What is the base for the weak-$\star$ topology on X and $B^{\star\star}$?

Question

1. What is the base for the weak-$\star$ topology on $X$?

   I understand that a base for the weak topology on $X^\star$ is: $$N_{\epsilon,x^{\star\star}_1,...x^{\star\star}_n}(\psi) = \{\psi' \in X^\star:|x^{\star\star}_i(\psi)-x^{\star\star}(\psi')|<\epsilon\}$$ This is a basis that makes all linear functionals on $X^{\star}$ continuous over $X^{\star}$. However, I'm having trouble understanding how this topology extends to $X$. In particular, its certainly the case that $x^{\star\star}$ can be an evaluation functional, in which case one may write: $$N_{\epsilon,x^{\star\star}_1,...x^{\star\star}_n}(\psi) = \{\psi' \in X^\star:|\psi(x_i)-\psi'(x_i)\}|<\epsilon\}$$ However this only provides a weak-${\star}$ basis for $X$ if $X$ is reflexive.

2. What is the base for the weak-$\star$ topology on $B^{\star\star}$? Here $B^{\star\star}$ is the closed unit ball in $X^{\star\star}$. Closure means closed in the norm of $X^{\star\star}$.

   Basically, using the insight gained from the first question, how would one repeat the same exercise with $B^{\star\star}$ instead of $X$?

Answer 1

For a normed space $X$ and a subset $\mathcal{ F}\subseteq X^\star$ we define $\sigma(X,\mathcal{ F})$ to be the smallest topology on $X$ such that all $f\in\mathcal{ F}$ are continous. Then one can show that the sets of the form

$$N_{\epsilon, f_1,...,f_n}(x)=\{y\in X : |f_i(x)-f_i(y)|<\epsilon,\space i=1,... ,n\}$$

where $\epsilon >0$, $n\in\mathbb N _0$, $f_i\in \mathcal{ F}$ and $x\in X$ form a base for the topology $\sigma(X,\mathcal{ F})$.

Now $\sigma(X,X^\star)$ is the weak topology on $X$ and $\sigma(X^\star,J(X))$ is the weak-$\star$ topology on $X^\star$, where $J:X\to X^{\star\star}$ is the canonical embedding.

Then by the above a basis for the weak-$\star$ topology on $X^\star$ is given by the sets

$$N_{\epsilon, J(x_1),...,J(x_n)}(\psi)=\{\psi'\in X^\star : |\psi(x_i)-\psi'(x_i)|<\epsilon,\space i=1,... ,n\}$$

where $\epsilon >0$, $n\in\mathbb N _0$, $x_i\in X$ and $\psi\in X^\star$.

Now if you want to show that the unit ball $B^\star$ is weak-$\star$ closed in $X^\star$ you can do it in two ways:

1. Using the basis:

For $\psi\in (B^\star)^c$ there is $x\in B$ and $\epsilon>0$ such that $|\psi(x)|=1+\epsilon$. Then all $\psi'\in N_{\epsilon,J(x)}(\psi)$ satisfy $|\psi(x)-\psi'(x)|<\epsilon$ and hence $|\psi'(x)|>1$ which implies $\psi'\in (B^\star)^c$.Thus $(B^\star)^c$ is weak-$\star$ open.

2. Directly from the definition:

$B^\star=\{\psi\in X^\star: \sup_{x\in B}|\psi(x)|\leq 1\}=\bigcap_{x\in B}\{\psi\in X^\star: |J(x)(\psi)|\leq 1\}$ and as each $J(x)$ is continous with respect to the weak-$\star$ topology this is a intersection of weak-$\star$ closed sets.

Hope this helps!

$\textbf{Edit:}$ Replacing $X$ by $X^\star$ then also yields that $B^{\star\star}$ is weak-$\star$ closed in $X^{\star\star}$.