Theorem 1 (Markoff-Kakutani)

— Let E be a Hausdorff locally convex vector space over the field $\mathbf{R}$, and K a non-empty compact convex subset of E. Let $\Gamma$ be a set of continuous affine transformations on K, pairwise permutable. Then there exists a point a in K such that $u(a) = a$ for all $u \in \Gamma$.

For every $u \in \Gamma$, let $K_u$ be the set of all $x \in K$ such that $u(x) = x$. We shall show that $K_u$ is non-empty. Let $x$ be a point of K; for every integer $n \geq 1$, let $x_n$ be the element $\frac{1}{n} \sum_{i=0}^{n-1} u^i(x)$ of E. Since K is convex and stable under $u$, the points $x_n$ belong to K and since K is compact, there exists a limit point $a$ of the sequence $(x_n)_{n \geq 1}$. The mapping $y \mapsto u(y) - y$ from K into E is continuous, hence $u(a) - a$ is a limit point of the sequence $(u(x_n) - x_n)_{n \geq 1}$. But we have $u(x_n) - x_n = \frac{1}{n} (u^n(x) - x)$.

Since K is compact, hence also bounded (III, p. 3, prop. 2), the sequence $(u^n(x) - x)_{n \geq 1}$ is bounded; consequently, the sequence $\left( \frac{1}{n}(u^n(x) - x) \right)_{n \geq 1}$ tends to 0 (III, p. 4, prop. 3), and since E is Hausdorff, we have $u(a) - a = 0$. Therefore $a \in K_u$.

Each of the sets $K_u$ is a closed and convex subset of the compact space K, and we shall prove that the intersection $\bigcap_{u \in \Gamma} K_u$ is non-empty. For this it is enough to prove that, for $n \geq 1$, and $u_1, ..., u_n$ in $\Gamma$, the set $K_{u_1} \cap ... \cap K_{u_n}$ is non-empty. The case $n = 1$ having been considered, we argue by induction on $n$. Suppose $n \geq 2$ and put $L = K_{u_1} \cap ... \cap K_{u_{n-1}}$. By the hypothesis of induction, L is a non-empty compact convex subset of E. Since $u_n$ commutes with $u_1, ..., u_{n-1}$, we have $u_n(L) \subset L$. Applying the first part of the proof to the affine transformation induced by $u_n$ on L, we conclude that there exists a point $a$ in L such that $u_n(a) = a$; then $a$ belongs to $K_{u_1} \cap ... \cap K_{u_n}$, which is then non-empty.