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

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

\end{conjecture}

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/}

Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)


* indicates original appearance(s) of problem.

Solution

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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