Let \(D_r(0)=\{z\in\mathbb C:|z|<r\}\) and \(\Delta=D_1(0)\). Use the following subordination result: if \(F:\Delta\to\mathbb C\) is injective and holomorphic, and \(g:\Delta\to F(\Delta)\) is holomorphic with \(g(0)=F(0)\), then \(g(D_r(0))\subset F(D_r(0))\) for \(0\leq r<1\). Prove that if \(g\) is holomorphic on \(\Delta\), \(g(0)=0\) and \(|\operatorname{Re}g(z)|<1\) for every \(z\in\Delta\), then
\[ | g ( z ) | \leq { \frac { 2 } { \pi } } \log \left\{ { \frac { 1 + | z | } { 1 - | z | } } \right\} \]
for all \(z \in D _ { 1 } ( 0 )\)
Solution
Let \[ S=\{w\in\mathbb C:|\operatorname{Re}w|<1\}. \]
1. The function\[ F(\zeta)=\frac{2i}{\pi}\log\frac{1+\zeta}{1-\zeta} \]is a biholomorphism from the unit disk onto \(S\) and satisfies \(F(0)=0\).
The Möbius map \[ M(\zeta)=\frac{1+\zeta}{1-\zeta} \] maps the unit disk biholomorphically onto the right half-plane. On that half-plane take the logarithm with argument in \((-\pi/2,\pi/2)\). Its image is the horizontal strip \[ \{u+iv:|v|<\pi/2\}. \] Multiplication by \(2i/\pi\) maps that strip biholomorphically onto \(S\), because \[ \operatorname{Re}\left(\frac{2i}{\pi}(u+iv)\right)=-\frac{2v}{\pi}. \] The composition is the displayed \(F\), and \(F(0)=0\).
2. Pointwise subordination uses the closed disk of radius \(|z|\).
The hypothesis \(|\operatorname{Re}g(z)|<1\) says exactly that \(g:\Delta\to S=F(\Delta)\), and \(g(0)=F(0)\). Equivalently, \[ h=F^{-1}\circ g:\Delta\to\Delta \] is holomorphic with \(h(0)=0\). Schwarz’s lemma [@SS03] gives \[ |h(z)|\le|z|. \] Thus, for \(r=|z|\), put \(K_r=\{\zeta\in\mathbb C:|\zeta|\leq r\}\). Then \(h(z)\in K_r\) and \[ g(z)=F(h(z))\in F(K_r). \] This includes \(r=0\), where \(K_0=\{0\}\) and \(g(0)=0\). The open-disk subordination statement in the problem follows from the same estimate whenever \(|z|<r\); the pointwise estimate at \(r=|z|\) uses the closed disk \(K_r\).
3. The explicit strip map has the required radial bound.
2. Then\[ |g(z)|=|F(h(z))| \le\frac2\pi\log\frac{1+|z|}{1-|z|}, \]which is the required estimate.