# Seminorms that define the weak* topology

Let $$E$$ be a normed space, and $$E'$$ its dual space. The weak* topology on $$E'$$ can be defined by the family of seminorms $$\{p_x\,|\, x\in E\}$$, where for any $$f\in E'$$, $$p_x(f)=|f(x)|$$.

My question is, can this topology be defined by fewer seminorms? For example, if we have a countable dense set $$\{x_1, x_2, ...\}\subset E$$, does the smaller family $$\{p_{x_n}\,|n\geq 1\}$$ of seminorms define the same topology on $$E'$$?

One of the difficulties is that open sets in weak* topology are not bounded in the original $$\|\cdot\|$$ norm.

• Topology generated by a countable number of semi-norms is metrizable but the weak* topology is not, if the space is infinite dimensional. Commented Dec 11, 2021 at 5:41
• @Kavi Rama Murthy Thanks! In the end of the post math.stackexchange.com/q/626599 it was claimed that, for a normed space X and its completion $\hat X$, the dual spaces $X'$ and $(\hat X)'$ are isomorphic and have the same weak* topology. However, the weak* topology on the dual $(\hat X)'$ has more seminorms (coming from $\hat X-X$) than on the dual $X'$. Is that claim true? Commented Dec 11, 2021 at 6:06
• @Kavi Rama Murthy Thanks! Can you provide some hint, or, where I can find a proof? This is a somehow similar situation, where one wants to check the topology remain the same if some extra seminnorms are added into the family... Commented Dec 11, 2021 at 6:23
• I am sorry. I am deleting my second comment because it was wrong. I have provided an answer to the question in your comment. Commented Dec 11, 2021 at 8:44

Let $$X$$ be the space of all finitely non-zero sequences with the sup norm. Its completion is $$c_0$$ and the dual is $$\ell^{1}$$. The sequence $$(ne_n)$$ tends to $$0$$ in the weak* topology induced by $$X$$ but not in the weak* topology induced by $$c_0$$ (since $$x_n \to 0$$ does not imply $$nx_n \to 0$$).