Schottky's theorem

By [L1], choose $h$ with $e^{2\pi ih}=f$. Since $f$ omits $1$, the function $h$ omits every integer. In particular $h$ and $h-1$ are nowhere zero, so [L2] gives holomorphic $u,v$ on $\mathbb{D}$ with $u^2=h$ and $v^2=h-1$. Then $(u-v)(u+v)=1$, so $u-v$ is nowhere zero, and [L2] gives a holomorphic $g$ with $e^g=u-v$.

Using [L4] and $(u-v)(u+v)=1$, one gets $u+v=e^{-g}$ and hence $2u=e^g+e^{-g}=2\cosh g$. Therefore $u=\cosh g$, $h=u^2={\cosh }^{2}g=(1+\cosh (2g))\mathrm{/}2$, and $f=e^{2\pi ih}=-e^{\pi i\cosh (2g)}$.

Put $R_0:=\max \{2,R\}$. First suppose $R_0^{-1}\le \mathrm{\mid }f(0)\mathrm{\mid }\le R_0$. In step 1.1 choose the logarithm $h$ so that $\mathrm{\mid }\mathrm{Re}h(0)\mathrm{\mid }\le 1\mathrm{/}2$, which is possible because $h$ is determined up to an integer. Since $\mathrm{\mid }f(0)\mathrm{\mid }=e^{-2\pi \mathrm{Im}h(0)}$, this also gives $\mathrm{\mid }\mathrm{Im}h(0)\mathrm{\mid }\le (\log R_0)\mathrm{/}(2\pi )$ and hence $\mathrm{\mid }h(0)\mathrm{\mid }\le H(R_0)$ for a fixed bound $H(R_0)$. Now $\mathrm{\mid }u(0)\mathrm{\mid }=\sqrt{\mathrm{\mid }h(0)\mathrm{\mid }}$ and $\mathrm{\mid }v(0)\mathrm{\mid }=\sqrt{\mathrm{\mid }h(0)-1\mathrm{\mid }}$, so for $P(R_0):=\sqrt{H(R_0)}+\sqrt{H(R_0)+1}$ one has $\mathrm{\mid }u(0)-v(0)\mathrm{\mid }\le P(R_0)$. Because $(u-v)(u+v)=1$ and $\mathrm{\mid }u(0)+v(0)\mathrm{\mid }\le P(R_0)$, one also has $\mathrm{\mid }u(0)-v(0)\mathrm{\mid }\ge P(R_0{)}^{-1}$. Finally choose the logarithm $g$ of $u-v$ so that $\mathrm{\mid }\mathrm{Im}g(0)\mathrm{\mid }\le \pi$; then $\mathrm{\mid }\mathrm{Re}g(0)\mathrm{\mid }\le \log P(R_0)$ and therefore $\mathrm{\mid }g(0)\mathrm{\mid }\le A(R_0):=\log P(R_0)+\pi$.

Let ${\alpha }_n:=\frac{1}{2}\mathrm{arcosh}(2n+1)$ for $n\ge 0$, and set $E:=\{\pm {\alpha }_n+m\pi i:n\ge 0,m\in \mathbb{Z}\}\cup \{\pm {\alpha }_n+(m+\frac{1}{2})\pi i:n\ge 0,m\in \mathbb{Z}\}$. If $g(z)=\zeta \in E$, then $\cosh (2\zeta )$ is an odd integer, so step 2.1 gives $f(z)=1$, impossible. Hence $g(\mathbb{D})\cap E=\mathrm{\varnothing }$. The horizontal gaps between consecutive ${\alpha }_n$ are less than $1$, and the two vertical translates reduce the vertical gap to $\pi \mathrm{/}2$, so every open Euclidean disc of radius $2$ in $\mathbb{C}$ meets $E$.

Fix $w\in \mathbb{D}$ and rescale $g$ from the disc $D(w,1-\mathrm{\mid }w\mathrm{\mid })$ to the unit disc. If $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }>2\mathrm{/}(b(1-\mathrm{\mid }w\mathrm{\mid }))$, then [L3] would produce a schlicht disc of radius greater than $2$ inside $g(\mathbb{D})$, contradicting step 3.1. Therefore $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }\le 2\mathrm{/}(b(1-\mathrm{\mid }w\mathrm{\mid }))$ for every $w\in \mathbb{D}$.

If $\mathrm{\mid }f(0)\mathrm{\mid }\ge R_0^{-1}$, integrate step 4.1 radially and use step 2.2 to get $\mathrm{\mid }g(z)\mathrm{\mid }\le A(R_0)+(2\mathrm{/}b)\log \text{\hspace{0.17em}⁣}(1\mathrm{/}(1-r))$ for $\mathrm{\mid }z\mathrm{\mid }\le r$. The formula in step 2.1 then bounds $\mathrm{\mid }f(z)\mathrm{\mid }$ by a constant depending only on $R$ and $r$. If $\mathrm{\mid }f(0)\mathrm{\mid }<R_0^{-1}\le 1\mathrm{/}2$, apply the same construction to $1-f$: its center value lies between $1\mathrm{/}2$ and $3\mathrm{/}2$, so the preceding case with center parameter $2$ uniformly bounds $\mathrm{\mid }1-f(z)\mathrm{\mid }$ and hence $\mathrm{\mid }f(z)\mathrm{\mid }\le 1+\mathrm{\mid }1-f(z)\mathrm{\mid }$. Taking the larger of the two bounds gives the required $C(R,r)$.