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Let $ f : D \rightarrow D $ be holomorphic with $ f(0) = \frac{1}{2}$ and $ f(\frac{1}{2}) = 0 $, where $ D = \{ z : |z|\leq 1 \} $. Please suggest which of the following can be correct ..

  1. $ |f'(0)|\leq \frac{3}{4}$.

  2. $ |f'(1/2) |\leq \frac{4}{3}$.

  3. $ |f'(0)|\leq \frac{3}{4}$ and $ |f'(1/2)|\leq \frac{4}{3}$.

  4. $ f(z)=z$, $ z\in D$

Please help.

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    @RobertIsrael: Thanks $f$or pointing out the mistake. I have edited now.2012-06-10

3 Answers 3

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Hint: Try some polynomials....

Hints to edited question: note that $f(f(0)) = 0$. Schwarz's Lemma may be useful. Also consider what fractional linear transformations take $D \to D$ with $0 \to 1/2$ and $1/2 \to 0$.

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    $f(f(0))=0$ then by scwarz lemma $|f(f(z))|\leq |z|$ and $|(f\circ f)'(0)|\leq 1$..... $(f\circ f)'(x)=f'(f(x))f'(x)\Rightarrow |(f\circ f)'(0)|=|f'(\frac{1}{2})f'(0)|\leq 1$ this says that if $|f'(\frac{1}{2})|\leq \frac{3}{4}$ then $|f'(0)|\leq \frac{4}{3}$ Infact for a,b >0 we have if $|f'(\frac{1}{2})|\leq \frac{a}{b}$ then $|f'(0)|\leq \frac{b}{a}$ but then how do i say that $|f'(\frac{1}{2})|\leq \frac{3}{4}$ :O please help me in this case...2014-01-24
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First you should recall the following result: Suppose that $\displaystyle f$ is analytic on the unit disc ∆=$\{z:|z|<1\}$ and satisfies the following conditions $|f(z)| \leq 1$
$f(a)=b$ for some $a,b \in $∆ then $|f'(a)| \leq (1-|f(a)|^2)/(1-|a|^2 )$. In our problem first you take $a=0$ and $b=1/2$ and apply the above result you will get the first option in your problem. For the second option take $a=1/2$ and $b=0$. For your problem options (a), (b) and (c) are true

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Schwarz Picks Lemma says $|f'(z)|\le {1-|f(z)|^2\over 1-|z|^2}$ says $1,2,3$ are Correct