MHB Is an analytic map with a certain condition injective?

  • Thread starter Thread starter Euge
  • Start date Start date
  • Tags Tags
    2017
Click For Summary
An analytic map f: D → C is considered, defined by a power series with coefficients a_n. The condition for injectivity is established as the convergence of the series sum of n|a_n| being less than |a_1|. The discussion highlights that under this condition, f is proven to be injective. Kokuhaku provided a solution to the problem, demonstrating the necessary proof. The topic emphasizes the relationship between the coefficients of the series and the injectivity of the analytic map.
Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
Here is this week's POTW:

-----
Consider an analytic map $f : \Bbb D \to \Bbb C$ such that $f(z) = \sum\limits_{n = 0}^\infty a_n z^n$ for all $z\in \Bbb D$. Prove that $f$ is injective, provided

$$\sum_{n = 2}^\infty n\lvert a_n\rvert < \lvert a_1\rvert.$$

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
This week's problem was solved by Kokuhaku. You can read his solution below.
Note that $f'(z)-a_1 = \sum_{n=2}^\infty na_n$ for $z \in \mathbb{D}$, so we actually have $|f'(z)-a_1|<|a_1|$ for $z \in \mathbb{D}$ (use triangle inequality).

Suppose now that $z_1 \neq z_2$ (from $\mathbb{D}$). Then using above inequality and estimation lemma (M-L inequality) we have
\begin{align*}
|f(z_2)-f(z_1)| &= \Bigl|\int_{z_1}^{z_2} f'(z)\, dz\Bigr| = \Bigl|a_1 \int_{z_1}^{z_2} dz + \int_{z_1}^{z_2} (f'(z) - a_1) \, dz\Bigr| \\
&\ge |a_1(z_2-z_1)| - \Bigl|\int_{z_1}^{z_2} (f'(z) - a_1) \, dz\Bigr| > |a_1||z_2-z_1| -|z_2-z_1||a_1| = 0,
\end{align*}
so $f(z_1) \neq f(z_2)$, which gives us that $f$ is injective function.