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 {\displaystyle X} be a locally convex topological vector space, with a compact convex subset K {\displaystyle K} . Let S {\displaystyle S} be a family of continuous mappings of K {\displaystyle K} to itself which commute and are affine, meaning that T ( λ x + ( 1 − λ ) y ) = λ T ( x ) + ( 1 − λ ) T ( y ) {\displaystyle T(\lambda x+(1-\lambda )y)=\lambda T(x)+(1-\lambda )T(y)} for all λ {\displaystyle \lambda } in ( 0 , 1 ) {\displaystyle (0,1)} and T {\displaystyle T} in S {\displaystyle S} . Then the mappings in S {\displaystyle S} share a fixed point.
Proof for a single affine self-mapping Let T {\displaystyle T} be a continuous affine self-mapping of K {\displaystyle K} . For x {\displaystyle x} in K {\displaystyle K} define a net { x ( N ) } N ∈ N {\displaystyle \{x(N)\}_{N\in \mathbb {N} }} in K {\displaystyle K} by
x ( N ) = 1 N + 1 ∑ n = 0 N T n ( x ) . {\displaystyle x(N)={1 \over N+1}\sum _{n=0}^{N}T^{n}(x).}
Since K {\displaystyle K} is compact, there is a convergent subnet in K {\displaystyle K} :
x ( N i ) → y . {\displaystyle x(N_{i})\rightarrow y.\,}
To prove that y {\displaystyle y} is a fixed point, it suffices to show that f ( T y ) = f ( y ) {\displaystyle f(Ty)=f(y)} for every f {\displaystyle f} in the dual of X {\displaystyle X} . (The dual separates points by the Hahn-Banach theorem; this is where the assumption of local convexity is used.) Since K {\displaystyle K} is compact, | f | {\displaystyle |f|} is bounded on K {\displaystyle K} by a positive constant M {\displaystyle M} . On the other hand
| f ( T x ( N ) ) − f ( x ( N ) ) | = 1 N + 1 | f ( T N + 1 x ) − f ( x ) | ≤ 2 M N + 1 . {\displaystyle |f(Tx(N))-f(x(N))|={1 \over N+1}|f(T^{N+1}x)-f(x)|\leq {2M \over N+1}.}
Taking N = N i {\displaystyle N=N_{i}} and passing to the limit as i {\displaystyle i} goes to infinity, it follows that
f ( T y ) = f ( y ) . {\displaystyle f(Ty)=f(y).\,}
Hence
T y = y . {\displaystyle Ty=y.\,}
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