On convergence of infinite products of convex combinations of mappings in CAT(0) spaces

Abstract

We study the weak convergence of infinite products of convex combinations of operators in complete CAT(0) spaces. We provide a new approach to this problem by considering a constructive selection of convex combinations in CAT(0) spaces that does not depend on the order of the involved elements and retain continuity properties with respect to them.

1 Introduction

A problem with numerous applications to mathematics and other sciences consists in finding a common fixed point to a given family of mappings. This problem has been extensively studied in the context of Hilbert spaces, where, given a collection of nonexpansive mappings $T_1, T_2,\ldots ,T_n$ defined from a closed convex and nonempty subset of the Hilbert space into itself, common fixed points of these mappings have been found through iterative procedures involving products of them. Some initial references for this topic are [1, 7, 9, 10].

CAT(0) are the natural nonlinear version of Hilbert spaces; see, for instance, [6, 8]. Therefore, it is natural to wonder about these results on existence of common fixed points when the ambient space is a CAT(0) space. This has already been done by several authors. In Ref. [18], Reich and Salinas gave a collection of results on weak convergence of infinite products of operators in complete CAT(0) spaces in a work following an earlier one by the same authors [17]. These results find their origin in a work from Bac̆ák et al. [4] where the authors proved that, for each $x\in X$, where X is a complete CAT(0) space, $B_1$ and $B_2$ two nonempty, closed, and convex subsets of X and $P_{B_1}$ and $P_{B_2}$ the corresponding nearest point projections, the sequence $((P_{B_1}P_{B_2})^n x)_n$ is weakly convergent to a point in $B_1\cap B_2$ depending on x. This result was extended to more than two nearest point projections in a larger class of metric spaces by Ariza-Ruiz et al. in Ref. [3]. In the last section of [18], the authors extended their own results from previous sections to infinite products of convex combinations of mappings.

Dealing with convex combinations in CAT(0) is not an easy task. The convex combination of two points is always clear as geodesic are unique in CAT(0) spaces, but once we consider more than two points the situation dramatically changes. In particular, when paralleling the usual convex combination of a finite collection of points in a linear space, the convex combination may depend on the order we align the elements involved in it and so, what we think as a same convex combination may find several representations. In this way, we lack any possible property about continuity of the convex combination with respect to coefficients or points at least we keep some sort of control on the procedure to obtain the convex combination. With the purpose to guarantee some continuity property, several tries have been done to define particular convex combination out of a finite collection of points in a geodesic space; see, for instance, [2, 13, 15]. For the case of Busemann spaces, a particular convex combination selection was studied in Ref. [12]. With the help of this construction, we propose in this paper a new approach to results given in Ref. [18] for convex combinations of mappings in complete CAT(0) spaces.

2 Preliminaries

Let X be a metric space and $x,y \in X$. A geodesic path from x to y is a mapping $c:[0,l] \subseteq \mathbb {R} \rightarrow X$, such that $c(0) = x, c(l) = y$ and $d\left( c(t),c(t^{\prime })\right) = \left| t - t^{\prime }\right|$ for every $t,t^{\prime } \in [0,l]$. The image of c forms a geodesic segment which joins x and y and we denote it by [x, y] when it is unique. X is a (uniquely) geodesic space if every two points in X can be joined by a (unique) geodesic path. A point $z\in X$ belongs to a geodesic segment joining x and y if there exists $t\in [0,1]$, such that $d(z,x)= td(x,y)$ and $d(z,y)=(1-t)d(x,y)$, and we write $z=(1-t)x \oplus ty$. In this case, $z = c(tl)$, where $c: [0,l] \rightarrow X$ is a geodesic path from x to y whose image is the geodesic segment in question. A subset A of X is convex if it contains any geodesic segment joining each two points in A.

CAT(0) spaces are usually defined by comparing their triangles with Euclidean ones. We will get a more analytic definition as they are equivalent. The interested reader may check [6] or [8] for a more detailed introduction to CAT(0) spaces. A geodesic metric space (X, d) is said to be CAT(0) if, for any points $x,y,z\in X$ and $t\in [0,1]$, the following CN-inequality [6] holds:

$$\begin{aligned} d(x,(1-t)y\oplus tz)^2\le (1-t)d(x,y)^2+td(x,z)^2-t(1-t)d(y,z)^2. \end{aligned}$$

As it is well known, Hilbert spaces, Hadamard manifolds, the Hilbert ball, and ${\mathbb R}$-trees are examples of CAT(0) spaces. In particular, CAT(0) spaces are examples of the so-called Busemann spaces, since they satisfy the inequality

$$\begin{aligned} d((1-t)w\oplus tx, (1-t)y \oplus tz)\le (1-t)d(w,y)+td(x,z) \end{aligned}$$

(1)

for any w, x, y and $z\in X$ and $t\in [0,1]$.

If B is a closed and convex subset of a complete CAT(0) space, then the operator $P_B:X\rightarrow B$ which satisfies

$$\begin{aligned} d(x,P_Bx)=\inf _{y\in B}d(x,y):=\mathrm{dist\;}(x,B) \end{aligned}$$

for each $x\in X$ is well defined and nonexpansive (see [6, pg. 176]). This operator is called the nearest point projection or metric projection from X onto B.

Definition 2.1

If (X, d) is a metric space, then $T:X\rightarrow X$ is said to be nonexpansive if

$$\begin{aligned} d(Tx,Ty)\le d(x,y) \end{aligned}$$

for all $x,y\in X$.

There exists a notion of weak convergence in CAT(0) spaces given through asymptotic centers of sequences. Given a bounded sequence $(x_n)$ in a CAT(0) space, we say that $x_0\in X$ is the asymptotic center of $(x_n)$ if

$$\begin{aligned} r(x_n,x_0):=\limsup _{n\rightarrow \infty } d(x_n,x_0)=\inf _{x\in X}\limsup _{n\rightarrow \infty }d(x_n,x)=: r(x_n). \end{aligned}$$

The point $x_0$ is the weak limit of $(x_n)$ if it is the asymptotic center of each subsequence of it. Asymptotic centers of bounded sequences exist and are unique in complete CAT(0) spaces. Properties of asymptotic centers of bounded sequences of CAT(0) spaces have been studied by several authors; see, for instance, [11, 14].

Convex combinations in unique geodesic spaces are clear when dealing with two points, but when we consider more than just two points, the situation changes. How to understand a convex combination of three points is not that trivial and basic questions regarding convex hulls in these spaces remain open nowadays (see, for instance, [5, 16]). The natural way to define the convex combination of three points problem $x,y,z\in X$ and $a_1, a_2,a_3$ convex coefficients would be as follows:

$$\begin{aligned} a_1x\oplus a_2y\oplus a_3z:= a_1 x\oplus (1-a_1)\left( \frac{a_2}{1-a_1}y\oplus \frac{a_3}{1-a_1}z\right) . \end{aligned}$$

However, this may differ from

$$\begin{aligned} a_1x\oplus a_2y\oplus a_3z:= a_2 y\oplus (1-a_2)\left( \frac{a_1}{1-a_2}x\oplus \frac{a_3}{1-a_2}z\right) . \end{aligned}$$

The problem of choosing a convex combination that does not depend on the order of the points have been addressed by several authors [2, 11, 13]. In Ref. [11], ideas from [2], where such a selection was proposed through an inductive construction, were taken farther. The resulting convex combination was shown to exhibit certain continuity properties with respect to points and coefficients that we will of interest for us. We show next the construction of the convex combination of three points proposed in Ref. [2].

For $k = 2$, one has the standard definition $\dot{\oplus }_{i=1}^2 a_ix_i = a_1 x_1 \oplus a_2 x_2$. Now let $k = 3$ and consider, for $1\le j\le 3$, the sequences $(x_j^n)_n$ inductively defined as

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1^n= a_1x_1^{n-1}{\oplus } (1-a_1)\Big (\frac{a_2}{1-a_1}x_2^{n-1}{\oplus } \frac{a_3}{1-a_1}x_3^{n-1}\Big ),\\ x_2^n= a_2x_2^{n-1}{\oplus } (1-a_2)\Big (\frac{a_1}{1-a_2}x_1^{n-1}{\oplus } \frac{a_3}{1-a_2}x_3^{n-1}\Big ),\\ x_3^n= a_3x_3^{n-1}{\oplus } (1-a_3)\Big (\frac{a_1}{1-a_3}x_1^{n-1}{\oplus } \frac{a_2}{1-a_3}x_2^{n-1}\Big ). \end{array}\right. } \end{aligned}$$

It is shown in Ref. [2] that the sequences $(x_j^n)$, for $j\in \{1,2,3\}$, converge to a same point which defines the (by definition) convex combination $\dot{\oplus }_{i=1}^3 a_ix_i$. Clearly, by the construction method, this limit is independent of the order the points $x_i$ are arranged in the 3-tuple (assuming the correspondence between points and coefficients is maintained). The following result was shown in Ref. [12].

Theorem 2.2

Let (X, d) be a complete Busemann space. Let $x_1,\ldots , x_k$ and $a_1,\ldots ,a_k$ be $k(\ge 3)$ points in X and associated convex coefficients, respectively. Then, there exists a point which is limit of sequences $(x_j^n)_n$ for $1\le j\le k$ where

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1^n= a_1x_1^{n-1}{\oplus } (1-a_1)\Big (\dot{\oplus }_{i=2}^k\frac{a_i}{1-a_1}x_i^{n-1}\Big ),\\ x_2^n= a_2x_2^{n-1}{\oplus } (1-a_2)\Big (\dot{\oplus }_{\underset{i \ne 2}{i=1}}^k\frac{a_i}{1-a_2}x_i^{n-1}\Big ),\\ \vdots \\ x_k^n= a_kx_k^{n-1}{\oplus } (1-a_k)\Big (\dot{\oplus }_{i=1}^{k-1}\frac{a_i}{1-a_k}x_i^{n-1}\Big ). \end{array}\right. } \end{aligned}$$

This point will be denoted as $\dot{\oplus }_{i=1}^k a_ix_i$ and satisfies that

$$\begin{aligned} d(\dot{\oplus }_{i=1}^ka_ix_i,\dot{\oplus }_{i=1}^ka_iy_i)\le \sum _{i=1}^k a_id(x_i,y_i) \end{aligned}$$

(2)

for any other k-tuple $(y_1,\ldots ,y_k)$ of points in X.

We will also need some facts about orbits. Let (X, d) be a metric space and $T:X\rightarrow X$ a mapping. Given $x\in X$, the sequence $(x_n)$ given by $x_1=x$ and $x_n=Tx_{n-1}$ is called an exact orbit of T with initial point x. A sequence $(x_n)$ is an inexact orbit of T with summable errors if $\sum _{n=1}^\infty d(x_{n+1}, Tx_n)< \infty$. The following result was shown in Ref. [17].

Theorem 2.3

Let (X, d) be a complete CAT(0) space and consider a nonexpansive mapping $T:X\rightarrow X$. If $\mathrm{Fix\;} (T)\ne \emptyset$, then the following are equivalent:

• (i) All exact orbits of T are weakly convergent to fixed points of T.

• (ii) All inexact orbits of T with summable errors are weakly convergent to fixed points of T.

3 Weak convergence of infinite products of convex combinations of operators in CAT(0) spaces

Our results will need the notion of strongly nonexpansive mapping.

Definition 3.1

Let (X, d) be a metric space, then $T:X\rightarrow X$ is said to be strongly nonexpansive if it is nonexpansive itself, Fix $(T)\ne \emptyset$ and for any bounded sequence $(x_n)$ in X and $z\in \mathrm{Fix\;} (T)$, such that $\lim _{n\rightarrow \infty } (d(x_n,z)-d(Tx_n,z))=0$, it must be the case that $\lim _{n\rightarrow \infty } d(x_n,Tx_n)=0$.

As it is shown in Ref. [18], nearest point projections onto closed and convex subsets of complete CAT(0) spaces are strongly nonexpansive. The next result on strongly nonexpansive mappings [18, Theorem 3.8] will also be needed.

Theorem 3.2

Let (X, d) be a compete CAT(0) space and $T:X\rightarrow X$ strongly nonexpansive. Then, for each $x\in X$, the sequence $(T^nx)$ converges weakly to a fixed point of T.

The process to define the mapping which results as convex combination of a finite family of mappings is straightforward, as we show next. Let $T_1,\ldots ,T_k$ be a finite family of self-mappings from a complete CAT(0) space X into itself and $a_1,\ldots ,a_k$ corresponding convex coefficients. Then, $\dot{\oplus }_{i=1}^ka_iT_i:X\rightarrow X$ is defined as

$$\begin{aligned} (\dot{\oplus }_{i=1}^ka_iT_i)x= \dot{\oplus }_{i=1}^ka_iT_i(x) \end{aligned}$$

for each $x\in X$, where the right-side term is the convex combination given in Theorem 2.2 with points $T_1(x),\ldots , T_k(x)$ and coefficients $a_1,\ldots ,a_k$.

Our first goal is to show that the convex combination of strongly nonexpansive mappings is strongly nonexpansive when they have common fixed points. We begin with the next lemma.

Lemma 3.3

Suppose (X, d) is a complete CAT(0) space and $T_i:X \rightarrow X$, for $i=1, \ldots , k$ are strongly nonexpansive mappings, such that $\bigcap _{i=1}^k {{\,\textrm{Fix}\,}}(T_i) \ne \emptyset$. Then

$$\begin{aligned} {{\,\textrm{Fix}\,}}(\dot{\oplus }_{i=1}^{k}a_i T_i) = \bigcap _{i=1}^k \, {{\,\textrm{Fix}\,}}(T_i). \end{aligned}$$

Proof

We consider $z \in \bigcap _{i=1}^k \, {{\,\textrm{Fix}\,}}(T_i)$, and hence, $z \in {{\,\textrm{Fix}\,}}(T_i)$ for any $i=1, \ldots , k$. We know that $(\dot{\oplus }_{i=1}^{k}a_i T_i)z$ is the limit of the sequence $(T^n_1 z)_n$, where

$$\begin{aligned} T_1^n z = a_1 T_{1}^{n-1} z \oplus (1-a_1) \bigg ((\dot{\oplus }_{i=2 }^k \frac{a_i}{1-a_1}T_i^{n-1})z \bigg ). \end{aligned}$$

Now, since z is fixed for any $T_i$, it trivially follows that:

$$\begin{aligned} T_1^n z= a_1 z \oplus (1-a_1) \bigg (\dot{\oplus }_{i=2 }^k \frac{a_i}{1-a_1}z \bigg ) =z \end{aligned}$$

for any $n\in \mathbb N$. We take limit on n and, therefore, $z \in {{\,\textrm{Fix}\,}}(\dot{\oplus }_{i=1}^{k}a_i T_i)$.

To prove the opposite implication, we assume that $z \in {{\,\textrm{Fix}\,}}(\dot{\oplus }_{i=1}^{k}a_i T_i)$ and fix $y \in \bigcap _{i=1}^k \, {{\,\textrm{Fix}\,}}(T_i)$. It follows from (2) that:

$$\begin{aligned} d(z,y) \le \sum _{i=1}^k a_i d(T_i z,y) \le \sum _{i=1}^k a_i d(z,y) = d(z,y). \end{aligned}$$

Then

$$\begin{aligned} \sum _{i=1}^k a_i [d(y,z) - d(T_i z,y)] =0. \end{aligned}$$

From the above and the strongly nonexpansivity of mappings $T_i$, $i = 1, \ldots , k$, we have $T_i z = z$ for every $i=1, \ldots , k$. Then, $z \in \bigcap _{i=1}^k \, {{\,\textrm{Fix}\,}}(T_i)$. $\square$

Remark 3.4

Notice that for the previous result to hold, it only takes the space to be a complete Busemann space.

Lemma 3.5

Suppose (X, d) is a complete CAT(0) space and $T_i:X \rightarrow X$, for $i=1, \ldots , k$, are strongly nonexpansive mappings, such that $\bigcap _{i=1}^k {{\,\textrm{Fix}\,}}(T_i) \ne \emptyset$, then mapping $\dot{\oplus }_{i=1}^{k}a_i T_i$ is strongly nonexpansive too.

Proof

From Lemma 3.3, we know that $\mathrm{Fix\;} (\dot{\oplus }_{i=1}^{k}a_i T_i)$ is nonempty. Let us show next that $\dot{\oplus }_{i=1}^{k}a_i T_i$ is nonexpansive. For a point $x\in X$, we know that $(\dot{\oplus }_{i=1}^{k}a_i T_i)x$ is the limit of an inductively defined sequence $(T^n_1x)_n$ where

$$\begin{aligned} T^n_1x= a_1 T_{1}^{n-1} x \oplus (1-a_1) \Bigg (\dot{\oplus }_{i=2 }^k \frac{a_i}{1-a_1}T_i^{n-1}x \Bigg ). \end{aligned}$$

Therefore, for x and y in X, we consider $(\dot{\oplus }_{i=1}^{k}a_i T_i)x$, $(\dot{\oplus }_{i=1}^{k}a_i T_i)y$ and corresponding sequences $(T^n_1x)_n$ and $(T^n_1y)_n$. Using the nonexpansivity of mapping $T_1$, inequality (1), and directly applying the inductive hypothesis on k, we obtain that

$$\begin{aligned}&d(T^n_1x,T^n_1y) \\&\quad \le a_1 d(T_1^{n-1}x,T_1^{n-1}y) + (1-a_1) d\Bigg (\dot{\oplus }_{i=2 }^k \frac{a_i}{1-a_1}T_i^{n-1}x,\dot{\oplus }_{i=2 }^k \frac{a_i}{1-a_1}T_i^{n-1}y \Bigg ) \le \\&\quad \le a_1 d(x,y) + (1-a_1)d(x,y) = d(x,y). \end{aligned}$$

Then, by taking limit on n

$$\begin{aligned} d((\dot{\oplus }_{i=1}^{k}a_i T_i)x, (\dot{\oplus }_{i=1}^{k}a_i T_i)y) \le d(x,y), \end{aligned}$$

(3)

and nonexpansivity is proved. To finish the proof that $\dot{\oplus }_{i=1}^{k}a_i T_i$ is strongly nonexpansive, we fix $z \in {{\,\textrm{Fix}\,}}(\dot{\oplus }_{i=1}^{k}a_iT_i)$ and consider a bounded sequence $\{x_n\}$, such that

$$\begin{aligned} \lim _{n \rightarrow \infty } \, d(x_n,z) - d(z, (\dot{\oplus }_{i=1}^{k}a_iT_i) x_n) =0. \end{aligned}$$

Since $T_i$, for $i = 1, \ldots , k$, is nonexpansive, and z is a common fixed point for all $T_i$ and inequality (2), we obtain that

$$\begin{aligned} d(z, (\dot{\oplus }_{i=1}^{k}a_i T_i) x_n) \le \sum _{i=1}^k a_i d(T_i z, T_i x_n) \le \sum _{i=1}^k a_i d(z,x_n) = d(z,x_n). \end{aligned}$$

From here, we have that

$$\begin{aligned} 0 \le \sum _{i=1}^k a_i [d(x_n,z) - d(T_i x_n, z)] \le d(x_n,z) - d(z, \dot{\oplus }_{i=1}^{k}a_iT_i x_n) \rightarrow 0, \end{aligned}$$

(4)

as n goes to infinity. Therefore

$$\begin{aligned} 0 \le \sum _{i=1}^k a_i [d(x_n,z) - d(T_i x_n,z)] \rightarrow 0, \quad n \rightarrow \infty . \end{aligned}$$

Therefore

$$\begin{aligned} \lim _{n \rightarrow \infty } d(x_n,z) - d(T_i x_n,z) = 0, \end{aligned}$$

for every $i = 1, \ldots , k$. Now, since every map $T_i$, for $i=1, \ldots , k$, is strongly nonexpansive

$$\begin{aligned} \lim _{n \rightarrow \infty } d(x_n, T_i x_n) = 0, \quad i=1, \ldots , k. \end{aligned}$$

Using (2), we have

$$\begin{aligned} 0 \le d(x_n, \dot{\oplus }_{i=1}^{k}a_i T_i x_n) \le \sum _{i=1}^k a_i d(x_n, T_i x_n) \rightarrow 0, \end{aligned}$$

as $n \rightarrow \infty$. Therefore

$$\begin{aligned} \lim _{n \rightarrow \infty } d(x_n, \dot{\oplus }_{i=1}^{k}a_i T_i x_n) =0, \end{aligned}$$

from where the lemma follows. $\square$

Now, the combination of Lemmas 3.3 and 3.5, as well as Theorems 2.3 and 3.2, leads to the counterpart of Lemma 5.1 in [18] that we state next.

Lemma 3.6

Suppose (X, d) is a complete CAT(0) space and $\{P_{B_i}: 1 \le i \le k\}$ are the nearest point projections of X onto closed and convex subsets $\{B_i \subset X: 1 \le i \le k \}$, where $B_1 \cap \ldots \cap B_k \ne \emptyset$. Let $T=\dot{\oplus }_{i=1}^{k}a_i P_{B_i}$, where $a_1, \ldots , a_k \in (0,1)$ satisfy that $a_1 + \cdots +a_k =1$. Suppose further that there exist a sequence $(x_n)_n$ and a convergent series of positive numbers $\sum _{n \in \mathbb {N}} \gamma _n$ satisfying the inequality

$$\begin{aligned} d(x_{n+1}, T x_n) \le \gamma _n \end{aligned}$$

(5)

for all $n\in \mathbb N$. Then, there exists a point $\overline{x} \in B_1 \cap \cdots \cap B_k$, such that $\lim _{n \rightarrow \infty } x_n = \overline{x}$ weakly.

As in [18], we will give a weak convergence result of infinite products of mappings (possible discontinuous) that approximate nearest point projections. For that, we need to impose the same extra condition on CAT(0) spaces imposed in [18] and which is satisfied in certain CAT(0) spaces as the Hilbert ball.

Definition 3.7

A metric space (X, d) is said to be metrically homogeneous if, for any $x,y\in X$, there exists an onto isometry I on X, such that $I(x)=y$.

An immediate property about isometries is that the images of segments of unique geodesic metric spaces are determined by the images of the extremes of the segments. Furthermore, we have the following immediate lemma.

Lemma 3.8

Let (X, d) be a complete CAT(0) space and $I:X\rightarrow X$ an isometry, then, for any $x_1,\ldots ,x_k$ in X and $a_1, \ldots , a_k \in (0,1)$, such that $a_1 + \cdots +a_k =1$

$$\begin{aligned} I(\dot{\oplus }_{i=1}^{k}a_i x_i)=\dot{\oplus }_{i=1}^{k}a_iI(x_i). \end{aligned}$$

Now, we go with the main result of this work which is a counterpart of Theorem 5.2 in Ref. [18].

Theorem 3.9

Let (X, d) be a complete and metrically homogeneous CAT(0) space. Let $\{P_{B_i}: 1 \le i \le k\}$ be the nearest point projections of X onto closed and convex subsets $\{B_i \subset X: 1 \le i \le k \}$, where $B_1 \cap \ldots \cap B_k \ne \emptyset$. Let given mappings $A^i_n$, $i=1\ldots , k$, $n\in \mathbb N$ and a point $x_0\in X$ be such that

$$\begin{aligned} d(A^i_nx,P_{B_i}x)\le \gamma _nd(x_0,x) \end{aligned}$$

for each $x\in X$ and where $\gamma _n$, for $n\in \mathbb N$, are positive numbers, such that $\sum _{n\in {\mathbb N}} \gamma _n < \infty$. Then, for each $x\in X$, there exists a point $\bar{x}=\bar{x}(x)\in B_1\cap \cdots \cap B_k$, such that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Bigg ( \prod _{j=1}^n (\dot{\oplus }_{i=1}^k a_iA^i_j)\Bigg )x=\bar{x}\text { weakly,} \end{aligned}$$

where $a_1, \ldots , a_k \in (0,1)$ are numbers satisfying $a_1 + \cdots +a_k =1$ and $\prod$ stands for the composition of mappings.

Proof

We follow the proof of Theorem 5.2 in [18]. Take $z\in B_1\cap \ldots \cap B_k$ and let I be an onto isometry on X, such that $I(z)=x_0$. For each $i=1,\ldots , k$, make $\tilde{B}_i=I(B_i)$ and $\tilde{A}_n^i(I(x))=I(A_n^i(x))$. Make $w=I(x)$. Since I is an isometry, Lemma 3.8 implies that $P_{\tilde{B}_i}(w)=I(P_{B_i}x)$. Notice that $P_{\tilde{B}_i}$ is the nearest point projection onto $\tilde{B}_i$ for each i, too. In particular, $x_0\in \tilde{B}_i$ for all i, and so

$$\begin{aligned} d(x_0,P_{\tilde{B}_i}w)= d(P_{\tilde{B}_i}x_0,P_{\tilde{B}_i}w)\le d(x_0,w) \end{aligned}$$

for all $w\in X$. Now, from the hypothesis of the theorem, we have that

$$\begin{aligned} d(\tilde{A}_n^i w,P_{\tilde{B}_i}w)&\le \gamma _n d(x_0,x)\le \gamma _n (d(x_0,z)+d(z,x)) \nonumber \\&= \gamma _n (d(x_0,z)+d(Iz,Ix))= \gamma _n (d(x_0,z)+d(x_0,w)). \end{aligned}$$

(6)

Putting the two above inequalities together, we have that

$$\begin{aligned} d(x_0,\tilde{A}_n^i w)\le (1+\gamma _n)d(x_0,w)+\gamma _nd(x_0,z) \end{aligned}$$

(7)

for all $w\in X$ and $i=1, \ldots , k$. Therefore, from inequality (2)

$$\begin{aligned} d(x_0, (\dot{\oplus }_{i=1}^{m+1} b_i\tilde{A}^i_n )w)\le (1+\gamma _n)d(x_0,w)+\gamma _nd(x_0,z) \end{aligned}$$

(8)

for any numbers $b_1,\ldots , b_{m+1}\in (0,1)$ with $b_1+\cdots +b_{m+1}=1$ and $m=1,\ldots , k-1$. Moreover, from (2)

$$\begin{aligned} d((\dot{\oplus }_{i=1}^{m+1} b_i\tilde{A}^i_n )w, (\dot{\oplus }_{i=1}^{m+1} b_iP_{\tilde{B}_i} )w)\le \sum _{i=1}^{m+1}b_i d(\tilde{A}^i_n w, P_{\tilde{B}_i}w). \end{aligned}$$

(9)

Now, given $x\in X$, let $(x_n)_n$ be the sequence given by $x_1=x$ and $x_{n+1}= (\dot{\oplus }_{i=1}^{k}a_i A^i_n)x_n$ for $n\ge 1$. Consider also $(w_n)_n$, such that $w_n=I(x_n)$. Now, applying (8) in an inductive way, we have that

$$\begin{aligned} d(x_0,w_{n+1})\le \prod _{j=1}^n(1+\gamma _j)d(x_0,w)+\Bigg (\sum _{i=1}^{n-1} \gamma _i\prod _{j=i+1}^n (1+\gamma _j)+\gamma _n \Bigg ) d(x_0,w). \end{aligned}$$

Since $\sum _{j\in {\mathbb N}}\gamma _j <\infty$, it follows that $\prod _{j\in {\mathbb N}} (1+\gamma _j)<\infty$ and so, from the above, there exists $M>0$, such that:

$$\begin{aligned} d(x_0,w_{n})\le M(d(x_0,w)+d(x_0,z)) \end{aligned}$$

for all $n\in \mathbb N$. Let $\tilde{T} =\dot{\oplus }_{i=1}^{k}a_iP_{\tilde{B}_i}$, then, from (6) and (9)

$$\begin{aligned} d(w_{n+1}, \tilde{T}w_n)\le \sum _{i=1}^k d(\tilde{A}^i_n w_n, P_{\tilde{B}_i}w_n)\le \gamma _n d(x_0,z)+\gamma _n d(x_0,w_n). \end{aligned}$$

Therefore

$$\begin{aligned} d(w_{n+1}, \tilde{T}w_n)\le \gamma _n (M+1)(d(x_0,w)+d(x_0,z)) \end{aligned}$$

for all $n\in \mathbb N$. Consequently, $(w_n)_n$ is an inexact orbit of ${\tilde{T}}$ with summable errors. Then, applying Lemma 3.6, $(w_n)_n$ weakly converges to a point $\bar{w}$ which is in $\tilde{B}_1\cap \ldots \cap \tilde{B}_k$. Now, since I is weakly continuous (see [17, Lemma 6.2]), it follows that there exists $\bar{x}=I^{-1} (\bar{w})\in B_1\cap \ldots \cap B_k$ that is the weak limit of $(\prod _{j=1}^n (\dot{\oplus }_{i=1}^{k}a_i A^i_j) x)_n$, which completes the proof. $\square$

Remark 3.10

In this work, we have shown that the direct convex combination construction exhibited in the last section of [18] may be replaced with the one studied in Ref. [12], and denoted as $\dot{\oplus }$, to obtain counterparts to the results given in the last section of [18]. Main features of $\dot{\oplus }$ are that this convex combination does not depend on the way the terms of the convex combination are ordered, and still, it keeps good continuity and convex properties as inequality (2) or Theorem 3.2 in Ref. [12] shows. Recall that this theorem states that if $(x_1, \ldots , x_k)$ is a k-tuple of points in a complete Busemann space X and $(a_1,\ldots , a_k)$ is a k-tuple of convex coefficients, then, if, for each $n \in \mathbb {N}$, we consider the k-tuples $(z_1^n,\ldots ,z_k^n)$ of points in X and $(b_1^n,\ldots ,b_k^n)$ of convex coefficients, such that

$$\begin{aligned}(z_1^n,\ldots , z_k^n)\rightarrow (x_1,\ldots , x_k) \quad \text {and} \quad (b_1^n,\ldots ,b_k^n)\rightarrow (a_1,\ldots , a_k),\end{aligned}$$

as n goes to infinity, then

$$\begin{aligned} \dot{\oplus }_{i=1}^k b_i^n z_i^n\rightarrow \dot{\oplus }_{i=1}^ka_ix_i, \end{aligned}$$

as n goes to infinity.