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

Proof for a single affine self-mapping
Let $$T$$ be a continuous affine self-mapping of $$K$$.

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


 * $$ x(N)={1\over N+1}\sum_{n=0}^N T^n(x). $$

Since $$K$$ is compact, there is a convergent subnet in $$K$$:


 * $$ x(N_i)\rightarrow y. \,$$

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

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


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

Taking $$N = N_i$$ and passing to the limit as $$i$$ goes to infinity, it follows that


 * $$ f(Ty) = f(y). \, $$

Hence


 * $$Ty = y. \, $$

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


 * $$ K^S=\{y\in K\mid Ty=y, \, T\in S\}=\bigcap_{T\in S} K^T \,$$

is non-empty (and compact and convex).