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?

  • $\begingroup$ A starting point could be the definition of a locally convex topological vector space. As far as I know, one does not define "locally convex" in the setting of a general topology (since convex combinations require vector space operations). It would improve your Question to share what connections you are already aware of between the parts of the problem, as well as some context for what you consider "an elementary proof". Topological vector spaces are typically introduced at a graduate level of analysis. $\endgroup$
    – hardmath
    Oct 29, 2019 at 0:17
  • $\begingroup$ I gave some details, and I hope that now it is okay. $\endgroup$
    – serenus
    Oct 29, 2019 at 15:50
  • $\begingroup$ For a locally convex topology you need a directed family of seminorms. The seminorms you need here appear in your definition of $U_n $. $\endgroup$
    – Jochen
    Oct 29, 2019 at 22:12
  • 1
    $\begingroup$ Directed families of semi-norms have e.g. the advantage that the induced topology is verbatim as in metric spaces. $\endgroup$
    – Jochen
    Jan 27, 2020 at 5:48
  • 1
    $\begingroup$ I do not want to belabor the obvious. But another nice thing is that (semi-) metrizability is equivalent to having an increasing sequence of semi-norms. $\endgroup$
    – Jochen
    Jan 27, 2020 at 15:17

1 Answer 1


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 $q\le r$).


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