Inequality for square summable complex series

Importance: Medium ✭✭

Author(s):

Retkes, Zoltan

Subject:

Analysis

Keywords:

Inequality

Recomm. for undergrads: yes

Prize: Ł10

Posted	by:	tigris35711
	on:	December 25th, 2012

Conjecture For all $\alpha=(\alpha_1,\alpha_2,\ldots)\in l_2(\cal{C})$ the following inequality holds $\sum_{n\geq 1}|\alpha_n|^2\geq \frac{6}{\pi^2}\sum_{k\geq0}\bigg| \sum_{l\geq0}\frac{1}{l+1}\alpha_{2^k(2l+1)}\bigg|^2$

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Solution

On November 3rd, 2013 Anonymous says:

It's a simple application of the Shwartz inequality:

$\sum_{k}\left|\sum_{l} \frac{1}{l+1}a_{2^k(2l+1)}\right|^2 \le$ $\le \sum_{k}\left|\sum_{l} \frac{1}{l+1}\left|a_{2^k(2l+1)}\right|\right|^2 \le$ Shwartz: $\le \sum_{k} \left(\sum_{l}\frac{1}{(l+1)^2}\right)\left(\sum_{h}|a_{2^k(2l+1)}|^2\right) =$ $= \sum_{k} \frac{\pi^2}{6}\sum_{h}|a_{2^k(2l+1)}|^2 =$ $= \frac{\pi^2}{6} \sum_{k}\sum_{h}|a_{2^k(2l+1)}|^2 =$ $= \frac{\pi^2}{6} \sum_{n}|a_n|^2$ because $A_k:=\{ 2^k(2l+1)| l\in \mathbb N\}$ is a partition of $\mathbb N^+$ .