$\{x_n\}$ a sequence in NLS $X$ s.t. $\sum f(x_n)$ converges for all $f \in X^*$ , is the function $f \in X^* \to \sum f(x_n)$ continuous ?

Question

Let $X$ be a NLS , $X^*$ be the set of all bounded real valued functions on $X$ ( i.e. the topological dual of $X$) , let $\{x_n\}$ be a sequence in $X$ such that $\sum_{n=1}^{\infty} f(x_n)$ converges for all $f \in X^*$ ; then is the mapping $T:X^* \to \mathbb R$ , defined as $T(f):=\sum_{n=1}^{\infty} f(x_n)$ continuous ?

Answer 1

Let $T_i:X^*\to\mathbb{R}$ be defined as $T_if=\sum_{n=1}^if(x_n)$. Then $\|T_i\|\leq\sum_{n=1}^i\|x_n\|$ so that $(T_i)_{i=1}^\infty$ forms a sequence of continuous linear operators from $X^*$ to $\mathbb{R}$. Furthermore, for each $f\in X^*$ we have $\sup_{i\in\mathbb{N}}|T_if|<\infty$. Hence, by uniform boundedness principle (AKA Banach-Steinhaus), $C:=\sup_{i\in\mathbb{N}}\|T_i\|<\infty$. It follows easily that $\|T\|\leq C$.