The Krein-Milman Theorem
by Nirakar Neo
1. The Krein-Milman theorem in Locally Convex Spaces
My project work this semester focuses to understand the paper the Krein-Milman Theorem in Operator Convexity by Corran Webster and Soren Winkler, which appeared in the Transactions of the AMS [Vol 351, #1, Jan 99, 307-322]. But before reading the paper, it is imperative to understand the (usual) Krein-Milman theorem which is proved in the context of locally convex spaces. My understanding of this part follows the book A Course in Functional Analysis by J B Conway. To begin with we shall collect the preliminaries that we shall need to understand the Krein-Milman theorem.
Let denote the real() or the complex() number fields. Let be a vector space over . A subset of a vector space is called convex if for any two points in the subset, the line segment joining them lies completely in the subset. We make this idea more precise.
If , the line segment from to is given by the set
By an open line segmentwe shall mean that the end points should be excluded, that is
We shall call a line segment proper if . A subset is called convexif for any , the line segment . It is easy to see that intersection of arbitrary number of convex subsets is again convex. We characterize a convex set as –
Proposition 1.1 : is convex if and only if whenever , and with , then .
Proof: If the latter condition holds, then it is easy to see that is convex. For the converse part, we use induction on the number of points, . For , this is trivially true, while the case follows from the definition. Moreover note that if , the theorem holds. So we shall assume . Suppose the theorem holds for . Then, if there are points, say , and , with and then we can write
Now by the induction hypothesis, and by the convexity of , we get that . Thus the proof is complete.
If , the convex hull of , denoted by , is the intersection of all convex sets containing . Clearly this definition is meaningful since itself a convex set containing , and that convexity is preserved under intersection. is also convex. Using proposition 1.1, we have the following characterization of the convex hull of .
The convex hull of is the set of all convex combinations of elements of , that is
As an example, the convex hull of three points is the closed triangular region formed by those points.
Definition 1. Let be a convex subset of a vector space . A point is called an extreme point of , if there does not exist any proper open line segment containing and which lies completely in . We denote the set of extreme points of by .
Convex sets may or may not have extreme points. For example the set , does not have any extreme point, whereas the extreme points of the set are all the boundary points. The extreme points of a closed square region in are its four corners.
We have following equivalent characterizations.
Proposition 1.3 : Let is convex subset of a vector space , and let . Then the following are equivalent –
- (a) .
- (b) If and and , then either or or .
- (c) If and , then for some .
- (d) is a convex set.
Proof:The equivalence of and is straightforward from the definition of extreme point and convexity of .
: Let . Then by proposition 1.2, we get that , where , and . If for some , we get that . So suppose that for each . Then we can write
This happens only if which means .
: Suppose . Then there are points such that , where , and . So . Then dictates that or . But then .
: We take two points , then , since for any . Thus is convex.
: Suppose that is not an extreme point. Then there exists and with , and that . But this means that does not completely in , which contradicts its convexity.
We shall return to convexity once again after we have introduced the concept of a topological vector space.
1.2. Locally Convex Spaces
For doing analysis on vector spaces we need to have a topological structure on the space. Usually we deal with normed spaces, which already have a topology determined by the metric induced by the norm. We generalise to topological vector spaces which contain normed spaces as a subclass. We note, however, that the topological structure should be compatible with operations on the vector space. Thus we have the following definition.
Definition 2 : A topological vector space (TVS) is a vector space with a topology such that relative to this topology the maps defined by and are continuous for all , and .
Note that inherits the product topology. Addition is continuous means that if and if is a neighborhood of , there should exist neighborhoods of and of such that . Similarly, if , and , and is a neighborhood of then for some , and for some neighborhood of , we have whenever . We also note that translation by a vector and multiplication by a scalar are homeomorphisms on .
Every normed space is a TVS, since the norm satisfies and . Now that we have a topology on the vector space, we can talk about open and closed sets. Let be a TVS. We define the closed convex hull of , which is defined as the intersection of all closed convex subsets of containing , and is denoted by . Then is closed and convex. One may ask – whether ? This proposition answers that in affirmative.
Proposition 1.4 Let be a TVS and let be a convex subset of . Then
- (a) is convex, and
- (b) if and , then
- (a) Let . Choose an element , and let . Let be a net in converging to . Then since addition and scalar multiplication is continuous, we get
that is . Now let be a net in converging to , then for each . Since is closed, we have
Thus . Hence is closed.
- (b) For this part, we fix , and , where and . Since translation is a homeomorphism, there is an open set , such that . Then for any
It only remains to show that there is an element , such that for then this set becomes a neighborhood of 0, and hence its translation to will mean that . Now, means . But , and this set is open. Since , we can always find .
Corollary – If , then .
Proof: Since is convex, is convex too. Since is the intersection of all closed convex sets containing , we have . Now let , and let be any closed convex set containing . Then clearly, . And since is closed, we also have . Thus , and hence .
Definition 3. Let be a vector space over . A seminorm is a function satisfying –
- (a) and
for all , and .
Note that, from (b), we have , and thus . It may happen that is zero for elements which are not zero. A seminorm is called a norm if if and only if . Thus, every norm is a seminorm. Let be the set of all continuous functions . Fix . Define , by . Then
Also . Thus is a seminorm. But, does not mean that .
Let be a vector space and be a family of seminorms on . Let be the topology on that has as subbase, the sets of the form
where and . Since the basis determined by a subbase is the collection of all finite intersection of the elements of the subbase, we see that a subset is open if and only if for every there are and such that
Proposition 1.6 – with the above mentioned topology is a TVS.
Proof: We need to show that the maps and are continuous for all , and .
To show the continuity of addition, let , and let be a neighborhood of . Then we need to find neighborhoods of and of such that . Since is open there exist and , such that the basis element
Let , and define
as basis neighborhoods of and respectively. Let and , then
for each . Thus . Thus . The continuity of multiplication is handled in a similar manner.
Definition 4 – A locally convex space (LCS) is a topological vector space whose topology is defined by a family of seminorms such that
The condition imposed above ensures that the topology so defined is Hausdorff. To show that this is Hausdorff, let , with . Then there is a for which . For if for all , then , which contradicts that only 0 belongs to that set. Let , and
Then . For if , then
which again yields a contradiction.
There is another characterization of a LCS, which explains why the name locally convex is given. A set is called balanced if , whenever , and .
Theorem 1 Let be a TVS, and let be the collection of all open convex balanced subsets of . Then is locally convex if and only if is a basis for the neighborhood system at 0.
1.3. The Krein-Milman Theorem
Roughly speaking, the Krein-Milman theorem is a generalization of the fact that in , one can recover the shape of a convex polygon once its extreme points are known.
We shall be using the following crucial separation theorem.
Theorem 2 Let be a complex LCS and let and be disjoint closed convex subsets of . If is compact, then there is continuous linear functional , an , and an such that for , and we have
Theorem 3 If is a non-empty compact convex subset of a locally convex space , then and .
Proof: We shall prove the theorem in steps. First note that if , a singleton set, then the theorem trivially holds. Therefore, we’ll assume that contains at least two points.
Step 1. According to Proposition 1.3, is an extreme point if and only if is convex. Every locally convex space is Hausdorff, and since is a compact subset of , we get that is closed. Hence is open relative to . Thus, is an extreme point if and only if is relatively open convex subset of . In this first step of the proof, we shall establish the existence of a maximal relatively open convex proper subset of (using Zorn’s lemma).
Let denote the collection of all relatively open convex proper subsets of . This collection is not empty. For, let , then is open and non-empty (since we assumed that contains at least two points); let . Since is locally convex, there is an open convex subset containing . Then is a relatively open convex proper subset of . We introduce a partial order relation on by inclusion, and let be a totally ordered subset of . Put
Clearly, is open and since is totally ordered, is convex. We only have to show that is proper. If , then is an open cover for , and by compactness of , we have that for some , which contradicts that . Thus this totally ordered set has a maximal element, and hence by Zorn’s lemma, has a maximal element . This completes our first step.
Step 2. In this step we shall prove that consists of a single point, thereby proving that ext is not empty. If , and , let be defined by
Clearly, is continuous and it preserves convex combinations, that is,
where and for and . If we take , and , then by convexity of , we have , which implies . Since pull back of an open set by a continuous function is open, we get that is open in , and it is convex too. If then by proposition 1.4 we have
Thus , and hence by maximality of , we get which translates into .
Now, let be any open convex subset of . Let . If both points lie in , their convex combinations also lie in , and hence in . Suppose and , then from what we just proved above, all the linear combinations (where ) lies in . Thus which proves that is an open convex subset of . But by maximality of , we must have either or . Using this we complete our claim that is singleton.
Let , and . Let and be disjoint open convex subsets of such that and . Since , we have (from the previous paragraph). This implies that . But this is contradiction since does not lie in as well as . Thus is singleton, and hence ext is not empty.
Step 3. In this last step we prove that is the closed convex hull of ext , the last claim of the theorem. We first prove that if is an open convex subset of , and ext , then . If this does not hold, there will be an open convex subset of such that ext but . Then , and is contained in a maximal element . Since for some , this contradicts that ext .
Finally, let . Suppose , and therefore let . Then and are disjoint closed convex sets with being compact. Then by theorem 1.8, there exists a continuous linear functional , an , an such that , and . Let . Then is open, convex and which implies . But as we proved in the previous paragraph, we should have . But this contradicts that .
Hence , and this completes the proof.