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Let $(X,s)$ be a topological vector space over $\mathbb{F}$ with linear topology $s$, which we will henceforth refer to as the strong topology. Then, as usual we can construct the continuous dual space $(X,s)'$, which consists of all linear maps $X \rightarrow \mathbb{F}$ which are continuous with respect to the strong topology. Now, we can construct the weak topology $w$, the smallest topology on $X$ such that every map $\Lambda \in (X,s)'$ remains continuous.

So now here's my question. The weak topology on $X$ yields a new topological vector space $(X,w)$, so we can construct a continuous dual for that too. So what does $(X,w)'$ look like? It must at the very least contain $(X,s)'$, of course. Other possible questions:

  • For what types of TVS's can we infer something about the structure of $(X,w)'$ with respect to $(X,s)'$? What happens in a normed vector space?
  • Can somebody give me an explicit construction indicating that there exists a linear map $X \rightarrow \mathbb{F}$ which is not continuous in the strong topology but is continuous in the weak topology?
  • I suppose we can repeat this process to produce a "doubly weak" topology $ww$ on $X$ that is the weakest topology such that every $\Lambda \in (X,w)'$ remains continuous. Then we can produce yet another continuous dual space $(X,ww)'$. Does this process repeat forever? Does it terminate?
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In a Banach space we should have $(X, w)'=(X, s)'$. That is, by weakening the topology the dual stays the same. –  Giuseppe Negro Feb 13 '13 at 2:03

1 Answer 1

up vote 6 down vote accepted

If $F$ is a family of linear functionals on a vector space $X$ then it induces a weak topology $w_F$ on $X$. The dual space of $(X,w_F)$ is the linear span of $F$ in the algebraic dual of $X$. In particular $(X,s)' = (X,w)'$ always holds.

This fully answers the first bullet. It also shows that there is no example as in the second bullet and that the procedure in the third bullet stabilizes at $(X,w)$ after one step.

The point is that if $\varphi \colon X \to \mathbb{F}$ is $w_F$-continuous then $U = \{x \in X \mid \lvert \varphi(x)\rvert \lt 1\}$ is a $w_F$-open neighborhood of $0$. By definition of the weak topology, there are $f_1,\dots,f_n \in F$ and $\varepsilon \gt 0$ such that $V = \bigcap_{i=1}^n \left\{x \in X \mid \lvert f_i(x)\rvert \lt \varepsilon\right\} \subseteq U$ and in particular $\bigcap_{i=1}^n \ker f_i \subseteq \ker \varphi $. Therefore $\varphi$ is a linear combination of the $f_i$ (a proof of this last assertion is here).

You can find a detailed proof in pretty much every book on topological vector spaces. A good reference is chapter 3 of Rudin's Functional Analysis, see in particular Lemma 3.9 and Theorem 3.10 (the Hausdorff assumption is only there because Rudin assumes locally convex spaces to be Hausdorff, as many authors do).

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See also this closely related answer of mine. –  Martin Feb 13 '13 at 2:53
    
Ah, this is fantastic. I was fundamentally "disturbed" that one might be able to generate an infinitely increasing sequence of weaker topologies. I'm glad this cannot be. I apologize if this is some sort of well-known fact. Despite appearances I'm not at all a theorist =P –  Christopher A. Wong Feb 13 '13 at 21:39
    
Sorry, by no means did I intend to be condescending or imply "you should know this". I only tried to be as concise as possible by isolating the main ingredients. It's a good and very natural question and I don't think its answer is obvious. –  Martin Feb 15 '13 at 11:35

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