Question: Let $\{x_i\}_{i\in \mathbf{N}}$ be a sequence in a normed space $\mathscr{X}$ such that for any $f\in \mathscr{X}^\ast$, \begin{equation} \sum_{i=1}^{\infty}|f(x_i)|<\infty. \end{equation} Now we want to show that $\exists M<\infty$ such that \begin{equation} \sum_{i=1}^{\infty}|f(x_i)|\leqslant M\|f\|. \end{equation} Here is my attempt: We define on $\mathscr{X}^\ast$ that \begin{equation} T\colon \mathscr{X}^\ast\longrightarrow \mathbf{F},\,\,\, T(f)\colon\!=\sum_{i=1}^{\infty}f(x_i). \end{equation} Then \begin{equation} |T(f)|=\left|\sum_{i=1}^{\infty}f(x_i)\right|\leqslant\sum_{i=1}^{\infty}|f(x_i)|<\infty,\,\,\, \forall f\in\mathscr{X}^\ast. \tag{$\ast$} \end{equation} Also $T$ is a linear map, so $T\in \mathscr{X}^{\ast\ast}$. Then by ($\ast$), $\|T\|<\infty$.
I am stuck at here: The best we can show now is \begin{equation} \left|\sum_{i=1}^{\infty}f(x_i)\right|\leqslant \|T\|\cdot \|f\|. \end{equation} Is it possible to further show that \begin{equation} \sum_{i=1}^{\infty}|f(x_i)|\leqslant \|T\|\cdot \|f\|. \end{equation} If not, how can I adjust my proof?
You're on the right track; the ideas are the same as yours just with a little tweak to account for the absolute value.
Here $$\ell_1= \left\{ (z_i)_{i=1}^\infty: \sum_{i=1}^\infty |z_i|<\infty\right\}, \qquad \| (z_i)_i\|_{\ell_1} =\sum_{i=1}^\infty |z_i|.$$
Consider the map $S: \mathscr{X}^* \to \ell_1$ given by $f\mapsto (f(x_i))_i$. By hypothesis this is well defined (i.e. $S(f)\in \ell_1$ for every $f\in \mathscr{X}^*$), so it's enough to show $S$ is bounded. Consider the auxiliary $S_N:\mathscr{X}^*\to \ell_1$ with $S_N(f):=(f(x_i))_{i=1}^N$. Since $\sup_N \| S_N(f)\|_{\ell_1} = \sup_N\sum_{i=1}^N |f(x_i)|<\infty$, by the uniform boundedness principle we get that there exists $M>0$ such that $\| S_N(f)\|_{\ell_1}\leq M\| f\|_{\mathscr{X}^*}$. This is what you want after taking the limit.