Fixed points of groups of affine transformations

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.

By the definition of a solvable group (A, I, § 6, No. 4) there exists a finite decreasing sequence $(G_i)_{0 \leq i \leq n}$ of distinct subgroups of G, such that $G_0 = G$, $G_n = \{e\}$ and such that the group $G_{i-1}/G_i$ is commutative for $1 \leq i \leq n$. Let $K_i$ denote the set of fixed points of $G_i$ in K. Then $K_n = K$. Moreover, for $1 \leq i \leq n$, every element of $G_i$ induces the identity transformation on $K_i$; we thus deduce an action of the abelian group $G_{i-1}/G_i$ on K if $K_i$ is non-empty; it follows from th. 1 that the set $K_{i-1}$ of fixed points of $G_{i-1}/G_i$ in $K_i$ is non-empty. By descending induction on $i$, we conclude that $K_0$ is non-empty, hence the corollary.

Let $\mu$ be a linear form on E, such that $\mu(1) = 1$. For every function $f \in E$, we define the function $f' \in E$ by $f'(x) = \|f\| - f(x)$ ($x \in X$). First assume that $\mu$ is a mean; for every $f \in E$, we have $f' \geq 0$, hence $\mu(f') \geq 0$, i.e. $\mu(f) \leq \|f\|$; therefore $\|\mu\| \leq 1$.

Conversely, suppose $\mu$ belongs to $E'$, and that $\| \mu \| \leq 1$; for every positive function $f \in E$, we have $\mu(f') \leq \| f' \|$, hence

and finally $\mu(f) \geq 0$; consequently, $\mu$ is a mean.

It is clear that $K$ is convex; that it is compact for $\sigma(E', E)$ follows from cor. 3 of III, p. 17.

Q.E.D.

Let $\Gamma$ be a set of continuous mappings from $X$ into $X$ which commute pairwise. Let $\gamma \in \Gamma$. For every function $f \in E$, we have $f \circ \gamma \in E$; hence we can define an affine transformation $u_{\gamma}$ on the set $K$ of means on $X$, by

If $K$ is assigned the topology induced by $\sigma(E', E)$, the mapping $u_{\gamma}$ is continuous. If $\gamma$ is a homeomorphism, $u_{\gamma} \mu$ can be deduced from $\mu$ by transport of structure. Finally, we have $u_{\gamma} u_{\gamma'} = u_{\gamma' \gamma} u_{\gamma}$ for all $\gamma, \gamma'$ in $\Gamma$. By the Markoff-Kakutani th. (IV, p. 39, th. 1), there exists a mean $\mu$ on $X$, such that $u_{\gamma} \mu = \mu$ for all $\gamma \in \Gamma$; in other words, $\mu$ satisfies the relation $\mu(f) = \mu(f \circ \gamma)$ for $f \in E$ and $\gamma \in \Gamma$.

In an analogous way, the corollary of th. 1 (IV, p. 40) implies the following result:

It is enough to apply prop. 1 to the solvable group $G \times G$ acting on $G$ by $(g.g').x = {gxg'}^{-1}$.

a) $K_1$ is convex and compact ; b) we have $\| x_1 - x_2 \| < \varepsilon$ for every $x_1$ and $x_2$ in $K_2$.

Let $L$ be the closure of the set of all extremal points of $K$. By the Krein-Milman th. (II, p. 55, th. 1), $K$ is the closed convex envelope of $L$. Since $K$ contains at least two points, so does $L$. For every $x \in L$, let $A_x$ be the set of all $y \in L$ such that $\| x - y \| \leq \varepsilon / 4$. By the hypothesis, on $K$, there exists a countable subset $D$ of $L$ such that $L = \bigcup_{x \in D} A_x$. Since the norm is lower semi-continuous, each of the sets $A_x$ is closed. Applying Baire’s th. (GT, IX, § 5, No. 3, th. 1) to the compact space $L$, we see that there exists a point $a$ in $D$ and an open subset $U$ in $E$ such that $L \cap U$ is non-empty and is contained in $A_a$. Since $L$ contains at least two points, and since $E$ is Hausdorff, we can choose $U$ in such a way that $L \notin U$.

Let $M$ be the closed convex envelope of $L \cap \complement U$. For every real number $t$ such that $0 < t < 1$, let $M_t$ be the set of all vectors of the form $tx_1 + (1 - t) x_2$ with $x_1 \in M$ and $x_2 \in K$; this is a non-empty, compact convex subset of $K$. We shall prove that $M_t \neq K$ by reductio ad absurdum. Suppose that $M_t = K$; then every extremal point $x$ in $K$ belongs to $M_t$, hence can be written in the form $x = tx_1 + (1 - t) x_2$ with $x_1 \in M$ and $x_2 \in K$. This implies that $x = x_1 = x_2$, and so $x \in M$. By Krein-Milman th. (II, p. 55, th. 1), we have $K = M$, and $K$ is the closed convex envelope of $L \cap \complement U$. By II, p. 56, corollary, this implies that $L \subset L \cap \complement U$, which contradicts the relation $L \cap U \neq \varnothing$.

Put $d = \sup_{x \in K, y \in K} \| x - y \|$ and choose a real number $t$ such that $0 < t < 1$ and $t < \varepsilon / 4d$. Put $K_1 = M_t$ and $K_2 = K - M_t$. By the preceding argument, the sets $K_1$ and $K_2$ are non-empty, and $K_1$ is convex and compact. Let $M'$ be the closed convex envelope of $L \cap U$. Since $K$ is the closed convex envelope of the set $L = (L \cap \complement U) \cup (L \cap U)$, it is also the closed convex envelope of $M \cup M'$. Let $x_1$ and $x_2$ be two points in $K_2$; for $i = 1, 2$, there exist $y_i \in M$, $z_i \in M'$ and a real number $\alpha_i$ such that $0 \leq \alpha_i \leq 1$ and $x_i = \alpha_i y_i + (1 - \alpha_i) z_i$. If $\alpha_i \geq t$, then

; this contradicts the assumption that $x_i \notin M_i$. Hence $\alpha_i < t$, for $i = 1, 2$, and so For every point $z$ in $M'$, we have $\| z - a \| \leq \varepsilon / 4$, since $L \cap U \subset A_a$, and so, in particular $\| z_i - a \| \leq \varepsilon / 4$. Thus This completes the proof.

Let $\mathfrak{J}$ be the family of non-empty subsets of K which are closed convex and stable for G. If $(L_\alpha)_{\alpha \in I}$ is a family of elements of $\mathfrak{J}$ which is totally ordered by inclusion, then the set $L = \bigcap_{\alpha \in I} L_\alpha$ belongs to $\mathfrak{J}$. Consequently (S, III, § 3, No. 4, th. 2), there exists an element L in $\mathfrak{J}$ which is minimal for the relation of inclusion. We shall prove that L reduces to a point.

We argue by reductio ad absurdum, assuming that L contains at least two distinct points $x_1$ and $x_2$, put $x = (x_1 + x_2)/2$ and $\varepsilon = \|x_1 - x_2\|/2$. The convex and compact set L satisfies the first axiom of countability for the distance defined by the norm (GT, IX, § 2, No. 8). Hence we can apply lemma 2 and find a compact and convex subset $L_1$ of L, distinct from $\varnothing$ and from L, having the following property :

(A) For every $y_1$ and $y_2$ in $L - L_1$, we have $\|y_1 - y_2\| < \varepsilon$.

We shall prove by reductio ad absurdum that $gx \in L_1$ for all $g \in G$. Let $g \in G$ be such that $gx \in L - L_1$ then we have

for $i = 1, 2$. By property (A), we have $gx_i \in L_1$. Since $L_1$ is convex, we conclude that $gx = (gx_1 + gx_2)/2$ belongs to $L_1$, which contradicts the assumption.

Let $L'$ be the closed convex envelope of the orbit $Gx$ of x. The set $L'$ belongs to $\mathfrak{J}$. By the preceding argument, we have $L' \subset L_1$, hence $L' \subset L, L' \neq L$. This contradicts the minimal character of L and the proof is complete.

For every $g \in G$, let $K_g$ denote the set of all points x in K such that $gx = x$; let K be assigned the weakened topology; each set $K_g$ is convex and closed in the compact space K. We shall prove that the intersection $\bigcap_{g \in G} K_g$ is non-empty; for this, it is enough to prove that the set $K_{g_1} \cap ... \cap K_{g_n}$ is non-empty for every $g_1, ..., g_n$ in G. Fix $g_1, ..., g_n$ and let H be the subgroup of G generated by $\{ g_1, ..., g_n \}$. Choose a point a in K and let L denote the closed convex envelope of the orbit Ha of a. Let D be the countable set of elements of the form $\lambda_1 h_1 a + \cdots + \lambda_m h_m a$, where $\lambda_1, ..., \lambda_m$ are positive rational numbers such that $\lambda_1 + \cdots + \lambda_m = 1$, and $h_1, ..., h_m$ are elements in H. The closure $\overline{D}$ of D for the strong topology, is convex, hence it is closed for $\sigma(E, E')$ (IV, p. 4, prop. 2); therefore $\overline{D} = L$ and this proves that L is a metric space satisfying the first axiom of countability for the distance $(x, y) \mapsto \|x - y\|$. We can now apply lemma 2. There exists a point b in L which is invariant under H, hence $b \in K_{g_1} \cap ... \cap K_{g_n}$.

Since E is reflexive, K is compact for $\sigma(E, E')$ (IV, p. 15, th. 1). Moreover, every element of G belongs to $\mathscr{L}(E)$.