TITLE: Weak convergence on a dense subset of test functions
QUESTION [8 upvotes]: 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?

REPLY [8 votes]: This isn't true-- here's a counter-example.  Let $B=c_0$ and $S=c_{00}$ the sequences which are eventually zero.  Then $B^*=\ell^1$ and let $(\delta_n)$ be the canonical basis.  Let $\varphi_n = n^2\delta_n$ and $\varphi=0$.  Then $\varphi_n(g) \rightarrow \varphi(g)=0$ for any $g\in c_{00}$ because for each $g$, eventually $\varphi_n(g)=0$.  However, if $f=(1,1/2,1/3,1/4,\cdots)\in c_0$ then $\varphi_n(f)  = n \not\rightarrow \varphi(f)=0$.
(The problem with the idea of using uniform boundeness is that you need to apply it to maps from a Banach space to a normed space (see http://en.wikipedia.org/wiki/Principle_of_uniform_boundedness ) and so here we'd want to map from $S$, which isn't Banach.