show that $x$ is in $\ell^1$ if $\sum x_ny_n$ is convergent for all $y=(y_n)\in{c_0}$ If $x=(x_n)$ is a sequence of complex numbers such that the series $\sum  x_ny_n$ is convergent for all $y=(y_n)\in{c_0}$. Then prove that $x\in{\ell^1}$. Can anyone tell me what is the meaning of $c_0$  in this question and how to proceed on proving it.
 A: I would upvote Omran Kouba's answer if I could, since it is probably the one you were looking for. Here is an alternative direct proof, if one is willing to use more tools from functional analysis.
Equip $c_0$ with the sup norm $\|y\|_\infty = \sup |y_n|$. This a Banach space [exercise].
Now let us take your sequence $y=(y_n)$ and consider, for every $k$, the linear functional
$$
L_k:x\longmapsto \sum_{n=0}^k x_ny_n.
$$
On $c_0$, its norm is [exercise]
$$\|L_k\|=\sum_{n=0}^k|y_n|.$$
By assumption $\sup_k |L_kx|<\infty$ for every $x\in c_0$.
Therefore, by the Uniform Boundedness Principle, $\sum_{n=0}^\infty|y_n|=\sup_k \|L_k\|<\infty$.
A: Suppose that $x=(x_n)_n\notin \ell^1$, that is $\sum_{n=1}^\infty|x_n|=+\infty$. Let $S_n=\sum_{k=1}^n|x_k|$. We will define the sequence $(y_n)_n$ by setting $y_n=0$ if $x_n=0$ and
$$
y_n=\frac{1}{x_n}\log\left(\frac{S_n}{S_{n-1}}\right)\qquad\hbox{if $x_n\ne0$}
$$
(with $S_0=1$.)
Now, clearly, if $x_n\ne 0$, we have
$$
 |y_n|=\frac{1}{|x_n|}\log\left(1+\frac{|x_n|}{S_n}\right)\leq \frac{1}{S_n}.
$$
This allows us to conclude that $\lim_{n\to\infty}y_n=0$, ${\it i.e.}$ $(y_n)_{n}\in c_0$. On the the other hand, if $x_{n_0}\ne 0$ then for $n>n_0$ we have
$$0\leq x_ny_n=\log(S_n)-\log(S_{n-1})$$
and this proves that $\sum x_ny_n$ is divergent.
We have proved that if $x=(x_n)_n\notin \ell^1$ then there is a sequence $y=(y_n)_{n}\in c_0$, such that $\sum x_ny_n$ is divergent. This proves the desired result by contraposition.
A: Suppose that
$$
\sum_{k=1}^\infty|x_k|=\infty\tag1
$$
The divergence in $(1)$ implies that there is a sequence $\{n_m\}_{m=1}^\infty$ so that $n_1=1$ and
$$
\sum_{n_m\le k\lt n_{m+1}}\!\!\!\!|x_k|\ge m\tag2
$$
Let the sequence $\{y_k\}_{k=1}^\infty$ be given by
$$
y_k=\left\{\begin{array}{cl}
\frac1m\frac{|x_k|}{x_k}&\text{if $n_m\le k\lt n_{m+1}$ and $x_k\ne0$}\\
\frac1m&\text{if $n_m\le k\lt n_{m+1}$ and $x_k=0$}
\end{array}\right.\tag3
$$
Then, $\lim\limits_{k\to\infty}y_k=0$, and
$$
\begin{align}
\sum_{k=1}^\infty y_kx_k
&=\sum_{m=1}^\infty\sum_{n_m\le k\lt n_{m+1}}x_ky_k\tag{4a}\\
&=\sum_{m=1}^\infty\frac1m\sum_{n_m\le k\lt n_{m+1}}\!\!\!\!|x_k|\tag{4b}\\
&\ge\sum_{m=1}^\infty\frac1m\cdot m\tag{4c}
\end{align}
$$
and $\text{(4c)}$ diverges.
Therefore, if $\sum\limits_{k=1}^\infty|x_k|$ diverges, then there is a sequence $y\in c_0$ so that $\sum\limits_{k=1}^\infty y_kx_k$ also diverges.
