Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer 1

up vote 4 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 http://en.wikipedia.org/wiki/Principle_of_uniform_boundedness ) and so here we'd want to map from $B$, which isn't Banach.

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.