A Note on Geodesically Bounded -Trees
© W. A. Kirk. 2010
Received: 4 March 2010
Accepted: 10 May 2010
Published: 8 June 2010
It is proved that a complete geodesically bounded -tree is the closed convex hull of the set of its extreme points. It is also noted that if is a closed convex geodesically bounded subset of a complete -tree and if a nonexpansive mapping satisfies then has a fixed point. The latter result fails if is only continuous.
Recall that for a metric space a geodesic path (or metric segment) joining and in is a mapping of a closed interval into such that and for each Thus is an isometry and An -tree (or metric tree) is a metric space such that:
(i)there is a unique geodesic path (denoted by ) joining each pair of points
From (i) and (ii), it is easy to deduce that
(iii)if then for some
(Here denotes the usual Euclidean distance and denotes the origin.) For the river metric on , if two points , and are on the same vertical line, define Otherwise define where and More subtle examples of -trees also exist, for example, the real tree of Dress and Terhalle .
It is shown in  that -trees are hyperconvex metric spaces (a fact that also follows from Theorem B of  and the characterization of ). They are also CAT spaces in the sense of Gromov (see, e.g., [6, page 167]). Moreover, complete and geodesically bounded -trees have the fixed point property for continuous maps. This fact is a consequences of a result of Young  (see also ), and it suggests that complete geodesically bounded -trees have properties that one often associates with compactness. The two observations below serve to affirm this.
2. A Krein-Milman Theorem
In  Niculescu proved that a nonempty compact convex subset of a complete CAT space (called a global NPC space in ) is the convex hull of the set of all its extreme points. Subsequently, in , Borkowski et al. proved (among other things) that compactness is not needed in the special case when is a complete and bounded -tree. Here we show that in complete -trees even the boundedness assumption may be relaxed.
Let be a complete and geodesically bounded -tree. Then is the convex hull of its set of extreme points.
Let and let We will show that lies on a segment joining to some other element of We proceed by transfinite induction. Let denote the set of all countable ordinals, let let and assume that for all with has been defined so that the following condition holds:
There are two cases.
(1) If there is nothing to prove because . Otherwise, there are elements such that lies on the segment and At least one of these points say does not lie on the segment . Set and observe that lies on the segment
(2) is a limit ordinal. Since is geodesically bounded, it must be the case that This implies that is a Cauchy net. Since is complete, it must converge to some
Therefore, is defined for all Since is geodesically bounded, But since is uncountable, it is not possible that for each Hence this transfinite process must terminate, and for some It now follows from (i) that and lies on the segment
The above proof shows that in fact each point of is on a segment joining any given extreme point to some other extreme point.
3. A Fixed Point Theorem
then has a fixed point (see [11, Theorem ] also [12, Corollary ]). This fact carries over to -trees since -trees are also CAT spaces. However, we note here that if is an -tree, then again boundedness of can be replaced by the assumption that is merely geodesically bounded. In fact, we prove the following. (In the following theorem, we assume is nonexpansive relative to the Hausdorff metric on the bounded nonempty closed subsets of
Suppose is a closed convex and geodesically bounded subset of a complete -tree and suppose is a nonexpansive mapping taking values in the family of nonempty bounded closed convex subsets of Suppose also that Then there is a point for which
Proof of Theorem 3.1.
Thus – a contradiction. Therefore, there exists such that
Suppose is a closed convex and geodesically bounded subset of a complete -tree and suppose is a nonexpansive mapping for which Then has a fixed point.
In view of the fact that continuous self-maps of have fixed points, it is natural to ask whether Corollary 3.3 holds for continuous mappings. The answer is no, even when is bounded. Let be the Euclidean plane with the radial metric. Let be a sequence of distinct points on the unit circle, and let We now define a continuous fixed-point free map for which . First move each point of the segment to the right onto a segment where and is on the ray which extends (Thus For each let denote the point on the segment which has distance from . It is now clearly possible to construct a continuous (even lipschitzian) fixed point-free map (a shift) of the segment onto the segment for which Thus for all
Corollary 3.3 for bounded is also a consequence of Theorem of .
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