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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?

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It's worth noting that, as you showed, this does hold under the additional assumption that $\sup_n \|\varphi_n\|_{B^*} < \infty$. It happens quite often that the sequence we study is a priori known to be bounded in norm, and this shows that in that case, to check weak convergence we need only look at a dense subspace of $B$. – Nate Eldredge Jun 15 '12 at 22:33
up vote 7 down vote accepted

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 ) and so here we'd want to map from $S$, which isn't Banach.

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Thanks Matthew, for this fantastically simple counter-example! You've convinced me that I should reacquaint myself with sequence spaces. I don't understand the second paragraph of your answer though, I thought $B = c_0$ is a Banach space under the $\sup$-norm? – mmunoz Jan 19 '12 at 21:42
Sorry, the remark was a bit cryptic-- you only know what's happening on $S$, so it would be natural to apply UBP to $S$, but that's not complete. – Matthew Daws Jan 19 '12 at 22:01
Thank you very much. – mmunoz Jan 19 '12 at 22:06

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