Application of Banach-Steinhaus theorem Let  $(x_n)$ be a sequence in a Banach space $E$ such that $\sum_{j=1}^{\infty} |\varphi (x_j) |<\infty$, $\forall \varphi \in E'.$ Then $\sup \limits_{\|\varphi\| \leq 1} \sum_{j=1}^{\infty}|\varphi (x_j)| <\infty $.
My attempt:
For all $n \in \mathbb{N}$, define $f_n: E' \to \mathbb{K}$, $f_n(\varphi) = \sum_{j=1}^{n} \varphi (x_j)$.
($\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$)
Each $f_n$ is a continuous linear functional, since:
$$|f_n(\varphi)| = \bigg| \sum_{j=1}^{n} \varphi (x_j) \bigg| \leq 
\sum_{j=1}^{n} |\varphi (x_j)| \leq \sum_{j=1}^{n} \|\varphi\|\|x_j\| = 
(n \max \{\|x_j\|\} ) \|\varphi\|$$
For each $\varphi \in E'$, $(|f_n(\varphi)|)$ is bounded since
$$|f_n(\varphi)|  \leq \sum_{j=1}^{n} |\varphi (x_j)| \leq 
\sum_{j=1}^{\infty} |\varphi (x_j)| = M_\varphi \in \mathbb{R}  $$
By Banach-Steinhaus theorem, there exists $M>0$ such that $\sup \limits_{n \in \mathbb{N}}  \|f_n\|<M$.
For all $n \in \mathbb{N}$, we have:
$$M> \|f_n\| = \sup \limits_{\|\varphi\| \leq 1} |f_n(\varphi)| = 
\sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{n}\varphi (x_j)\bigg| $$
Then
$$\sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{\infty}\varphi (x_j)\bigg| \leq M $$
Unfortunately, this is not  what we want to prove. 
How can I fix it? 
 A: Consider the collection $\mathcal C$ of functionals on $E'$ of the following form: 
$$
f(\varphi) = \sum_{k=1}^N u_n \varphi(x_n),
$$
where $N\ge 1$ and $|u_n|=1$, $n= 1,\dots, N$. 
Each $f\in\mathcal C$ is a bounded linear functional on $E'$, moreover, for any $\varphi\in E'$ 
$$
\sup_{f\in\mathcal C}|f(\varphi)| = \sum_{k=1}^\infty |\varphi(x_n)|<\infty.
$$
Therefore,
$$
\sup_{f\in\mathcal C}||f|| = \sup_{f\in\mathcal C}\sup_{||\varphi||\le 1}|f(\varphi(x))| = \sup_{||\varphi||\le 1}\sup_{f\in\mathcal C}|f(\varphi(x))| = \sup_{||\varphi||\le 1}\sum_{k=1}^\infty |\varphi(x_n)|<\infty.
$$
A: Hint: You have the good idea, you just have to enlarge the space of functions that you are using. Here ar5e the modifications needed for the field of reals.
Consider $C$ the space of function such that an element $f_{n,x_i^j}:E'\rightarrow R  , i=1,...,n$, $j=0,1$ such that $x_i^0=x_i, x_i^1=-x_j$ and $f_{n,x_i^j}(\phi)=\sum_{i=1}^{i=n}(\phi(x_i^j)$. You can repeat your argument, and obtain the result since you have $(x_i^j)$ such that $\sum_{i=1}^{i=n}f_{n,x_i^j}(\phi)=\sum_{i=1}^{i=n}\mid \phi(x_i)\mid$.
For the complex field, you can define $C$ the set of $f_{n,u_i,x_i}(\phi)=\sum u_i\phi(x_i)$ where $u_i$ is a complex number of module $1$.
