Suppose Ball $X^*$ (closed unit ball of $X^*$) is weak-star metrizable. To show $F^\perp\subseteq\bigcap\limits^{\infty}_{n=1}U_n$.

Question

Let $X$ be a normed space. Suppose Ball $X^*$ (closed unit ball of $X^*$) is weak-star metrizable. Then there exists a metric $d$ on ball $X^*$ such that the weak-star topology on ball $X^*$ is induced by $d$. For each $n\in\mathbb{N}$, let $U_n=\{x^*\in\text{ball $X^*$}:d(x^*,0)<\frac{1}{n}\}.$ Then $U_n$ is open in (ball $X^*$, $wk^*$) such that $0\in U_n$ for each $n\in\mathbb{N}$ and $\bigcap\limits^{\infty}_{n=1}U_n=\{0\}$. For each $n$, since $0\in U_n$, by the definition of the weak-star topology on ball $X^*$, there exist $x_{n_i}\in X$ and $\epsilon_{n_i}>0$ ($i=1,\cdots$, $m_n$) such that $$\bigcap\limits_{i=1}^{m_n}\{x^*\in\text{ball $X^*$},|\langle x_{n_i},x^*\rangle|<\epsilon_{n_i}\}\subseteq U_n.$$ For each $n$, let $\epsilon_n=\min\{\epsilon_{n_1},\cdots$, $\epsilon_{n_{m_n}}\}$ and let $F_n=\{x_{n_1},\cdots,x_{n_{m_n}}\}$. Then for each $n$, \begin{align*} \{x^*\in\text{ball }X^*:|\langle x,x^*\rangle|<\epsilon_n \text{ for all }x\in F_n\}&\subseteq\bigcap\limits_{i=1}^{m_n}\{x^*\in\text{ball $X^*$},|\langle x_{n_i},x^*\rangle|<\epsilon_{n_i}\}\\&\subseteq U_n. \end{align*} Let $F=\bigcup\limits_{n=1}^{\infty}F_n$.

And my question is:

1. Let $x^*\in\bigcap\limits^{\infty}_{n=1}U_n$. What will $x^*$ satisfy? Is $x^*$ satisfy $|\langle x,x^*\rangle|<\epsilon_{n}$ for all $x\in F_n$ and for all n?

2. How to show $F^\perp\subseteq\bigcap\limits^{\infty}_{n=1}U_n$?

Any help will be appreciate!

Answer 1

I assume that $\langle x,x^*\rangle$ denotes $x^*(x)$ for each $x\in X$, $x^*\in X^*$ and $$F^\perp=\{x^*\in X^*: \langle x,x^*\rangle=0\mbox{ for all }x\in F\}.$$

1. Is $x^*$ satisfy $|\langle x,x^*\rangle|<\epsilon_{n}$ for all $x\in F_n$ and for all n?

It would be naturally satisfied provided $x^*\in \{x^*\in\text{ball }X^*:|\langle x,x^*\rangle|<\epsilon_n \text{ for all }x\in F_n\}\subseteq U_n$ for each $n$.

What will $x^*$ satisfy?

But I have to confess that $x^*=0$, because $\bigcap\limits^{\infty}_{n=1}U_n=\{0\}$.

2. Let $y^*\in F^\perp$. Then there exists $\lambda>0$ such that $\lambda y^*$ belongs to Ball $X^*$. Then $\lambda y^*\in \{x^*\in\text{ball }X^*:|\langle x,x^*\rangle|<\epsilon_n \text{ for all }x\in F_n\}\subseteq U_n$ for each $n$. So $\lambda y^*\in \bigcap\limits^{\infty}_{n=1}U_n=\{0\}$ and $y^*=0$.