Does there exist a holomorphic function$f: D \to D$ with $f(\frac{1}{2}) =-\frac{1}{2}$ and $f '(\frac{1}{4}) =1$ where $ D= \{ z \in \mathbb{C} : |z|<1\}$.
I cannot use any of Schwarz lemma or Schwarz-Pick theorem here.
Any help would be appreciated. Thanks in advance.
We apply Schwarz-Pick twice - let $f(\frac{1}{4})=a$
$1=|f'(\frac{1}{4})| \le \frac{1-|a|^2}{1-\frac{1}{16}}$ means $|a| \le \frac{1}{4}$
$|\frac{f(1/2)-f(1/4)}{1-\bar f(1/2) f(1/4)}| \le \frac{1/2-1/4}{1-1/8}$ means
$|\frac{2a+1}{a+2}| \le \frac{2}{7}$
But $|\frac{2a+1}{a+2}| = |2-\frac{3}{a+2}| \ge 2-\frac{3}{|a+2|} \ge 2-\frac{12}{7}=\frac{2}{7}$ means that we must have equality in Schwarz-Pick both times, so $a=-\frac{1}{4}$ and $f$ is a Mobius transform and by inspection $f(z)=-z$ (or noting that any Mobius transform cannot have more than one fixed point inside the unit disc unless it is the identity and $-f$ has two such), but then $f'=-1$ so contradiction!
(assume we have equality in $|2-\frac{3}{a+2}| \ge 2-\frac{3}{|a+2|} \ge 2-\frac{12}{7}=\frac{2}{7}$; it follows from the last inequality that $-\frac{3}{|a+2|}=-\frac{12}{7}$ which means $|a+2|=7/4$; the triangle inequality $|2+a| \ge 2-|a| \ge 7/4$ must be an equality, so $2,a$ must be opposite, hence $a$ is a negative real)
(quick proof that a Mobius automorphism of the disc $g(z)=\gamma \frac{z-c}{1-\bar c z}, |\gamma|=1$ with two fixed points inside the disc is the identity: $g(z)=z$ means $z-\bar c z^2=\gamma (z-c)$; if $c=0, g(z)= \gamma z$ and that has only $0$ as fixed point unless $\gamma=1$; if $c \ne 0$ we get a quadratic with the product of roots having absolute value $|\gamma \frac{c}{\bar c}|=1$ by Viete, hence if a root has $|z_1| <1$ the other must have $|z_2| >1$)
Hence there is no such $f$