Banach–Alaoglu theorem

Remarks

Since $U$ is absorbing, each $x \in E$ lies in $r_x U$ for some $r_x > 0$, so $\abs{f(x)} \le r_x$ for $f \in U^\circ$. Hence $U^\circ$ sits inside the product $P = \prod_{x \in E} \set{c \in \mathbb{K} : \abs{c} \le r_x}$, which is compact by Tychonoff's theorem, and the weak-* topology is the topology induced by the product topology. The functions in $P$ that are linear and bounded by $1$ on $U$ form a closed subset of $P$. Each such function is continuous, since it is bounded on a neighbourhood of $0$. So this closed subset is $U^\circ$, and it is compact. Example: for $E = c_0$, the dual is $\ell^1$, and the closed unit ball of $\ell^1$ is weak-* compact, although it is not compact in norm. The weak-* topology is needed: in an infinite-dimensional normed space $E$, the closed unit ball of $E'$ is never compact in the norm topology. The theorem gives compactness, not sequential compactness. If $E$ is a separable normed space, the closed unit ball of $E'$ is weak-* metrizable, and every bounded sequence in $E'$ has a weak-* convergent subsequence.

Going deeper

For a student

The Banach–Alaoglu theorem is the standard source of compactness in functional analysis. It is used to prove existence results, for instance minimizers in the calculus of variations, weak solutions of partial differential equations, and invariant means and measures. With Goldstine's theorem, it characterizes reflexive Banach spaces by the weak compactness of their unit ball. With the Krein–Milman theorem, it shows that the dual unit ball has many extreme points. In duality theory, it says that polars of neighbourhoods of $0$ are weak-* compact, which is a key step of the Mackey–Arens theorem.