Let \[ f(z)=z+a_2z^2+a_3z^3+\cdots \] be univalent on \(\Delta\), and suppose \[ \sup_{z\in\Delta}|f(z)|\le M. \] Show that \[ |a_2|\le2\sqrt{1-\frac1M}. \]
Hint: the normalized exterior map obtained from \(f(z^{-2})^{-1/2}\) omits the disk of radius \(M^{-1/2}\); apply the area theorem.
Solution
Construct the normalized exterior map suggested in the hint. Its first negative Laurent coefficient is \(-a_2/2\), while boundedness of \(f\) forces a disk of area \(\pi/M\) into the omitted set. The area theorem then gives the desired coefficient estimate.
1. There is a holomorphic nonvanishing function \(h\) on \(\Delta\) with\(h(0)=1\) such that\[ f(w)=w h(w)^2. \]
Since \(f\) is univalent and \(f(0)=0\), the origin is the only zero of \(f\) in \(\Delta\), and it is simple because \(f'(0)=1\). Thus \[ q(w)=\frac{f(w)}{w} \] extends holomorphically to \(\Delta\), satisfies \(q(0)=1\), and never vanishes. Because \(\Delta\) is simply connected, \(q\) admits a holomorphic square root \(h\) with \(h(0)=1\). Hence \[ f(w)=w h(w)^2. \]
2. For \(|z|>1\), define\[ G(z)=\frac{z}{h(z^{-2})}. \]Then \(G\) is a normalized univalent exterior map, so \(G\in\Sigma\).
The function \(G\) is holomorphic for \(|z|>1\) and near infinity satisfies \(G(z)=z+O(z^{-1})\), because \(h(0)=1\).
Moreover, by 1, \[ G(z)^2 =\frac{z^2}{h(z^{-2})^2} =\frac1{f(z^{-2})}. \] Suppose \(G(z_1)=G(z_2)\). Squaring and using the preceding identity gives \[ f(z_1^{-2})=f(z_2^{-2}). \] Since \(f\) is univalent, \[ z_1^{-2}=z_2^{-2}, \] so \(z_2=\pm z_1\). But the definition shows \(G(-z)=-G(z)\), and \(G\) never vanishes on \(|z|>1\). Thus \(G(z_1)=G(-z_1)\) is impossible, so \(z_2=z_1\). Hence \(G\) is univalent and normalized at infinity, i.e. \(G\in\Sigma\).
3. The Laurent expansion of \(G\) at infinity begins\[ G(z)=z-\frac{a_2}{2z}+O(z^{-3}). \]
From \[ f(w)=w+a_2w^2+a_3w^3+\cdots, \] we obtain \[ f(z^{-2}) =z^{-2}\left(1+a_2z^{-2}+a_3z^{-4}+\cdots\right). \] Using the branch fixed in 2, \[ \begin{aligned} G(z) &=f(z^{-2})^{-1/2}\\ &=z\left(1+a_2z^{-2}+a_3z^{-4}+\cdots\right)^{-1/2}\\ &=z\left(1-\frac{a_2}{2}z^{-2}+O(z^{-4})\right)\\ &=z-\frac{a_2}{2z}+O(z^{-3}). \end{aligned} \] Thus, in the notation \[ G(z)=z+\sum_{n\ge1}\frac{b_n}{z^n}, \] we have \[ b_1=-\frac{a_2}{2}. \]
4. The omitted compact set \(K(G)=\mathbb C\setminus G(\widehat{\mathbb C}\setminus\overline\Delta)\) contains the disk\[ B(0,M^{-1/2}). \]
For \(|z|>1\), put \(w=z^{-2}\in\Delta\). By hypothesis, \[ |f(w)|\le M. \] Using \(G(z)^2=1/f(w)\) from 2, \[ |G(z)|=|f(w)|^{-1/2}\ge M^{-1/2}. \] Thus no point with modulus less than \(M^{-1/2}\) belongs to the image of \(G\), so \[ B(0,M^{-1/2})\subset K(G). \] Consequently \[ \operatorname{area}K(G)\ge\frac{\pi}{M}. \]
5. The area theorem gives\[ \operatorname{area}K(G) \le \pi\left(1-\frac{|a_2|^2}{4}\right). \]
The area theorem says that for \[ G(z)=z+\sum_{n\ge1}\frac{b_n}{z^n}\in\Sigma, \] \[ \operatorname{area}K(G) =\pi\left(1-\sum_{n\ge1}n|b_n|^2\right). \] Discarding all terms except \(n=1\) gives \[ \operatorname{area}K(G) \le\pi(1-|b_1|^2). \] By 3, \(|b_1|=|a_2|/2\), proving the claim.
6. Therefore\[ |a_2|\le2\sqrt{1-\frac1M}. \]
Combining 4 and 5, \[ \frac\pi M \le\pi\left(1-\frac{|a_2|^2}{4}\right). \] Canceling \(\pi\) and rearranging gives \[ \frac{|a_2|^2}{4}\le1-\frac1M. \] Taking square roots yields \[ |a_2|\le2\sqrt{1-\frac1M}. \]