A weak version of Markov-Kakutani fixed point theorem 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). 
 A: 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, $$\mathcal{F} \subset \mathcal{A} \text{ finite} \Rightarrow \bigcap_{A \in \mathcal{F}} K_A \neq \varnothing.$$
Therefore,
$$K := \bigcap_{A \in \mathcal{A}} K_A \neq \varnothing.$$
$K$ 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, $X_r = X_{r-1} \cap \varphi_r^{-1}(0) \neq \varnothing$.
