Lemma 3

Let G be a group of continuous (for $\mathscr{T}$) affine transformations on K. Suppose that K is non-empty and that $\|gx - gy\| = \|x - y\|$ for all x, y in K and all g in G. Then there exists a point in K which is invariant under G.

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.