Let $S$ be a (norm-)dense subset of a Banach space $B$.
Suppose $\varphi_n$ is a sequence in the dual space $B^*$ and that for some $\varphi \in B^*$ we have $\varphi_n(g) \to \varphi(g)$ for all $g \in S$. Is it true that $\varphi_n(f) \to \varphi(f)$ for all $f \in B$?
Here are some of my thoughts:
Fix $f \in B$. For any $\epsilon > 0$ choose $g \in S$ with $\| f - g \|_B < \epsilon$. By the triangle inequality we have \begin{align} | \varphi_n(f) - \varphi(f) | & \le | \varphi_n(f) - \varphi_n(g) | + | \varphi_n(g) - \varphi(g) | + | \varphi(g) - \varphi(f) | \\ & \le \| \varphi_n \|_{B^*} \| f - g \|_B + | \varphi_n(g) - \varphi(g) | + \| \varphi \|_{B^*} \| f - g \|_B \end{align}
By hypothesis, the middle term can be made smaller than $\epsilon$ for sufficiently large $n$, hence
$| \varphi_n(f) - \varphi(f) | \le \| \varphi_n \|_{B^*} \epsilon + (1 + \| \varphi \|_{B^*}) \epsilon $
The second term can be made arbitrarily small, but the first term may be unbounded if $\| \varphi_n \|_{B^*}$ get arbitrarily large. Is there any reason to believe that these norms are uniformly bounded?