Markov–Kakutani fixed-point theorem

In mathematics, the Markov–Kakutani fixed-point theorem, named after Andrey Markov and Shizuo Kakutani, states that a commuting family of continuous affine self-mappings of a compact convex subset in a locally convex topological vector space has a common fixed point. This theorem is a key tool in one of the quickest proofs of amenability of abelian groups.

Statement

Let ${\displaystyle X}$ be a locally convex topological vector space, with a compact convex subset ${\displaystyle K}$. Let ${\displaystyle S}$ be a family of continuous mappings of ${\displaystyle K}$ to itself which commute and are affine, meaning that ${\displaystyle T(\lambda x+(1-\lambda )y)=\lambda T(x)+(1-\lambda )T(y)}$ for all ${\displaystyle \lambda }$ in ${\displaystyle (0,1)}$ and ${\displaystyle T}$ in ${\displaystyle S}$. Then the mappings in ${\displaystyle S}$ share a fixed point.^[1]

Proof for a single affine self-mapping

Let ${\displaystyle T}$ be a continuous affine self-mapping of ${\displaystyle K}$.

For ${\displaystyle x}$ in ${\displaystyle K}$ define a net ${\displaystyle \{x(N)\}_{N\in \mathbb {N} }}$ in ${\displaystyle K}$ by

${\displaystyle x(N)={1 \over N+1}\sum _{n=0}^{N}T^{n}(x).}$

Since ${\displaystyle K}$ is compact, there is a convergent subnet in ${\displaystyle K}$:

${\displaystyle x(N_{i})\rightarrow y.\,}$

To prove that ${\displaystyle y}$ is a fixed point, it suffices to show that ${\displaystyle f(Ty)=f(y)}$ for every ${\displaystyle f}$ in the dual of ${\displaystyle X}$. (The dual separates points by the Hahn-Banach theorem; this is where the assumption of local convexity is used.)

Since ${\displaystyle K}$ is compact, ${\displaystyle |f|}$ is bounded on ${\displaystyle K}$ by a positive constant ${\displaystyle M}$. On the other hand

${\displaystyle |f(Tx(N))-f(x(N))|={1 \over N+1}|f(T^{N+1}x)-f(x)|\leq {2M \over N+1}.}$

Taking ${\displaystyle N=N_{i}}$ and passing to the limit as ${\displaystyle i}$ goes to infinity, it follows that

${\displaystyle f(Ty)=f(y).\,}$

Hence

${\displaystyle Ty=y.\,}$

Proof of theorem

The set of fixed points of a single affine mapping ${\displaystyle T}$ is a non-empty compact convex set ${\displaystyle K^{T}}$ by the result for a single mapping. The other mappings in the family ${\displaystyle S}$ commute with ${\displaystyle T}$ so leave ${\displaystyle K^{T}}$ invariant. Applying the result for a single mapping successively, it follows that any finite subset of ${\displaystyle S}$ has a non-empty fixed point set given as the intersection of the compact convex sets ${\displaystyle K^{T}}$ as ${\displaystyle T}$ ranges over the subset. From the compactness of ${\displaystyle K}$ it follows that the set

${\displaystyle K^{S}=\{y\in K\mid Ty=y,\,T\in S\}=\bigcap _{T\in S}K^{T}\,}$

is non-empty (and compact and convex).