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The Krein–Milman TheoremA Project in Functional Analysis
Samuel Pettersson
November 29, 2016
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
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
Outline
1. An informal example
2. Extreme points
3. The Krein–Milman theorem
4. An application
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
‖x‖1≤ 1
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
‖x‖1≤ 1
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
‖x‖1≤ 1 ‖x‖2≤ 1
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
‖x‖1≤ 1 ‖x‖2≤ 1
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
‖x‖1≤ 1 ‖x‖2≤ 1 ‖x‖∞≤ 1
Convex sets and their “corners”
ObservationSome convex sets are the convex hulls of their “corners”.
‖x‖1≤ 1 ‖x‖2≤ 1 ‖x‖∞≤ 1
Convex sets and their “corners”
ObservationSome convex sets are not the convex hulls of their “corners”.
x1, x2 ≥ 0
Convex sets and their “corners”
ObservationSome convex sets are not the convex hulls of their “corners”.
x1, x2 ≥ 0
Convex sets and their “corners”
ObservationSome convex sets are not the convex hulls of their “corners”.
x1, x2 ≥ 0
Convex sets and their “corners”
ObservationSome convex sets are not the convex hulls of their “corners”.
x1, x2 ≥ 0 ‖x‖∞< 1
Objectives
I Formalize the notion of a corner of a convex set (extremepoint).
I Find a sufficient condition for a convex set to be the closedconvex hull of its extreme points (Krein–Milman).
Objectives
I Formalize the notion of a corner of a convex set (extremepoint).
I Find a sufficient condition for a convex set to be the closedconvex hull of its extreme points (Krein–Milman).
Objectives
I Formalize the notion of a corner of a convex set (extremepoint).
I Find a sufficient condition for a convex set to be the closedconvex hull of its extreme points (Krein–Milman).
Outline
1. An informal example
2. Extreme points
3. The Krein–Milman theorem
4. An application
Definition of an extreme point
DefinitionAn extreme point of a convex set K ⊆ E in a vector space E is apoint 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
RemarkIn 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 arealways extremal
Definition of an extreme point
DefinitionAn extreme point of a convex set K ⊆ E in a vector space E is apoint 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
RemarkIn 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 arealways extremal
Definition of an extreme point
DefinitionAn extreme point of a convex set K ⊆ E in a vector space E is apoint 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
RemarkIn 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 arealways extremal
Definition of an extreme point
DefinitionAn extreme point of a convex set K ⊆ E in a vector space E is apoint 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
RemarkIn 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 arealways extremal
Definition of an extreme point
DefinitionAn extreme point of a convex set K ⊆ E in a vector space E is apoint 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
RemarkIn 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 arealways extremal
Definition of an extreme point
DefinitionAn extreme point of a convex set K ⊆ E in a vector space E is apoint 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
RemarkIn 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 arealways extremal
Examples of extreme points
Extreme points of the closed unit ball:
Space Extreme points
`1 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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
±ei = (0, . . . , 0,
ith term︷︸︸︷±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 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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 ±ei = (0, . . . , 0,
ith term︷︸︸︷±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
Statement of Krein–Milman
Theorem (Krein–Milman)
A compact convex set K ⊆ E in a normed space coincides with theclosed convex hull of its extreme points:
K = conv(extK )
Reminder
convA := convA
= 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 theclosed convex hull of its extreme points:
K = conv(extK )
Reminder
convA := convA
= the smallest closed and convex set containing A.
Preparation for the proof: Extreme sets
DefinitionGiven a compact convex set K ⊆ E in a normed space, an extremeset is a subset M ⊆ K that is
I non-empty
I closed
I such that any line segment in K whose interior intersects Mhas endpoints in M:
∃t ∈ (0, 1) : (1− t)x + ty ∈ M =⇒ x , y ∈ M, ∀x , y ∈ K
Preparation for the proof: Extreme sets
DefinitionGiven a compact convex set K ⊆ E in a normed space, an extremeset is a subset M ⊆ K that is
I non-empty
I closed
I such that any line segment in K whose interior intersects Mhas endpoints in M:
∃t ∈ (0, 1) : (1− t)x + ty ∈ M =⇒ x , y ∈ M, ∀x , y ∈ K
Preparation for the proof: Extreme sets
DefinitionGiven a compact convex set K ⊆ E in a normed space, an extremeset is a subset M ⊆ K that is
I non-empty
I closed
I such that any line segment in K whose interior intersects Mhas endpoints in M:
∃t ∈ (0, 1) : (1− t)x + ty ∈ M =⇒ x , y ∈ M, ∀x , y ∈ K
Preparation for the proof: Extreme sets
DefinitionGiven a compact convex set K ⊆ E in a normed space, an extremeset is a subset M ⊆ K that is
I non-empty
I closed
I such that any line segment in K whose interior intersects Mhas endpoints in M:
∃t ∈ (0, 1) : (1− t)x + ty ∈ M =⇒ x , y ∈ M, ∀x , y ∈ K
Preparation for the proof: Extreme sets
DefinitionGiven a compact convex set K ⊆ E in a normed space, an extremeset is a subset M ⊆ K that is
I non-empty
I closed
I such that any line segment in K whose interior intersects Mhas endpoints in M:
∃t ∈ (0, 1) : (1− t)x + ty ∈ M =⇒ x , y ∈ M, ∀x , y ∈ K
Preparation for the proof: Extreme sets
LemmaFor A ⊆ K an extreme set and f ∈ E ?,
B := {x ∈ A : 〈f , x〉 = maxy∈A
〈f , y〉}
= {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
LemmaFor A ⊆ K an extreme set and f ∈ E ?,
B := {x ∈ A : 〈f , x〉 = maxy∈A
〈f , y〉}
= {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
LemmaFor A ⊆ K an extreme set and f ∈ E ?,
B := {x ∈ A : 〈f , x〉 = maxy∈A
〈f , y〉}
= {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
LemmaFor A ⊆ K an extreme set and f ∈ E ?,
B := {x ∈ A : 〈f , x〉 = maxy∈A
〈f , y〉}
= {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
LemmaFor A ⊆ K an extreme set and f ∈ E ?,
B := {x ∈ A : 〈f , x〉 = maxy∈A
〈f , y〉}
= {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 theclosed convex hull of its extreme points:
K = conv(extK )
Proof of Krein–Milman
For conv(extK ) ⊆ K ,
K compact, convex, and K ⊇ extK
=⇒ K closed, convex, and K ⊇ extK
=⇒ conv(extK ) ⊆ K
Proof of Krein–Milman
For conv(extK ) ⊆ K ,
K compact, convex, and K ⊇ extK
=⇒ K closed, convex, and K ⊇ extK
=⇒ conv(extK ) ⊆ K
Proof of Krein–Milman
For conv(extK ) ⊆ K ,
K compact, convex, and K ⊇ extK
=⇒ K closed, convex, and K ⊇ extK
=⇒ conv(extK ) ⊆ K
Proof of Krein–Milman
For conv(extK ) ⊆ K ,
K compact, convex, and K ⊇ extK
=⇒ K closed, convex, and K ⊇ extK
=⇒ conv(extK ) ⊆ K
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Proof of Krein–Milman
For K ⊆ conv(extK ),
K = ∅ =⇒ K ⊆ conv(extK )
Otherwise, argue by contradiction:
∃x ∈ K \ conv(extK )
=⇒ ∃f ∈ E ? : f (conv(extK )) < f (x) (Hahn–Banach)
=⇒ ∃f ∈ E ? : f (extK ) < f (x)
=⇒ ∃B ⊆ K extreme set without extreme points (Lemma)
=⇒ Contradiction! (Proposition)
Hence, K ⊆ conv(extK )
Outline
1. An informal example
2. Extreme points
3. The Krein–Milman theorem
4. An application
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point
Not dual spaces
Example
c0 ⊆ `∞ and L1(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(extBE?) (generalized Krein–Milman)
=⇒ BE? has an extreme point