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.

A locally convex space is called Barrelled if each closed absorbing convex set is 0-neighborhood See. But i doubt that every absorbing set contains zero. Then is every LCV is barreled. I think, i am confusing with something...

Every absorbing set contains zero. Because if $A$ is absorbing set then there is some $t>0$ such that $0\in tA$, that is we have $0= t.a$ for some $a\in A$. This gives $a=0\in A$.

So My question is:

If every absorbing set contains zero, then it means that every LCV is barreled.. so what is need to defined barreled in LCV.. can i get some topological vector space which is not barreled

share|improve this question
    
Sorry, but I do not get what the actual question is. –  Michael Greinecker Jan 7 '13 at 11:09
    
Sorry.. I am editing the question.. My question is if every absorbing set contains zero, then it means that every LCV is barreled.. so what is need to defined barreled in LCV.. can i get some topological vector space which is not barreled.. –  zapkm Jan 7 '13 at 11:12
3  
In order for $A$ to be a $0$-neighborhood you need more than $0 \in A$. You need an entire open set $U \subset A$ such that $0 \in U$. –  Martin Jan 7 '13 at 11:16
    
@Martin. Thanks i got it... condition is CLOSED absorbing set contains zero Neighborhood.. –  zapkm Jan 7 '13 at 11:19

1 Answer 1

up vote 5 down vote accepted

Take the subspace $X$ of $C[0,1]$ consisting of continuous functions $f \colon [0,1] \to \mathbb C$ vanishing in a neighborhood of $0$. That is, for every $f \in X$ there exists $\delta \gt 0$ such that $f(t) = 0$ for all $t \in [0,\delta)$.

Equip $X$ with the supremum norm, so $X$ is locally convex. [Note that $X$ is not closed in $C[0,1]$: it is dense in the subspace of functions vanishing at $0$, but $\sqrt{x} \notin X$. It is necessary to have a non-complete space for an example involving a normed space, since Banach spaces (and more generally, Baire locally convex spaces) are barreled.]

The set $$B = \{g \in X : |g(1/n)| \leq 1/n \text{ for all } n \in \mathbb{N}\} \subset X$$ is a barrel: By definition it is closed, convex and circled and $0 \in B$. Moreover, $B$ is absorbent since every function $f \in X$ vanishes in a neighborhood of zero, so multiplying $f$ by a suitably small constant $\lambda$ ensures $\lambda f \in B$.

On the other hand, $B$ is not a neighborhood of zero:

For every $\varepsilon \gt 0$ there is $N \in \mathbb{N}$ such that $1/N \lt \varepsilon$; take $f_N\colon [0,1] \to [0,1/N]$ to be piecewise linear such that $f = 0$ on $[0,1/3N]$ and $f= 1/N$ on $[1/2N,1]$. This a function $f_N \in X$ which is in the $\varepsilon$-neighborhood of $0$, but $f(1/2N) = 1/N > 1/2N$, so $f_N \notin B$.

share|improve this answer
    
@ martin,I find these concepts of tvs very slippery just like this.can you suggest where I can find a clear exposition of this. –  Koushik Jan 7 '13 at 12:22
1  
@K.Ghosh I like Trèves's book on topological vector spaces. Other good references include Rudin's functional analysis, or the books on topological vector spaces by Grothendieck, Kelley-Namioka, Köthe or Schaefer. –  Martin Jan 7 '13 at 12:34

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.