Bound for a holomorphic function on the unit disk to a 3-punctured plane

Question

Let $\mathbb{Y}$ be the complex plane with the three points $\{1-i,1,1+i\}$ excluded, $\mathbb{Y} = \mathbb{C}\setminus\{1-i,1,1+i\}$. Let $F(z)$ be a holomorphic function defined on the open unit disk $\mathbb{D}$, $F(\mathbb{D})\subseteq \mathbb{Y}$, such that $F(0)=0$.
How close can $|F(\frac{1}{2})|$ get to $1$? Namely, can we, for any $\varepsilon>0$ construct a function $F$ with the properties indicated above such that $|F(\frac{1}{2})-1|<\varepsilon$?

Answer 1

The answer is no by Schottky theorem.

Consider $G$ the Mobius transform that sends $1-i \to 0, 1 \to \infty, 1+i \to 1$; an easy computation shows that

$G(z)=\frac{z-1+i}{2(z-1)}$ while $G(1+c)=\frac{c+i}{2c}$ so $|G(1+c)| \ge \frac{1}{4|c|}$ if $|c| < 1/2$ say since then $|c+i| >1/2$ by the reverse triangle inequality.

Assume there is such $F$ as above, then $H=G\circ F$ maps the unit disc to the plane excluding $0,1$ and $H(0)=\frac{1-i}{2}, |H(0)|=\sqrt 2 /2$ while if $F(1/2)=1+c$ with $|c|<1/2$ we have that $|H(1/2)| \ge \frac{1}{4|c|}$ by the above.

But by Schottky theorem we have $\log |H(z)| \le 7 \frac{1+|z|}{1-|z|}$ so $\log |H(1/2)| \le 21$ which means that $|c|$ cannot be too small ($|c| \ge \frac{1}{4e^{21}}$) so $F(1/2)$ cannot get closer than that to $1$.