A written fixed-point proof for affine isometries

Programme proof selection, CC0-1.0.

This note retains Sections 1–5 of the earlier programme lesson Weakly compact convex sets and fixed points, with exact local prerequisite links and two explicit topology details. It proves the full group-of-affine-isometries theorem needed by VT10, including uncountable groups and nonseparable Banach spaces.

The earlier written inputs are NP1–NP2: Hahn–Banach, norming, strict separation and equality of convex closures, AB02: completeness of the dual, and AB03: uniform boundedness. Finite products of compact spaces and the needed subnet compactness are proved in AB04. The maximal principle and real-number completeness are foundational assumptions. The category and extreme-point assertions are proved below.

The free human comparison is Jacob Lurie's Math 261y, Lecture 26, Theorem 1, Proposition 2 and Lemma 3, pages 1–3. The retained proof specifies affine actions, proves its auxiliary compact-face claims, and keeps the coefficient of the large set at least the positive threshold inside the retained subset. This corrects the mismatched inequality in part (ii) on page 3 of the notes. No conclusion below is supplied by a citation alone.

1. Compactness, faces and extreme points

A weakly compact subset of a Banach space is norm bounded. Indeed every $f\in X^*$ is bounded on that compact set. Apply uniform boundedness to its canonical images in $(X^*)^*$; their operator norms are their original norms.

We will use the category property of a compact Hausdorff space $Z$: if $Z$ is nonempty and is a countable union of closed sets, one of them has nonempty interior. Here is the argument. If closed sets $D_n$ all had empty interior, begin with a nonempty open set and successively choose nonempty open sets $V_n$ such that $\overline V_n\subseteq V_{n-1}\setminus D_n. \tag{1.1}$ Regularity of a compact Hausdorff space allows these choices. The nonempty compact sets $\overline V_n$ are nested, so their intersection is nonempty. A point in it lies in none of the $D_n$, contradicting the proposed covering.

The regularity used here has a short compactness proof. If $z$ lies in an open set $O\subseteq Z$, separate $z$ from each point of the compact set $Z\setminus O$ by disjoint open neighborhoods. Finitely many of the latter neighborhoods cover that compact set. The intersection $V$ of the corresponding neighborhoods of $z$ has closure disjoint from their union, and hence $\overline V\subseteq O$. This supplies the choices in (1.1).

Let $K$ be a nonempty compact convex set in a Hausdorff locally convex space. A face of $K$ is a convex subset $F$ with the following property: if a point of an open segment between two points of $K$ lies in $F$, then both endpoints lie in $F$. A point is extreme when its singleton is a face.

Lemma 1.1. Every nonempty closed face of $K$ contains an extreme point of $K$. Moreover $K$ is the closed convex hull of its extreme points.

Proof. Among the nonempty closed faces contained in a given one, a descending chain has a nonempty intersection by compactness. The intersection is again a closed face. The maximal principle therefore gives a minimal such face $F$. If $F$ contained distinct points, a continuous real linear functional would distinguish them. Its maximizing set on $F$ would be a proper nonempty closed face of $F$, hence a face of $K$. This contradicts minimality. Thus $F$ is a singleton.

Write $E=\operatorname{ext}K$ and $C=\overline{\operatorname{co}}E$. The first assertion makes $E$ nonempty. If some point of $K$ were outside $C$, separation would give a continuous real linear functional $f$ with $\max_K f>\sup_C f.$ The maximizing face of $K$ contains an extreme point by the first assertion. This point also belongs to $C$, a contradiction. $\square$

The next observation explains why an extreme point cannot be reconstructed entirely from a compact set that omits it.

Lemma 1.2. If $A\subseteq K$ is compact and $p\in\operatorname{ext}K$ belongs to $\overline{\operatorname{co}}A$, then $p\in A$.

Proof. Suppose $p\notin A$. Choose an open balanced convex neighborhood $V$ of zero whose closure is disjoint from $A-p$. Such a neighborhood exists by local convexity, compactness of $A$, and separation of $p$ from each point of $A$. Cover $A$ by finitely many sets $a_j+V$, with $a_j\in A$. Put $A_j=A\cap(a_j+\overline V),\qquad C_j=\overline{\operatorname{co}}A_j.$ Each $C_j$ is a compact convex subset of $K\cap(a_j+\overline V)$: it is closed inside the compact set $K$, and the translated neighborhood closure is closed and convex.

The convex join of the finitely many $C_j$ is compact, being the image of their product with a finite-dimensional probability simplex. It is therefore closed and equals $\overline{\operatorname{co}}A$. Express $p$ as a convex combination of points in the $C_j$. Extremality forces every point with a positive coefficient to equal $p$. Thus $p\in a_j+\overline V$ for some $j$. Balancedness gives $a_j\in p+\overline V$, contrary to the choice of $V$. $\square$

In particular, if $A$ is a compact subset of the weak closure of $\operatorname{ext}K$ and omits one of those extreme points, its closed convex hull omits that point too.

2. Removing everything except a small part

We now combine the category property with the preceding two lemmas. Separability is needed only in this auxiliary result.

Lemma 2.1. Suppose $K\subseteq X$ is weakly compact, convex and norm separable. Given $\varepsilon>0$ with $\operatorname{diam}K>\varepsilon,$ there is a nonempty proper weakly compact convex subset $C\subset K$ such that $\operatorname{diam}(K\setminus C)<\varepsilon. \tag{2.1}$

Proof. Let $Z=\overline{\operatorname{ext}K}^{\,w}$. Lemma 1.1 makes $Z$ nonempty and gives $K=\overline{\operatorname{co}}Z^{\,w}$. It is compact Hausdorff. Choose a countable norm-dense set in $K$. The closed norm balls of radius $\varepsilon/8$ about those points cover $Z$, and each intersection with $Z$ is weakly closed. Closed norm balls are weakly closed by Hahn–Banach. The category property gives a nonempty relatively weakly open set $U\subseteq Z$ lying in one of these balls, say $B$.

Set $A=Z\setminus U,\qquad D=Z\cap B,\qquad K_A=\overline{\operatorname{co}}A^{\,w},\qquad K_D=\overline{\operatorname{co}}D^{\,w}. \tag{2.2}$ If $A$ were empty, then $Z\subseteq B$ and hence $K\subseteq B$, which would give $\operatorname{diam}K\leq\varepsilon/4$. Thus $A$ is nonempty. Both $A$ and $D$ are compact; both convex hull closures in (2.2) are weakly compact because they are closed subsets of $K$. Also $K_D\subseteq B$, so $\operatorname{diam}K_D\leq\varepsilon/4. \tag{2.3}$

Because $A\cup D=Z$, compactness of the convex join gives $K=\{t a+(1-t)d:a\in K_A,\ d\in K_D,\ 0\leq t\leq1\}. \tag{2.4}$ The relatively open set $U$ meets $\operatorname{ext}K$, since the latter is dense in $Z$. Choose an extreme point $p\in U$. Lemma 1.2 applied to the compact set $A$ gives $p\notin K_A$.

Write $R=\operatorname{diam}K$; it is finite and positive. Choose $0<\delta<1$ with $2\delta R<\varepsilon/2$, and define $C=\{t a+(1-t)d:a\in K_A,\ d\in K_D,\ \delta\leq t\leq1\}. \tag{2.5}$ This set is nonempty and weakly compact. It is convex: in a convex combination of two displayed expressions, the new coefficient of $K_A$ is a convex combination of their coefficients and remains at least $\delta$; regroup the $K_A$ and $K_D$ terms using their convexity. A vanishing coefficient of $K_D$ causes no difficulty.

The point $p$ is outside $C$. Indeed a representation in (2.5), together with extremality, would force $p=a\in K_A$ because $t>0$. Thus $C$ is proper.

Every $y\in K\setminus C$ has a representation in (2.4) with $t<\delta$; otherwise it would belong to $C$. Its distance from the corresponding $d$ is at most $tR<\delta R$. Consequently for $y,y'\in K\setminus C$, using (2.3), $\|y-y'\|<2\delta R+\varepsilon/4<\varepsilon.$ This proves (2.1). $\square$

The conclusion does not say that $C$ is invariant under any action. Its purpose is to make two orbit points outside $C$ necessarily close in norm.

3. One affine map has a fixed point

Lemma 3.1. A weakly continuous affine map $T:K\to K$ on a nonempty weakly compact convex subset of a Banach space has a fixed point.

Proof. Fix $z\in K$ and consider its successive averages $z_n=\frac1n\sum_{j=0}^{n-1}T^jz\in K. \tag{3.1}$ Affineness gives $Tz_n-z_n=\frac{T^nz-z}{n}. \tag{3.2}$ Since $K$ is norm bounded, the right side tends to zero in norm. A subnet of $(z_n)$ converges weakly to some $z_\infty\in K$. Weak continuity of $T$ and (3.2) give $Tz_\infty=z_\infty$. $\square$

No isometry hypothesis was used here. For several maps that do not commute, their average can have a fixed point without an immediate reason for each map to fix it. The next argument supplies that reason when the maps belong to a group of isometries.

4. An averaged fixed point is fixed by each isometry

Proposition 4.1. Let a group $G$ act on a nonempty weakly compact convex set $K\subseteq X$ by weakly continuous affine bijections preserving norm distances. For $g_1,\ldots,g_m\in G$ and positive numbers $\lambda_i$ summing to one, a point $x\in K$ satisfying $x=\sum_{i=1}^m\lambda_i g_i x \tag{4.1}$ is fixed by every $g_i$.

Proof. Suppose some $g_i$ moves $x$. Remove the indices which fix $x$ from (4.1), subtract their terms, and divide by the sum of the remaining coefficients. This leaves an equation of the same form with positive coefficients, and now every listed map moves $x$.

Let $H$ be the subgroup generated by these finitely many maps. It is countable. Its orbit $Hx$ is countable, and $L=\overline{\operatorname{co}}^{\,w}(Hx) \tag{4.2}$ is a nonempty weakly compact convex subset of $K$. It is invariant under $H$, using weak continuity and the inverse of each group element. It is norm separable: weak and norm closures of a convex set agree by NP2, and rational convex combinations of the countable orbit are norm dense in its convex hull closure. To apply NP2, a weakly continuous linear functional is norm continuous because the weak topology is weaker; conversely every norm-continuous linear functional is weakly continuous by definition. In the complex case apply the real-part conversion proved in NP1. Thus the two topologies have the same continuous real dual, as NP2 requires. This uses no separability assumption on $X$ or on the original $K$.

Choose $0<\varepsilon<\min_i\|g_i x-x\|. \tag{4.3}$ The diameter of $L$ exceeds $\varepsilon$. Lemma 2.1 gives a proper weakly closed convex subset $C\subset L$ with $\operatorname{diam}(L\setminus C)<\varepsilon$. Since the closed convex hull of $Hx$ is $L$, the orbit cannot be contained in $C$. Choose $h\in H$ with $hx\notin C$. Apply this affine map to (4.1): $hx=\sum_i\lambda_i h g_i x. \tag{4.4}$ At least one $hg_i x$ is outside $C$, since otherwise convexity would put $hx$ in $C$. Both these points belong to $L\setminus C$. The isometry property therefore gives $\|g_i x-x\|=\|hg_i x-hx\|<\varepsilon,$ contradicting (4.3). Thus no listed map moves $x$. $\square$

5. The common fixed point theorem

Theorem 5.1 (Ryll-Nardzewski, group form). Under the hypotheses of Proposition 4.1, there is a point of $K$ fixed by all of $G$.

Proof. Given any finite list $g_1,\ldots,g_m$, its average $T(y)=\frac1m\sum_i g_i y$ is a weakly continuous affine map from $K$ into $K$. Lemma 3.1 supplies a fixed point of $T$; Proposition 4.1 says this point is fixed by each $g_i$. Each set $\operatorname{Fix}_K(g)=\{y\in K:gy=y\}$ is weakly closed. They have the finite intersection property, so compactness of $K$ makes their total intersection nonempty. $\square$

The group may be uncountable. The countable group and separable convex set appeared only inside the proof for a finite list. Bounded affine isometries of a Banach space restrict to actions of the type used here: their linear parts are bounded and hence weakly continuous, and translations are weakly continuous too.