Krein–Milman theorem

Throughout, ${\displaystyle X}$ will be a real or complex vector space.

For any elements ${\displaystyle x}$ and ${\displaystyle y}$ in a vector space, the set ${\displaystyle [x,y]:=\{tx+(1-t)y:0\leq t\leq 1\}}$ is called the closed line segment or closed interval between ${\displaystyle x}$ and ${\displaystyle y.}$ The open line segment or open interval between ${\displaystyle x}$ and ${\displaystyle y}$ is ${\displaystyle (x,x):=\varnothing }$ when ${\displaystyle x=y}$ while it is ${\displaystyle (x,y):=\{tx+(1-t)y:0<t<1\}}$ when ${\displaystyle x\neq y}$;[1] it satisfies ${\displaystyle (x,y)=[x,y]\setminus \{x,y\}}$ and ${\displaystyle [x,y]=(x,y)\cup \{x,y\}.}$ The points ${\displaystyle x}$ and ${\displaystyle y}$ are called the endpoints of these interval. An interval is said to be non-degenerate or proper if its endpoints are distinct.

The intervals ${\displaystyle [x,x]=\{x\}}$ and ${\displaystyle [x,y]}$ always contain their endpoints while ${\displaystyle (x,x)=\varnothing }$ and ${\displaystyle (x,y)}$ never contain either of their endpoints. If ${\displaystyle x}$ and ${\displaystyle y}$ are points in the real line ${\displaystyle \mathbb {R} }$ then the above definition of ${\displaystyle [x,y]}$ is the same as its usual definition as a closed interval.

For any ${\displaystyle p,x,y\in X,}$ the point ${\displaystyle p}$ is said to (strictly) lie between ${\displaystyle x}$ and ${\displaystyle y}$ if ${\displaystyle p}$ belongs to the open line segment ${\displaystyle (x,y).}$[1]

If ${\displaystyle K}$ is a subset of ${\displaystyle X}$ and ${\displaystyle p\in K,}$ then ${\displaystyle p}$ is called an extreme point of ${\displaystyle K}$ if it does not lie between any two distinct points of ${\displaystyle K.}$ That is, if there does not exist ${\displaystyle x,y\in K}$ and ${\displaystyle 0<t<1}$ such that ${\displaystyle x\neq y}$ and ${\displaystyle p=tx+(1-t)y.}$ In this article, the set of all extreme points of ${\displaystyle K}$ will be denoted by ${\displaystyle \operatorname {extreme} (K).}$[1]

For example, the vertices of any convex polygon in the plane ${\displaystyle \mathbb {R} ^{2}}$ are the extreme points of that polygon. The extreme points of the closed unit disk in ${\displaystyle \mathbb {R} ^{2}}$ is the unit circle. Every open interval and degenerate closed interval in ${\displaystyle \mathbb {R} }$ has no extreme points while the extreme points of a non-degenerate closed interval ${\displaystyle [x,y]}$ are ${\displaystyle x}$ and ${\displaystyle y.}$

A set ${\displaystyle S}$ is called convex if for any two points ${\displaystyle x,y\in S,}$ ${\displaystyle S}$ contains the line segment ${\displaystyle [x,y].}$ The smallest convex set containing ${\displaystyle S}$ is called the convex hull of ${\displaystyle S}$ and it is denoted by ${\displaystyle \operatorname {co} S.}$ The closed convex hull of a set ${\displaystyle S,}$ denoted by ${\displaystyle {\overline {\operatorname {co} }}(S),}$ is the smallest closed and convex set containing ${\displaystyle S.}$ It is also equal to the intersection of all closed convex subsets that contain ${\displaystyle S}$ and to the closure of the convex hull of ${\displaystyle S}$; that is,

$${\displaystyle {\overline {\operatorname {co} }}(S)={\overline {\operatorname {co} (S)}},}$$

where the right hand side denotes the closure of ${\displaystyle \operatorname {co} (S)}$ while the left hand side is notation. For example, the convex hull of any set of three distinct points forms either a closed line segment (if they are collinear) or else a solid (that is, "filled") triangle, including its perimeter. And in the plane ${\displaystyle \mathbb {R} ^{2},}$ the unit circle is not convex but the closed unit disk is convex and furthermore, this disk is equal to the convex hull of the circle.

Krein–Milman theorem[1] — Suppose ${\displaystyle X}$ is a Hausdorff locally convex topological vector space (for example, a normed space) and ${\displaystyle K}$ is a compact and convex subset of ${\displaystyle X.}$ Then ${\displaystyle K}$ is equal to the closed convex hull of its extreme points:

$${\displaystyle K~=~{\overline {\operatorname {co} }}(\operatorname {extreme} (K)).}$$

Moreover, if ${\displaystyle B\subseteq K}$ then ${\displaystyle K}$ is equal to the closed convex hull of ${\displaystyle B}$ if and only if ${\displaystyle \operatorname {extreme} K\subseteq \operatorname {cl} B,}$ where ${\displaystyle \operatorname {cl} B}$ is closure of ${\displaystyle B.}$

The convex hull of the extreme points forms a subset of ${\displaystyle K}$ so the main burden of the proof is to show that there are enough extreme points so that their convex hull covers all of ${\displaystyle K.}$

A corollary to the above theorem, which is also some called "the Krein–Milman theorem", is that every non-empty compact convex subset of a Hausdorff locally convex TVS has extreme points; that is, the set of its extreme points is not empty.[1]

A particular case of this theorem, which can be easily visualized, states that given a convex polygon, the corners of the polygon are all that is needed to recover the polygon shape. The statement of the theorem is false if the polygon is not convex, as then there can be many ways of drawing a polygon having given points as corners.