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Suppose that $f$ is analytic in the unit disc D = {$z \in \mathbb{C}$ : |$z$| < 1} and $|$f($z$)$| \le 1/(1-|$z$|)$ for all $z\in D$.

Let $f($z$)= \sum _{n=0}^{\infty } a_nz^n$ be the power series expansion of f about $0$.

Prove that $$|a_n| \le (n+1)(1+1/n)^{n} < e(n+1)$$

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I've been working on this problem for a while now, but can't seem to get anywhere. I've tried expressing 1/(1-$|$z$|$ as the geometric series $\sum _{n=0}^{\infty} |z|^n$ and writing $|f($z$)|= \sum _{n=0}^{\infty } |a_n||z|^n$ $\le$ $\sum _{n=0}^{\infty} |z|^n$ –  matt Oct 14 '12 at 0:32
    
Also, I know that $$e=\lim_{n \to \infty} (1 + 1/n)^n.$$ My approach was to prove the $\le$ inequality first and then use this fact to show the last inequality –  matt Oct 14 '12 at 0:33
    
Please avoid using $$ in titles. –  Pedro Tamaroff Oct 14 '12 at 0:43
    
k, thanks. sorry! –  matt Oct 14 '12 at 0:45
    
The second inequality is really easy, use $(1+\frac{1}{n})^n <e<(1+\frac{1}{n})^{n+1}$ –  Alex Oct 14 '12 at 0:50

2 Answers 2

up vote 0 down vote accepted

Recall that Cauchy integral formula gives us $$ a_n= \frac{1}{2\pi i}\int\limits_{\partial B(0,r)}\frac{f(z)}{z^{n+1}}dz \frac{1}{2\pi i}\int\limits_{0}^{2\pi}\frac{f(r e^{it})}{(re^{it})^{n+1}}d(r e^{it})= \frac{1}{2\pi}\int\limits_{0}^{2\pi}\frac{f(r e^{it})}{r^n e^{int}}dt $$ hence $$ |a_n|\leq \frac{1}{2\pi}\int\limits_{0}^{2\pi}\frac{|f(r e^{it})|}{r^n |e^{int}|}dt\leq \frac{1}{2\pi}\int\limits_{0}^{2\pi}\frac{dt}{r^n (1-|r e^{it}|)}= \frac{1}{r^n (1-r)} $$ As the consequence $$ a_n\leq\min\limits_{r\in(0,1)}\frac{1}{r^n (1-r)}=(n+1)\left(1+\frac{1}{n}\right)^n $$ I suggest you to fill the gaps.

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thanks for the help! –  matt Oct 14 '12 at 1:32
    
sorry, i'm still not sure how you get the equality on the right hand side of $$a_n\leq\min\limits_{r\in(0,1)}\frac{1}{r^n (1-r)}=(n+1)\left(1+\frac{1}{n}\right)^n$$ –  matt Oct 14 '12 at 1:50
    
use calculus to find point of minimum. Find derivative, find stationary points, substitute into the function –  Norbert Oct 14 '12 at 1:59
    
thanks again :D –  matt Oct 14 '12 at 2:01

Here are some hints.

  1. Cauchy's theorem can help you recover $a_n$ from $f$.
  2. Use the inequality $|\int f| \leq \int |f|$
  3. What is the minimum of $\frac{1}{z^{n+1}(1-z)}$ when $z\in (0,1)$?
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thanks for the help! –  matt Oct 14 '12 at 1:32

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