The Markov-Kakutani fixed point theorem for abelian affine actions

For $g\in G$ and $n\in \mathbb{N}$, define $g^{0}x=x$, $g^{i+1}x=g(g^{i}x)$, and $A_n(g)x:=\frac{1}{n+1}{\sum }_{i=0}^n{g}^{i}x$, with vector-valued finite sums as in [F5]. Since $1\mathrm{/}(n+1)>0$ and the sum of the $n+1$ equal coefficients is $(n+1)\mathrm{/}(n+1)=1$, convexity makes $A_n(g)$ a self-map of $X$. The action iterates are continuous and affine by induction. For any finite convex combination $y={\sum }_{\mathrm{\ell }}{t}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$, their affine identities and finite sum distributivity give $A_n(g)y=\frac{1}{n+1}{\sum }_i{\sum }_{\mathrm{\ell }}{t}_{\mathrm{\ell }}{g}^i{y}_{\mathrm{\ell }}={\sum }_{\mathrm{\ell }}{t}_{\mathrm{\ell }}{A}_n(g)y_{\mathrm{\ell }}$; hence $A_n(g)$ is affine. Continuity follows from the TVS addition and scalar-multiplication maps and the continuity of the action iterates.

Let $\mathrm{\Gamma }$ be the monoid of finite compositions of the maps $A_n(g)$, including the identity. For $g,h\in G$, affinity and the action law give $A_n(g)A_m(h)x=\frac{1}{(n+1)(m+1)}{\sum }_{i=0}^n{\sum }_{j=0}^m{g}^i{h}^{j}x$ and $A_m(h)A_n(g)x=\frac{1}{(n+1)(m+1)}{\sum }_{j=0}^m{\sum }_{i=0}^n{h}^j{g}^{i}x$. Because $G$ is abelian, $g^i{h}^j=h^j{g}^i$; associativity and commutativity of vector addition and scalar distributivity reorder the finite sums, so the two maps commute. Therefore $\mathrm{\Gamma }$ is abelian, and every member is a continuous self-map of $X$.

For every $\gamma \in \mathrm{\Gamma }$, the image $\gamma (X)$ is nonempty and compact by [F2], hence closed in $X$ by [F2]. Given a nonempty finite list ${\gamma }_1,\dots ,{\gamma }_k\in \mathrm{\Gamma }$, their composition $\gamma :={\gamma }_1\cdots {\gamma }_k$ lies in $\mathrm{\Gamma }$; commutativity lets us write $\gamma ={\gamma }_i\circ {\gamma }_i^{\mathrm{\prime }}$ for each $i$ with ${\gamma }_i^{\mathrm{\prime }}$ the composition of the other factors, using the identity when $k=1$. Thus the nonempty set $\gamma (X)$ lies in every ${\gamma }_i(X)$. The empty finite intersection is $X\mathrm{\ne }\mathrm{\varnothing }$, so $\{\gamma (X):\gamma \in \mathrm{\Gamma }\}$ has the finite intersection property, and compactness of $X$ gives $x_0\in {\bigcap }_{\gamma \in \mathrm{\Gamma }}\gamma (X)$.

Fix $g\in G$ and suppose $v:=x_0-g{x}_0\mathrm{\ne }0$. By [A1] and [F3], choose a continuous linear functional $\phi \in V^{\mathrm{\prime }}$ whose real part $\psi :=\mathrm{Re}\phi$ satisfies $\psi (v)\mathrm{\ne }0$. The continuous real-valued function $\psi$ is bounded on compact $X$ by [F4]; fix $C\ge 0$ with $\mathrm{\mid }\psi (y)\mathrm{\mid }\le C$ for every $y\in X$.

For every $n\in \mathbb{N}$, membership $x_0\in A_n(g)(X)$ gives some $x\in X$ with $x_0=A_n(g)x$. Affinity of the action and real-linearity of $\psi$ yield $\psi (v)=(\psi (x)-\psi (g^{n+1}x))\mathrm{/}(n+1)$ by telescoping, so $\mathrm{\mid }\psi (v)\mathrm{\mid }\le 2C\mathrm{/}(n+1)$. This bound is valid for every $n$; no sequence of preimages is chosen. If $C=0$, the bound gives $\psi (v)=0$ directly. If $C>0$, then for any $\epsilon >0$, [F6] applied to $\epsilon \mathrm{/}(2C)>0$ gives $n\ge 1$ with $1\mathrm{/}n<\epsilon \mathrm{/}(2C)$. Since $n+1>n>0$, [F5] gives $1\mathrm{/}(n+1)<1\mathrm{/}n$, hence $\mathrm{\mid }\psi (v)\mathrm{\mid }\le 2C\mathrm{/}(n+1)<\epsilon$. As this holds for every positive $\epsilon$, $\psi (v)=0$, contradicting step 4.1. Therefore $g{x}_0=x_0$. Since $g$ was arbitrary, $x_0$ is fixed by all of $G$.