The Krein–Milman Theorem A Project in

Samuel Pettersson

November 29, 2016 2. Extreme points

3. The Krein–Milman theorem

4. An application

Outline

1. An informal example 3. The Krein–Milman theorem

4. An application

Outline

1. An informal example

2. Extreme points 4. An application

Outline

1. An informal example

2. Extreme points

3. The Krein–Milman theorem Outline

1. An informal example

2. Extreme points

3. The Krein–Milman theorem

4. An application Outline

1. An informal example

2. Extreme points

3. The Krein–Milman theorem

4. An application Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”. Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”.

kxk1≤ 1 Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”.

kxk1≤ 1 Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”.

kxk1≤ 1 kxk2≤ 1 Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”.

kxk1≤ 1 kxk2≤ 1 Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”.

kxk1≤ 1 kxk2≤ 1 kxk∞≤ 1 Convex sets and their “corners”

Observation Some convex sets are the convex hulls of their “corners”.

kxk1≤ 1 kxk2≤ 1 kxk∞≤ 1 x1, x2 ≥ 0

Convex sets and their “corners”

Observation Some convex sets are not the convex hulls of their “corners”. Convex sets and their “corners”

Observation Some convex sets are not the convex hulls of their “corners”.

x1, x2 ≥ 0 Convex sets and their “corners”

Observation Some convex sets are not the convex hulls of their “corners”.

x1, x2 ≥ 0 Convex sets and their “corners”

Observation Some convex sets are not the convex hulls of their “corners”.

x1, x2 ≥ 0 kxk∞< 1 I Formalize the notion of a corner of a ().

I Find a sufficient condition for a convex set to be the closed of its extreme points (Krein–Milman).

Objectives I Find a sufficient condition for a convex set to be the closed convex hull of its extreme points (Krein–Milman).

Objectives

I Formalize the notion of a corner of a convex set (extreme point). Objectives

I Formalize the notion of a corner of a convex set (extreme point).

I Find a sufficient condition for a convex set to be the closed convex hull of its extreme points (Krein–Milman). Outline

1. An informal example

2. Extreme points

3. The Krein–Milman theorem

4. An application I points are never extremal

I points may be extremal

I boundary points (inside the set) of strictly convex sets are always extremal

Definition An extreme point of a convex set K ⊆ E in a E is a point z ∈ K not in the interior of any segment in K:

z 6= (1 − t)x + ty, ∀t ∈ (0, 1), ∀x, y ∈ K, x 6= y

Remark In a normed space,

Definition of an extreme point I interior points are never extremal

I boundary points may be extremal

I boundary points (inside the set) of strictly convex sets are always extremal

Remark In a normed space,

Definition of an extreme point

Definition An extreme point of a convex set K ⊆ E in a vector space E is a point z ∈ K not in the interior of any in K:

z 6= (1 − t)x + ty, ∀t ∈ (0, 1), ∀x, y ∈ K, x 6= y I interior points are never extremal

I boundary points may be extremal

I boundary points (inside the set) of strictly convex sets are always extremal

Definition of an extreme point

Definition An extreme point of a convex set K ⊆ E in a vector space E is a point z ∈ K not in the interior of any line segment in K:

z 6= (1 − t)x + ty, ∀t ∈ (0, 1), ∀x, y ∈ K, x 6= y

Remark In a normed space, I boundary points may be extremal

I boundary points (inside the set) of strictly convex sets are always extremal

Definition of an extreme point

Definition An extreme point of a convex set K ⊆ E in a vector space E is a point z ∈ K not in the interior of any line segment in K:

z 6= (1 − t)x + ty, ∀t ∈ (0, 1), ∀x, y ∈ K, x 6= y

Remark In a normed space,

I interior points are never extremal I boundary points (inside the set) of strictly convex sets are always extremal

Definition of an extreme point

Definition An extreme point of a convex set K ⊆ E in a vector space E is a point z ∈ K not in the interior of any line segment in K:

z 6= (1 − t)x + ty, ∀t ∈ (0, 1), ∀x, y ∈ K, x 6= y

Remark In a normed space,

I interior points are never extremal

I boundary points may be extremal Definition of an extreme point

Definition An extreme point of a convex set K ⊆ E in a vector space E is a point z ∈ K not in the interior of any line segment in K:

z 6= (1 − t)x + ty, ∀t ∈ (0, 1), ∀x, y ∈ K, x 6= y

Remark In a normed space,

I interior points are never extremal

I boundary points may be extremal

I boundary points (inside the set) of strictly convex sets are always extremal Space Extreme points ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) Entire unit sphere `∞ (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit : ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) Entire unit sphere `∞ (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points ith term z}|{ ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) Entire unit sphere `∞ (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points

`1 `p (1 < p < ∞) Entire unit sphere `∞ (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) Entire unit sphere `∞ (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) `∞ (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) Entire unit sphere (±1, ±1,... )

Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) Entire unit sphere `∞ Examples of extreme points

Extreme points of the closed unit ball:

Space Extreme points ith term 1 z}|{ ` ±ei = (0,..., 0, ±1 , 0,... ) `p (1 < p < ∞) Entire unit sphere `∞ (±1, ±1,... ) Outline

1. An informal example

2. Extreme points

3. The Krein–Milman theorem

4. An application Reminder conv A := conv A = the smallest closed and convex set containing A.

Statement of Krein–Milman

Theorem (Krein–Milman) A compact convex set K ⊆ E in a normed space coincides with the closed convex hull of its extreme points:

K = conv(ext K) Statement of Krein–Milman

Theorem (Krein–Milman) A compact convex set K ⊆ E in a normed space coincides with the closed convex hull of its extreme points:

K = conv(ext K)

Reminder conv A := conv A = the smallest closed and convex set containing A. I non-empty

I closed

I such that any line segment in K whose interior intersects M has endpoints in M:

∃t ∈ (0, 1): (1 − t)x + ty ∈ M =⇒ x, y ∈ M, ∀x, y ∈ K

Definition Given a compact convex set K ⊆ E in a normed space, an extreme set is a M ⊆ K that is

Preparation for the proof: Extreme sets I non-empty

I closed

I such that any line segment in K whose interior intersects M has endpoints in M:

∃t ∈ (0, 1): (1 − t)x + ty ∈ M =⇒ x, y ∈ M, ∀x, y ∈ K

Preparation for the proof: Extreme sets

Definition Given a compact convex set K ⊆ E in a normed space, an extreme set is a subset M ⊆ K that is I closed

I such that any line segment in K whose interior intersects M has endpoints in M:

∃t ∈ (0, 1): (1 − t)x + ty ∈ M =⇒ x, y ∈ M, ∀x, y ∈ K

Preparation for the proof: Extreme sets

Definition Given a compact convex set K ⊆ E in a normed space, an extreme set is a subset M ⊆ K that is

I non-empty I such that any line segment in K whose interior intersects M has endpoints in M:

∃t ∈ (0, 1): (1 − t)x + ty ∈ M =⇒ x, y ∈ M, ∀x, y ∈ K

Preparation for the proof: Extreme sets

Definition Given a compact convex set K ⊆ E in a normed space, an extreme set is a subset M ⊆ K that is

I non-empty

I closed Preparation for the proof: Extreme sets

Definition Given a compact convex set K ⊆ E in a normed space, an extreme set is a subset M ⊆ K that is

I non-empty

I closed

I such that any line segment in K whose interior intersects M has endpoints in M:

∃t ∈ (0, 1): (1 − t)x + ty ∈ M =⇒ x, y ∈ M, ∀x, y ∈ K B := {x ∈ A : hf , xi = maxhf , yi} y∈A = {maxima of f on A}

is an extreme subset of K. Proposition Every extreme set A ⊆ K contains an extreme point of K. Proof. Use Zorn’s lemma and the above lemma (details omitted).

Preparation for the proof: Extreme sets

Lemma For A ⊆ K an extreme set and f ∈ E ?, is an extreme subset of K. Proposition Every extreme set A ⊆ K contains an extreme point of K. Proof. Use Zorn’s lemma and the above lemma (details omitted).

Preparation for the proof: Extreme sets

Lemma For A ⊆ K an extreme set and f ∈ E ?,

B := {x ∈ A : hf , xi = maxhf , yi} y∈A = {maxima of f on A} Proposition Every extreme set A ⊆ K contains an extreme point of K. Proof. Use Zorn’s lemma and the above lemma (details omitted).

Preparation for the proof: Extreme sets

Lemma For A ⊆ K an extreme set and f ∈ E ?,

B := {x ∈ A : hf , xi = maxhf , yi} y∈A = {maxima of f on A}

is an extreme subset of K. Proof. Use Zorn’s lemma and the above lemma (details omitted).

Preparation for the proof: Extreme sets

Lemma For A ⊆ K an extreme set and f ∈ E ?,

B := {x ∈ A : hf , xi = maxhf , yi} y∈A = {maxima of f on A}

is an extreme subset of K. Proposition Every extreme set A ⊆ K contains an extreme point of K. Preparation for the proof: Extreme sets

Lemma For A ⊆ K an extreme set and f ∈ E ?,

B := {x ∈ A : hf , xi = maxhf , yi} y∈A = {maxima of f on A}

is an extreme subset of K. Proposition Every extreme set A ⊆ K contains an extreme point of K. Proof. Use Zorn’s lemma and the above lemma (details omitted). Reminder

Theorem (Krein–Milman) A compact convex set K ⊆ E in a normed space coincides with the closed convex hull of its extreme points:

K = conv(ext K) K compact, convex, and K ⊇ ext K =⇒ K closed, convex, and K ⊇ ext K =⇒ conv(ext K) ⊆ K

Proof of Krein–Milman

For conv(ext K) ⊆ K, =⇒ K closed, convex, and K ⊇ ext K =⇒ conv(ext K) ⊆ K

Proof of Krein–Milman

For conv(ext K) ⊆ K,

K compact, convex, and K ⊇ ext K =⇒ conv(ext K) ⊆ K

Proof of Krein–Milman

For conv(ext K) ⊆ K,

K compact, convex, and K ⊇ ext K =⇒ K closed, convex, and K ⊇ ext K Proof of Krein–Milman

For conv(ext K) ⊆ K,

K compact, convex, and K ⊇ ext K =⇒ K closed, convex, and K ⊇ ext K =⇒ conv(ext K) ⊆ K ∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

Proof of Krein–Milman

For K ⊆ conv(ext K), ∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

Otherwise, argue by contradiction:

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K) ∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction: =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K)

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) Hence, K ⊆ conv(ext K)

Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition) Proof of Krein–Milman

For K ⊆ conv(ext K),

K = ∅ =⇒ K ⊆ conv(ext K)

Otherwise, argue by contradiction:

∃x ∈ K \ conv(ext K) =⇒ ∃f ∈ E ? : f (conv(ext K)) < f (x) (Hahn–Banach) =⇒ ∃f ∈ E ? : f (ext K) < f (x) =⇒ ∃B ⊆ K extreme set without extreme points (Lemma) =⇒ Contradiction! (Proposition)

Hence, K ⊆ conv(ext K) Outline

1. An informal example

2. Extreme points

3. The Krein–Milman theorem

4. An application ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki)

=⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman)

=⇒ BE ? has an extreme point

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. Proposition

The closed unit ball BE ? has an extreme point. Proof.

Not dual spaces ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki)

=⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman)

=⇒ BE ? has an extreme point

Proposition

The closed unit ball BE ? has an extreme point. Proof.

Not dual spaces

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki)

=⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman)

=⇒ BE ? has an extreme point

Proof.

Not dual spaces

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. Proposition

The closed unit ball BE ? has an extreme point. ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki)

=⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman)

=⇒ BE ? has an extreme point

Not dual spaces

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. Proposition

The closed unit ball BE ? has an extreme point. Proof. =⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman)

=⇒ BE ? has an extreme point

Not dual spaces

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. Proposition

The closed unit ball BE ? has an extreme point. Proof. ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki) =⇒ BE ? has an extreme point

Not dual spaces

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. Proposition

The closed unit ball BE ? has an extreme point. Proof. ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki)

=⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman) Not dual spaces

Example ∞ 1 c0 ⊆ ` and L (R) are not dual spaces. Proposition

The closed unit ball BE ? has an extreme point. Proof. ? BE ? is weakly compact (Banach–Alaoglu–Bourbaki)

=⇒ BE ? = conv(ext BE ? ) (generalized Krein–Milman)

=⇒ BE ? has an extreme point