Sep. Crit: Comp. of the Closed Unit Ball of X* in the Weak* Topology

# Separable Criterion for the Compactness and Sequential Compactness of the Closed Unit Ball of X* in the Weak* Topology

Before we look at the main theorem for this page we will first need to recall some basic definitions.r

Definition: Let $X$ be a topological space. A subset $A \subseteq X$ is Compact if every open cover of $A$ has a finite subcover. |

Definition: Let $X$ be a topological space. A subset $A \subseteq X$ is Sequentially Compact if every infinite sequence has a convergent subsequence. |

Important Note

In general topological spaces, compactness does not imply sequential compactness AND sequential compactness does not imply compactness.

Theorem 1: If $X$ is a metric space and $A \subseteq X$ then $A$ is compact if and only if $A$ is sequentially compact. |

Theorem 2: Let $X$ be a normed linear space. If $X$ is separable then the closed unit ball of the topological dual $X^*$ is compact and sequentially compact in the weak* topology. |

**Proof:**Since $X$ is separable it has a countable and dense subset of the closed unit ball $B_*$ of the topological dual $X^*$. Let $\{ x_n : n \in \mathbb{N} \}$ be such a set. Define a function $d : X^* \times X^* \to [0, \infty)$ for all $\varphi, \psi \in B^*$ by:

\begin{align} \quad d(\varphi, \psi) = \sum_{n=1}^{\infty} \frac{1}{2^n} | \varphi(x_n) - \psi(x_n) | \end{align}

- We note that $d$ is well-defined as if $\varphi, \psi \in B^*$ then $\| \varphi \|, \| \psi \| \leq 1$.

- We first show that $d$ is a metric on $X^*$.

**(1) Showing that $d(\varphi, \psi) = 0$ if and only if $\varphi = \psi$:**Suppose that $d(\varphi, \psi) = 0$. Then $\varphi(x_n) = \psi(x_n)$ for all $n \in \mathbb{N}$. Let $x \in X$ and let $\epsilon > 0$ be given.. Since $\{ x_n : n \in \mathbb{N} \}$ is dense in $X$, there exists $n \in \mathbb{N}$ such that:

\begin{align} \quad | \varphi(x) - \varphi(x_n) | < \frac{\epsilon}{2} \quad \mathrm{and} \quad | \psi(x) - \psi(x_n) | < \frac{\epsilon}{2} \end{align}

- Therefore we have that:

\begin{align} \quad | \varphi(x) - \psi(x) | \leq \underbrace{| \varphi(x) - \varphi(x_n) |}_{< \frac{\epsilon}{2}} + \underbrace{| \varphi(x_n) - \psi(x_n) |}_{=0} + \underbrace{| \psi(x_n) - \psi(x) |}_{< \frac{\epsilon}{2}} < \epsilon \end{align}

- Since $\epsilon$ is arbitrary we see that $\varphi(x) = \psi(x)$ for all $x \in X$. So $\varphi = \psi$.

- On the other hand, if $\varphi = \psi$ then trivially $d(\varphi, \psi) = 0$.

**(2) Showing that $d(\varphi, \psi) = d(\psi, \varphi)$:**This is clear since:

\begin{align} \quad d(\varphi, \psi) = \sum_{n=1}^{\infty} \frac{1}{2^n} |\varphi(x_n) - \psi(x_n)| = \sum_{n=1}^{\infty} \frac{1}{2^n}|\psi(x_n) - \varphi(x_n)| = d(\psi, \varphi) \end{align}

**(3) Showing that $d(\varphi, \psi) \leq d(\varphi, \eta) + d(\eta, \psi)$:**

\begin{align} \quad d(\varphi, \psi) &= \sum_{n=1}^{\infty} \frac{1}{2^n} |\varphi(x_n) - \psi(x_n)| \\ &= \sum_{n=1}^{\infty} \frac{1}{2^n} |\varphi(x_n) - \eta(x_n) + \eta(x_n) - \psi(x_n)| \\ & \leq \sum_{n=1}^{\infty} \frac{1}{2^n} | \varphi(x_n) - \eta(x_n)| + \sum_{n=1}^{\infty} \frac{1}{2^n} |\eta(x_n) - \psi(x_n)| \\ & \leq d(\varphi, \eta) + d(\eta, \psi) \end{align}

- So indeed, $d$ is a metric on $B_*$. Furthermore, this metric induces the weak* topology on $B_*$.

- By Helly's Theorem, every sequence of continuous linear functionals on $B_*$ (which are bounded) has a weak* convergent subsequence, and so $B_*$ is sequentially compact. Since $B_*$ is also a metric space, we have by Theorem 1 that $B_*$ is also compact. $\blacksquare$