Let $b=(b_n)$ be a real sequence, suppose that for any $a\in\ell^2$, $\sum_j a_j b_j<\infty$, then $b\in\ell^2$. [duplicate]

Let $x = (x_n)$ be a sequence of complex numbers with the property that for every $y = (y_n) \in \ell^2$ we have that the sequence $(S_N(y))_{N\geq1}$ with $$S_N(y) =\sum_{n=1}^N x_ny_n $$ converges. Show that $x\in \ell^2$ I think we have to use Riesz representation theorem somehow. Can we identify the $y_n$ with some $\ell_n$ and then use uniform boundedness? But how do we know that $\ell_n$ can act on our x?

For each $N\ge 1$, the map $$S_N:\ell^2\to\mathbb{C},\quad y\mapsto S_N(y)$$ defines a bounded linear functional on $\ell^2$, and evidently its norm $\|S_N\|=\left(\sum_{n=1}^N|x_n|^2\right)^{\frac{1}{2}}$. Since $\lim_{N\to\infty}S_N(y)$ exists for every $y\in\ell^2$, in particular $\{S_N(y)\}_{N\ge 1}$ is a bounded sequence in $\mathbb{C}$ for every $y\in\ell^2$. Then by uniform boundedness principle, $\sup_{N\ge 1}\|S_N\|<\infty$, i.e. $\|x\|<\infty$.