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## Convex Geometry

Csaba Vincze (2013)

University of Debrecen

7.2 Krein-Milman's theorem

## 7.2 Krein-Milman's theorem

The theorem belongs to the basis of the theory of convex sets. The presented version for convex sets in the finite dimensional coordinate space was proved by H. Minkowski. The generalization of this result to infinite dimensional topological vector spaces involves an additional closure operator too. It is due to Krein and Milman (1940).

Figure 48: Hermann Minkowski, 1864-1909.

Definition Let K be a convex subset in the space. The point p in K is called an extreme point if the punctured set K - {p} is also convex. Ext K denotes the set of the extreme points or, in an equivalent terminology, the profile of K.

Remark We have some simple examples for extreme points like the endpoints of segments. Another type of examples are related to convex closed domains in the plane bounded by smooth curves with curvature having no zeros. In this case all of the boundary points are extreme points.

Theorem 7.2.1 (Krein-Milman) If K is a non-empty compact convex set then it is the convex hull of its extreme points,i.e.

 $K:=convextK.$

Proof Let K be a non-empty compact convex set in the coordinate space of dimension n. The inclusion

 $convextK\subset K$

is trivial and, consequently, it is enough to prove that each element q in K can be expressed as a convex combination of extreme points of K. We use an induction on the dimension of the space. If n=1 then K is a closed bounded interval with the endpoints as extreme points. Therefore q can be obviously expressed as a convex combination of them. Suppose that the statement is true for thecoordinate spaces of dimension at most n - 1. If q is an extreme point then the proof is finished. Otherwise K can not be punctured at q without loosing the convexity. Therefore there exists a segment

 $s\left({v}_{1},{v}_{2}\right)\subset K$

such that

 $q\in s\left({v}_{1},{v}_{2}\right)butq\ne {v}_{i}\left(i=1,2\right).$

Using that K is compact we can suppose that both endpoints of this segment are on the boundary of K. Consider the supporting hyperplanes

Then

 ${K}_{i}={H}_{i}\cap K\left(i=1,2\right)$ (7.3)

are convex compact sets of dimension

 $\mathrm{d}\mathrm{i}\mathrm{m}{H}_{i}\cap K\le n-1\left(i=1,2\right).$

By the inductive hypothesis each endpoint of the segment can be expressed as a convex combination of the extreme points of 7.3. To finish the proof we are going to prove that the extreme points of the intersections 7.3 are extreme points of the set K as well. If (for example) i=1 and z is an extreme point of the set

 ${H}_{1}\cap K$ (7.4)

then segments containing z in their relative interiors could not run in the hyperplane because z is an extreme point of 7.4. But they could not intersect the hyperplane because it is a supporting hyperplane for K. Therefore we have no such a segment and z is an extreme point of K as was to be proved. ▮

Corollary 7.2.2 The set of extreme points of a compact convex set is non-empty.

Remark An alternative argumentation: let K be a compact convex set and consider the point where the norm function attains its maximum at (the furthest point of K from the origin). We can easily prove that it must be an extreme point of K, see also excercise 7.4.7.