1
$\begingroup$

$\textbf{Theorem(Schottky)}$ 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 $z \in \mathbb{D_r}=\{z \in \mathbb{C}: |z|<r \}$.

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?

$\endgroup$
  • $\begingroup$ 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 $z\in\mathbb{D}_r$. $\endgroup$ – user10354138 Jun 15 at 12:26
  • $\begingroup$ Oh yes, that was a typo in the theorem, I fixed it, $\endgroup$ – EinStone Jun 15 at 12:40
  • $\begingroup$ 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.) $\endgroup$ – Lukas Geyer Jun 17 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.