The Klein-Millman Theorem

Definition 4.0.1   Let $X$ be a vector space, $K \subset X$ be convex. Call $a \in K$ an extreme point of $K$ if $a = t k_1 + (1-t) k_2$ for $k_1, k_2 \in K, t \in (0,1)$, then $a = k_1 = k_2$. The set of such points is denoted by ext $K$.

More generally, we can define a face of $K$ to be $F \subset K$ such that if

$$tk_1 + (1-t) k_2 \in F, k_1, k_2 \in K, t \in (0,1)$$

then $k_1, k_2 \in F$.
An extreme point is a one-point face; $x \in$   ext $K$ iff $\{ x \}$ is a face of $K$.

Notice (1) a face is convex, (2) if $F$ is a face of $K$ and $G$ is a face of $F$, then $G$ is a face of $K$.

Monday, February 20, 2006:

More Examples:
Let $XC$ be a normed space.

$$\ext(\overline{B_1(0)} ) \subset \{ x \in X : \Vert x \Vert = 1 \}$$

It is possible for $\ext(\overline{B_1(0)}) = \emptyset$. Let $X = L^1([0,1])$. If $f \in L^1([0,1])$, then $\Vert f\Vert _1 \leq 1$.
Choose $x_0 \in [0,1]$ such that

$$\int_0^{x_0} \vert f\vert = \frac12$$

Let

$$g_1(x) = \begin{cases}2 f(x), & x \leq x_0 \ 0, & x > x_0 \end{cases}$$

$$g_2(x) = \begin{cases}0, & x \leq x_0 \ 2 f(x), & x > x_0 \end{cases}$$

Then, $g_1, g_2$ have norm $1$ and $f = \frac12( g_1 + g_2 )$. Clearly, $f \neq g_1$, $f \neq g_2$, so $\ext( \overline{B_1(0)} ) = \emptyset$.

Let $Z$ be an infinite compact Hausdorff space. Then ball $C(Z)$ is not compact (exercise). Nonetheless, $\ext($   ball $C(Z) )$ is all $f \in C(Z)$ such that $\vert f(z)\vert\vert = 1$ for all $z \in Z$.
If $\mathbb{F}= \mathbb{R}$, this is two functions - $f \equiv 1$, $f \equiv -1$. If $\mathbb{F}= \mathbb{C}$, this is uncountably many functions. In fact, the convex hull of this set is (uniformly) dense in ball $C(Z)$.

Proposition 4.0.2   Let $X$ be a vector space, $K \subset X$ convex, and $a \in K$. The following are equivalent:

1. $a \in$   ext $K$
2. $x,y \in X$, $a = \frac{x + y}2$ implies $x \notin K$, $y \notin K$, or $x=y=a$.
3. $x,y \in X$, $\lambda \in (0,1)$, $a = \lambda x + (1-\lambda)y$ implies $x \notin K, y \notin K$, or $x=y=a$
4. If $a \in$   co $\{ x_1, \ldots, x_n \}, x_i \in K$, then $a = x_i$ for some $i$.
5. $K \setminus \{ a \}$ is convex.

Proof. Left as an exercise $\qedsymbol$

Theorem 4.0.3   Krein-Milman Theorem
Let $X$ be an LCS, $K \subset X$ nonempty, convex, compact set. Then $K = \overline{\text{co }}(\text{ext }K)$. In particular, ext $K \neq \emptyset$.

Proof. WLOG, $K$ is not a singleton. By $(5)$ from the previous proposition, $a$ is an extreme point iff $K \setminus \{ a \}$ is a relatively open convex subset of $K$. We use Zorn's Lemma to find a maximal relatively open convex proper subset and then show its complement is a singleton.

Let $U$ be the relatively open convex proper subsets, ordered by inclusion. As $K$ is not a singleton, $U$ is not the empty set. Let $U_0 \subset U$ is a totally ordered subset.

Let $A = \bigcup_{w \in U_0} W$. Then it is easy to show that $A$ is a relatively open convex subset of $K$. If $A = K$, then $U_0$ is an open cover of compact sets, and so there is a finite subcover. Then, picking the biggest set in the finite subcover, $K \in U_0$. This is a contradiction, thus $A$ is a proper subset.

Let $U$ be a maximal element (by Zorn). For $x \in K$, $\lambda \in [0,1]$, define $T_{x,\lambda}:K \rightarrow K$ by $T_{x,\lambda}(y) = \lambda y + (1-\lambda) x$. Clearly, $T_{x,\lambda}$ is continuous.

Claim 1: $T_{x,\lambda}(K) \subset U$ for $x \in U$, $\lambda \in [0,1)$.
If $y_1, \ldots, y_n \in K$, $\alpha_1, \ldots, \alpha_n \geq 0$, $\sum_{i=1}^n \alpha_i = 1$, then

$$T_{x,\lambda}(\sum_{i=1}^n \alpha_i y_i ) = \sum_{i=1}^n \alpha_i( T_{x,\lambda}(y_i) )$$

By the convexity of $U$, $T_{x,\lambda}(U) \subset U$ for $x \in U$, $\lambda < 1$.
Thus, $T_{x,\lambda}^{-1}(U)$ is a relatively open convex subset of $K$.
If $y \in \bar U \setminus U$, $T_{x,\lambda}(y) \in [x,y) \subset U$ (by IV.1.11).
Thus, $\bar U \subset T_{x,\lambda}^{-1}(U)$. By maximality of $U$, $T_{x,\lambda}^{-1}(U) = K$, which finishes the claim.

Wednesday, February 22, 2006:

Claim 2: $K \setminus U$ is a singleton.
If not, there are $a, b \in K \setminus U$, $a \neq b$. Pick $V_a$, $V_b$ open convex disjoint sets with $a \in V_a$, $b \in V_b$. As $a \neq U$, $V_a \cup U = K$. As $b \notin U$, we must have $b \in V_a$.

Claim 3: If $V$ is an open convex subset of $X$ with ext $K \subset V$, then $K \subset V$
If not, there is such a $V$ with ext $K \subset V$ but $K \not\subset V$. That is, $V \cap K \neq K$, so $K \cap V \in U$. Hence, $K \cap V$ is contained in a maximal element, $U$. .
Now, $K \setminus U = \{ a\}$ and since $U$ is convex, by (5) of the previous proposition, $a \in$   ext $K$, so ext $K \not\subset V$.

Finally, let $E = \overline{\text{co }}(\text{ext }K)$. If $f \in X^*$, $\alpha \in \mathbb{R}$ such that

$$E \subset \{ x \in X :$$    Re $$f(x) \leq \alpha \} =: V$$

Then by claim 3, $K \subset V$. By the Hahn-Banach separation theorem, $E = K$.

$\qedsymbol$

Proof. Let $\mathcal{F}$ be the closed faces of $K$, ordered by reverse inclusion. Let $\mathcal{F}_0$ be a totally ordered set of faces. By compactness,

$$F_0 := \displaystyle \bigcap_{F \in \mathcal{F}_0} F \neq \emptyset$$

Notice that $F_0$ is a face of $K$ (there are a few details here). By Zorn, we have a minimal face of $F \in \mathcal{F}$.

Claim: $F = \{ a \}$ for some $a \in K$.
Let $f \in X^*$ and $s = \inf \{$    Re $f(x) : x \in F \}$. $F$ is weakly compact and Re $f$ is weakly-continuous so $f$ attains its minimal value of $F$.
Let $E := \{ x \in F :$    Re $f(x) = s \} \neq \emptyset$. A moment's thought shows that $E$ is a face of $F$.
By minimality, $E = F$ so each $f \in X^*$ is constant on $F$. Since $X^*$ separates points, $F$ is a singleton.

Since every $G \in \mathcal{F}$ contains a minimal face of $K$, $G \cap$   ext $K \neq \emptyset$ for all $G \in \mathcal{F}$. If $x \in K \setminus \overline{\text{co }}(\text{ext }K)$, then there is $U$ open such that $x \in U$ and $U \cap \overline{\text{co }}(\text{ext }K) = \emptyset$.
By Hahn-Banach separation theorem (open), there is $f \in X^*$, $\alpha \in \mathbb{R}$ such that for all $k \in \overline{\text{co }}(\text{ext }K)$,

    Re $$f(x) \in$$    Re $$f(U) < \alpha \leq$$    Re $$f(k)$$

Then,

$$s := \inf \{$$    Re $$f(k) : k \in K \} < \alpha$$

and so

$$\{ k \in K :$$    Re $$f(k) = s \} \cap \overline{\text{co }}(\text{ext }K ) = \emptyset$$

As before, $\{ k \in K :$    Re $f(k) = s \}$ is a closed face and by this intersection, it contains no extreme points. $\qedsymbol$

Friday, February 24, 2006:

Corollary 4.0.4   Let $X$ be an LCS, $K \subset X$ convex, compact, and nonempty. If $f \in X^*$, there is $x \in$   ext $K$ such that

    Re $$f(k) \leq$$    Re $$f(x)$$

for all $k \in K$.

Proof. Let $s = \sup \{$    Re $f(k) : k \in K \}$ and $F = \{ k \in K :$    Re $f(k) = s \} \neq \emptyset$.
Then $F$ is a compact face of $K$. If $a \in F$ and $a = \frac{b+c}2$, $b,c \in K$, then

$$s =$$    Re $$f(a) = \frac{\text{ Re }f(b) + \text{ Re }f(c)}2$$

and so Re $f(b) =$    Re $f(c) = s$. Thus, $b, c \in F$.
By KM, there is $x \in$   ext $F$. But ext $F \subset$   ext $K$. By the choice of $s$, the desired condition holds.

$\qedsymbol$

Applications:
For a Banach space $X$, we know ball $X^*$ is wk $^*$-compact and so has extreme points.

Corollary 4.0.5  

1. $L^1([0,1])$ is not the dual of any Banach space.
2. $c_0$ is not the dual of any Banach space.

Proof.

1. ball $L^1([0,1])$ has no extreme points.
2. We claim ball $c_0$ has no extreme points. If $(x_n) \in$   ball $c_0$, then $\lim_{n \rightarrow \infty} x_n = 0$. So, there exists an $N$ such that $\vert x_N\vert \leq \frac12$. Let
   $$y_n = \begin{cases}x_n, & n \neq N \ x_N + \frac12, & n = N \end{cases}$$
   $$z_n = \begin{cases}x_n, & n \neq N \ x_N - \frac12, & n = N \end{cases}$$
   Then, $y, z \in$   ball $c_0$ and $x = \frac{y+z}2$ and $y \neq x \neq z$.

$\qedsymbol$

Theorem 4.0.6   Let $X$ be an LCS, $K \subset X$ is compact and convex. If $F \subset K$ satisfies $F$ closed and $\overline{\text{co }}(F) = K$, then ext $K \subset F$.

Proof. Suppose $x_0 \in$   ext $K \setminus F$. Pick $p$, a continuous seminorm such that

$$F \cap \{ x \in X : p(x - x_0) < 1 \} = \emptyset$$

Trick: find $p_1, \ldots, p_n$ such that

$$F \cap \{ x \in X : p_i(x - x_0) < 1, i = 1, \ldots, n \} = \emptyset$$

Let $p = p_1 + p_2 + \cdots + p_n$.
Let $U_0 = \{ x \in X : p(x) < \frac13 \}$. Then,

$$(x_0 + U_0) \cap (F + U_0) = \emptyset$$

Thus, $x_0 \notin \overline{F + U_0}$. As $F$ is compact, there exists $y_1, \ldots, y_n \in F$ such that

$$F \subset \bigcup_{k=1}^n y_k + U_0$$

Let $K_k = \overline{\text{co }}(F \cap (y_k + U_0))$. Then,

$$K_k \subset y_k + \overline U_0$$

as $y_k + \bar U_0$ is a closed convex set containing $F \cap (y_k + U_0)$, and

$$K_k \subset K$$

as $K$ is a closed convex set containing $F \cap (y_k + U_0)$.
Next, by the selection of the $y_k$ and $\overline{\text{co }}(F) = K$, $\overline{\text{co }}( K_1 \cup K_2 \cup \cdots \cup K_n) = K$.

Lemma 4.0.7   As $K_i$ are compact,

$$\overline{\text{co }}(K_1 \cup \cdots \cup K_n) = \text{co }( K_1 \cup \cdots \cup K_n )$$

Proof. Let $x \in \overline{\text{co }}(K_1 \cup \cdots \cup K_n)$. Then, there is a net $(x_\lambda)$ such that $x_\lambda \rightarrow x$ and each $x_\lambda \in$   co $( K_1 \cup \cdots \cup K_n)$. That is, there are $k_{\lambda,1} \in K_i$, $\alpha_{\lambda,i} \geq 0$ such that

$$\sum_{i=1}^n \alpha_{\lambda,i} = 1$$

and

$$x_\lambda = \alpha_{\lambda,1} k_{\lambda,1} + \cdots + \alpha_{\lambda,n} k_{\lambda,n}$$

Looking at each net $(\alpha_{\lambda,i})_\lambda, (k_{\lambda,i})_\lambda$ in turn and choose convergent subnets. Thus, we may assume that there is $\beta_i$ and $l_i$ such that

$$\alpha_{\lambda,i} \rightarrow \beta_i$$

$$k_{\lambda,i} \rightarrow l_i$$

for $i = 1, \ldots, n$. By continuity of vector space operations,

$$x = \lim_{\lambda} \sum_{i=1}^n \alpha_{\lambda,i} k_{\lambda,i}$$

$$= \sum_{i=1}^n ( \lim_\lambda \alpha_{\lambda,i} ) (\lim_\lambda k_{\lambda,i})$$

$$= \sum_{i=1}^n \beta_i l_i \in$$   co $$( K_1 \cup \cdots \cup K_n)$$

$\qedsymbol$

As $x_0 \in K =$   co $( K_1 \cup \cdots \cup K_n)$,

$$x_0 = \sum_{i=1}^n \alpha_i x_i, x_i \in K_i$$

$$\sum_{i=1}^n \alpha_i = 1, \alpha_i \geq 0$$

As $x_0 \in$   ext $K$, there is $k$ in $1, \ldots, n$ such that $x_0 = x_k$. Thus,

$$x_0 \in K_k \subset y_k + \bar U_0 \subset \overline{ F + U_0 }$$

$\qedsymbol$

Monday, February 27, 2006:
Recall: If $X, Y$ are vector spaces, then $T:X \rightarrow Y$ is called affine if for a convex combination $\sum_{i=1}^n \lambda_i x_i$,

$$T(\sum_{i=1}^n \lambda_i x_i ) = \sum_{i=1}^n \lambda_i T(x_i)$$

Example: If $X = Y = \mathbb{R}$, then $T$ affine means $T(x) = mx + b$.

An affine map is a linear map plus a translate. That is, if $T:X \rightarrow Y$ is affine, then

$$L(x) = T(x) - T(0)$$

is linear.

Proposition 4.0.8   Let $X, Y$ be LCS's, $K \subset X$, compact and convex. If $T:X \rightarrow Y$ is a continuous affine map,, then $T(K)$ is a compact convex set. If $y \in$   ext $T(K)$, then $T^{-1}(y)$ contains an element of ext $K$.

Example: If $x \in$   ext $K$, then $T(x)$ need not be extreme. $T:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ $(x,y,z) \mapsto (x,y)$.

Proof. By compactness, $T(K)$ is compact. By convexity, a convex combination of elements of $T(K)$ is in $T(K)$. Let $y \in$   ext $T(K)$. Then, $T^{-1}(y)$ is compact and convex.
By KM, $T^{-1}(y)$ has an extreme point, $x$.

Claim: $x \in$   ext $K$.
If $x = \frac{a+b}2$, $a, b \in K$,

$$y = T(x) = \frac{T(a) + T(b)}2$$

As $y \in$   ext $T(K)$, $T(a), T(b) \in T(K)$, $T(a) = T(b) = y$. So, $a, b \in T^{-1}(y)$.
As $x \in$   ext $T^{-1}(y)$, $a = b = x$. $\qedsymbol$

Further improvements on KM: Choquet's Theorem (Lax):

Theorem 4.0.9   Let $X$ be an LCS, $K \subset X$ a compact, convex, nonempty set. For all $k \in K$, there is a probability measure $\mu_k$ on $\overline{\text{ext }K}$ so that, for all $f \in X^*$,

$$f(k) = \int_{\overline{\text{ext }K}} f(x) d\mu_k(x)$$

That is,

$$k = \int_{\overline{\text{ext }K}} x d\mu_k(x)$$

holds weakly.

$\mu_k$ is a ``continuous" version of a convex combination. If ext $K$ is a finite set, then we can say that $k$ is a convex combination of the extreme points.

We need a preliminary result about extending positive linear functionals.

Definition 4.0.10   An ordered vector space is a vector space $X$ over $\mathbb{R}$ with a reflexive and transitive order $\leq$ such that

1. $x \leq y$, $z \in X$, then $x + z \leq y = z$
2. $x \leq y$, $\alpha \geq 0$, then $\alpha x \leq \alpha y$.

Recall $A \subset X$ is cofinal if for all $x \geq 0$ then there is a $a \in A$ such that $a \geq x$. For ordered vector spaces, $X, Y$, we say $T:X \rightarrow Y$ is positive if $x \geq 0$ implies $Tx \geq 0$. This is denoted $T \geq 0$.

Theorem 4.0.11   Let $(X,\leq)$ be an ordered vector space over $\mathbb{R}$ and $Y$ a linear subspace that is cofinal. A positive linear functional extends to a positive linear functional on $X$.

Proof. Let $f:Y \rightarrow \mathbb{R}$ be a positive linear functional. Let

$$P := \{ x \in X : x \geq 0 \}$$

Let $X_1 = Y - P + P$. Then $X_1$ is a linear subspace of $X$. If $\hat f$ is a positive extension of $f$ to $X_1$ and $g$ is any extension of $\hat f$ to $X$ then, for $x \geq 0$,

$$g(x) = \hat f (x) \geq 0$$

as $x \in P$. So, WLOG, $X = Y - P + P$.

Claim: $X = Y + P = Y - P$.
Let $x \in X$. Then, $x = y + p_1 - p_2$ for some $y \in Y$, $p_i \in P$. As $Y$ is cofinal, there is $y_1 \in Y$ such that $y_1 \geq p_1$. Thus,

$$p_1 = y_1 - (y_1 - p_1 \in Y - P.$$

So,

$$x = y + y_1 - (p_2 + (y_1-p_1)) \in Y - P$$

Now,

$$X = -X = -(Y-P) = -Y + P = Y + P.$$

If $x \in X$, then

$$x = y_1 - p_1 = y_2 + p_2$$

for some $y_i \in Y$, $p_i \in P$. Thus,

$$y_2 \leq x \leq y_1$$

Define $q$ on $X$ by $q(x) = \int \{ f(y) : y \in Y, y \geq x \}$. This is well-defined by the previous sentence.
If $\beta \geq 0$, $x \in X$,

$$q(\beta x) = \inf \{ f(y) : y \in Y, y \geq \beta x \}$$

$$= \inf \{ f(\beta z) : z \in Y, \beta z \geq \beta x \}$$

$$= \beta q(x)$$

Similarly, $q(x + y) \leq q(x) + q(y)$, so $q$ is sublinear. Since $f$ is positive, if $x \leq y$, then $f(x) \leq f(y)$ for all $x,y \in Y$. Thus, $f(x) \leq q(x)$.

Letting $g$ be the Hahn-Banach extension of $f$, dominated by $q$, we claim $g$ is positive.

If $x \geq 0$, then $-x \leq 0$, $0 \in y$, $q(-x) \leq f(0)$. So,

$$-g(x) = g(-x) \leq q(-x) \leq 0$$

So, $q(x) \geq 0$ and $g(x) \geq 0$. $\qedsymbol$

Example:
Let $Y \subset C_b(X)$, $X$ locally compact. Observe $C_b(X)$ is an ordered vector space under the natural order ($f \leq g$ if $f(x) \leq g(x)$ for all $x \in X$).
If $\vec {1}\in Y$, then $Y$ is cofinal.
Let $f \in C_b(X)$. So, there is a $M > 0$ such that $f(x) \leq M$ for all $x \in X$. Then, $f \leq M \vec {1}$ and $M \vec {1}\in Y$.

Proof. (of Choquet's Theorem):
Let $X$ be an LCS over $\mathbb{R}$, $K \subset X$ compact, convex, and nonempty. Let $k \in K$.
Goal: Find $\mu_k$ on $\overline{\text{ext }K}$ such that

$$\forall f \in X^*, \, f(k) = \int_{\overline{\text{ext }K}} f(x) d\mu k(x)$$

Define $\tau : X^* \rightarrow C( \overline{\text{ext }K} )$ by $\tau(f) = f\vert _{\overline{\text{ext }K}}$.
Claim: $\tau$ is one-to-one.
Suppose $f(x) = g(x)$ for all $x \in \overline{\text{ext }K}$, $f,g \in X^*$. Then, $(f - g)(x) = 0$ for all $x \in \overline{\text{ext }K}$. By the corollary of KM,

$$\sup{x \in K} (f - g)(x)$$

occurs at some $x \in$   ext $K$. Thus, $(f-g)(x) \leq 0$ for $x \in K$. Similarly, looking $\sup x \in K (g-f)(x)$ shows $(g-f)(x) \leq 0$ for $x \in K$. Thus, $g(x) = f(x)$ for all $x \in K$. Thus, $g - f$ vanishes on $\overline{\text{span}} K$. So $g - f$ is in $\{ h \in X^*, h(\overline{\text{span}} K) \equiv 0\} =: K^\perp$.
For the other inclusion, $f \in K^\perp$, $f\vert _K = 0$, $f\vert _{\overline{\text{ext }K}} = 0$, $\tau(f) = 0$.

Fix $k \in K$. Define $\sigma_k : \tau(X^*) \rightarrow \mathbb{R}$ by $\sigma_k(f) = g(k)$ where $g \in \tau^{-1}(f)$.
Suppose $g_1, g_2 \in \tau^{-1}(f)$. Well, $g_1 - g_2 \in \ker \tau = K^\perp$, so $(g_1 - g_2)(k) = 0$ and $g_1(k) = g_2(k)$, and $\sigma_k$ is well-defined.
Observe that $\sigma_k$ is a positive linear functional. Let $f \in \tau(X^*)$ with $f(\ell) \geq 0$ for all $\ell \in \overline{\text{ext }K}$. Why is $\sigma_k(f) \geq 0$? Let $g \in \tau^{-1}(f)$. Then, $g(\ell) \geq 0$ for all $\ell \in \overline{\text{ext }K}$. But,

$$\inf_{x \in K} g(x) = \inf_{x \in \overline{\text{ext }K}} g(x)$$

by the corollary of KM. So, $g(x) \geq 0$ for all $x \in K$. In particular, $\sigma_k(f) = g(k) \geq 0$.
To extend $\sigma_k$ to a positive functional on $C(\overline{\text{ext }K})$, we need to have a cofinal subspace.
Let $Y$ be the linear subspace containing $\tau(X^*)$ and $\vec {1}$. Using the corollary of KM again,

$$\inf_{x \in \overline{\text{ext }K}} f(x) \leq \sigma_k(f) \leq \sup_{x \in \overline{\text{ext }K}} f(x)$$

So, if $f \leq \vec {1}$, $f \in \tau(x^*)$, then $\sigma_k(f) \leq 1$.
Extend $\sigma_k$ by defining $\sigma_k( \vec {1}) = 1$ and extend by linearity. Using the fact that if $f \leq \vec {1}$, $f \in \tau(X^*)$, then $\sigma_k(f) \leq 1$, one can show the extension is still positive. By the extension theorem for positive linear functionals, $\sigma_k$ extends to a positive linear functional on $\mathcal{C}(\overline{\text{ext }K})$.
Observe that $f \in$   ball $\mathcal{C}(\overline{\text{ext }K})$ iff $-\vec {1}\leq f \leq \vec {1}$. So, by positivity,

$$-1 \leq \tilde \sigma_k(f) \leq 1$$

where $\tilde \sigma_k$ is the extension of $\sigma_k$.
So, $\sigma_k$ is bounded and has norm $1$.
By Riesz, there is a signed measure $\mu_k$ such that

$$\tilde \sigma_k(f) = \int_{\overline{\text{ext }K}} f(x) d\mu_k(x)$$

for $f \in \mathcal{C}(\overline{\text{ext }K})$.
Let $g \in X^*$. Notice that $\tau(g) \in C(\overline{\text{ext }K})$ and $\sigma_k(\tau(g)) = g(k)$. So, $\tau(g)(e) = g(e)$ for all $e \in \overline{\text{ext }K}$.

So, the formula becomes:

$$g(k) = \int_{\overline{\text{ext }K}} g(x) d\mu_k(x)$$

$\qedsymbol$