## Weak* and product topology are the same on $B_1((\ell^1)^*) = D^N$ (p19 (C))

Let $\mathbb{D} = \{ x \in \mathbb{C} : |x| \le 1 \}$. Show that $B_1(\ell^\infty = (\ell^1)^*) = \mathbb{D}^\mathbb{N}$ and that the weak*-topology on $B_1(\ell^\infty)$ is the same as the product topology on $\mathbb{D}^\mathbb{N}$.

Remarks
Without further comment, I will interchange between $S \in (\ell^1)^*$ and $(s_n) \in \ell^\infty$.  Also, I will sometimes abbreviate $B_1( (\ell^1)^* )$ to simply $B_1$.  A basic open neighborhood of the weak*-topology on $B_1$ is
$V_{a_1, \ldots, a_n ; \epsilon ; T} = \{ S \in B_1( (\ell^1)^* ) : |S(a_1) - T(a_1)| < \epsilon, \ldots,|S(a_n) - T(a_n)| < \epsilon\} ,$
for some $a_i = (a_{in}) \in \ell^1$, $\epsilon > 0$, and $T \in B_1$. Also, $e_n \in \ell^\infty$ is defined by $m \neq n \Rightarrow e_n(m) = 0,$ and $e_n(n) = 1$.

Proof.
$B_1(\ell^\infty) = \{x \in \ell^\infty : \| x \| \le 1 \} = \mathbb{D}^\mathbb{N}$.

Let $U$ be a basic open neighborhood of $(t_n)$ in the product topology on $\mathbb{D}^\mathbb{N}$. That is, $U = \prod_{i = 0}^\infty U_i$, such that $U_i = \mathbb{D}$ if $i \notin \{i_1, \ldots, i_n \}$ and
$U_{i_j} = \{ x \in \mathbb{D} : | x - t_{i_j} | < \epsilon_j \}.$
Note that $|(S - T)(e_{i_j})| = |s_{i_j} - t_{i_j}|$. Let $\epsilon = \min \{ \epsilon_j \}$.  Then $V_{e_{i_1}, \ldots, e_{i_n} ; \epsilon ; T } \subset U$.

Then, let $V$ be a basic open neighborhood of $B_1$ in the weak*-topology, that is $V = V_{a_1, \ldots, a_n ; \epsilon ; T}$. Now, $\exists M > 0$, such that $\forall i$, $\sum_{n = M}^\infty |a_{in}| < \epsilon / 4$.  Let $K > 0$, such that $K \ge \max \{ \| a_1 \|_1, \ldots, \| a_n \|_1 \}$.  Let $\eta = \frac{\epsilon}{2MK}$. Choose $(s_n) \in \mathbb{D}^\mathbb{N}$, such that $n < M \Rightarrow |s_n - t_n| < \eta$.
Then
$|(S - T)(a_i)| = \left| \sum_{n = 0}^\infty (s_n - t_n)a_{in} \right| \le \sum_{n = 0}^\infty |(s_n - t_n)a_{in}|$
$= \sum_{n = 0}^{M - 1} |(s_n - t_n)a_{in}| + \sum_{n=M}^\infty |(s_n - t_n)a_{in}|$
$< \epsilon / 2 + \epsilon / 2 = \epsilon.$

Thus, if $U = \prod_{i = 0}^\infty U_i$, where $i < M \Rightarrow U_i = \{x \in \mathbb{D} : |x - t_i| < \eta \}$ and $i \ge M \Rightarrow U_i = \mathbb{D}$, then $U \subset V$.