Multiple Choice Question: Let f be holomorphic on D with $ f(0) = \frac{1}{2}$ and $ f(\frac{1}{2}) = 0 $, where $ D = \{ z : |z|\leq 1 \}$.
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 ..
$ |f'(0)|\leq \frac{3}{4}$.
$ |f'(1/2) |\leq \frac{4}{3}$.
$ |f'(0)|\leq \frac{3}{4}$ and $ |f'(1/2)|\leq \frac{4}{3}$.
$ f(z)=z$, $ z\in D$
Please help.
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 $1/2 \to 0$.
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
Schwarz Picks Lemma says $|f'(z)|\le {1-|f(z)|^2\over 1-|z|^2}$ says $1,2,3$ are Correct