A convergence result on random products of mappings in metric trees

Abstract

Let X be a metric space and {T₁, ..., T _N } be a finite family of mappings defined on D ⊂ X. Let r : ℕ → {1,..., N} be a map that assumes every value infinitely often. The purpose of this article is to establish the convergence of the sequence (x _n ) defined by

$$x_0\in D;\mathsf{\text{and}}{x}_{n+1}=T_{r\left(n\right)}\left(x_n\right),\mathsf{\text{for}}\mathsf{\text{all}}n\ge 0.$$

In particular we prove Amemiya and Ando's theorem in metric trees without compactness assumption. This is the first attempt done in metric spaces. These type of methods have been used in areas like computerized tomography and signal processing.

Mathematics Subject Classification 2000: Primary: 06F30; 46B20; 47E10.

1. Introduction

Many problems in mathematics [1] and physical sciences [2–4] uses a techniques known as search for a common fixed point. Indeed, let X be a metric space and suppose T₁,..., T _N are pairwise distinct self-mappings of some nonempty and closed subset D of X. Suppose further that the fixed point set, Fix(T _i ) = {x ∈ D; T _i (x) = x}, of each mapping T _i is nonempty and that $C=\mathsf{\text{Fix}}\left(T_1\right)\cap \cdots \cap \mathsf{\text{Fix}}\left(T_N\right)\ne \varnothing$. The aim is to find a common fixed point of these mappings. One frequently employed approach is the following.

Let r be a random mapping for {1,..., N}, i.e., a surjective mapping from ℕ onto {1,...,N} that takes each value in {1,...,N} infinitely often. Then generate a random sequence (x _n ) by x₀ ∈ D arbitrary, and

$$x_{n+1}=T_{r\left(n\right)}\left(x_n\right),foralln>0,$$

and hope that this sequence converges to a point in C. We call it a random or unrestricted product (resp. iteration). For products generated by using control sequence, there are many results: for instance, cyclic control arises when r(n) = n + 1 mod N (see, for example, [5]).

In general, this random product fails to have well convergence behavior. The first positive results were done in the case when D = X is a Hilbert space and each mapping T _i , is the projection onto some nonempty, closed and convex subset C _i of X ; hence Fix(T _i ) = C _i , i = 1, ..., N. The problem of finding a common fixed point is then the well-known convex feasibility problem (see, for example, [5]). Combettes article [6] proposed several interesting applications of this problem. Some of the early known results in this case are.

• (1) Amemiya and Ando [7]: If each set C _i , is a closed subspace, then the random product converges weakly to the projection onto C.

• (2) Bruck [8]: If some set C _i , is compact, then the random product converges in norm to a point in C. If N = 3 and each set C _i , is symmetric, then the random product converges weakly to a point in C (see also [[9], Theorem 2]).

• (3) Aharoni and Censor [10], Flam and Zowe [11], Tseng [12], Eisner et al. [13]: If X is finite dimensional, then the random product converges in norm to a point in C.

The authors in [14] were successful in their extension of Amemiya and Ando's [7] results from Hilbert spaces to Banach spaces. In this study, we investigate Amemiya and Ando's result and [14]'s results in metric spaces. Such extension is the first attempt so far.

The main difficulty faced in such extensions is the heavy use of the linearity structure of either the Hilbert space in [7] or Banach spaces in [14]. Indeed when one tries to extend concepts from linear functional analysis, one has to pay attention to look deep into the supporting basic ideas and what intrinsic interrelations exist between them. Most of the main theorems in nonlinear functional analysis were done in the framework of linear Banach spaces. So it was interesting to investigate the extension of these fundamental results in nonlinear structures like metric spaces. As an example of this research is Kirk's fixed point theorem [15]. Many researchers have tried to do it but the best approach is the one given by Penot [16]. The impact of this approach went beyond what it intended to do initially. This research follows the same motivations. In particular we investigate the concept of weak convergence in metric spaces which is central, for instance, in [14]. We consider the case of metric trees to illustrate some of these ideas.

2. Basic definitions and results

Metric trees were first introduced by Tits [17] in 1977. A metric tree is a metric space (M, d) such that for every x, y in M there is a unique arc between x and y and this arc is isometric to an interval in ℝ. For example, a connected graph without loop is a metric tree. One basic property of metric trees is their one dimensionality. Again in the late seventies, while studying t-RNA molecules of the E. Coli bacterium Eigen raised several questions which led Dress [18, 19] to construct metric trees, (named as T-theory). Metric trees also arise naturally in the study of group isometries of hyperbolic spaces. For metric properties of trees we refer to [20].

Since a metric tree is a space in which there is only one path between two points x and y, this would imply that if z is a point between x and y, by which we mean if d(x, z) + d(z, y) = d(x, y) then we know that z is actually on the path between x and y. This will motivate the next concept of a metric interval.

Definition 2.1. A metric interval or metric segment [x, y] in a metric space M is defined by

$$\left[x,y\right]:=\left\{z\in M:d\left(x,z\right)+d\left(z,y\right)=d\left(x,y\right)\right\}.$$

First let us give the definition of a metric tree.

Definition 2.2. A metric tree is a nonempty metric space M satisfying:

• (a) Any two points x and y in M are the endpoints of a metric segment [x, y].

• (b) If x,y, z ∈ M then [x, y]∩[x, z] = [x, w] for some w ∈ M. (i.e., if we have two metric segments with a common endpoint, then their intersection is a metric segment.)

• (c) If x, y, z ∈ M and [x, y] ∩ [y, z] = {y} then [x, y] ∪ [y, z] = [x, z] (i.e., If two metric segments intersect in a single point, then their union is a metric segment.)

Metric trees are very special. They enjoy properties which are shared by l^∞ and Hilbert space. In particular, Kirk [21] showed that complete metric trees are hyperconvex. Since the weak topology has an intimate relationship with convexity, let us define convex subset in this setting.

Definition 2.3. Let M be a metric tree and C ⊂ M. We say that C is convex if for all x, y ∈ C we have [x, y] ⊂ C.

Clearly a metric tree M and the empty set ∅ are convex. Also any closed ball B(a, r) = {z ∈ M: d(a, z) ≤ r} in a metric tree is also convex. Let $\mathcal{C}\left(M\right)$ denotes the collection of all closed and convex subsets of M, we set:

$$\mathsf{\text{conv}}\left(A\right)=\cap \left\{B:B\mathsf{\text{is}}\mathsf{\text{a}}\mathsf{\text{convex}}\mathsf{\text{subset}}\mathsf{\text{of}}M\mathsf{\text{such}}\mathsf{\text{that}}A\subseteq B\right\}.$$

Note that $\mathcal{C}\left(M\right)$ is invariant by intersection, i.e. the intersection of any family of convex subsets of M is convex. We need the following result of Baillon [22] in order to prove our first fact about $\mathcal{C}\left(M\right)$.

Theorem 2.1. [22]Let M be a bounded metric space and let {H _β }_β∈Γbe a decreasing family of nonempty hyperconvex subsets of M then$\bigcap _{\beta \in \Gamma }{H}_{\beta }\ne \varnothing$and is hyperconvex.

Since convex subsets of a metric tree are metric trees, then they are hyperconvex by [21]. This combined with Baillon's result we get the following theorem.

Theorem 2.2. Let M be a bounded complete metric tree and let {C _β }_β∈Γbe a family of nonempty, closed and convex subsets of M such that$\bigcap _{\beta \in {\Gamma }_f}{C}_{\beta }\ne \varnothing$, where Γ _f is any finite subset of Γ, then$\bigcap _{\beta \in \Gamma }{C}_{\beta }\ne \varnothing$and is convex.

This is known as compactness of $\mathcal{C}\left(M\right)$ according to Penot's formulation [16]. Note the slight difference between the statements of the two theorems. Indeed the intersection of two convex sets is convex while the intersection of two hyperconvex sets may not be hyperconvex.

Next we discuss the nearest point projections in metric trees. Let C be a nonempty, closed and convex subset of a complete metric tree M. For any x ∈ M, denote

$$P_C\left(x\right)=\left\{c\in C;d\left(x,c\right)=\mathsf{\text{dist}}\left(x,C\right)=\underset{y\in C}{\mathsf{\text{inf}}}d\left(x,y\right)\right\}.$$

In a Hilbert space, the metric projections on closed and convex subsets are nonexpansive. In uniformly convex spaces, the metric projections are uniformly Lipschitzian. In fact, they are nonexpansive if and only if the space is Hilbert. In what follows we will show that the metric projections in metric trees are nonexpansive. This result is not true in hyperconvex metric spaces.

Lemma 2.1. [23, 24]If C is a nonempty, closed and convex subset of a complete metric tree M, then for any x ∈ M there exists a unique c _x ∈ C such that dist(x,C) = d(x,c _x ), which means that P _C is single valued. Moreover if c ∈ C we have

$$d\left(x,P_C\left(x\right)\right)+d\left(P_C\left(x\right),c\right)=d\left(x,c\right),$$

and

$$d\left(P_C\left(x\right),P_C\left(y\right)\right)=d\left(x,y\right)-d\left(x,P_C\left(x\right)\right)-d\left(y,P_C\left(y\right)\right)$$

or P _C (x) = P _C (y), for any x, y ∈ M. In particular, P _C is nonexpansive.

Next we prove another property of the mapping P _C .

Proposition 2.1. If C is a nonempty, closed and convex subset of a complete metric tree M, then for any x ∈ M we have

$$P_C\left(\left[x,P_C\left(x\right)\right]\right)=\left\{P_C\left(x\right)\right\}.$$

In other words, P _C is a sunny nonexpansive mapping[25, 26].

Proof. Let x ∈ M and y ∈ [x,P _C (x)]. Since M is a metric tree, there exists w ∈ M such that [y, P _C (x)] ∩ [y, P _C (y)] = [y, w]. We have w ∈ [P _C (x), P _C (y)]. Since C is convex, we get that w ∈ C. Also the definition of w implies

$$d\left(y,w\right)+d\left(w,P_C\left(y\right)\right)=d\left(y,P_C\left(y\right)\right).$$

The properties of P _C will force P _C (y) = w which will imply P _C (y) ∈ [y, P _C (x)]. Since y ∈ [x, P _C (x)] we get

$$d\left(x,P_C\left(x\right)\right)=d\left(x,y\right)+d\left(y,P_C\left(x\right)\right)=d\left(x,y\right)+d\left(y,P_C\left(y\right)\right)+d\left(P_C\left(y\right),P_C\left(x\right)\right)$$

In particular we get d(x, P _C (y)) ≤ d(x, P _C (x)) which implies P _C (y) = P _C (x).

3. Amemiya and Ando's theorem in metric trees

In 1965 Amemiya and Ando [7] proved the astonishing result.

Theorem 3.1. [7]Let H be a Hilbert space and {P₁,..., P _N } be a finite family of orthogonal linear projections defined on H. Let r : ℕ → {1,..., N} be a map that assumes every value infinitely often. The sequence (x _n ) defined by

$$x_0\in H,and{x}_{n+1}=P_{r\left(n\right)}\left(x_n\right),foralln\ge 0.$$

converges weakly in H.

Today, 46 years later, it is still not known whether (x _n ) converges strongly, even for N = 3. There is doubt expressed in the literature as to whether this sequence does converge strongly (cf. [[27], Example 4]) for an interesting example of possible relevance. In general, strong convergence may be obtained when some kind of compactness is assumed. Next, we show that in the case of metric trees, we have strong convergence without any compactness assumption. The Amemiya and Ando's theorem was preceded by von Neumann [28] for alternating products of two projections (with strong convergence as the conclusion).

Theorem 3.2. Let C₁, ..., C _N be a finite family of nonempty, closed and convex subsets of a complete metric tree M such that$C=\bigcap _{1\le i\le N}{C}_i\ne \varnothing$. Let r : ℕ → {1, ..., N} be a map that assumes every value infinitely often. The sequence (x _n ) defined by

$$x_0\in X,and{x}_{n+1}=P_{C_{r\left(n+1\right)}}\left(x_n\right),foralln\ge 0.$$

converges strongly in M. Moreover we have

$$\underset{n\to \infty }{\mathsf{\text{lim}}}{x}_n=\underset{n\to \infty }{\mathsf{\text{lim}}}{P}_{C_{r\left(n+1\right)}}\circ \cdots \circ P_{C_{r\left(1\right)}}\left(x_0\right)=P_C\left(x_0\right).$$

Fix

Proof. c ∈ C. Using Lemma 2.1 we have

$$d\left(x_n,x_{n+1}\right)=d\left(x_n,c\right)-d\left(x_{n+1},c\right),$$

for any n ≥ 0. In particular we have

$$\sum _{k=n}^{k=n+h}d\left(x_k,x_{k+1}\right)=d\left(x_n,c\right)-d\left(x_{n+h},c\right),$$

for any n ≥ 0 and h ≥ 0. Since $P_{C_{r\left(n\right)}}\left(c\right)=c$, we get d(x_n+1,c) ≤ d(x _n , c), for any n ≥ 0. In other words, the sequence (d(x _n ,c)) is decreasing. Hence $\underset{n\to \infty }{\mathsf{\text{lim}}}d\left(x_n,c\right)$ exists. Therefore the sequence (x _n ) is Cauchy. Since M is complete, there exists c ∈ M such that $\underset{n\to \infty }{\mathsf{\text{lim}}}{x}_n=c$. For any i ∈ {1,..., N}, there exists a subsequence of (x _n ) which belongs to C _i . Since C _i is closed, we conclude that c ∈ C _i , for any i ∈ {1,..., N}. Hence c ∈ C. Next we show that c = P _C (x₀). For any n ≥ 0, we get

$$d\left(x_0,x_n\right)\le \sum _{k=0}^{k=n}d\left(x_k,x_{k+1}\right)=d\left(x_0,c_0\right)-d\left(x_n,c_0\right).$$

If we let n → ∞, we get

$$d\left(x_0,c\right)\le d\left(x_0,c_0\right)-d\left(c,c_0\right).$$

The definition of c₀ implies

$$d\left(x_0,c\right)\le d\left(x_0,c_0\right)-d\left(c,c_0\right)\le d\left(x_0,c\right)-d\left(c,c_0\right)$$

which implies d(c, c₀) = 0, or c = c₀.

Remark 3.1. In[14]the authors made heavy use of the property that in smooth reflexive Banach spaces X, if E is a closed subspace of X, then there is at most one nonexpansive retraction of X onto E[26]. In the case of metric trees, we have a similar result. Indeed, let C be a nonempty, closed and convex subset of a metric tree M. Then P _C is a sunny nonexpansive retract of M onto C. Let Q : M → C be another sunny mapping. Let x ∈ M. There exists w ∈ M such that [x, P _C (x)] ∩ [x, Q(x)] = [x, w]. Since C is convex, then w ∈ C. Also since d(x,w) + d(w, P _C (x)) = d(x, P _C (x)), the definition of P _C (x) will force P _C (x) = w. Hence P _C (x) ∈ [x, Q(x)]. Since Q is sunny, we must have Q(P _C (x)) = Q(x), which implies P _C (x) = Q(x). In other words, P _C is the only sunny retract from M onto C.

In the next section we investigate the behavior of the random product of mappings other than the nearest point projections.

4. Random product of mappings in metric trees

As the authors did in [14], one inspires itself from the Amemiya and Ando's work in Hilbert spaces to extend it to other underlying spaces. In particular the authors in [14] introduced the concepts of (W) and (S) properties. Since the (W) property is strongly linked to the weak-topology, we are not able to extend such property to metric trees.

Definition 4.1. Let M be a metric space. Let T : M → M be a nonexpansive map with a nonempty fixed point set Fix(T). We will say that T satisfies the property (S) if and only if for any c ∈ Fix(T) and any sequence (v _n ) such that$\underset{n\to \infty }{\mathsf{\text{lim}}}\left[d\left(v_n,c\right)-d\left(T\left(v_n\right),c\right)\right]=0$, we have$\underset{n\to \infty }{\mathsf{\text{lim}}}d\left(v_n,T\left(v_n\right)\right)=0$.

Remark 4.1. Note that if T₁,...,T _N are nonexpansive mappings with a common fixed point and satisfy the property (S), then we have

$$\mathsf{\text{Fix}}\left(T_1\circ \cdots \circ T_N\right)=\bigcap _{1\le i\le N}\mathsf{\text{Fix}}\left(T_i\right).$$

Indeed let c₀ ∈ M be a common fixed point of T₁, ...,T _N . Let us only prove that$Fix\left(T_1\circ \cdots \circ T_N\right)\subset \bigcap _{1\le i\le N}Fix\left(T_i\right)$. Let c ∈ Fix(T₁ ○ ⋯ ○ T _N ). Then we have T₁ ○ ⋯ ○ T _N (c) = c.

Since each mapping is nonexpansive, we get

$$d\left(c_0,c\right)=d\left(c_0,T_1\circ \cdots \circ T_N\left(c\right)\right)\le d\left(c_0,T_N\left(c\right)\right)\le d\left(c_0,c\right).$$

Since T _N satisfies the property (S), we get T _N (c) = c. Similarly one will show that T _i (c) = c, for i = 1,...,N.

Another property discovered by Caristi [29] (see also [30]) and extensively used to obtain some beautiful results extending Banach contraction principle is the following definition.

Definition 4.2. Let M be a metric space. Let T : M → M be a mapping. We will say that T satisfies the property (C)-λ if and only if there exists a map λ : M → [0, ∞) such that

$$d\left(x,T\left(x\right)\right)\le \lambda \left(x\right)-\lambda \left(T\left(x\right)\right),$$

for any x ∈ M.

It is easy to check that if T satisfies the (C)-λ property, then any orbit (Tⁿ(x)) is a Cauchy sequence for any x ∈ M. In particular if M is complete and T is continuous, then $P\left(x\right)=\underset{n\to \infty }{\mathsf{\text{lim}}}{T}^n\left(x\right)$ is a retraction from M into Fix(T) which is nonempty.

Example 4.1. Let M = [0,2] is a metric tree being an interval of the metric tree ℝ. Define the mapping T : [0,2] → [0,1] by

$$T\left(x\right)=\left\{\begin{array}{ccc}\hfill x\hfill & \hfill if\hfill & \hfill 0\le x\le 1\hfill \\ \hfill 2-x\hfill & \hfill if\hfill & \hfill 1\le x\le 2.\hfill \end{array}\right.$$

Note that T is a nonexpansive retraction and Fix(T) = [0,1]. In particular (Tⁿ(x)) is convergent and its limit is T(x). But the nearest point projection on [0,1] is the map

$$P\left(x\right)=\left\{\begin{array}{ccc}\hfill x\hfill & \hfill if\hfill & \hfill 0\le x\le 1\hfill \\ \hfill 1\hfill & \hfill if\hfill & \hfill 1\le x\le 2\hfill \end{array}\right.$$

which is different from T. Moreover one can easily show that

$$d\left(x,T\left(x\right)\right)\le \lambda \left(x\right)-\lambda \left(T\left(x\right)\right)$$

for any x ∈ [0,2], where λ(t) = t. Therefore T satisfies the (C)-λ property.

In the next result, we show how Theorem 3.2 extends to the family of mappings satisfying the (C)-λ property.

Theorem 4.1. Let M be a complete metric space. Let T₁,...,T _N be a finite family of pairwise distinct self-mappings of some nonempty and closed subset D of M. Suppose further that each map T _i , i = 1,..., N, is continuous and satisfies the (C)-λ property, with the same function λ. Let r be a random mapping for {1,...,N}, i.e., a surjective mapping from ℕ onto {1,...,N} that takes each value in{1,...,N} infinitely often. Then the random sequence (x _n ) defined by x₀ ∈ D arbitrary, and

$$x_{n+1}=T_{r\left(n\right)}\left(x_n\right),foralln>0,$$

is convergent. Its limit is a common fixed point of the mappings T₁,...,T _N .

Proof. Let x ∈ D. Our assumptions on the mappings T _i imply

$$d\left(x_n,x_{n+1}\right)\le \lambda \left(x_n\right)-\lambda \left(x_{n+1}\right),$$

for any n ≥ 0. In particular we have

$$\sum _{k=n}^{k=n+h}d\left(x_k,x_{k+1}\right)\le \lambda \left(x_n\right)-\lambda \left(x_{n+h}\right),$$

for any n ≥ 0 and h ≥ 0. On the other hand, we have λ(x_n+1) ≤ λ(x _n ), for any n ≥ 0. Therefore the positive sequence (λ(x _n )) is convergent. Clearly this will imply that the sequence (x _n ) is Cauchy. Since M is complete, there exists c ∈ M such that $\underset{n\to \infty }{\mathsf{\text{lim}}}{x}_n=c\in D$ since D is closed. For any i ∈ {1,..., N}, there exists a subsequence of (x_ϕ(n)) such that x_ϕ(n)+1= T _i (x_ϕ(n)). Hence T _i (c) = c, for any i = 1,..., N.

Note that the limit defines a retraction on the common fixed point set of the mappings T₁,...,T _N . But this retraction may not be equal to the nearest point projection even in the case of a metric tree as the Example 4.1 shows.

The next result investigates the extension of some of the results discovered in [14]. Before we do this, we need to discuss the weak-topology in the nonlinear setting of metric spaces. Indeed, let (x _n ) be a bounded sequence in the metric tree M. Define the real-valued function

$${\phi }_{\mathcal{U}}\left(x\right)=\underset{n,\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,x\right)$$

where is a nontrivial ultrafilter [31]. We have the following theorem which will play a central role in our work.

Theorem 4.2. Let M be a complete metric tree. Let (x _n ) be a bounded sequence in M. Then for any nontrivial ultrafilter, there exists a unique$z_{\mathcal{U}}\in M$such that

$${\phi }_{\mathcal{U}}\left(x\right)={\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right)+d\left(x,z_{\mathcal{U}}\right),$$

for any x ∈ M.

Proof. Let

$$r=\mathsf{\text{inf}}\left\{{\phi }_{\mathcal{U}}\left(x\right);x\in M\right\}.$$

For any ε > 0, consider the set

$$C_{\epsilon }=\left\{x\in M;{\phi }_{\mathcal{U}}\left(x\right)\le r+\epsilon \right\}.$$

Using the properties of metric trees, we know that C _ε is a nonempty, bounded and convex subset of M. Since ${\phi }_{\mathcal{U}}$ is continuous, then it is also closed. Using the compactness of $\mathcal{C}\left(M\right)$, then

$$\bigcap _{\epsilon >0}{C}_{\epsilon }=\left\{x\in M;{\phi }_{\mathcal{U}}\left(x\right)=r\right\}\ne \varnothing .$$

Now, we will show that this intersection is reduced to one point. Indeed, let us Fix $z_{\mathcal{U}}\in M$ such that ${\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right)=r$. Let x be any point in M. Using the properties of metric trees, for any n ≥ 1, there exists $w_n\in \left[x,z_{\mathcal{U}}\right]$ such that $\left[x_n,x\right]\cap \left[x_n,z_{\mathcal{U}}\right]=\left[x_n,w_n\right]$. Since $\left[x,z_{\mathcal{U}}\right]$ is compact, then there exists $w\in \left[x,z_{\mathcal{U}}\right]$ such that $\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(w_n,w\right)=0$. Since $d\left(x_n,w_n\right)+d\left(w_n,z_{\mathcal{U}}\right)=d\left(x_n,z_{\mathcal{U}}\right)$ for any n ≥ 1 then we have

$$\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,w_n\right)+\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(w_n,z_{\mathcal{U}}\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,z_{\mathcal{U}}\right)={\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right).$$

Hence

$$\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,w\right)+d\left(w,z_{\mathcal{U}}\right)={\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right),$$

that is, ${\phi }_{\mathcal{U}}\left(w\right)+d\left(w,z_{\mathcal{U}}\right)={\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right)$. Obviously this will imply $d\left(w,z_{\mathcal{U}}\right)=0$ or $\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(w_n,z_{\mathcal{U}}\right)=0$. Also since d(x _n , w _n ) + d(w _n ,x) = d(x _n ,x) for any n ≥ 1, then we have

$$\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,w_n\right)+\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(w_n,x\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,z_{\mathcal{U}}\right)={\phi }_{\mathcal{U}}\left(x\right).$$

Hence

$$\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_n,z_{\mathcal{U}}\right)+d\left(z_{\mathcal{U}},x\right)={\phi }_{\mathcal{U}}\left(x\right),$$

that is, ${\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right)+d\left(z_{\mathcal{U}},x\right)={\phi }_{\mathcal{U}}\left(x\right)$. This latest identity, also known as Uniform Opial condition, will easily show that $z_{\mathcal{U}}$ is unique.

Definition 4.3. Let M be a metric tree and (x _n ) be a bounded sequence in M. For any nontrivial ultrafilter, the unique point$z_{\mathcal{U}}$found in Theorem 4.2 is called the weak-limit of (x _n ) along. We will say that (x _n ) is weakly convergent if and only if$z_{\mathcal{U}}=z_{\mathcal{V}}$, for any nontrivial ultrafiltersand.

It is because of the absence of a dual space that we used Opial behavior to try to catch the weak-limit of a bounded sequence. In the next result we show some close similarities between the classical weak-limit point in Banach spaces and the one introduced above.

Proposition 4.1. Let M be a complete metric tree and (x _n ) be a bounded sequence in M. For any nontrivial ultrafilter, then

$$z_{\mathcal{U}}\in \Omega \left(x_n\right)=\bigcap _{n\ge 1}\overline{\mathsf{\text{conv}}}\left\{x_n,x_{n+1},\dots \right\}.$$

Fix

Proof. i ≥ 1. Set P _n the nearest point projection on $\overline{\mathsf{\text{conv}}}{\left(x_i\right)}_{i\ge n}$. Since is nontrivial then

$$\left\{i;d\left(x_i,P_n\left(z_{\mathcal{U}}\right)\right)\le d\left(x_i,z_{\mathcal{U}}\right)\right\}\in \mathcal{U}$$

where we used the nonexpansiveness of P _n . This obviously implies

$$\underset{i,\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_i,P_n\left(z_{\mathcal{U}}\right)\right)\le \underset{i,\mathcal{U}}{\mathsf{\text{lim}}}d\left(x_i,z_{\mathcal{U}}\right)$$

which implies $P_n\left(z_{\mathcal{U}}\right)=z_{\mathcal{U}}$ or $z_{\mathcal{U}}\in \overline{\mathsf{\text{con}}}{\left(x_i\right)}_{i\ge n}$ for any n ≥ 1. So $z_{\mathcal{U}}\in \Omega \left(x_n\right)$.

Next we discuss the behavior of mapping which satisfies the property (S).

Theorem 4.3. Let M be a complete metric tree. Let T : C → C be a nonexpansive mapping which satisfies the property (S), where C is a nonempty, bounded, closed, and convex subset of M. Then the sequence (Tⁿ(x)) converges weakly to a fixed point of T.

Proof. Let and be any nontrivial ultrafilters. Let $z_{\mathcal{U}}$ and $z_{\mathcal{V}}$ be the minimum point of ${\phi }_{\mathcal{U}}\left(z\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z\right)$ and ${\phi }_{\mathcal{V}}\left(z\right)=\underset{\mathcal{V}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z\right)$, respectively. Proposition 4.1 implies that $z_{\mathcal{U}}$ and $z_{\mathcal{V}}$ are in Ω(Tⁿ(x)) ⊂ C. Next we will prove that $z_{\mathcal{U}}$ and $z_{\mathcal{V}}$ are fixed point of T. It is enough to prove that $T\left(z_{\mathcal{U}}\right)=z_{\mathcal{U}}$. Since C is hyperconvex and bounded, we know that T has a nonempty fixed point set (see [32, 33]). Let c ∈ Fix(T). The sequence (d(Tⁿ(x),c)) is a decreasing sequence of positive numbers. Since T satisfies the property (S), we deduce that $\underset{n\to \infty }{\mathsf{\text{lim}}}d\left(T^n\left(x\right),T^{n+1}\left(x\right)\right)=0$. Hence

$${\phi }_{\mathcal{U}}\left(z\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(T^{n+1}\left(x\right),z\right),$$

which implies ${\phi }_{\mathcal{U}}\left(T\left(z\right)\right)\le {\phi }_{\mathcal{U}}\left(z\right)$, for any z ∈ C. The properties of $z_{\mathcal{U}}$ will force the identity $T\left(z_{\mathcal{U}}\right)=z_{\mathcal{U}}$, i.e., $z_{\mathcal{U}}\in \mathsf{\text{Fix}}\left(T\right)$. Note that the sequence $\left(d\left(T^n\left(x\right),z_{\mathcal{U}}\right)\right)$ is decreasing which implies

$${\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{U}}\right)=\underset{n\to \infty }{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{U}}\right)=\underset{n\ge 1}{\mathsf{\text{inf}}}d\left(T^n\left(x\right),z_{\mathcal{U}}\right).$$

Hence

$${\phi }_{\mathcal{U}}\left(z_{\mathcal{V}}\right)=\underset{\mathcal{U}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{V}}\right)=\underset{n\to \infty }{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{V}}\right)=\underset{\mathcal{V}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{V}}\right),$$

because z _V ∈ Fix(T). Since

$${\phi }_{\mathcal{V}}\left(z_{\mathcal{V}}\right)=\underset{\mathcal{V}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{V}}\right)\le {\phi }_{\mathcal{V}}\left(z_{\mathcal{U}}\right)=\underset{\mathcal{V}}{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{U}}\right)=\underset{n\to \infty }{\mathsf{\text{lim}}}d\left(T^n\left(x\right),z_{\mathcal{U}}\right),$$

which implies ${\phi }_{\mathcal{U}}\left(z_{\mathcal{V}}\right)\le {\phi }_{\mathcal{U}}\left(z_{\mathcal{U}}\right)$. The properties of $z_{\mathcal{U}}$ imply $z_{\mathcal{V}}=z_{\mathcal{U}}$. This completes the proof of Theorem 4.3.