Analytic function on an open disc.

Solution 1:

For 1. Incorrect. Since Using Schwarz Lemma we must have $|f'(0)|\leq 1$.

For 2. Correct. consider $f(z) = \frac{3}{4}z + \frac{3}{16}$

For 3. Correct. consider $f(z) = -\frac{3}{4}z - \frac{3}{16}$

For 4. Incorrect. Using Schwarz Lemma (derivative version) i.e

$|f'(z)| \leq \frac{1-|f(z)|^2}{1-|z|^2}$. Here given that $f'(\frac{1}{4})=1$. So we have $|f(\frac{1}{4})| \leq \frac{1}{4}$.

Also using Schwarz Pick Lemma we have $$\frac{|f(z) -f(w)|}{|1-f(z)\overline{f(w)}|} \leq \frac{|z-w|}{|1-z\bar{w}|}$$

Using above inequality and the fact that $f(\frac{1}{2}) =-\frac{1}{2}$, one will get \begin{align} \frac{|f(\frac{1}{4}) + \frac{1}{2}|}{|1+ \frac{1}{2}f(\frac{1}{4})| } &\leq \frac{2}{7} \\ \implies |2 - \frac{3}{f(\frac{1}{4})+2}|&\leq \frac{2}{7} \\ \implies \frac{12}{7} \leq \frac{3}{|f(\frac{1}{4})+2|} &\leq \frac{16}{7}\\ \implies \frac{21}{16}\leq |f(\frac{1}{4})+2| &\leq \frac{21}{12} \end{align} So $f(\frac{1}{4})$ belong to Annulus with centre at $-2$ and with inner radius $\frac{21}{16}$ and outer radius $\frac{21}{12}$.

we also have $|f(\frac{1}{4})| \leq \frac{1}{4}$ which gives us $f(\frac{1}{4})$ lies in disk with centre $0$ and radius $\frac{1}{4}$. This two domain intersect at $-\frac{1}{4}$. So, one must have $f(\frac{1}{4}) = -\frac{1}{4}$.

Then we have $|f(\frac{1}{4})|= \frac{1}{4}$ and $f'(\frac{1}{4}) =1$. we have equality in the inequality $|f'(z)| \leq \frac{1-|f(z)|^2}{1-|z|^2}$ for $z = \frac{1}{4}$. Hence $f$ has to be scalar times identity function i.e $f(z) = cz$ for all $z$ in unit disk with $c$ is an unimodulur constant. But $f'(\frac{1}{2}) =1 $ gives us $c=1$. But that contradict $f(\frac{1}{2}) = - \frac{1}{2}$.