Proving that sequence is in $\ell^\infty$ Let $(y_k)_{k\in\mathbb{N}}$ be a sequence in $\mathbb{K}$ with the property that for all sequences $(x_k)_{k\in\mathbb{N}}\in\ell^1$ the sequence $\sum_{k=1}^{n} x_k y_k$ has a limit in $\mathbb{K}$ for $n\to\infty$. Prove that $y\in \ell^\infty$.
I tried to define an operator $T_y: \ell^1\rightarrow \ell^1, (x_k)_{k\in\mathbb{N}}\mapsto  (x_ky_k)_{k\in\mathbb{N}}$, but as I do not know about absolute convergence of $(x_k y_k)_{k\in\mathbb{N}}$, I am not sure if it really maps to $\ell^1$.
 A: Suppose the sequence $(y_n)_{n\in\mathbb{N}}$ to be unbounded.
Then for each $n>0$, there is $k_n$ such that $|y_{k_n}|\ge n$. Without loss of generality we can replace the original $y_n$ by this subsequence.
Let $x_n:=\frac{\overline{y}_n}{|y_n|n^2}$; the $n^2$ is introduced so $(x_n)\in\ell^1$. Then
$$
 \sum_nx_ny_n=\sum_n\frac{|y_n|^2}{|y_n|n^2}=\sum_n\frac{|y_n|}{n^2}\ge\sum_n\frac{1}{n}
$$
Hence, going the other way, if $\sum_nx_ny_n$ converges for all $(x_n)\in\ell^1$ then $(y_n)\in\ell^\infty$.
A: Your argument can work. If $(x_k)_k \in \ell^1$, then also
$\left(x_k \operatorname{sgn} x_k\operatorname{sgn} y_k\right)_k \in \ell^1$ where $\operatorname{sgn} z_k = \frac{\overline{z_k}}{|z_k|^2}$ if $z_k \ne 0$ or $0$ if $z_k=0$.
By the assumption $$\sum_{k=1}^\infty x_k (\operatorname{sgn} x_k)(\operatorname{sgn} y_k)y_k = \sum_{k=1}^\infty |x_ky_k| < +\infty$$
so $(x_ky_k) \in \ell^1$. Hence your operator $T_y : \ell^1\to \ell^1$, $(x_k)_k\mapsto  (x_ky_k)_k$ is well-defined. Moreover, it is the limit in strong operator topology of the sequence of bounded operators $T_y^n : \ell^1 \to \ell^1$ given by
$$T^n_y(x_k)_k = (x_1y_1, \ldots, x_ny_n,0,0, \ldots)$$
so by the uniform boundedness principle, $T_y$ is also bounded, or
$$\sum_{k=1}^\infty |x_ky_k| \le \|T_y\| \sum_{k=1}^\infty |x_k|$$
By plugging in canonical vectors $e_n$, we get that $|y_n| \le \|T_y\|$ and hence $(y_k)_k \in \ell^\infty$.
