145 lines
9.6 KiB
TeX
145 lines
9.6 KiB
TeX
\section{Compact Convex Sets}
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\label{section:compact-convex}
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\begin{theorem}[Mazur]
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\label{theorem:convex-hull-complete}
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Let $E$ be a locally convex space over $K \in \RC$ and $A \subset E$ be compact, then
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\begin{enumerate}
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\item $\conv(A)$ is totally bounded in $E$.
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\item If $E$ is complete, then $\conv(A)$ is relatively compact.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(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
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\[
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\conv(A) \subset \conv(B + U) = \conv(B) + U \subset C + U
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\]
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which yields a finite covering of $\conv(A)$ using $U$.
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(2): By \autoref{proposition:compact-uniform}.
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\end{proof}
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\begin{definition}[Extreme Point]
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\label{definition:extreme-point}
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Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K \setminus \bracsn{x}$ such that $x \in (y, z) \subset K$.
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\end{definition}
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\begin{definition}[Extreme Subset]
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\label{definition:extreme-subset}
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Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.
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\end{definition}
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\begin{lemma}
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\label{lemma:extreme-subset-extreme-point}
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Let $E$ be a vector space over $\real$, $K \subset E$ be convex, $A \subset K$ be extreme, and $x \in A$ be an extreme point of $A$, then $x$ is an extreme point of $K$.
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\end{lemma}
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\begin{proof}
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Let $y, z \in K$ such that $x \in (y, z)$. Since $A$ is extreme, $y, z \in A$. As $x$ is extremal in $A$, $x = y = z$, so $x$ is extremal in $K$.
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\end{proof}
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\begin{lemma}
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\label{lemma:extremal-face}
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Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$.
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\end{lemma}
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\begin{proof}
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Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by \autoref{proposition:compact-extensions}. 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.
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\end{proof}
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\begin{theorem}[Krein-Milman]
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\label{theorem:krein-milman}
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Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 1.12.5]{Bogachev}}}. ]
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Assume without loss of generality that $K \ne \emptyset$.
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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.
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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 \autoref{lemma:extremal-face}, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $A$ 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 $K_0$.
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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.
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Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, 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 \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ 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 $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$.
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\end{proof}
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\begin{theorem}[Markov-Kakutani Fixed Point Theorem]
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\label{theorem:markov-kakutani}
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Let $E$ be a separated topological vector space over $\real$, $K \subset E$ be a compact convex set, and $\cf \subset C(K; K)$ such that:
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\begin{enumerate}[label=(\alph*)]
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\item For any $f, g \in \cf$, $f \circ g = g \circ f$.
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\item For each $f \in \cf$, $x, y \in K$, and $t \in [0, 1]$,
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\[
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f(tx + (1 - t)y) = tf(x) + (1 - t)f(y)
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\]
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\end{enumerate}
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then there exists $x_0 \in K$ such that $f(x_0) = x_0$ for all $f \in \cf$.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 1.12.10]{Bogachev}}}. ]
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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:
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\begin{enumerate}
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\item For any $f \in \cf$ and $n \in \natp$, $f^{(n)} \in \cf$.
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\item For any $f, g \in \cf$, $f \circ g \in \cf$.
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\end{enumerate}
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Let $\bracsn{f_j}_1^N \subset \cf$ and $\bracsn{n_j}_1^N \subset \natp$, then
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\[
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\bigcap_{j = 1}^n f_j^{(n_j)}(K) \supset \braks{\prod_{j = 1}^n f_j^{(n_j)}}(K) \ne \emptyset
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\]
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thus any finite intersections of elements in
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\[
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\mathcal{K} = \bracsn{f^{(n)}(K)| f \in \cf, n \in \natp}
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\]
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is non-empty. Since $E$ is separated, by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}, $\mathcal{K}$ is a family of closed sets satisfying the finite intersection property. Hence $\bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K) \ne \emptyset$.
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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
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\[
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x = \frac{1}{N}\sum_{j = 0}^{N-1}f^j(y) \quad f(x) = \frac{1}{N}\sum_{j = 1}^{N}f^j(y)
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\]
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In which case,
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\[
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f(x) - x = \frac{1}{N}(f^N(y) - y) \in \frac{1}{N}(K - K) \subset U - U
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\]
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As this holds for all $U \in \cn_E(0)$ and $E$ is separated, $f(x) = x$.
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\end{proof}
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\begin{lemma}
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\label{lemma:compact-null-auxiliary}
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Let $E$ be a Banach space over $K \in \RC$ and $A \subset E$ be compact, then:
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\begin{enumerate}
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\item There exists a null sequence $\seq{x_n} \subset E$ such that $A \subset \ol{\conv}(\seq{x_n})$.
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\item There exists a compact, convex, and circled set $B \subset E$ such that $A$ is compact in $E_B$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Lemma III.9.1]{SchaeferWolff}}}. ]
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(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 $A \subset \bigcup_{x \in A_n}B_E(x, r_n)$.
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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 $B_n = \bracsn{z_j}_1^k$.
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By the above construction, $A_n \subset \sum_{j = 0}^n \lambda_j B_j$ for all $n \in \natz$. As $\sum_{n \in \natz}\lambda_n = 1$,
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\[
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A \subset \ol{\conv}\braks{\bigcup_{n \in \natz}A_n} \subset \ol{\conv}\braks{\bracs{0} \cup \bigcup_{n \in \natz}B_n}
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\]
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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.
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(2): Using (1) and \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, 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 $\seq{x_n}$.
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Since $\seq{x_n} \subset E$ is a null sequence, there exists $\seq{\lambda_n} \subset [1, \infty)$ such that:
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\begin{enumerate}[label=(\roman*)]
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\item $\lambda_n \to \infty$ as $n \to \infty$.
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\item $\lambda_n x_n \to 0$ as $n \to \infty$.
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\end{enumerate}
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Let $B = \ol{\aconv}(\seq{\lambda_n x_n})$, then by \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, $B$ 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.
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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 \autoref{proposition:compact-extensions}. As such, $A' = A$ by \autoref{proposition:closure-of-image} and \autoref{proposition:compact-closed}. Therefore $A$ is a compact subset of $E_B$.
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\end{proof}
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