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
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 $f_N \notin B$.