Advanced Analysis, Notes 13: Banach spaces (convex hulls and the Krein–Milman theorem)
by Orr Shalit
It would be strange to disappear for a week without explanations. This blog was not working for the past week because of the situation in Israel. The dedication in the beginning of the previous post had something to do with this, too. We are now back to work, with the modest hope that things will remain quiet until the end of the semester. We begin our last chapter in basic functional analysis, convexity and the Krein–Milman theorem.
1. Closed convex sets and convex hulls
Let be a Banach space. Throughout this post we will consider either with its norm topology or with a weak topology where separates the points of .
Definition 1: Let be a set. The convex hull of , denoted , is the smallest convex set containing . The closed convex hull of , denoted , is the smallest closed convex set containing .
Since arbitrary intersections of convex sets are convex, one sees that the convex hull of a set exists simply by taking the intersection of all convex sets containing . Similar remarks hold for the closed convex hull.
Exercise A: Let . Then
Lemma 2: Let be convex and let be a subspace that separates points. Then and are convex.
Proof: One can give a proof with nets that works in both strong and weak topologies. I prefer to give a proof using neighborhoods just to get a chance to use this language a bit.
Let and . Put . Let be a basic neighborhood of . We must show that . Now because are in the closure of , then there are points and which are in . Put . Then , and also for all , so . Thus for any such , so .
Corollary 3: .
Proof: Lemma 2 tells us that the right hand side is convex and closed, so it must contain the left hand side. On the other hand is contained in the closed , so the closure must also be contained.
Definition 4: Let be convex. A non-empty set is said to be a face of if is convex and if
If , and if is a face of , then is said to be an extreme point of . The set of extreme points of is denoted .
- Consider the extreme points and faces of the closed and open unit balls in w.r.t the norms, . We see here that , and this is what happens in general.
- Let . Then . If , let be such that . Then one easily constructs supported on and supported on such that .
Exercise B: The intersection of faces is again a face (so long as it is non–empty). If we have convex sets , then if is a face of and is a face of , then is a face of . In particular, .
Lemma 5: Let be convex and a face. Suppose that , where and are non-negative and sum to . Then .
Proof: Suppose . If then all other s are zero, so . Otherwise put . is in . Define . Then and . Thus , from the definition of face.
Lemma 6: Let be a non-empty, compact convex subset of . Let be a continuous (w.r.t the topology at hand) functional and define . Then the set is a compact face in .
Proof: is continuous on a compact set, so attains it maximum, so . is a closed subset of a compact Hausdorff space, so is compact. To see that is a face, first note that is convex. If , and , then
while and . It follows that , and .
2. The Krein–Milman theorem
Theorem 7 (Krein–Milman theorem): Let be a Banach space with either the normed topology or the –topology, where is a subspace of that separates points. Let be compact and convex (and assume it is non-empty, to avoid worrying about the silly situation that it is). Then .
Remark: In particular, the result implies that is not empty. Reconsidering Example 2 above, we see that the unit ball of is not compact with respect to any weak topology, and in particular it is not closed with respect to the weak topology. This also implies that is not the dual of any Banach space, since otherwise the closed unit ball would be weak* compact by Alaoglu’s theorem.
Proof: The first step is to show that is not empty. Let be the family consisting of all compact faces of . is not empty because . Order by inclusion. An application of Zorn’s lemma (the intersection of a totally ordered chain in is a non–empty face) shows that has a minimal element, which we shall denote as . We shall show that is a singleton, and hence that .
Assume for contradiction that there are two distinct elements in . Then there is a functional , continuous w.r.t. the relevant topology, such that
Now let , and define as in Lemma 6. The lemma then implies that is a compact face of , therefore it is a compact face of . But (*) implies that , which contradicts the minimality of . Therefore, , and .
We shall now prove that . Assume (again, for the sake of obtaining a contradiction) that there is some which is not in . By the Hahn–Banach Theorem (more precisely, Theorem 8 in Notes 11) there is a continuous functional and some such that
for all . Denote and . By Lemma 6, is a compact face of . By the first half of the proof, has an extreme point . But then is an extreme point of , so by (*) we have that . On the other hand, , so this is a contradiction. It follows that , as required.
3. Minimality of Ext(C)
Theorem 8 (Milman’s Theorem): Let be a non–empty, compact and convex subset of . If is a closed set such that , then .
Proof: Let . Let be a basic open neighborhood of . We will be done once we show that . Indeed, if for any basic neighborhood of , then .
For all , denote . Then the open sets cover , and since is compact there are such that . Denote and .
We claim that is compact. We leave this as an exercise.
We have that . Whence . In particular, , so , where and for all , and . Since is an extreme point there is some such that . Thus
for . It follows that , so , and the proof is complete.