# Topological vector space generated by weak topology

Q1. Let $(X, \|.\|)$ be a real Banach space and $\tau$ is the weak topology on $X$. I would like to ask whether $(X,\tau)$ is a topological vector space?

Q2. Let $(X, \|.\|)$ be a real Banach space and its dual space $X^*$. Let $\tau^*$ is the weak$^*$ topology on $X^*$. I would like to ask whether $(X^*,\tau^*)$ is a topological vector space?

Thank you for all construction and helping.

## 1 Answer

Both are true. For your specific question, one might note both the weak and $weak^*$ topologies are generated by a separating family of semi-norms. For the weak topology the seminorms are \begin{equation} p_{\ell}(x,y)=|\ell(x)-\ell(y)|, \end{equation} where $\ell\in X^{*}$ and $x,y\in X$,

and for the $weak^*$ topology the seminorms are \begin{equation} \rho_{x}(\ell_1,\ell_2)=|\ell_1(x)-\ell_2(x)|, \end{equation}where $x\in X$ and $\ell_1,\ell_2\in X^{*}$. And whenever you have a separating family of seminorms their initial topology is the locally convex vector space topology, this is not difficult and a good reference is Rudin's Functional Analysis.

Actually, you can see that we do not actually use the fact that $X$ is a normed space, only that $X^{*}$ separates points in $X$, and a general theorem is that if you have a separating family of linear functionals you can use them to define a locally convex vector space topology on the original space.