The Krein–Milman Theorem A Project in Functional Analysis
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 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).
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 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 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,
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 line segment 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 ball: 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 subset 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