# Bound in the proof of Schottky's Theorem

$$\textbf{Theorem(Schottky)}$$ Let $$M > 0$$ and $$r\in (0,1)$$ be given. Then there exists a constant $$C > 0$$ such that the following holds: If F is holomorphic in in the unit disk $$\mathbb{D}$$, omits 0 and 1 from its range, and if $$|F(0)| ≤ M$$, then $$|F(z)| ≤ C$$ for all $$z \in \mathbb{D_r}=\{z \in \mathbb{C}: |z|.

I am trying to understand the proof and here is how far I got: We define

\begin{align*} A &= \frac{\log F}{2\pi i} \\ B &= \sqrt{A} - \sqrt{A-1} \\ H &= \log{B} \end{align*}

One can show that these are well defined and both logs are chosen to have an argument in $$[-\pi,\pi]$$. We see that the definition $$\exp(H)=\sqrt{\frac{\log F}{2\pi i}} - \sqrt{\frac{\log F}{2\pi i}-1}$$ implies that $$\frac{\exp(H(z)) + \exp(-H(z))}2 = \sqrt{\frac{\log F}{2\pi i}}.$$ From here we get the estimate $$|F(z)| \leq \exp(\pi e^{2|H(z)|}).$$ Further one can show that $$|H(z)| \le |H(0)| -130 \log(1-r)$$ which shows that $$|F(z)|\leq C$$, where $$C$$ only depends on $$|H(0)|$$ and $$r$$.

So it remains to show that $$|H(0)|\leq C_1$$ where $$C_1$$ only depends on $$M$$. We have $$|\textrm{Im} (H(0))|\leq \pi$$ by construction of $$H$$.

We can assume that $$|F(0)| \geq \frac12$$ (otherwise we work with $$1-F$$). Since then $$|F(0)|$$ is bounded we get that $$C_2 \geq \left|\sqrt{\frac{\log F(0)}{2\pi i}}\right| = \left|\frac{\exp(H(0)) + \exp(-H(0))}2\right| \geq \sinh(\textrm{Re} (H(0)))$$ so that $$\textrm{Re} (H(0)) \leq \sinh^{-1}(C_2)$$ for some constant $$C_2$$ that depends only on M.

To complete the proof we need a lower bound for $$\textrm{Re} (H(0))$$, which is supposed to work similarly, but I don't see how. Can anyone help?

• You can't bound it on the entire $z\in\mathbb{D}$, because we can take $F$ to be the conformal equivalence $\mathbb{D}\to\mathbb{H}$ for example. You can only bound it on $z\in\mathbb{D}_r$. – user10354138 Jun 15 at 12:26
• Oh yes, that was a typo in the theorem, I fixed it, – EinStone Jun 15 at 12:40
• I don't see how you can bound the imaginary part of $\log F$ in any way. E.g., if $F(z) = e^{az-b}$ with $0<a<b$, then $F$ omits $0$ and $1$, and $\log F(z) = a z - b$ has imaginary part ranging in $(-a,a)$, so you can make that range arbitrarily big. (Geometrically, this function wraps the unit circle many times around $0$ in the punctured unit disk.) – Lukas Geyer Jun 17 at 23:08