Let $X$ be a separable normed space with a countable dense subset $S:=(x_n)_{n=1}^\infty$ and let $X'$ be the topological dual of $X$.
How can we prove existence of a vectorial topology on $X'$, say $\sigma(X',S)$, with which $X'$ is locally convex (that is, $(X', \sigma(X',S))$ is a locally convex topological vector space) and $\sigma(X',S)$ admits as a basis of neighborhoods of zero the sets of the form
$$U_n=\{x'\in X': \ \displaystyle\sup_{1\leq j\leq n} |x'(x_j)|<1/n \},$$
and, moreover, $\sigma(X',S)$ is weaker than the usual weak$^*$ topology $\sigma(X',X)$ on $X'$ ?
(Here, by a vectorial topology we mean a Hausdorff topology which is compatible with the vector space structure, that is, the operations of vector addition and scalar multiplication are continuous.)
In other words, how can we show that:
$i)$ first of all, existence of a vectorial topology on $X'$ associated with the countable dense subset $S$ of $X$ (which we denote by $\sigma(X',S)$); such that
$ii)$ the topological dual $X'$ (as a vector space) endowed with the topology $\sigma(X',S)$ is locally convex, that is, $(X', \sigma(X',S))$ is a locally convex (topological vector) space;
$iii)$ the sets $(U_n)_{n=1}^\infty$ form a basis of neighborhoods of zero for the topology $\sigma(X',S)$;
$iv)$ and finally, $\sigma(X',S)\leq \sigma(X',X)$ on $X'$?
Can any body provide a proof of these facts stated above in $i), ii), iii)$ and $iv)$ ?
I think we should use the fact that the set $S$ separates points of $X$. But how?
Best Answer
Given a directed family $\mathcal P$ of seminorms on a vector space $Y$ (in your case, $Y=X'$ and $\mathcal P=\{p_n:n\in\mathbb N\}$ with $p_n(x')=\max\{|x'(x_j)|:1\le j\le n\}$) a set $A\subseteq Y$ is called $\mathcal P$-open if, for each $a\in A$ there are $p\in \mathcal P$ and $r>0$ with $B_p(a,r)\subseteq A$, where $B_p(a,r)=\{y\in Y: p(a-y)<r\}$. It is easy to check that the system of all $\mathcal P$-open sets is a topology (for the stability with respect to intersections you need that $\mathcal P$ is directed, i.e., for all $p,q\in \mathcal P$ there is $r\in \mathcal P$ with $p\le r$ and $q\le r$).