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The following theorem is due to Dieudonné:

Let $p(z) = a_0 + a_1 z + \cdots + a_n z^n$, a complex polynomial. $p$ is univalent on $D$, the unit disk, if and only if $f_\theta(z) = \sum_{k=1}^n a_k \dfrac{\sin(k\theta)}{\sin(\theta)} z^{k-1}$ has no zeros in $D$. ($\theta \in (0, \dfrac{\pi}{2}]$).

The idea of the proof is as follows:

1) $p$ is univalent on $D$ if and only if $t\mapsto p(re^{it})$ is injective on $t\in [0,2\pi)$ for each $0< r < 1$.

2) $f_\theta$ has no zero in $D$ for $\theta \in (0,\dfrac{\pi}{2}]$ if and only if $p(re^{i(t+\theta)}) \ne p(re^{i(t-\theta)})$ for $t \in [0,2\pi)$, $0<r<1$ and $\theta \in (0,\dfrac{\pi}{2}]$.

The second pars of 1) and 2) are saying the exact same thing, so this would prove the theorem.

What (I think) I don't really get is is the statement 1). It seems like the proof for $p$ polynomial should work for any analytic function.

So it seems like the theorem should be true for a general analytic function, where the sum becomes an infinite sum.

What am I missing?

EDIT: Basically, it seems like nowhere in the proof of Dieudonné the fact that $p$ is a polynomial is used.

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See if you can come up with an analytic function yourself for which the hypotheses hold but the conclusion of the theorem fails. That is what happens next, you alternate trying to prove the theorem under weakened hypotheses with trying to find a counterexample under weakened hypotheses. Eventually you learn a little. –  Will Jagy Jan 3 '12 at 0:54
    
Can you give me a hint? I haven't been able to come up with a counterexample. –  Braindead Jan 3 '12 at 1:14
    
I did not expect this. I am sorry that I gave the impression that I know an answer to this. I was pleased with your curiosity, and hoped you could proceed on your own. Meanwhile, given any constant $a$ with $|a| \leq 1,$ the linear fractional transformation $$ h(z) = (z -a) / (1 - \bar{a}z) $$ is univalent and takes the disk to itself. How would you characterize the behavior of $h_\theta (z)?$ –  Will Jagy Jan 3 '12 at 3:27
    
Have you seen Grunsky? en.wikipedia.org/wiki/Grunsky%27s_theorem –  Will Jagy Jan 3 '12 at 3:53
    
$h$ is univalent in $D$ implies $h_\theta$ has no zero in $D$, no question. The fact that $h_\theta$ has no zero in $D$ for $\theta$ in $(0,\pi/2]$ is equivalent to the fact that $h$ restricted to circles $|z|=r$ is injective for each $0<r<1$. I mean, $h_t(z) = (h(z e^{it}) - h(z e{-it}))/(z \sin t)$. In the case of the Mobius tranformation, it just gives $h_t(z) = \dfrac{2i(1-|a|^2)}{(e^{i t}-\bar{a}z)(1-b e^{it}z)}$. –  Braindead Jan 3 '12 at 4:03

1 Answer 1

up vote 3 down vote accepted

You do, in fact, have the following statement:

Lemma Let $H$ be a open, simply connected domain bounded by a Jordan curve $\partial H$. If $f$ is an analytic function on $H$ that extends continuously to $\partial H$ and is injective on $\partial H$, then $f$ is univalent in $H$ and $f(H)$ is the inner domain of the Jordan curve $f(\partial H)$.

(See C. Pommerenke, Univalent Functions; Lemma 1.1 and Corollary 9.5.)

This implies that step (1) in the proof you wrote down carries over directly to the case of arbitrary analytic functions.


Step (2) can be formulated thus:

By step one it suffices to consider the zero sets of $P(z e^{i\theta}) - P(z e^{-i\theta})$, when $z \neq 0$ and $\theta \in (0,\pi/2]$. This is equivalent, in the region concerned, to looking at the zero sets of the function

$$ Q(\theta;z) = \frac{P(z e^{i\theta}) - P(z e^{-i\theta})}{z (e^{i\theta} - e^{-i\theta})} $$

using that the denominator is non-zero in the region concerned. Since the numerator, a difference of two holomorphic functions (in $z$), vanishes at $z = 0$, we have that $Q(\theta;z)$ is also holomorphic in $z$. Using that $\lim_{\theta\to 0}Q(\theta;z) = P'(z)$ and $Q(\theta;0) = Q(\theta';0)$ (so locally univalent implies that $Q(\theta;0) \neq 0$), you get that $P$ being univalent is equivalent to $Q$ having no zeroes.


Indeed, if you look at the Zbl review of Dieudonné's 1931 paper (I don't have the paper handy to check, unfortunately), the "infinite sum" version of what you want is described. (I assume that is how the theorem is stated in the paper, but I am not sure. Most citations to the criterion, however, cites this other slightly earlier paper which only deals with the polynomials.)


Thanks to the French national library, I found a digital copy of the paper in question ("Sur les fonctions univalentes"), from which I quote

La condition nécessaire et suffisante pour que la fonction (1) soit univalente dans le cercle unité, est que l'équation en $x$ $$ > \phi(x,\theta) \equiv 1 + a_2 x \frac{\sin 2\theta}{\sin \theta} + > \ldots + a_n x^{n-1} \frac{\sin n\theta}{\sin \theta} + \ldots = 0 $$ n'ait aucune racine dans le cercle unité, quelle que soit la valeur de $\theta$ comprise entre $0$ et $2\pi$.

where he defined the function in expression (1) to be

$$ f(z) = z + a_2 z^2 + \ldots + a_n z^n + \ldots $$

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Wow, thank you for such a complete and detailed response! –  Braindead Jan 3 '12 at 16:55
    
Question: Is there a simpler proof of step (1) for polynomials that doesn't use the stated lemma? –  Braindead Jan 3 '12 at 17:01
    
There may be? The stated lemma is more or less a direct application of the argument principle, which is fairly elementary (and its proof for polynomials, if you allow the fundamental theorem of algebra, is a direct computation). Off the top of my head I don't know of a more direct way of showing step (1) for polynomials though. –  Willie Wong Jan 4 '12 at 10:08

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