We have geometric Hahn Banach theorem in standard functional analysis course, saying that there exists a separating hyperplane separates two special sets in usual topology. However this theorem can be used for some special sets in weak* topology. This is the problem 9 in the functional analysis book by Prof. Brezis.

Theorem 1Let be a Banach space, be two nonempty disjoint convex subsets. Assume is open in weak* topology. Then there exist some and a constant such that the hyperplane separates and .

*Proof:* Since is open in weak* topology, which is coarser than the norm topology for , so is open in norm topology. By geometric Hahn-Banach theorem, there exist and a constant such that

Fix an element , since is open in weak* topology, we can find a weak* open symmetric neighborhood around 0, say . Then we have

Since and is symmetric, we have

Therefore defines a bounded linear functional on and is continuous in weak* topology. Hence there exists an such that .

There is a more applicable version of this theorem, in which case is closed and is compact in weak* topology. Before proving it, we need the following lemma:

Lemma 2Assume that be two nonempty convex subsets, is closed and is compact in weak* topology. Then is convex and closed in weak* topology.

*Proof:* To see is convex. Since ; . Then .

To see is weak* closed, we only need to show is weak* open. Fixed a point , since is weak* closed, for any , there exists a weak* open, convex neighborhood around 0, say , such that

Clearly we have , and since is weak* compact, we can find a finite coverings that cover , i.e. . Define , I want to show

Assume such that . Then for some . Since and is convex (intersection of convex sets is convex), . This implies there exists a such that , which leads to a contradiction.

Theorem 3Let be a Banach space, be two nonempty disjoint convex subsets. Assume is closed in weak* topology and is compact in weak* topology. Then there exist some and a constant such that the hyperplane strictly separates and .i.e. there exists an such thatand

*Proof:* By the lemma we know is convex and closed in weak* topology. Since and are disjoint, does not contain 0. We can find a weak* open, convex neighborhood around 0, say , that does not intersect with . Then by Theorem 1, there exists a hyperplane that separates and .

Since is convex, we can take , which implies . Therefore we have

Define , . Then we have the desired separating property.