proof of Krein-Milman theorem

proof of Krein-Milman theorem

The proof is consist of three steps for good understanding.
We will show initially that the set of

extreme points

of

K

,

E

⁢

x

⁢

(

K

)

is non-empty,

E

⁢

x

⁢

(

K

)

≠

∅

.
We consider that

𝒜

=

{

A

⊂

K

:

A

⊂

K

,

extreme

}

.

Step1

The family set

𝒜

ordered by

⊂

has a

minimal element

, in other words there exist

A

∈

𝒜

such as

∀

B

∈

𝒜

,

B

⊂

A

we have that

B

=

A

.

Proof1

We consider

A

<

B

⇔

B

⊂

A

,

∀

A

,

B

∈

𝒜

. The

ordering relation

<

is a partially relation
on

𝒜

. We must show that

A

is maximal element for

𝒜

.
We apply Zorn’s lemma.We suppose that

𝒞

=

{

A

i

:

i

∈

I

}

is a chain of

𝒜

.Witout loss of
generality we take

A

=

⋂

i

∈

I

A

i

and then

A

≠

∅

.

𝒞

has the property of finite

intersections

and it is consist of

closed sets

. So we have that

⋂

i

∈

I

A

i

≠

∅

. It is easy to see that

A

∈

𝒜

.
Also

A

⊂

A

i

, for any

i

∈

I

, so we have that

A

>

A

i

, for any

i

∈

I

.

Step2

Every minimal element of

𝒜

is a set which has only one point.

Proof2

We suppose that there exist a minimal element

A

of

𝒜

which has at least two points,

x

,

y

∈

A

. There exist

x

*

∈

X

*

such as

x

*

⁢

(

x

)

≠

x

*

⁢

(

y

)

, witout loss of
generality we have that

x

*

⁢

(

x

)

<

x

*

⁢

(

y

)

.

A

is

compact set

(closed subset of the compact

K

). Also there
exist

α

∈

ℝ

such that

α

=

sup

z

∈

A

⁡

x

*

⁢

(

z

)

and

B

=

{

z

∈

A

:

x

*

⁢

(

z

)

=

α

}

≠

∅

.
It is obvious that

B

is an extreme subset of

A

,

B

is an extreme subset of

K

,

B

∈

𝒜

.

x

∉

B

since

B

∈

𝒜

and

B

⊊

A

that contradicts to the fact that

A

is

minimal

extreme subset of

𝒜

.

From the above two steps we have that

E

⁢

x

⁢

(

K

)

≠

∅

.

Step3

K

=

c

¯

⁢

o

⁢

(

E

⁢

x

⁢

(

K

)

)

where

c

¯

⁢

o

⁢

(

E

⁢

x

⁢

(

K

)

)

denotes the closed

convex hull

of extreme points of

K

.

Proof3

Let

L

=

c

¯

⁢

o

⁢

(

E

⁢

x

⁢

(

K

)

)

. Then

L

is closed subset of

K

, therefore it is compact, and convex clearly by the definition.
We suppose that

L

⫋

K

. Then there exist

x

∈

K

-

L

. Let use

Hahn-Banach theorem

(

geometric form

).
There exist

x

*

∈

X

*

such as

sup

w

∈

L

⁡

x

*

⁢

(

w

)

<

x

*

⁢

(

x

)

. Let

α

=

sup

⁡

{

x

*

⁢

(

y

)

:

y

∈

K

}

,

B

=

{

y

∈

K

:

x

*

⁢

(

y

)

=

α

}

.

Similar

to
Step2

B

is extreme subset of

K

.

B

is compact and from step1 and step2 we have that

E

⁢

x

⁢

(

B

)

≠

∅

. It is true that

E

⁢

x

⁢

(

B

)

⊂

E

⁢

x

⁢

(

K

)

⊂

L

. Now let

y

∈

E

⁢

x

⁢

(

B

)

then

x

*

⁢

(

y

)

=

α

and if

y

∈

L

⁢

x

*

⁢

(

y

)

<

x

*

⁢

(

x

)

≤

α

. That is a

contradiction

.

Title

proof of Krein-Milman theorem

Canonical name

ProofOfKreinMilmanTheorem

Date of creation

2013-03-22 15:24:40

Last modified on

2013-03-22 15:24:40

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

6

Author

georgiosl (7242)

Entry type

Proof

Classification

msc 46A03

Classification

msc 52A07

Classification

msc 52A99

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