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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.

Sriti Mallick
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preeti
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3 Answers3

5

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 $1/2 \to 0$.

Robert Israel
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    @preeti: Liouville's theorem does not apply, here. Entire means holomorphic on $\C$, not just on a function's domain of definition. – Cameron Buie Jun 10 '12 at 20:20
  • @CameronBuie: Thanks for the help. – preeti Jun 10 '12 at 20:37
  • Further help needed. $f(z)={2\over 3}z(z-{1\over 2})$ is one such with $|f'(0)|=|f'({1\over 2})|={1\over 3}$. No conclusion can be made from here. – Sugata Adhya Dec 15 '12 at 14:11
  • One what? The fractional linear transformation $\phi(z) = \dfrac{2z-1}{z-2}$ takes $D$ onto $D$ and interchanges $0$ and $1/2$. So (with $f$ as in the question) $f \circ \phi$ takes $D$ into $D$, $0 \to 0$ and $1/2 \to 1/2$. By Schwarz's lemma, $f \circ \phi(z) = z$, i.e. $f = \phi^{-1} = \phi$. – Robert Israel Dec 16 '12 at 03:19
  • $f(f(0))=0$ 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... –  Jan 24 '14 at 13:59
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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

learner
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S. Veeramani
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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

Myshkin
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