Suppose that $a_n$ is a sequence of real numbers such that $\sum_na_nb_n$ converges whenever $\sum_n b_n^2 \lt \infty$. Show that $\sum_{n=1}^{\infty}a_n^2 \lt \infty$.
My try: I defined $T: l^2 \to \mathbb{R}$ by sending $(b_1,b_2,\ldots,b_n,\ldots,) \to \sum_{n=1}^{\infty}a_nb_n$. Then $T$ is linear. I want to show that $T$ is bounded and then that will give us the result for $b^n=(a_1,a_2,\ldots,a_n,0,\ldots,0)$, $$T\left(\frac{b^n}{\sqrt{\sum_{j=1}^n a_j^2}}\right)=\frac{a_1^2+a_2^2+\ldots a_n^2}{\sqrt{\sum_{j=1}^n a_j^2}}=\sqrt{\sum_{j=1}^n a_j^2} \le \|T\|, \forall n \in \mathbb{N}$$
which implies that $$\sum_{j=1}^{\infty} a_j^2 =\lim_{n \to \infty} \sum_{j=1}^n a_j^2 \le \|T\|^2 \lt \infty$$
The only thing which remains to be shown now is that $T$ is bounded for which I tried to evoke the Closed Graph Theorem. Suppose that $b^n=(b^n(1),b^n(2),\ldots,b^n(j),\ldots) \in l^2 $ converge to $0$ and $T(b^n) \to y$. Let $\epsilon \gt 0$. Then there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, we have $\|b^n\|_2 \lt \frac{\epsilon}{2}$ which in particular implies that $|b^n(j)| \lt \frac{\epsilon}{2}$ for all $j$ and all $n \ge n_0$. Since $T(b^n) \to y$, there exists $n_1 \in \mathbb{N}$ such that for all $n \ge n_1$, $|T(b^n)-y| \lt \frac{\epsilon}{2}$.
Then for all $n \ge \max{(n_0,n_1)}$$$|y| \le |y-T(b^n)|+|T(b^n)| \lt \epsilon+\frac{\epsilon}{2}(|a_1|+|a_2|+\ldots+|a_k|)$$
(Note: Since $T(b^n) \lt \infty$, the tail of the series goes to $0$ which means that $|\sum_{j \ge k} a_jb^n(j)| \lt \frac{\epsilon}{2}$ )
This is not what I intended to show. Can I conclude from here that $y=0$ since $\epsilon \gt 0$ is arbitrary? For me the problem is that apriori, I don't have a way to get away with the $|a_j|$'s, since they depend on the choice of $b^n$.
Note: There is an answer to this question here: If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$. But I wanted to know if I can go via this route and get to the answer and if not, why so.
Thanks for the help!!
Best Answer
EDIT: As pointed out in the comments, I was using a slightly different method. Rather, I was proving the map $(b_n)\mapsto(a_nb_n)$ is a bounded linear map from $\ell^2\to\ell^1$. The conclusion of course remains the same, but the method is not quite what the OP wanted.
It's possible to approach the question this way, although it is not the best method. We want to show if $b^n\to0$ in $\ell^2$ and $T(b^n)\to y$ in $\ell^1$ then $y=0$, so it is sufficient to show that $y_k=0$ for each $k$. Fix $\varepsilon>0$. Let $n_0$ be large enough that $\max\{\|b^n\|_2,\|T(b^n)-y\|_1\}<\varepsilon$ for all $n\ge n_0$. Observe that $$|a_kb^n(k)-y_k|\le\|T(b^n)-y\|_1<\varepsilon,$$ so if we can show that $(a_n)$ is a bounded sequence it will follow that $$|y_k|\le\varepsilon+|a_k|\|b^n\|_2<(1+M)\varepsilon$$ where $M=\sup_n|a_n|$, and so we will be done. Suppose for a contradiction that $(a_n)$ is not bounded. Then there is a subsequence $(n_k)$ such that $|a_{n_k}|\ge k$. Let $b=(b_n)$ be the sequence defined by
$$b_n=\begin{cases} \frac1k&\text{if }n=n_k,\\ 0&\text{otherwise.} \end{cases}$$
Clearly $b\in\ell^2$, but $|a_{n_k}b_{n_k}|\ge1$ for all $k$, so $a_nb_n\not\to0$ and in particular $\sum_na_nb_n$ cannot converge. This completes the proof.