A relation between the sequence spaces $c_0$ and $\ell^1$ I am reviewing my functional analysis exam and having trouble with one question. 

Let $y=(y_n)$ be a real sequence and assume that $\sum_{n=1}^{\infty}x_ny_n$ is convergent for every $x=(x_n)\in c_0$. Prove that $y\in \ell^1$. 

Can anyone give me some hints to attack this problem ? Thank you very much.
 A: Assume the contrary that $y\notin \ell^1$. Then 
$$\sum_{i=1}^n |y_i| \to \infty,\ \ \text{as } n\to \infty.$$
Then there is $n_1 < n_2 < n_3<\cdots $ so that 
$$\sum_{i=n_k+1}^{n_{k+1}} |y_i| \ge 1$$
for all $k$. Define 
$$x_n = \text{sgn}(y_n) \frac{1}{k},\ \ \ \text{if } n_k<n\le n_{k+1}.$$
Then $x =(x_n)\in c_0$ and 
$$\sum_{i=1}^\infty x_i y_i = \sum_{k=1}^\infty \sum_{i = n_k+1}^{n_{k+1}} x_i y_i = \sum_{k=1}^\infty \sum_{i=n_k+1}^{n_{k+1}} \frac 1k |y_i| \ge \sum_{k=1}^\infty \frac 1k$$
is divergent. Thus $y\in \ell^1$. 
A: Here is a slightly different proof that $(\mathcal{c}_0)^*=\ell_1$.
Define $\lambda(n):=\Lambda(\mathbf{e}_n)$ where $\mathbf{e}_n:\mathbb{N}\rightarrow\mathbb{C}$ such that $\mathbf{e}_n(m)=\mathbb{1}_{\{n\}}(m)$, Clearly $\mathcal{e}_n\in\mathcal{c}_0$ and $\|\mathbf{e}_n\|_\infty=1$.
For any $\lambda\in\ell_1$, the map $\Lambda:\mathcal{c}_0\rightarrow\mathbb{C}$ defined as
$$\Lambda f=\sum^\infty_{j=1}\lambda(j)f(j)$$
is a bounded linear functional. Furthermore, taking $f_n:\mathbb{N}\rightarrow\mathbb{C}$ as
$$\begin{align}
f_n(m)=\sum^n_{j=1}\frac{\overline{\lambda(j)}}{|\lambda(j)|}\mathbb{1}_{(0,\infty)}(|\lambda(j)|)\tag{1}\label{1}\mathbf{e}_j(m)
\end{align}$$
we obtain that a sequence in $\mathcal{c}_0$ with $\|f_n\|_1$ (for all $n$ large enough, unless $\Lambda=0$) such that
$$
\lim_n|\Lambda f_n|=\sum^\infty_{j=1}|\lambda(j)|$$
whence it follows that $\|\Lambda\|=\|\lambda\|_1$.
Conversely, let $\Lambda\in(\mathcal{c}_0)^*$, and set $\lambda(n)=\Lambda(\mathbf{e}_n)$. It suffices to show that

*

*$\lambda\in\ell_1$, and

*$\Lambda f=\sum^\infty_{j=1}\lambda(j)f(j)$ for all $f\in\mathcal{c}_0$.

to conclude that $(\mathcal{c}_0)^*$ and $\ll_1$ are isometrically isomorphic.
Notice that $\|\mathbf{e}_n\|_\infty=1$ for all $n\in\mathbb{N}$. For any $f\in c_0$, define the sequence $f_n=\sum^n_{j=1}f(j)\mathbf{e}_j$. It is obvious that $\{f_n:n\in\mathbb{N}\}\subset\mathcal{c}_0$, $\|f_n\|_\infty\leq\|f\|_\infty$, and that
$$\|f-f_n\|_\infty\xrightarrow{n\rightarrow\infty}0$$
By continuity of $\Lambda$,
$$|\Lambda f_n|\leq\|\Lambda\|\|f_n\|_\infty\leq\|\Lambda\|\|f\|_\infty,$$
and
$$\Lambda f=\lim_{n\rightarrow\infty}\Lambda f_n=\lim_{n\rightarrow\infty}\sum^n_{j=1}f(j)\lambda(j)=\sum^\infty_{j=1}f(j)\lambda(j)
$$
This gives (2). To show that $\lambda\in\ell_1$, define for each $n\in\mathbb{N}$ the function $f_n:\mathbb{N}\rightarrow\mathbb{C}$ as in \eqref{1}. Then, $f_n\in\mathcal{c}_0$, $\|f_n\|_\infty\leq 1$ (equality for all $n$ large enough, unless $\Lambda\equiv0$), and
$$\sum^n_{j=1}|\lambda(j)|=|\Lambda f_n|\leq\|\Lambda\|\|f_n\|_\infty\leq\|\Lambda\|<\infty$$
for all $n\in\mathbb{N}$. (1) follows immediately.
