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.

When saying two topological vector spaces $E$ and $F$ are in duality, does it mean that they are each other's continuous dual, i.e. $E = F^*$ and $F=E^*$, or just that one is the other's continuous dual, not necessarily true for the reverse?

If it is the former, when is $E = F^*$ and $F=E^*$ true?

For example,

a biorthogonal system is a pair of topological vector spaces $E$ and $F$ that are in duality, with a pair of indexed subsets $ \tilde v_i$ in $E$ and $\tilde u_i$ in $F$ such that $$ \langle\tilde v_i , \tilde u_j\rangle = \delta_{i,j} $$ with the Kronecker delta

Thanks and regards!

share|improve this question
2  
Yes, each is the dual of the other. But dual in the weak topology defined by the other. Which is automatically true as long as you have a bilinear pairing so that each space separates the points of the other. You can forget that last condition if you don't mind non-Hausdorff topologies. –  GEdgar Mar 31 '12 at 21:53
    
@GEdgar: +1 Thanks as always! (1) Do you mean that two TVSes are continuous dual of each other wrt their weak topologies, if there exists "a bilinear pairing so that each space separates the points of the other"? (2) What does "a bilinear pairing so that each space separates the points of the other" mean? (3) I just found a Wiki article about two vector spaces to be in dual pair wrt a bilinear form. I wonder in your comment, if you meant two TVSes are continuous dual wrt the evaluation, where the evaluation is the bilinear form? –  Tim Apr 1 '12 at 0:25
add comment

2 Answers 2

Let $E,F$ be two real vector spaces. A "pairing" is a map $E \times F \to \mathbb R$, written say $\langle x,y \rangle$, that is linear separately in each variable, and such that each space separates points in the other: for each nonzero $x \in E$ there is $y \in F$ with $\langle x,y \rangle\ne 0$ and for each nonzero $y \in F$ there is $x \in E$ with $\langle x,y \rangle \ne 0$. We say "$E$ and $F$ are in duality".

Now define the weak topology $\sigma(E,F)$ on $E$ as the weakest topology such that, for each fixed $y \in F$, the map $x \mapsto \langle x,y \rangle$ is continuous. Then we may identify $F$ with the set $E^*$ of all $\sigma(E,F)$-continuous linear functionals on $E$ as follows: continuous linear functional $\phi \in E^*$ corresponds to point $y \in F$ iff $\phi(x) = \langle x,y \rangle$ for all $x$. Similarly, define the weak topology $\sigma(F,E)$ on $F$, and then identify the $\sigma(F,E)$-dual $F^*$ of $F$ with $E$.

share|improve this answer
    
+1. Thanks! I wonder why there is a bijection between $F$ with $E^*$? I can understand there is an injective, but not sure why it is surjective? –  Tim Apr 1 '12 at 19:08
    
Perhaps get a textbook on locally convex spaces. This explanation of the "weak topology" should be found near the beginning. –  GEdgar Apr 1 '12 at 21:13
add comment

You might find the paragraph starting out with "Duality in the theory of topological vector spaces." helpful here:

Duality. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Duality&oldid=21977

share|improve this answer
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.