Suppose $f: \mathbb{D}\to \mathbb{C}$ is a univalent function with $$f(z)=z+a_2z^2+a_3z^3+\cdots.$$ The Bieberbach conjecture/de Branges' theorem asserts that $|a_n|\leq n$ with equality for the Koebe function, which has an unbounded image. Suppose we restrict to the class of univalent functions whose image is actually bounded. Is there a better bound than $|a_n|\leq n$ ?
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2 Answers
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The answer is yes, even the rate of increase of coefficients is much smaller. However even the precise rate is not known (not speaking of the exact estimate). The strongest results in this direction are here:
MR1162188 Carleson, Lennart; Jones, Peter W. On coefficient problems for univalent functions and conformal dimension. Duke Math. J. 66 (1992), no. 2, 169–206.
answered Jun 11, 2016 at 8:04
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1$\begingroup$ There are a couple of papers by Hedenmalm and Shimorin (Duke Math. J. '05, Proc AMS '07) on universal integral means spectra that may be relevant: as a byproduct of their results they get some further bounds for the decay rate of coefficients. $\endgroup$ Commented Jun 11, 2016 at 14:49
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$\begingroup$ This is a very interesting result, but it's not clear if this answer the OP's question. C-J prove that $|a_n|\le Cn^{-b}$, with $b>0$ (note that they don't normalize by requiring $a_0=0$, $a_1=1$, this would be pointless in their setting); this just tells us that for any given $f$, eventually $|a_n|\ll n$, but it could still be that for a given $n$, all one can say in general is $|a_n|\le n$. $\endgroup$ Commented Jun 11, 2016 at 16:00
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$\begingroup$ @Christian Remling: They do normalize their functions, and actually estimate $A_n$ which is the sup of $a_n$ over the normalized class. So the estimate is definitely better than $n$ for large $n$. $\endgroup$ Commented Jun 12, 2016 at 21:25
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$\begingroup$ @AlexandreEremenko: Yes, I have read the result very carefully, though not in the original source, but in Garnett's book. In any event, the point of my comment was to point out the difference between an asymptotic bound, valid for $n\ge N_0(f)$, and a bound valid for all $f$ for a given $n$ (addressed by Lasse's answer also). $\endgroup$ Commented Jun 12, 2016 at 21:26
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$\begingroup$ In the original paper, this is done slightly differently, by imposing a bound on $f$, but of course similar comments apply. $\endgroup$ Commented Jun 12, 2016 at 21:30
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This is a softer answer than Alex's, in line with Christian's comment to that answer.
If you are looking at the family of all bounded conformal maps, then you clearly do not get a better bound, as the Koebe function can be approximated by such.
On the other hand, if you look at all functions with the usual normalisations satisfying a fixed bound (say functions taking values in the unit disc) , this family is compact, and hence you do get a better bound for all coefficients. Of course this does not help you finding those bounds, which according to Alex's answer would seem to be currently hopeless.
answered Jun 12, 2016 at 8:54