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Problem of the Week #241 - Jan 10, 2017

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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
Here is this week's POTW:

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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

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

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
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.
 
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