If $f(z)=z+a_2 z^2 + a_3 z^3 + \cdots$ and $\sum_{n\geq 2} n|a_n| \leq 1$, then $f$ is univalent

Question

I am trying to understand the proof from this book of the following theorem:
$\textbf{Theorem}:$ If $f(z)= z+ \sum_{n\geq 2} a_n z^n$ is such that $\sum_{n\geq 2} n|a_n| \leq 1$, then $f$ is univalent in the unit disk $\mathbb{D}$.

In the proof, the authors first proved that the series converges and hence $f$ is an analytic function in $\mathbb{D}$. Till now I have no problem in understanding. The remaining part of the proof given in the book is as follows:

Let $|z_0|<1$, then \begin{equation} (f(z)-f(z_0))-(z-z_0) = \sum_{n\geq 2}a_n (z^n -z_{0}^{n}) = (z-z_0)\sum_{n\geq 2} a_n (z^{n-1} +z^{n-2}z_0 + \cdots + z_{0}^{n-1}). \end{equation} As $|z^{n-1} +z^{n-2}z_0 + \cdots + z_{0}^{n-1}|<n$ for $|z|<1$, we have $$|(f(z)-f(z_0))-(z-z_0)| < |z-z_0|\sum_{n\geq 2} n|a_n| \leq |z-z_0|.$$ According the Rouche's theorem, $f(z)-f(z_0)$ and $z-z_0$ have the same number of zeros in $\mathbb{D}$, that is $f(z)=f(z_0)$ has exactly one solution.

$\textbf{My doubt}:$ In Rouche's theorem, the inequality must hold on the boundary of the region, but in this proof $|z|<1$. So how can we apply Rouche's theorem here?

$\textbf{My explanation}: $ From the proof we can say that the inequality holds on any circle inside $\mathbb{D}$, as we can fix $|z|=\delta <1$, and $|z_0|<|z|$. We get the conclusion by allowing $\delta \rightarrow 1^-$.

$\textbf{My question}:$ Is my explpanation correct? Or I am missing out some basic fact?

$|z^{n-1} +z^{n-2}z_0 + \cdots + z_{0}^{n-1}|<n$ is also true when $|z|=1$ , so when $|z|=1$, you can get $|(f(z)-f(z_0))-(z-z_0)| < |z-z_0|$. — Riemann, Jun 21 '19 at 04:34
OK. But in that case don't we need $f$ to be analytic on a region containing the closed unit disk? — skylark, Jun 21 '19 at 04:53
From $\sum_{n\geq 2} n|a_n| \leq 1$, you can prove the convergent radius $R$ of $f(z)= z+ \sum_{n\geq 2} a_n z^n$ is not less than than $1$, that is to say $R\geq1$. — Riemann, Jun 21 '19 at 05:15
You can see the post https://math.stackexchange.com/questions/65298/general-form-of-rouches-theorem — Riemann, Jun 21 '19 at 05:16
Your explanation with $\delta$ (but with $\delta\to 1^-$) is fine — Hagen von Eitzen, Jun 21 '19 at 06:02

score 6 · Accepted Answer · answered Jun 21 '19 at 05:46

This is not an explanation of the steps in the book but I would like to point out that the proof in the book looks a bit silly to me. You don't require Rouche's Theorem. There is a very elementary 'High School' proof: suppose $f(z)=f(w)$. Then $z-w=a+a_2(w^{2}-z^{2})+a_3(w^{3}-z^{3})+\cdots$. Note that $|w^{n}-z^{n}| =|z-w|(|w^{n-1}+w^{n-2}z+\cdots+wz^{n-2}+z^{n-2}|< n|z-w|$ if $z \neq w$. Using the hypothesis we arrive at the contradiction $|z-w| <|z-w|$ if $z \neq w$.

If $f(z)=z+a_2 z^2 + a_3 z^3 + \cdots$ and $\sum_{n\geq 2} n|a_n| \leq 1$, then $f$ is univalent

1 Answers1

Linked