Home ยป Inequality for square summable complex series

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^+ $.

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