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de Branges has proved de Branges's theorem (the famous Bieberbach conjecture) that if a holomorphic function $f(z) = z+\sum_{n=2}^{\infty} a_nz^n$ in the unit disk $D = \{z\in \mathbb{C},|z| \leq 1\}$ is univalent, then we have $|a_n| \leq n,\forall n\geq 2$. Conversely, let's consider a holomorphic function $g(z) = z+\sum_{n=2}^{\infty} b_nz^n$ which is defined in $D$ and satifies $|b_n| \leq n$, then what are the general sufficient conditions(I've known some special conditions on this problems, such as Nehari's univalence criterion and other criterions, unfortunately, they are not in full generality) to ensure $g(z)$ is univalent. Any clues and facts are welcomed, best regards !


Updated question: necessary and sufficient conditions for a holomorphic function defined in the unit disk to be univalent (as far as I known, several conditions have be proposed, but all of them seem to be not practical), simple forms and only depend on function g(z) or its derivatives, integrals, their combinations, and so on. For example something like Milin's inequality. Unfortunately, I've tried several variants of this inequality (together with some additional conditons), but fails.

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    $\begingroup$ Bieberbech has conjectured de Branges' theorem. $\endgroup$
    – abx
    Sep 24 '20 at 8:25
  • $\begingroup$ @abx,fixed, thanks! :) $\endgroup$
    – Milin
    Sep 24 '20 at 8:51
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It is highly unlikely that anything "reasonable" can be said in terms of the coefficients.

Already the allowable region for $(a_2,a_3)$ (so they are the first non-trivial coefficients of $z+a_2z^2+a_3z^3+...$ univalent) is quite complicated as for example the sharpness (ie for each $0 \le \alpha \le 1$ there is an univalent functions for which the equality holds) of the Fekete-Szego inequalities $|a_3-\alpha a_2^2| \le 1+2e^{\frac{-2\alpha}{1-\alpha}}, 0 \le \alpha \le 1$ shows - here $\alpha =0$ corresponds to the highly non-trivial third coefficient bound $|a_3| \le 3$, while $\alpha=1$ corresponds to the easy $|a_3-a_2^2| \le 1$

See also Schaeffer Spencer book on the coefficient regions for Schlicht functions https://www.ams.org/books/coll/035/

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  • $\begingroup$ @ Conrad, many thanks for your kind reply. $\endgroup$
    – Milin
    Sep 25 '20 at 1:43
  • $\begingroup$ hopefully it was helpful - finding the coefficient space for $S$ seems hopeless as noted; finding analytic conditions in terms of the function, who knows, maybe something reasonable can be done $\endgroup$
    – Conrad
    Sep 25 '20 at 3:06
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You did not specify in what terms do you need a necessary and sufficient condition. Presumably in terms of the coefficients. One such condition can be obtained as follows: $f(z)$ is univalent if and only if $$F(z,w)=\frac{f(z)-f(w)}{z-w}$$ has no zeros in the unit polydisk, which is equivalent to $$\log F(z,w)$$ to be convergent in the unit polydisk. And for this, there is a formula generalizing the Cauchy Hadamard's formula for the radius of convergence. So this property is (in principle) expressed in terms of the coefficients. Further conditions of this sort can be found in the book of Goluzin, Geometric theory of functions of a complex variable.

Another criterion of univalence is Milin's criterion, see, for example https://iopscience.iop.org/article/10.1070/SM1967v003n01ABEH002364, Theorem 1. I believe that some version of Milin's criterion was used in de Branges original proof.

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  • $\begingroup$ @ Alexandre, thanks for your clarification and great answers, what I mean is that the conditions which satisfy: 1.simple forms. 2. only depend on function $g(z)$ or its coefficients, that's all. Actually, I have consideried the converse of a modified version of Milin's conjecture with some additional conditions, but seems do not work. $\endgroup$
    – Milin
    Sep 25 '20 at 1:46

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