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Suppose that $(X,\tau)$ is already a locally convex TVS. Let us denote by $X'$, the space of all $\tau$-continuous linear functionals on $X$, the topological dual of $X$. For each $f\in X'$, define $p_f(x)=|f(x)|, x\in X$. Then $$\Gamma=\{p_f: f\in X'\}$$ is a separating family of seminorms on $X$. By a theorem of Rudin, there exists a locally convex topology on $X$, commonly denoted by $\sigma:=\sigma(X,X')$, with the property that every member of $\Gamma$ is $\sigma$-continuous. Specifically, $\sigma$ consists of arbitrary unions of the translates of finite intersections of the sets $\{x\in X: |f(x)|<\epsilon\}$, $f\in X'$ and $\epsilon>0$.

Question. Is $\tau=\sigma$? What is only clear to me (if Im correct) is that $\sigma\subset\tau$. I need some help on this.

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  • $\begingroup$ Its already 12:30 am here. If somebody ask for clarification, my apology that I cannot entertain your question. I got to back @ this site by tomorrow. $\endgroup$ – Juniven Aug 17 '13 at 16:12
  • $\begingroup$ In general $\tau$ is strictly finer than $\sigma$. Think of normed spaces, then $\sigma$ is the weak topology. $\endgroup$ – timur Aug 18 '13 at 1:41
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As timur pointed out this topologies in general are different. Indeed, take arbitrary infinite dimensional normable space $(X,\tau)$. We know that in such spaces there are neighbourhhods of zero which does not contain linear subspaces, e.g. unit ball. But as for the weak topology $\sigma(X,X')$ it is not true, because all neighbourhoods of zero contain infinnite dimensional subspace (see remark before theorem 3.12 in Rudin's Functional analysis)

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