(Folland's page 169) Let $X$ be a normed vector space. Let $B^*:= \{ f \in X^* \,: \, ||f|| \le 1 \}$ be the unit ball in $V^*$ under the operator norm. $B^*$ is compact in $X^*$ in the weak$^*$ topology.
In his proof, he stated that
We may identify $B^*$ with $D:= \prod D_x$, where $D_x:= \{ z \in \mathbb{C} \, : \, |z| \le ||x|| \}$. Then the weak$^*$ topology inherited in $B^*$ coincides with the product topology.
So I tried to proved this; and I proved the following:
Claim: If $V^*$ is identified as a subset of $\prod D_x$, then the weak $^*$ topology is the same as the product topology inherited.
Proof:
- Let $\tau_1, \tau_2$ denote the weak $^*$ topology, product topology respectively.
A basis element of $V^*$ is of the form $U:=\bigcap_{i=1}^n B(x_i, f, \varepsilon) \cap V^*$ for some $f \in V^*$ where $$ B(x_i,f, \varepsilon) = \{ g \in V^* \, : \, |g(x_i)-f(x_i)|<\varepsilon \} $$ Hence $h \in U$ iff $$ |h(x_i)-f(x_i)| < \varepsilon \text{ for all } i \Leftrightarrow h \in V_* \cap h \in \prod_i B(f(x_i), \varepsilon) $$ So $\tau_1 \subseteq \tau_2$.
Conversely, let $U:=\prod B(\alpha_i, \varepsilon)$ (only finitely many $i$ are not whole space.) be any basis element of $\prod \mathbb{C}_x$. We can define a linear functional, $g \in V^*$, such that $g(x_i) = \alpha_i$ for finitely many elements $i$. This can be extended to $V$ by Hahn Banach. Hence, $U \cap V^*$ corresponds to the basis element $V^*$ $$ \bigcap_{i=1}^n B(x_i,g, \varepsilon) $$ Thus, $\tau_2 \subseteq \tau_1$.
I have not found any reference for the proof of these two topologies. So (i) is my proof correct? (ii) is Hahn Banach necessary?
In 3) you are proving that the restriction of product topology to $X^{*}$ is the weak* topology. So you have to take a basic open set in $\prod D_x$ and intersect it with $X^{*}$. This makes 3) evident and you do not have to construct any $x^{*}$ using Hahn - Banach Theorem.