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 $\ell^1$.

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 $(y_n)\in\ell^\infty$.
