# 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

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 $$$$(f(z)-f(z_0))-(z-z_0) = \sum_{n\geq 2}a_n (z^n -z_{0}^{n}) = (z-z_0)\sum_{n\geq 2} a_n (z^{n-1} +z^{n-2}z_0 + \cdots + z_{0}^{n-1}).$$$$ As $$|z^{n-1} +z^{n-2}z_0 + \cdots + z_{0}^{n-1}| 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 at 4:34
• OK. But in that case don't we need $f$ to be analytic on a region containing the closed unit disk? – skylark Jun 21 at 4:53
• 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 at 5:15
• You can see the post math.stackexchange.com/questions/65298/… – Riemann Jun 21 at 5:16
• Your explanation with $\delta$ (but with $\delta\to 1^-$) is fine – Hagen von Eitzen Jun 21 at 6:02

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 $$z \neq w$$.