TITLE: Convex Hull of univalent functions and Bieberbach Conjecture
QUESTION [7 upvotes]: Consider the class $S$ of univalent functions on the unit disk $D$ normalized so that $f(0)=0$ and $f'(0)=1$. 
Each function in $S$ satisfy the Bieberbach conjecture, that is the $n$-th coefficient in the power series  expansion is less or equal than $n$.  I am looking at the convex hull of class $S$. Any function there satisfies the Bieberbach conjecture by an easy computation.
Are there functions which do not belong to the convex hull of $S$ but satisfy the Bieberbach conjecture? 
More generally, how do you prove that a function is not in the convex hull of $S$ without using the Bieberbach conjecture? There are characterizations of convex hulls of subfamilies of $S$, for example starlike maps, in terms of integral representations.

REPLY [4 votes]: By extremality, a function of the form $f(z)= z+\sum_{n\geq 2} n a_n z^n$ with $|a_n|=1$ belongs to the convex hull of $S$ if and only if it belongs to $S$. 
According to wikipedia, the only univalent such functions are those for which $a_n= \omega^{n-1}$ for some $|\omega|=1$, so you get many examples of functions satisfying Bieberbach's conjecture but not in the convex hull of $S$. For example, $f(z) = 2z - \frac{z}{(1-z)^2}$.