Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.