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

Recall from the Helley's Theorem page that Helley's theorem states that if $X$ is a normed linear space and $X$ is separable then every bounded sequence in $X^*$ will weak-* converge.

We will now use Helley's theorem to show that if $X$ is a normed linear space and $X$ is separable then the closed unit ball of $X^*$ is weak-* compact and weak-* sequentially compact.

Theorem 1: 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. |

*In general the concepts of compactness and sequential compactness are different. If $X$ is a metric space, they are the same.*

**Proof:**Let $B_{X^*}$ denote the closed unit ball of $X^*$. That is:

- Since $X$ is separable, $B_{X^*} is separable. Let [[$ \{ x_n : n \in \mathbb{N} \}$ be a countable dense subset of $B_{X^*}$. Define a function $d : B_{X^*} \times B_{X^*} \to [0, \infty)$ for all $f, g \in B_{X^*}$ by:

- We note that $d$ is well-defined. If $f, g \in B_{X^*}$ then $\| f \|, \| g \| \leq 1$ and so:

- And so:

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

**(1) Showing that $d(f, g) = 0$ if and only if $f = g$:**Suppose that $d(f, g) = 0$. Then $f(x_n) = g(x_n)$ 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:

- Therefore we have that:

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

- On the other hand, if $f = g$ then trivially $d(f, g) = 0$.

**(2) Showing that $d(f, g) = d(g, f)$:**This is clear since:

**(3) Showing that $d(f, g) \leq d(f, h) + d(h, g)$:**

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

- By Helley's Theorem, every sequence of continuous linear functionals on $B_{X^*}$ (which are bounded) has a weak-* convergent subsequence, and so $B_{X^*}$ is sequentially compact. Since $B_{X^*}$ is also a metric space, we have sequential compactness is equivalent to compactness. So $B_{X^*}$ is compact.$\blacksquare$

#### Important Note

The theorem above tells us that if $X$ is a normed linear space and $X$ is separable then the closed unit ball $B_{X^*}$ of $X^*$ is weak-* compact and weak-* sequentially compact.

Actually, if $X$ is just a normed linear space (and we don't assume $X$ is separable) then we can conclude that $B_{X^*}$ is weak-* compact. This is Alaoglu's Theorem. The same cannot be said for weak-* sequential compactness though. That is, if $X$ is not separable then $B_{X^*}$ may be weak-* sequentially compact OR may not be.