### 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.

** 1.1. Convexity **

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 segment**we shall mean that the end points should be excluded, that is

We shall call a line segment **proper** if . A subset is called **convex**if 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 pointof , if there doesnotexist 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

*Proof:*

- (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

seminormis a function satisfying –

- (a) and
- (b)
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

Then

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

and

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 1Let 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 2Let 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 3If 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.

[…] post is a continuation of this previous post. The aim of this post is to introduce some common objects that are encountered in operator algebra. […]

Reblogged this on John Pappas's blog.