Let \[ f(z)=z+\sum_{n=2}^{\infty}a_nz^n \] be univalent on \(\Delta\). Give an explicit bound \[ |a_n|\le M_n \] valid for every such \(f\), using only elementary methods such as the area theorem or the distortion theorems.
Solution
For \(n\ge2\), one explicit bound is \[ \boxed{ |a_n| \le M_n := \frac{(n+1)^2}{4} \left(\frac{n+1}{n-1}\right)^{n-1} }. \]
1. The distortion theorem and Cauchy’s coefficient estimate give, for every \(0<r<1\),\[ |a_n| \le \frac{r^{1-n}}{(1-r)^2}. \]
The Koebe distortion theorem gives, on \(|z|=r\), \[ |f(z)|\le\frac{r}{(1-r)^2}. \] By Cauchy’s formula for the coefficient \(a_n\), \[ a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{n+1}}\,dz, \] and therefore \[ |a_n| \le \frac{1}{r^n}\max_{|z|=r}|f(z)| \le \frac{r^{1-n}}{(1-r)^2}. \]
2. Optimizing this elementary estimate gives the displayed \(M_n\).
For fixed \(n\ge2\), set \[ \Phi_n(r)=\frac{r^{1-n}}{(1-r)^2}. \] Then \[ \frac{d}{dr}\log\Phi_n(r) =\frac{1-n}{r}+\frac{2}{1-r}. \] The unique critical point in \((0,1)\) is \[ r_n=\frac{n-1}{n+1}. \] Since \(\Phi_n(r)\to\infty\) at both endpoints, this is its minimum. Substitution gives \[ \begin{aligned} \Phi_n(r_n) &=\left(\frac{n+1}{n-1}\right)^{n-1} \left(\frac{n+1}{2}\right)^2\\ &=\frac{(n+1)^2}{4} \left(\frac{n+1}{n-1}\right)^{n-1}. \end{aligned} \] Thus \(|a_n|\le M_n\) for every normalized univalent \(f\).