Explicit construction of a nested local base in the weak*-topology

Question

Consider the $C^*$-algebra $C([a,b])$ of complex continuous functions in a closed interval of $\mathbb{R}$. I want to construct a countable nested local base in the state space of the algebra with respect to the $weak*$-topology. Such a base exists since the algebra is separable. How can we construct it explicitly?

Answer 1

Theorem. Let $X$ be a normed space.

1. If the algebraic dimension of $X$ is uncountable then the origin of the dual space $X'$ does not admit a countable neighborhood base relative to the weak$^*$ topology.

2. If $X$ is separable then every bounded subset of $X'$ is metrizable relative to the weak$^*$ topology.

Proof. (1) Suppose by contradiction that $\{V_n\}_n$ is a countable neighborhood base of the origin. By definition of the weak$^*$ topology, for each $n$ there is a finite set $F_n\subseteq X$, and a positive number $\varepsilon _n$, such that the set $$ U(F_n, \varepsilon _n):= \{f\in X': |f(x)|<\varepsilon _n, \forall x\in F_n\} $$ is contained in $V_n$.

By hypothesis we have that $$ \text{span}\left(\bigcup_nF_n\right) \subsetneq X, $$ so we may pick some vector $y$ in $X$ not lying in the linear span of $\bigcup_nF_n$. Observing that $U(\{y\},1)$ is a neighborhood of the origin in $X'$, we will reach a contradiction by proving that $$ V_n\not\subseteq U(G,1), $$ for any given $n$. In fact, since $y\notin \text{span}(F_n)$, we may use Hahn-Banach to produce a continuous linear functional $f$ on $X$ vanishing on $F_n$, and such that $f(y)=1$. It then follows that any multiple of $f$ lies in $U(F_n, \varepsilon _n)$, and hence also in $V_n$, but $\lambda f$ is not in $U(G,1)$ if $|\lambda |\geq 1$.

(2) Assuming that $X$ is separable, choose a dense subset $\{x_n\}_n$ of the unit ball of $X$. Given any $f$ and $g$ in $X'$, define $$ d(f,g) = \sum_n 2^{-n}\min\{1, |f(x_n)-g(x_n)|\}. $$ It is easy to see that $d$ is a metric on $X'$ and moreover that a net $\{f_i\}_i$ converges to some $f$ according to this metric iff $$ f_i(x_n)\ {\buildrel i\to\infty \over \longrightarrow}\ f(x_n),\quad\forall n. \tag 1 $$

On the other hand, it is well known that a bounded net $\{f_i\}_i$ converges to $f$ in the weak$^*$ topology iff (1) holds.

So if $S$ is a bounded subset of $X'$, the nets within $S$ converge weak$^*$ iff they converge according to $d$, thus proving that the weak$^*$ topology on $S$ is metrizable by $d$. QED

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It is well known that the algebraic dimension of any infinite dimension Banach space is uncountable, so the Theorem above applies to $C([0, 1])$ and hence the origin in $C([0, 1])'$ does not admit a countable neighborhood base relative to the weak$^*$ topology.

On the other hand, the state space of $C([0, 1])$ is a bounded set, so part (2) of the above Theorem applies.