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There is a proof of this in: If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$

But I am wondering if there is a contrapositive proof.

Attempt

Given $\sum a_{n}^{2}=\infty$ we want $b_{n}\in l^{2}$ s.t. $\sum a_{n}b_{n}=\infty$.

Let $n_{k}$ be such that $\sum_{n_{k}}^{n_{k+1}}|a_{m}|>\sum_{n_{k}}^{n_{k+1}}a_{m}^{2}>k$, then for $b_{m}=\frac{sign(a_{m})}{k}$ for $n_{k}\leq m\leq n_{k+1}$ we get

$$\sum a_{n}b_{n}=\sum_{n_{k}}^{n_{k+1}}|a_{m}|\frac{1}{k}>\sum 1=\infty.$$

However, the problem is that $\sum b_{m}^{2}=\infty$ if say $n_{k+1}-n_{k}=k^{2}$.

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  • $\begingroup$ write that there exists $\frac{|a_n|^2}{|c_n|^2} \in l^1$, $\frac{|a_n|^2}{|c_n|} \not \in l^1$, and let $b_n = \frac{\overline{a_n}}{|c_n|}$ $\endgroup$ – reuns Oct 6 '16 at 23:26
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In line with what you wrote, assume that $\{ a_n \} \notin \ell^2$ and define $$ b_n = \frac{\overline{a_n}}{1+\sum_{m=0}^{n}|a_m|^2} $$ The integral test can be used to prove the convergence of $$ \sum_{n=0}^{\infty}|b_n|^2=\sum_{n=0}^{\infty}\frac{|a_n|^2}{(1+\sum_{m=0}^{n}|a_m|^2)^2}, $$ which follows because $\sum_{m=0}^{n}|a_m|^2\rightarrow \infty$ as $n\rightarrow\infty$. The tricky part is proving the divergence of $$ \sum_{n=0}^{\infty}a_nb_n= \sum_{n=0}^{\infty}\frac{|a_n|^2}{1+\sum_{m=0}^{n}|a_m|^2} $$ For the divergence, take a look at the really nice solution to this question I posted some time ago: If $a_n \ge 0$ and $\sum_n a_n = \infty$, must $\sum_n \frac{a_n}{1+\sum_{k=1}^{n}a_k}$ also diverge? . The person who posted the solution did a great job of making this look simple.

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I think the sequence $(n_k)_k$ isn't very well defined, because given $n_k=10000$ or whatever, you can't figure out where $n_{k+1}$ is, because it musn't be too far from $n_{k}$ nor from $n_{k+2}$, but if they are far from each other in the first place then you have a problem. But that again depends on where $n_{k+3}$....

Basically your last statement shows that your proof holds a fault, and I think that's where it is However I don't have a proof.

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Hint. May assume that $\lim a_n = 0$ and $a_1 \neq 0$. Let $s_n = |a_1|^2 + \cdots + |a_n|^2$ and $b_n = a_n^* / s_n$. If $(s_n)$ diverges, then we can check that

$$ \sum_{n=1}^{\infty} a_n b_n = \sum_{n=1}^{\infty} \frac{|a_n|^2}{s_n} = \infty \qquad \text{and} \qquad \sum_{n=1}^{\infty} |b_n|^2 = \sum_{n=1}^{\infty} \frac{|a_n|^2}{s_n^2} < \infty. $$

Since my iPad is crappy, I will complement when I am in front of my laptop. But this is essentially the $p$-test with $p=1$ and with $p=2$ and the proof is not hard!

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