Markov–Kakutani fixed-point theorem

Remarks

For $T \in \mathcal{F}$ and $n \ge 1$, let $A_{T,n} = \frac1n (I + T + \cdots + T^{n-1})$, an affine map of $K$ into $K$ by convexity. If $x = A_{T,n}(y)$, then $T(x) - x = \frac1n (T^n(y) - y)$ lies in $\frac1n(K - K)$. Since $K - K$ is compact, hence bounded, every point of $\bigcap_n A_{T,n}(K)$ is a fixed point of $T$, as $X$ is Hausdorff. Because the maps commute, the compact sets $A_{T_1,n_1} \cdots A_{T_k,n_k}(K)$ lie in each $A_{T_i,n_i}(K)$, so the family of all $A_{T,n}(K)$ has the finite intersection property, and its intersection is nonempty.

Example: an abelian group $G$ acts on the weak-$*$ compact convex set of means on $\ell^\infty(G)$ by commuting affine continuous maps. A common fixed point is an invariant mean, so $G$ is amenable.

The commutativity hypothesis cannot be dropped. The affine maps $x \mapsto x/2$ and $x \mapsto (x+1)/2$ of $[0,1]$ do not commute. Their only fixed points are $0$ and $1$, so they have no common fixed point.

Going deeper

For a student

The theorem is the abelian prototype of the fixed point characterizations of amenability. Applied to the set of means, it shows that abelian groups are amenable; Day's fixed point theorem extends this to all amenable groups. Compared with the Kakutani fixed-point theorem for set-valued maps, it needs affine maps but allows an arbitrary commuting family.