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 $$\sum_{n\geq 1}|\alpha_n|^2\geq \frac{6}{\pi^2}\sum_{k\geq0}\bigg| \sum_{l\geq0}\frac{1}{l+1}\alpha_{2^k(2l+1)}\bigg|^2 $$


* indicates original appearance(s) of problem.


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^+ $.

Oh Yes.

Where shall I send the £10 prize ?

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