Existence of an analytic function in a unit disc

Question

Let $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$.Then, is it true that there exists a holomorphic function $f:\mathbb{D} \to\mathbb{D}$ such that $f(\frac{1}{2})=-\frac{1}{2}$ and $f'(\frac{1}{4})=1$?
Can we use Schwarz's lemma to solve it?

Answer 1

This is a double application of Schwarz-Pick; let $f(1/4)=w$

first $1=|f'(1/4)| \le \frac{1-|w|^2}{1-|1/4|^2}$, so one gets $|w| \le 1/4$

then: $|\frac{f(1/2)-f(1/4)}{1-\overline {f(1/2)} f(1/4)}| \le |\frac{(1/2)-(1/4)}{1-\overline {(1/2)} (1/4)}| =\frac {2}{7}$, so:

$|\frac{2w+1}{2+w}| \le \frac {2}{7}$ with $|w| \le 1/4$

But (see below for a direct proof) $|\frac{2w+1}{2+w}| \ge |\frac{2(-1/4)+1}{2+(-1/4)}|=2/7$ when $|w| \le 1/4$ with equality iff $w=-1/4$ hence we get $f(1/4)=-1/4$ and $f$ is a disc automorphism as we have equality in Schwarz-PIck, while it is obvious that $f(z)=-z$ is the only such, but $f'(1/4)=-1$ so contradiction and no such function exists!

(for example $|\frac{2w+1}{2+w}| =|2-\frac{3}{2+w}| \ge 2-|\frac{3}{2+w}|$, while $-|\frac{3}{2+w}| \ge -|\frac{3}{2+(-1/4)}|=-12/7$ since $|2+w| \ge |2+(-1/4)|, |w| \le 1/4$)