to $g\circ f$ yields $|f'(0)|\leq \left|g'\left(\frac{2}{3}\right)\right|^{-1}$, and you can check that this upper bound is attained when $f=g$.

In general, if $f$ maps the disk to itself and $f(0)=a\neq 0$, you can compose with $\phi_a(z)=\frac{a-z}{1-\overline{a}z}$ to get a map $\phi_a\circ f$ from the disk to itself that sends $0$ to $0$, so that Schwarz's lemma can be applied. Notice that $\phi_a$ swaps $0$ and $a$, which is part of why $\phi_{2/3}$ is so helpful in this problem. More generally, if $f(b)=a$, then Schwarz's lemma can be applied to $\phi_a\circ f\circ \phi_{b}$.

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Schwarz-Pick Lemma:

$$|f^{\prime} (a)| \le \frac{ 1 -|f(a)|^2}{ 1 -|a|^2} $$

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