NEW ZEALAND JOURNAL OF MATHEMATICS Volume 34 (2005), 25–29

HYPERCONVEX ULTRAMETRIC SPACES AND FIXED POINT THEORY

Sara Bouamama and Driss Misane (Received May 2002)

Abstract. This paper is inspired by the work of Sine and Soardi [9], [10]. We introduce the notion of hyperconvexity in generalized ultrametric spaces i.e. where the distance has values in an arbitrary partially ordered set. Then we give a fixed point theorem in these spaces for contracting or nonexpansive mappings which leads to some corollaries.

1. Introduction The hyperconvexity is a recent notion introduced by Aronsjazn and Panitchpakdi [1] in 1956. It was intensively studied for nonexpansive maps. In 1976, Sine and Soardi [9], [10] proved that fixed point property holds for nonexpansive mappings in bounded hyperconvex spaces. Recently, Jawhari , Misane and Pouzet [3] showed that Sine and Soardi Fixed Point Theorem is equivalent to the classical Tarski’s theorem via the notion of generalized metric spaces. In this paper, the notion of hyperconvexity is defined on generalized ultrametric spaces and is also successfully used in fixed point theory. Generalized ultrametric spaces are a common generalization of Partially ordered sets and ordinary ultrametric spaces, Lawrence 1973, Rutten 1995. The topology for generalized metric spaces extend both the Scott topology for algebraic complete partial orders and the Alexandroff topology for ordinary metric spaces reduced to the ε–ball topology. A new fixed point theorem holds for nonexpansive mappings and leads to an analogue of Banach fixed point theorem in ultrametric spaces . We discuss in this context the consequences which have been made so far of this theorem and its corollaries.

2. Hyperconvex Ultrametric Spaces Let X be a set and let (Γ, ≤) be a complete lattice with a least element 0 and a greatest element 1 (a complete lattice is an ordered set where each non–empty subset has a supremum and an infinimum). Let d : X ∗ X −→ Γ be a mapping which satisfies the following properties: For all x, y, z ∈ X and γ ∈ Γ: (1) d(x, y) = 0 if and only if x = y (2) d(x, y) = d(y, x) (3) if d(x, y) ≤ γ and d(y, z) ≤ γ, then d(x, z) ≤ γ.

1991 Mathematics Subject Classification 06F30, 46A16, 46A19. Key words and phrases: Ultrametric spaces, hyperconvexity, nonexpansive mappings, fixed points. 26 SARA BOUAMAMA and DRISS MISANE

The triple (X, d, Γ) is called an ultrametric space. Let 0 < γ ∈ Γ and a ∈ X. The set Bγ (a) = {x ∈ X; d(x, a) ≤ γ} is called a ball. We can also use the notation B(a, γ) instead of Bγ (a). We note the following property for an ultrametric space (X, d, Γ).

(1.1) If 0 6= α ≤ β and Bα(a) ∩ Bβ(b) 6= ∅, then Bα(a) ⊆ Bβ(b).

Proof. Let x ∈ Bα(a), then d(x, a) ≤ α. On the other hand, there exists y ∈ Bα(a) ∩ Bβ(b), hence d(y, a) ≤ α and d(y, b) ≤ β. Thus by (1), (2), d(x, y) ≤ α ≤ β and therefore d(x, b) ≤ β. So x ∈ Bβ(b). 

Let E a set. A family F = (Ei)i∈I of subsets of E has the 2–Helly’s property if 0 T 0 for all sub–family F = (Ei)1∈I0 the intersection of this family ( {Ei : i ∈ I } is 0 non–empty if and only if for all i, j ∈ I the intersections Ei ∩ Ej are not empties. In the same way, a set F of subsets of E has the 2–helly’s property if for all subset F 0, the intersection of subsets in F 0 is non–empty if and only if the intersection of two arbitrary subsets in F 0 is non-empty. In fact, F has the 2–helly’s property if and only if each subset of elements of F has the 2–helly’s property. Definition 2.1. Let (X, d, Γ) be an ultrametric space. X is called hyperconvex if it satisfies the two following properties:
(i) the set of blls has the 2–Helly’s property: ie For each family B(xi, γi) i∈I of closed balls of X: T If B(xi, γi) ∩ B(xj, γj) 6= ∅ for all i, j, then i∈I B(xi, γi) 6= ∅.

(ii) Convexity: For all x, y ∈ X and γ1, γ2 ∈ Γ: If d(x, y) ≤ sup(γ1, γ2), then there exists z ∈ X such that d(x, z) ≤ γ1 and d(z, y) ≤ γ2. Proposition 2.2. Let (X, d, Γ) be an ultrametric space. X is hyperconvex if and
only if each family of closed balls B(xi, γi) i∈I of X satisfies the property P: \ B(xi, γi) 6= ∅ if and only if d(xi, xj) ≤ sup(γi, γj) for all i, j ∈ I. i∈I Proof. We first show that: T B(xi, γi) 6= ∅ implies that d(xi, xj) ≤ sup(γi, γj) for all i, j ∈ I. i∈I T Let z ∈ i∈I B(xi, γi) then for all i, j ∈ I d(xi, z) ≤ γi and d(xj, z) ≤ γj and so d(xi, xj) ≤ sup(γi, γj).
Now, let X be a hyperconvex space and B(xi, γi) i∈I a family of balls of X with d(xi, xj) ≤ sup(γi, γj) for all i, j ∈ I, then for all i, j ∈ I there exists z ∈ X such T that d(xi, z) ≤ γi and d(xj, z) ≤ γj (from the convexity). Thus i∈I B(xi, γi) 6= ∅ from 2–Helly. Conversely, suppose that all the families of closed balls of X satisfy the property
P and let B(xi, γi) i∈I a family of balls of X with B(xi, γi) ∩ B(xj, γj) 6= ∅ for all i, j ∈ I. Let z ∈ B(xi, γi) ∩ B(xj, γj) then d(xi, z) ≤ γi and d(xj, z) ≤ γj. So T d(xi, xj) ≤ sup(γi, γj) for all i, j ∈ I and thus i∈I B(xi, γi) 6= ∅ from the property P. Let x, y ∈ X and γ1, γ2 ∈ Γ such that d(x, y) ≤ sup(γ1, γ2) then B(x, γ1) ∩ B(y, γ2) 6= ∅ from P. Let z ∈ B(x, γ1) ∩ B(y, γ2) then d(x, z) ≤ γ1 and d(z, y) ≤ γ2.  HYPERCONVEX ULTRAMETRIC SPACES AND FIXED POINT THEORY 27

Proposition 2.3. The non–empty intersection of closed balls of an hyperconvex space is hyperconvex.

Proof. Let (X, d, Γ) be an hyperconvex ultrametric space and let B = B(x, r) be a ball of X.

Let Br(xi, ri) i∈I a family of balls of B Br(xi, ri) = B ∩B(xi, ri) which satis- fies: d(xi, xj) ≤ sup(ri, rj). Let i ∈ I, we have Br(xi, ri) 6= ∅. Let yi ∈ Br(xi, ri), then B(yi, ri) = B(xi, ri) from (1.1). Then we can assume with- out loss of generality that all xi are in B. The balls B(xi, ri) satisfy the property T P in X hyperconvex, hence i∈I B(xi, ri) 6= ∅. Let y in this intersection, then d(y, xi) ≤ ri ≤ sup(r, ri) for all i ∈ I. Hence y ∈ B. That shows that B is hyperconvex. We will proceede in the similar way for the intersection of two balls and so one for an arbitrary family of balls. 

Let (X, d, Γ) be an ultrametric space and T : X −→ X a mapping. T is called nonexpansive or contracting if dT (x),T (y)
≤ d(x, y) for all x, y ∈ X. If for all x, y ∈ X, x 6= y, dT (x),T (y)
< d(x, y) then T is strictly contracting. Let x ∈ X, the orbit of x by T is the set {x, T (x),T 2(x),...}. T is called strictly contracting on orbits when for every x ∈ X such that T (x) 6= x, we have dT 2(x),T (x)
< dT (x), x
. We define B as the set of the nonempty intersections of closed balls of X and we denote A = T{B ∈ B; A ⊆ B} for all A ⊆ X and δ(A) = sup{d(x, y); x, y ∈ A} the diameter of A.

Theorem 2.4. Let (X, d, Γ) be a hyperconvex ultrametric space and T : X −→ X be a nonexpansive mapping, then T has a fixed point or there exists a nonempty hyperconvex subset S of X such that T (S) ⊆ S, S = T (S) and δ(S) = dx, T (x)
for all x ∈ S.

Proof. Suppose that T doesn’t have a fixed point. Let BT = {B ∈ B; T (B) ⊆ B}. BT 6= ∅ as X = B(x, 1) for all x ∈ X, then X ∈ BT . Let us show that BT has a minimal element. From Zorn’s lemma, it suffies to show that BT is inductively ordered or equivalently that each chain of BT has a lower bound (a chain is a family of elements of an ordered set such that each two elements of the chain are comparative). Let (B ) a chain of B B = T B(xk, γk)
. k k∈K T k i∈Ik i i k k k k Let k ∈ K. Bk 6= ∅. Let z ∈ Bk then d(xi , z) ≤ γi and d(xj , z) ≤ γj for all k k k k i, j ∈ Ik. Thus d(xi , xj ) ≤ sup(γi , γj ) for all i, j ∈ Ik.
Let k1, k2 ∈ K (k1 6= k2); Suppose that Bk1 ⊆ Bk2 as (Bk) is a chain . Let k1 k1 k2 k2 z ∈ Bk1 then d(xi , z) ≤ γi and d(z, xj ) ≤ γj for all i ∈ Ik1 and j ∈ Ik2 . So k1 k2 k1 k2 d(xi , xj ) ≤ sup(γi , γj ). We conclude that the family B(xk, γk)
satisfies the property P given i i k∈K,i∈Ik in Proposition 2.2 in X hyperconvex. Thus T B = T T B(xk, γk) 6= k∈K k k∈K i∈Ik i i T T T k k T T k k
∅ and T ( k∈K Bk) = T ( k∈K i∈I B(xi , γi ) ⊆ k∈K T i∈I B(xi , γi ) = T T k T k k∈K T (Bk) ⊆ k∈K Bk as all the Bk are in BT , then k∈K Bk ∈ BT . Thus from Zorn’s lemma, BT has a minimal element S.  28 SARA BOUAMAMA and DRISS MISANE

S ∈ BT then T (S) ⊆ S from the definition of BT . So T (S) ⊆ S. It follows that
T T (S) ⊆ T (S), then T (S) ∈ BT . Thus T (S) = S as S is minimal. T S ∈ BT then S can be written as S = k∈K B(xk, γk), so S is hyperconvex as an intersection of balls in an hyperconvex space. Let x ∈ S, then d(x, xk) ≤ γk
for all k ∈ K. T (x) ∈ S, then d T (x), x ≤ γk as d is ultrametric. We take

Bx = B x, d x, T (x) . T (x) 6= x, then Bx ⊆ B(xk, γk) for all k ∈ K from (1.1)

and so Bx ⊆ S. Let y ∈ Bx then d T (y),T (x) ≤ d(y, x) ≤ d x, T (x) as T is nonexpansive and so dT (y), x
≤ dx, T (x)
as d is ultrametric. Then we have T (Bx) ⊆ Bx and so Bx ∈ BT . Thus S = Bx as S is minimal in BT . Let y, z ∈ S then d(y, x) ≤ dx, T (x)
and d(z, x) ≤ dx, T (x)
. So d(y, z) ≤ dx, T (x)
. It follows that δ(S) = sup{d(y, z); y, z ∈ S} ≤ dx, T (x)
≤ δ(S). From the Theorem 2.4, we obtain the following corollary: Corollary 2.5. Let (X, d, Γ) be a hyperconvex ultrametric space and T : X −→ X be a nonexpansive mapping. Assume furthermore that for every x ∈ X with T (x) 6= x, there exists an integer n ≥ 0 such that dT n+2(x),T n+1(x)
< dT n+1(x),T n(x)
we take T 0(x) = x
. Then T has a fixed point and the set of fixed points of T is hyperconvex. Proof. We can deduce easily from the theorem 2.4 that T has a fixed point as we can’t find such an hyperconvex subset S of X. Let Fix(T ) be the set of fixed
points of T and let BFix(T )(xi, γi) i∈I a family of balls of Fix(T ) with d(xi, xj) ≤
sup(γi, γj) for all i, j ∈ I BFix(T )(xi, γi) = B(xi, γi)∩Fix(T ) . We have B(xi, γi)∩ Fix(T ) 6= ∅. Let yi ∈ B(xi, γi) ∩ Fix(T ), so B(yi, γi) = B(xi, γi) from (1.1). Then, we can suppose without loss of generality that xi ∈ Fix(T ) for all i ∈ I. The balls B(xi, γi) satisfy the property P given in Proposition 2.2 on X hyperconvex, T then F = {B(xi, γi)/i ∈ I}= 6 ∅ and as an intersection of balls of a hyperconvex space, F is hyperconvex, moreover T (F ) ⊆ F since xi ∈ Fix(T ), so the theorem 2.4 applied to F gives an x ∈ Fix(T ) ∩ F . Thus the last intersection is not empty , and Fix(T ) is hyperconvex.  This result generalizes the following theorem. Theorem 2.6. Let (X, d, Γ) be a hyperconvex ultrametric space and let T : X −→ X be contracting and strictly contracting on orbits. Then T has a fixed point. If moreover T is strictly contracting on X, then the fixed point is unique

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Sara Bouamama Driss Misane Universit´emohammed V Universit´emohammed V Facult´edes sciences Facult´edes sciences D´epartement de Math´ematiqueset D´epartement de Math´ematiqueset Informatique Informatique Rabat Rabat MOROCCO MOROCCO [email protected] [email protected]