# hayman's result for $A^2(D)$

Consider injective homolomorphic functions $f:\mathbb D\to \mathbb C$ on the unit disk $|z|\leq 1$, normalized by the conditions $f(0)=0$ and $f'(0)=1$.

Thus for $|z|\leq 1$ we have $f(z)=\sum_{k=0}^{\infty} a_k z^k$ with $a_0=0$ and $a_1=1$. Ludwig Bieberbach conjectured in 1916 and Louis de Branges proved in 1984 that for all $k \in \mathbb N$ the inequality $|a_k|\leq k$ holds. The Koebe function $K(z)=z/(1-z)^2$ does not belong to $H^2(D)$ and $A^2(D)$ . for which the extremal condition holds. Is there any analogous result for restricting the univalent functions to lie in $H^2(D)$.

Although the above is settled there is another result, due to Hayman and independent of De Branges' theorem, that $\lim a_n/n \leq 1$ and equality holds only for Koebe function. I think it might be worthwhile to find an analogue of this in $A^2(D)$. In $H^2(D)$, it doesn't make sense as $\lim a_n=0$

$A^2(D)$ is the bergman space with p=2

I am not sure what "analogous" means but the estimate $|a_n|\leq n$ is best possible when you restrict to $H^2$, and even when you restrict to polynomials. For the simple reason that the extremal function can be approximated by polynomials injective in the unit disc.
• @Kuoshik: for each strictly smaller radius $r<1$ all Taylor's approximations $f_k$ are univalent for sufficiently large $k$. This is a direct consequence of the Argument Principle. Now modify them as $f_{k,1}(z)=f_k((1-1/k)z)$ to get univalent approximation over the disc. Oct 27 '13 at 17:50