Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\emptyset \not = X\subseteq \Bbb{R}^n$ be convex and compact and let $\cal{A}$ be a commuting family of affine maps from $\Bbb{R}^n$ into $\Bbb{R}^n$ such that $X$ is invariant under each element of $\cal A$. How can I show that $\cal{A}$ has a common fixed point.

Remark I know this is a consequence of some stronger theorems such as Markov-Kakutani fixed point theorem. But I have no access to its proof. Besides I am looking for a much elementary proof (which I think must exists for this special case).

share|improve this question
What is the original version of Markov-Kakutani fixed point theorem? –  Belle-tiantian Apr 13 at 16:57
add comment

1 Answer

up vote 1 down vote accepted

For each $A \in \mathcal{A}$, let $K_A =\{ x \in X : A(x) = x\}$ the set of fixed points of $A$ in $X$. Since $A$ is continuous, $K_A$ is closed (hence compact), and since $A$ is affine, $K_A$ is convex. By - for example - Brouwer's fixed point theorem(1), $K_A \neq \varnothing$.

For $B \in \mathcal{A}$ and $x \in K_A$, we have

$$A(B(x)) = B(A(x)) = B(x),$$

i.e. $B(K_A) \subset K_A$, and since $K_A$ is convex and compact, $B$ has a fixed point in $K_A$.

From that, we can deduce that the family of convex compact sets $\mathcal{K}_\mathcal{A} = \{ K_A : A \in \mathcal{A}\}$ has the finite intersection property, $$\mathcal{F} \subset \mathcal{A} \text{ finite} \Rightarrow \bigcap_{A \in \mathcal{F}} K_A \neq \varnothing.$$


$$K := \bigcap_{A \in \mathcal{A}} K_A \neq \varnothing.$$

$K$ is the set of common fixed points of all $A \in \mathcal{A}$.

(1) Here, we can also prove directly that each $A$ has a fixed point in $X$. Let $\pi_r \colon \mathbb{R}^n \to \mathbb{R}$ the coordinate projection. Let $m_r(K) = \min \{ \pi_r(x) : x \in K\}$ and $M_r(K) = \max \{ \pi_r(x) : x \in K\}$ for any compact nonempty $K$. Let $X_0 = X$, and for $1 \leqslant r \leqslant n$, let $X_r = \{ x \in X_{r-1} : \pi_r(A(x)) = \pi_r(x)\}$. Since $A$ is continuous, each $X_r$ is closed in $X_{r-1}$, hence compact, and since $A$ is affine, each $X_r$ is convex. It remains to see that $X_{r-1} \neq \varnothing \Rightarrow X_r \neq \varnothing$.

Let $\varphi_r(x) = \pi_r(A(x)) - \pi_r(x)$. On $X_{r-1} \cap \pi_r^{-1}(m_r(X_{r-1}))$, we have $\varphi_r \geqslant 0$, and on $X_{r-1} \cap \pi_r^{-1}(M_r(X_{r-1}))$ we have $\varphi_r \leqslant 0$. Since $\varphi_r$ is continuous and $X_{r-1}$ connected, $X_r = X_{r-1} \cap \varphi_r^{-1}(0) \neq \varnothing$.

share|improve this answer
Nice solution. But do we have to use Brouwer's theorem !? –  Drop Aug 2 '13 at 18:07
I would think you can prove that an affine map that leaves a compact convex set (in a finite-dimensional space) invariant has a fixed point there without Brouwer. But Brouwer was the quick no-brains-needed way to know that $K_A \neq \varnothing$. –  Daniel Fischer Aug 2 '13 at 18:12
That would be a nice problem ! –  Drop Aug 2 '13 at 18:40
Let $X_r = \{x \in X : A(x)_r = x_r\}$ (the $r$-th component/coordinate). $x \mapsto A(x)_r - x_r$ is continuous, $\geqslant 0$ where the $r$-th coordinate minimises in $X$, $\leqslant 0$ where it maximises. Thus $X_r \neq \varnothing$. $X_r$ is closed (thus compact) and convex ($A$ is affine). Iterate over all coordinates. –  Daniel Fischer Aug 2 '13 at 18:45
@Daniel Fischer What is the original version of Markov-Kakutani fixed point theorem? –  Belle-tiantian Apr 14 at 12:08
show 1 more comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.