Theorem 11.3.5 (Markov-Kakutani Fixed Point Theorem).label 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:
- (a)
For any $f, g \in \cf$, $f \circ g = g \circ f$.
- (b)
For each $f \in \cf$, $x, y \in K$, and $t \in [0, 1]$,
\[f(tx + (1 - t)y) = tf(x) + (1 - t)f(y)\]
then there exists $x_{0} \in K$ such that $f(x_{0}) = x_{0}$ for all $f \in \cf$.
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)
For any $f \in \cf$ and $n \in \natp$, $f^{(n)}\in \cf$.
- (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 5.16.2 and Proposition 5.16.3, $\mathcal{K}$ 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, $f(x) = x$.$\square$
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