If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$ I'm trying to prove the following:

If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n<\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2<\infty$, then $\sum_{n=1}^\infty a_n^2 <\infty$.

The context here is functional analysis homework, in the subject of Hilbert spaces.
Here's what I've thought:
Let $f=(a_n)>0$. Then the problem reads: if $\int f\overline{g}<\infty$ for all $g>0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.
I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle <\infty$ for all $g>0,g\in \ell^2$ then $f\in \ell^2$.
I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P
From there, I don't know what else to do. I thought of Hölder's inequality which in this context states:
$$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$
but it's not useful here.
 A: Put $l_n(b) =\sum_{k=1}^n a_kb_k$. $l_n$ is linear, continuous and $\lVert l_n\rVert = \left(\sum_{k=1}^na_k^2\right)^{\frac 12}$. For all $b\in \ell^2$ the sequence $\{l_n(b)\}$ is bounded. By the principle of uniform boundedness we get that the sequence $\left\{\lVert l_n\rVert\right\}$ is bounded.  
A: Following the idea left by david as an answer, I'll post a detailed solution.
Let $T_n:\ell^2 \to \mathbb{C}, T_n(c)=\sum_{j=1}^n a_j c_j$. Then clearly $T_n \in (\ell^2)^*$ for every $n$.
I claim that for every $c\in \ell^2$, the limit $\lim_n T_n(c)=\sum_{j=1}^\infty a_jc_j$ exists.
Indeed, we know it does for $c\in \ell^2$ such that $c(n)\geq 0$ for all $n$. But then, for an arbitrary $c\in \ell^2$, we can decompose $c$ as:
$c=(\mbox{Re }  c)^+ - (\mbox{Re } c)^- + i\left( (\mbox{Im }c)^+ - (\mbox{Im } c)^-\right)$
which proves the claim.
As a consequence we have that $\sup_n \lVert T_n(c) \rVert <\infty$ for all $c\in \ell^2$, but then by the uniform boundedness principle, $\sup_n \lVert T_n \rVert <\infty$.
Now, $T_n(c) = \langle c, a^{(n)}\rangle_{\ell^2}$ where $a^{(n)}(j)=\begin{cases} a_j & \mbox{if }j\leq n \\ 0 & \mbox{if } j>n \end{cases}$.
By Riesz' representation theorem on Hilbert spaces, we know that $\lVert T_n \rVert = \lVert a^{(n)}\rVert_2= \left( \sum_{j=1}^n a_j^2 \right)^{\frac{1}{2}}$.
To conclude, since $a_n\geq 0$ for all $n$, we have that $\left( \sum_{n=1}^\infty a_n^2 \right)^{\frac{1}{2}} = \sup_n \lVert a^{(n)} \rVert_2 = \sup_n \lVert T_n \rVert < \infty$, and then $\sum_{n=1}^\infty a_n^2 <\infty$.
A: It can be proven using the Hölder inequality — see lemma 1.8 (page 8) of these notes. (There may be mistakes; I typeset these from lecture notes.) There's no need to invoke heavy machinery like the uniform boundedness principle or even the Riesz representation theorem; it's a special case of a general result for $\ell^p$ spaces.
