showing that a sequence converges in the dual space of a normed vector space

Suppose that $S=\left\{s_\alpha: \alpha \in A\right\}$ is a set of points in a normed vector space $X$ such that $\overline{span}(S)=X$. If $\left\{f_n\right\}$ is a bounded sequence in $X^*$ and $\left\{f_n(s_\alpha)\right\}$ converges for all $\alpha \in A$, then there exists $f\in X^*$ such that $\displaystyle \lim_{n\rightarrow \infty} f_n(x)=f(x)$ for all $x\in X$.

I know that if $X$ is a normed vector space then $X^*=B(X, \mathbb{R})$ is a Banach space. Thus, if we can show that $\left\{f_n(x)\right\}$ is a Cauchy sequence in $X^*$, then $f_n\rightarrow f$ and $f\in X^*$. The problem is that I don't know how to show that $\left\{f_n(x)\right\}$ is a Cauchy sequence. How can I incorporate the fact that $\left\{f_n\right\}$ is a bounded sequence in $X^*$ and $\left\{f_n(s_\alpha)\right\}$ converges for all $\alpha \in A$?

Thanks!

By linearity, $\left\{f_n(s)\right\}_s$ converges for all $s\in\text{span}(S)$. Let $K=\sup_{n\in\mathbb{N}}\Vert f_n\Vert$. Given $x\in X$, we can find $s\in\text{span}(S)$ such that $\Vert{x-s\Vert}<\varepsilon/4K$. Notice that, for all $n,m\in\mathbb{N}$ sufficiently large (depending on $s$ and $\varepsilon$), we have $$|f_n(x)-f_m(x)|\leq\Vert f_n\Vert\Vert x-s\Vert+|f_n(s)-f_m(s)|+\Vert f_m\Vert\Vert s-x\Vert\leq 2K(\varepsilon/4K)+\varepsilon/2=\varepsilon.$$
Thus, $\left\{f_n(x)\right\}_n$ is Cauchy for every $x\in X$, as you wanted.
Therefore, $f_n\rightarrow f$ pointwise, for some linear $f$. It is a consequence of the Uniform Boundedness Principle/Banach-Steinhauss Theorem that $f\in X^*$.
• You do not need Bnach-Steinhaus for the continuity of the limit since you easily get $|f(x)|\le K \|x\|$. – Jochen Apr 1 '14 at 7:10