Let $D=\{z\in \Bbb C:|z|<1\}$.
Show that there exists a holomorphic function $f:D\to D$ such that
- $f(\frac{3}{4})=-\frac{3}{4}$ and $f^{'}{(\frac{3}{4})}=-\frac{3}{4}$
Show that there exists no holomorphic function $f:D\to D$ such that
- $f(\frac{1}{2})=-\frac{1}{2}$ and $f^{'}{(\frac{1}{4})}=1$
By Schwartz Lemma we can at best say that $|f^{'}(z)|\le \dfrac{1-|f(z)|^2}{1-|z|^2}$
If I substitute the values of the two given problems in the equation then I can say that $|f^{'}(z)|\le 1$ which is satisfied in every case.
How can I show existence/non-existence in two cases of the problem? Please help.
Let $$ \varphi (z)=\frac{4z-3}{4-3z}, \quad g(z)=kz ;(|k|<1) $$ and consider $f(z)=\varphi \circ g\circ \varphi (z).$ Then $f$ satisfies $f(\frac{3}{4})=-\frac{3}{4}$ and $f^\prime (\frac{3}{4})=\varphi^\prime(0)\cdot g^\prime(0)\cdot \varphi ^\prime(\frac{3}{4})=\frac{7}{16}\cdot k\cdot \frac{16}{7}=k.$ Take $k=-\frac{3}{4}$. – ts375_zk26 May 03 '16 at 04:56