I have the following exercise: Suppose $y=\left(y_{j}\right)_{j=1}^{\infty}$ is a sequence of complex numbers. Prove that if the series $\sum_{j=1}^{\infty} y_{j} \bar{x}_{j}$ is convergent for any sequence $x=\left(x_{j}\right)_{j=1}^{\infty} \in c_{0}$ then $y \in \ell_{1}.$
My attempt: Consider the following: \begin{align*} T_n(x) : & c_0 \longrightarrow \mathbb{R} \\ & x\longmapsto \sum_{j=1}^{n} \bar{x}_j y_j \end{align*} Note that, due to the hypotesis, $\sup_{n\in \mathbb{N}}\left|\left| T_n(x) \right|\right|= \left|\left| \sum_{j=1}^{\infty} \bar{x}_j y_j \right|\right| < \infty$. Now we can apply the uniform boundedness principle, that tells us that $\sup \nolimits _{{n\in \mathbb{N},\|x\|=1}}\|T(x)\|= \sup \nolimits _{n \in \mathbb{N}}\|T_n\|<\infty$.
In this part is where I am confused, because $\sup \nolimits _{{n\in \mathbb{N},\|x\|=1}}\|T(x)\| =\left|\left| \sum_{j=1}^{\infty} \bar{x}_j y_j\right|\right|\leq \sum_{j=1}^{\infty} |\bar{x}_j y_j|\leq \sum_{j=1}^{\infty} |y_j|$. This last sum is what we want to be finite in order to see that $y\in \ell_1$, but how can I finish this argument?
Besides observing that $T_n$ is continuous for obvious reasons, there is no need to compute its norm. Once the hypothesis guarantees that $$ T(x):= \lim_{n\mapsto \infty } T_n(x) $$ exists for every $x$, a well known Corollary of the Uniform Boundedness Principle, known as the Banach-Steinhauss Theorem (https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Corollaries), implies that $T$ is continuous.
Since we know that $c_0' =\ell^1$, it follows that there exists some $z = (z_j)_{j=1}^\infty $ in $\ell^1$, such that $$ T(x) = \sum_{j=1}^\infty x_jz_j, $$ for all $x$. By computing both sides on the standard basis vectors of $c_0$, one easily deduces that $y_j=z_j$, from where it follows that $y$ lies in $\ell^1$.