12.3 Compact Convex Sets

Proof. (1): Let $U \in \cn_{E}(0)$ be convex and circled, then since $A$ is compact, there exists $B \subset A$ finite such that $A \subset B + U$. As such, $\conv(A) \subset \conv(B + U) = \conv(B) + U$. Since $B$ is finite, $\conv(B)$ is compact. Thus there exists $C \subset \conv(B)$ finite such that

which yields a finite covering of $\conv(A)$ using $U$.

(2): By Proposition 6.4.3.$\square$

Proof. Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by Proposition 5.16.3. Let $x \in A$ and $y, z \in K$ such that $x \in (y, z)$. By definition of $\alpha$, $\dpn{y, \phi}{E}= \dpn{z, \phi}{E}= \alpha$. Thus $y, z \in A$ as well.$\square$

Proof, [Theorem 1.12.5, BS17]. Assume without loss of generality that $K \ne \emptyset$.

Let $K_{0} \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_{0}) \subset 2^{K}$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_{0}$. Since $K_{0} \in \mathcal{E}(K_{0})$, $\mathcal{E}(K_{0}) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C}\subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme.

By Zorn’s lemma, there exists a minimal element $A$ of $\mathcal{E}(K_{0})$. Let $x, y \in A$, $\phi \in E^{*}$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha}\cap K$ is a non-empty, closed, and extreme subset of $K$ by Lemma 12.3.4, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the Hahn-Banach Theorem, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_{0}$.

Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point.

Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the Hahn-Banach Theorem, there exists $x \in K$ and $\phi \in E^{*}$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E}< \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by Lemma 12.3.4, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E}< \alpha$.$\square$

Proof, [Theorem 1.12.10, BS17]. For each $f \in \cf$ and $n \in \natp$, define $f^{(n)}= \frac{1}{n}\sum_{k = 0}^{n - 1}f^{k}$, then $f^{(n)}\in C(K; K)$ as well. Via a closure operation, assume without loss of generality that:

1. (1)

   For any $f \in \cf$ and $n \in \natp$, $f^{(n)}\in \cf$.

2. (2)

   For any $f, g \in \cf$, $f \circ g \in \cf$.

Let $\bracsn{f_j}_{1}^{N} \subset \cf$ and $\bracsn{n_j}_{1}^{N} \subset \natp$, then

thus any finite intersections of elements in

is non-empty. Since $E$ is separated, by Proposition 5.16.3 and Proposition 5.16.4, $\mathcal{K}$ is a family of closed sets satisfying the finite intersection property. Hence $\bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K) \ne \emptyset$.

Now, let $x \in \bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K)$ and $U \in \cn_{E}(0)$, then there exists $N \in \natp$ such that $NU \supset K$. In which case, there exists $y \in K$ such that

In which case,

As this holds for all $U \in \cn_{E}(0)$ and $E$ is separated, $f(x) = x$.$\square$

Proof, [Lemma III.9.1, SW99]. (1): Assume without loss of generality that $A \ne \emptyset$. Let $\bracsn{\lambda_n}_{0}^{\infty} \subset (0, \infty)$ such that $\sum_{n \in \natz}\lambda_{n} = 1$. For each $n \in \natz$, let $r_{n} = \lambda_{n+1}^{2}$ and $A_{n} \subset A$ be finite such that $A \subset \bigcup_{x \in A_n}B_{E}(x, r_{n})$.

Define $B_{0} = \lambda_{0}^{-1}A_{0}$. For each $n \in \natp$, write $A_{n} = \bracsn{x_j}_{1}^{k}$. By definition of $A_{n-1}$, there exists $\bracsn{y_j}_{1}^{k} \subset A_{n-1}$ such that $d(x_{j}, y_{j}) < r_{n-1}$ for all $1 \le j \le k$. For every $1 \le j \le k$, let $z_{j} = (x_{j} - y_{j})/\lambda_{n}$, and define $B_{n} = \bracsn{z_j}_{1}^{k}$.

By the above construction, $A_{n} \subset \sum_{j = 0}^{n} \lambda_{j} B_{j}$ for all $n \in \natz$. As $\sum_{n \in \natz}\lambda_{n} = 1$,

Finally, for each $n \in \natp$, $B_{n}$ is finite with $\norm{z}_{E} \le r_{n-1}/\lambda_{n}= \lambda_{n}$ for all $z \in B_{n}$. As $B_{0}$ is finite as well, any enumeration of $\bracs{0}\cup \bigcup_{n \in \natz}B_{n}$ yields a null sequence.

(2): Using (1) and Mazur’s Theorem, assume without loss of generality that there exists a null sequence $\seq{x_n}\subset E$ such that $A$ is the closed convex hull of $\seq{x_n}$.

Since $\seq{x_n}\subset E$ is a null sequence, there exists $\seq{\lambda_n}\subset [1, \infty)$ such that:

1. (i)

   $\lambda_{n} \to \infty$ as $n \to \infty$.

2. (ii)

   $\lambda_{n} x_{n} \to 0$ as $n \to \infty$.

Let $B = \ol{\aconv}(\seq{\lambda_n x_n})$, then by Mazur’s Theorem, $B$ is a compact, convex, and circled subset of $E$ with $\seq{x_n}\subset B$. In addition, $\seq{x_n}\subset E_{B}$ with $\norm{x_n}_{E_B}\le \lambda_{n}^{-1}$ for all $n \in \natp$. Thus $\seq{x_n}$ is a null sequence in $E_{B}$ as well.

Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_{B}$. Since the inclusion $E_{B} \to E$ is continuous, $A'$ is a compact convex set in $E$ by Proposition 5.16.3. As such, $A' = A$ by Proposition 5.5.3 and Proposition 5.16.4. Therefore $A$ is a compact subset of $E_{B}$.$\square$