current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
up vote 16 down vote favorite
4

I've read the claim that Fréchet spaces that are not Banach spaces never have a dual that is a Fréchet space, but have not been able to find a proof of this statement. Is it trivial or does someone have a reference?

share|cite|improve this question

1 Answer 1

up vote 19 down vote accepted

For any locally convex and metrizable space $E$, its strong dual is metrizable if and only if $E$ is normable.

This and related properties of (F)-spaces are discussed in detail in Topological Vector Spaces I by Köthe (see §29.1, pp. 393-394 in the English edition).

share|cite|improve this answer
    
Thanks, so one has to combine §21.5 (3): "For a Fréchet space, the original topology is equal to the strong topology", with §29.1 (7), which is what you quoted. – Tim van Beek Apr 29 '11 at 7:52
1  
SO what does this mean for the question, for someone who doesn't have instant recall about the lattice of properties of topological vector spaces? – David Roberts Apr 29 '11 at 7:56
1  
@David Roberts: This implies that the claim stated by the OP holds true. The observation is probably due to Grothendieck. – Andrey Rekalo Apr 29 '11 at 8:02
12  
A nice way to think of this is as the observation that a LCTVS cannot be a (non-trivial) projective limit and an inductive limit of countably infinite families of Banach spaces at the same time. Either one family has to be uncountable, or both have to be finite. – Loop Space Apr 29 '11 at 8:45
    
@Andrew Stacey: That's really nice and intuitively appealing. – Andrey Rekalo Apr 29 '11 at 8:55

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.