On the Properties of Iterations Generated with Composition Maps of Cyclic Contractive Self-Mappings and Strict Contractions in Metric Spaces

Abstract

This paper studies the convergence of distances between sequences of points and that of sequences of points in metric spaces. This investigation is focused on the iterative processes built with composed self-mappings of a cyclic contraction, which can involve more than two nonempty closed subsets in a metric space, which are combined with compositions of a strict contraction with itself, which operates in each of the individual subsets, in any order and any number of mutual compositions. It is admitted, in the most general case, the involvement of any number of repeated compositions of both self-maps with themselves. It is basically seen that, if one of the best-proximity points in the cyclic disposal is unique in a boundedly compact subset of the metric space is sufficient to achieve unique asymptotic cycles formed by a best-proximity point per each adjacent subset. The same property is achievable if such a subset is strictly convex and the metric space is a uniformly convex Banach space. Furthermore, all the sequences with arbitrary initial points in the union of all the subsets of the cyclic disposal converge to such a limit cycle.

1. Introduction

There is an abundant and rich background literature available on contractive cyclic self-mappings defined on subsets of complete metric spaces. See, for instance, [1,2,3] and some of the references therein. In [4], multivalued cyclic self-mappings are investigated while, in [5], Jungck iterative cyclic iterative processes are focused on. In [6], some results for cyclic contractions are obtained, which apply for Banach spaces, geodesic metric spaces and metric spaces. In [7], relatively nonexpansive mappings are investigated on the union of two subsets of a Banach space. If the pair of subsets has a proximal normal structure and a relative nonexpansive mapping, $T$ on $A\cup B$ satisfies $T\left(A\right)\subset B$ and $T\left(B\right)\subset A$, where $A$ and $B$ are weakly compact and convex subsets of the Banach space $\left(X,‖ ‖\right)$, then there is some proximal point $z\in A\cup B$ such that $‖z-Tz‖=dist\left(A,B\right)$. The obtained results are also adapted, in particular, for Banach spaces with strictly convex norms and they are valid for uniformly convex Banach spaces. In [8], cyclic ${\varphi }_A$-contractions defined on partially ordered orbitally complete metric spaces are considered. Some fixed-point and best-proximity-point theorems are stated and certain links between points of coincidence and common best proximal points are investigated. It is also found that there are conditions under which points of coincidence, common best proximal points and common fixed points are coincident. A study on the best-proximity points for cyclic mappings in multiplicative metric spaces is performed in [9]. On the other hand, best-proximity results in b-metric spaces endowed with a graph are established in [10], while, in [11,12], some new results for cyclic mappings in complete b-metric-like space are established. Also, some connections on cyclic contractions and contractive self-mappings with Hardy–Rogers self-mappings are investigated in [13] while proximity results in Busemann convex metric spaces are given in [14]. In [15], a so-called BSS-type cyclic zapping is investigated, which is a pair semi-cycle mapping supported with a special contractive rule. On the other hand, some new best proximity and cyclic mapping results are obtained in [16,17] concerned with S-cyclic mappings.

It can be pointed out the importance of the concept of approximative compactness for the non-emptiness of the sets of best-proximity points of two nonempty subsets $A$ and $B$ of a metric space. In fact, one of the simplest such conditions is that both best proximity sets are nonempty if $A$ is compact and $B$ is approximatively compact with respect to $A$, that is, every sequence ${\left\{x_n\right\}}_{n=0}^{\infty }\subset B$ such that $d\left(y,x_n\right)\to d\left(y,B\right)$ as $n\to \infty$ for some $y\in A$ such that $dist\left(A,B\right)=d\left(y,B\right)$ such have a convergent subsequence [1,18,19].

This paper extends the formalism on cyclic contractions to consider the compositions of cyclic self-mappings with usual strict contractions which are integrated in more general iterative processes in the following way. The subsets of the cyclic disposals are assumed to be nonempty and closed and they can be disjointed or not. The contractive cyclic self-mapping $T$ operates over adjacent nonempty closed subsets of the cyclical disposal in the metric space $\left(X,d\right)$ as usual, while the strict contraction $S$ operates over each of the individual subsets. The iterative processes which govern the combinations of both self-mappings are built in general under any order of the contributions of each of the mappings and any number of compositions of the self-mapping $T$ with itself. Also, the iterative processes can include any number of compositions in any order of the sets of compositions of both self-mappings $T$ and $S$ with themselves. The subjacent idea of the proposal is to adequate the evolution of sequences in each of the subsets being governed by the self-mapping $S$ or by compositions of the self-mapping $S$ with itself, with switching actions from the current subset to its adjacent one in the cyclic disposal, governed by the self-mapping $T$. Each switching in-between adjacent subsets of the cyclic disposal can be addressed from the current subset to another subset distinct to the adjacent one if a number of compositions of the cyclic self-mapping $T$ with itself are combined before the next contribution to the iterative process of the contractive self-mapping $S$, again, with eventual compositions with itself. Some applications of interest, as exhibited in some of the given worked examples, rely on the solution of differential systems of equations in alternative domains which can also have different parameterizations and are subject to mutual switching actions. For instance, each of the subsets of the cyclic disposal can be associated with a particular parameterization of the differential system, while switching actions can govern the activations of other alternative parameterizations in a different set to the current one among those being integrated in the cyclic disposal. This situation may arise in some processes in the real word as, for instance, in atmospheric models, in distillation column towers, in switching processes in financial markets, etc., and these are also relevant in certain control processes [20,21,22,23].

This paper is organized as follows. Section 2 gives some preliminary definitions and results on the contractive cyclic self-mapping and the strict contraction to be composed in the iterative processes of the third section. It is basically seen that under the condition that just one of the best proximity sets in the cyclic disposal is a singleton together with the bounded compactness of such a set (or its strict convexity if the metric space is, in addition, a uniformly convex Banach space), this suffices to achieve unique asymptotic cycles formed by a best-proximity point per each adjacent subset. All the sequences with arbitrary initial points in the union of all the subsets converge to such a limit cycle. The results also hold for extended cyclic contractions where the product of the different individual constants (one per transition from each subset to its adjacent one) is less than unity which, in fact, only requires that one of such constants be contractive. Section 3 is devoted to the convergence of distances of sequences and sequences when the compositions of the cyclic contraction are combined with compositions of the strict contraction in any order of mutual compositions while possibly involving any number of repeated compositions of both maps with themselves. It is assumed that the subsets of the cyclic disposal are nonempty and closed and that they are subsets of a uniformly convex Banach space. In addition, it is assumed that at least one of the subsets of the cyclic disposal is boundedly compact with its best proximity set to the next adjacent subset being a singleton, or alternatively that, at least one of such subsets is strictly convex. It is also assumed that the unique fixed point of the contractive self-mapping $S$ at each subset belongs to the set of best-proximity points to its next adjacent subset in the cyclic disposal. Section 4 describes some worked examples which illustrate the formal results and, finally, the Conclusions section ends this paper.

2. Preliminary Definitions and Preparatory Results on Cyclic Contractive Self-Mappings and Strict Contractions

Through this paper, $\overline{p}=\left\{1,\dots ,2,\dots \dots , p\right\}$ and $x\vee y$ is the logic disjunction “or” of the logic propositions $x$ and $y$. Also, $R_{+}=R_{0+}\cup \left\{0\right\}$ and $R_{0+}=\left\{r\in R:r\ge 0\right\}$ are the respective sets of positive and the non-negative real numbers, and $Z_{+}=Z_{0+}\cup \left\{0\right\}$ and $Z_{0+}=\left\{z\in Z : \mathrm{z}\ge 0\right\}$ are the respective sets of positive and the non-negative integer numbers.

A positive matrix $A\in R^{n\times m}$ (referred to as $A\succ 0$) is the one with all its entries being non-negative with at least one positive. $A\succ 0$ is strictly positive (referred to a $A\succ \succ 0$) if all its entries are positive. A vector $v\in R^n$ is positive (referred to as $v\succ 0$) if all its components are non-negative with at least one being positive. Also, $v\succ 0$ is strictly positive (referred to as $v\succ \succ 0$) if all its components are strictly positive.

Consider self-mappings $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ such that $A_i\subset X$ for $i\in \overline{p}$ are non-empty closed subsets of a metric space $\left(X,d\right)$ with $S\left(A_i\right)\subseteq A_i$ for $i\in \overline{p}$ and $T\left(A_i\right)\subseteq A_{i+1}$ for $i\in \overline{p}$.

$$d\left(Tx, Ty\right)\le K_{T}d\left(x, y\right)+\left(1-K_T\right)D \mathrm{for} x\in A_i, y\in A_{i+1}, \mathrm{any} i\in \overline{p}$$

(1)

$$d\left(Sx, Sy\right) \left\{\begin{array}{c} \le K_{S}d\left(x, y\right)\hspace{1em}\mathrm{if} d\left(x, y\right)>K_S^{-1}D\\ =D\hspace{1em}\hspace{1em}\mathrm{if} d\left(x, y\right)\le K_S^{-1}D\end{array}\right. \mathrm{for} x\in A_i , y\in A_{i+1}, \mathrm{any} i\in \overline{p}$$

(2)

$$d\left(Sx, Sy\right)\le K_{S}d\left(x, y\right) \mathrm{if} x, y\in A_i \mathrm{for} \mathrm{any} i\in \overline{p}$$

(3)

for some real constants $K_T, K_S\in \left(0 , 1\right)$, such that the distances in-between any adjacent subsets $A_i$ and $A_{i+1}$ for $i\in \overline{p}$ and their respective diameters are, respectively, as below:

The distance in-between adjacent subsets is denoted with D which is zero if such subsets intersect and greater than zero if they do not intersect. More formally,

$$D=d\left(A_i, A_{i+1}\right)=inf \left(d\left(x, y\right):x\in A_i , y\in A_{i+1}\right) and \mathit{diam}\left(A_i\right) \ge D ; \forall i\in \overline{p}$$

(if $A_i\cap A_{i+1}=\varnothing$; $\forall i\in \overline{p}$ then $D=0$), with $A_{p+1}=A_1$ then, $A_{np+\mathcal{l}}=A_{\mathcal{l}}$ for any integer $\mathcal{l}\in \left[0, p-1\right]$. Unless otherwise be stated, we will refer to as the “adjacent” subset of $A_i$ to its next adjacent one $A_{i+1}$ in the cyclic disposal rather than to the preceding one $A_{i-1}.$

• Under (1)–(3),

(1) 

$T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is said to be a $p\left(\ge 2\right)$-cyclic contractive self-mapping on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$

• from (2).

(2) 

$S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is:

(1)

Non-expansive from (2)–(3) for any pair $\left(x,y\right)$, with $x,y\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$, allocated either in the same $A_i$ or in adjacent subsets $A_i$ and $A_{i+1}$ for $i\in \overline{p}$ of elements allocated in adjacent subsets;

(2)

A (strict) contraction from (3) for any pair $\left(x,y\right)$ of elements $x,y\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ allocated in the same subset $A_i$ for $i\in \overline{p}$;

(3)

Contractive for any pair $\left(x,y\right)$ allocated in adjacent subsets $A_i$ and $A_{i+1}$ for $i\in \overline{p}$ under the constraints of the first equation in (2), that is, if $d\left(x,y\right)>K_s^{-1}D$.

Remark 1. 

1.1. Note that the constraints $diam\left(A_i\right)\ge D$ for any $i\in \overline{p}$  are clearly needed for the coherency of (2). Note that, if $D=0$ , then the contractive condition (3) applies for $S:A_i\to A_i$  for any $i\in \overline{p}$  if $x,y\in A_i$ , while this contractive condition applies if $x\in A_i$, $y\in A_{i+1}$  or if $x\in A_{i+1}$  and $y\in A_i$for $D>0$  if $\mathrm{if} d\left(x,y\right)>K_S^{-1}D$  from (2). Also, if $D>0$ , then there exist nonempty closed subsets $A_{i0}\subseteq A_i$for each $i\in \overline{p}$  which contain the best approximation point(s) of each subset $A_i$ to its adjacent subset $A_{i+1}$. 

1.2. Note that $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is a $p\left(\ge 2\right)$ -cyclic contraction from (1), [1,2,3], and $S:\left({\displaystyle {\cup }_{j\in \overline{p}}{A}_j}\right) |A_i\to A_i$ is a (strict) contraction on $A_i$  for each $i\in p$  from (3).

1.3.  Note that the second condition of (2) implies that if $D<d\left(x,y\right)\le K_S^{-1}D$  for $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)}$ then $\left(Sx,Sy\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_{i0}\times A_{i+1 ,0}\cup A_{i+1,0}\times A_{i0}\right)}$ to accomplish with $d\left(Sx , Sy\right)=D$. In order for that constraint to be feasible, it is necessary that a) $A_{i0}\ne \varnothing$; $\forall i\in \overline{p}$ ; b) $S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ may be multi-valued at some points in order to accomplish with that constraint in the event that distances $d\left(x, y\right)$  for points of adjacent subsets that exceed the given threshold. That is, $d\left(x,y\right)\le K_S^{-1}D\Rightarrow d\left(Sx,Sy\right)=D$ if $x,Sx\in A_i$ and $y,Sy\in A_{i+1}$ (or vice versa) for any given $i\in \overline{p}$ but the above constraint is not requested if $x,y,Sx,Sy$ belong to the same subset $A_i$ for any given $i\in \overline{p}$.

Remark 2. 

Note that, since $S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is subject to the constraint $S:\left({\displaystyle {\cup }_{j\in \overline{p}}{A}_j}\right) |A_i\to A_i$; $\forall i\in \overline{p}$, this function may be defined by $p$functions $S_i:A_i\to A_i$; $\forall i\in \overline{p}$ which are all identical in ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$if such an intersection is non-empty. By virtue of the contraction condition (3), each $S_i$has a unique fixed point in $A_i$;$\forall i\in \overline{p}$ which are all coincident if ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }$. Note also that for any $x\in A_i ,y\in A_{i+1}$and $i\in \overline{p}$with $A_{p+1}=A_1$:

$$d\left(T^{n}x, T^{n}y\right)\le K_{T}d\left(T^{n-1}x ,T^{n-1}y\right)+\left(1-K_T\right)D$$

$$\le K_T\left(K_{T}d\left(T^{n-1}x ,T^{n-2}y\right)+\left(1-K_T\right)D\right)+\left(1-K_T\right)D$$

$$\le K_T^{2}d\left(T^{n-2}x ,T^{n-2}y\right)+\left(1-K_T^2\right)D$$

$$\le \dots \le K_T^{n}d\left(x ,y\right)+\left(1-K_T^n\right)D$$

(4)

so that $d\left(T^{n}x, T^{n}y\right)\to D$ as $k\to \infty$ and $T^{n}x\in A_j$ and $T^{n}y\in A_{j+1}$ for some $j\in \overline{p}$ and all $n\in Z_{0+}$. Since $D=inf\left(d\left(x,y\right):x\in A_i,y\in A_{i+1}\right)$ and $A_i\subset X$ for all $i\in \overline{p}$ are closed, then there exist nonempty closed subsets $A_{i0}\subseteq A_i$ for each $i\in \overline{p}$ which contain the best approximation point(s) of each subset $A_i$  to its adjacent subset $A_{i+1}$. Throughout this paper, it is of interest for the convergence properties the case when $A_{i0}=\left\{z_i\right\}$  and $S_i{z}_i=z_i$ ; $\forall i\in \overline{p}$ , that is, there is a unique best-proximity point at each $A_i$  which is also the unique fixed point of $S_i$. Note that since $d\left(T^{np}x, T^{np+1}x\right)\to D$ as $n\to \infty$ with ${\left\{T^{np}x\right\}}_{n=0}^{\infty }\subset A_i$ and ${\left\{T^{np+1}x\right\}}_{n=0}^{\infty }\subset A_{i+1}$ if $A_{i0}=\left\{z_i\right\}$  and $S_i{z}_i=z_i$ ; $\forall i\in \overline{p}$ then the best-proximity point $z_i$  through $T$ is jointly the unique fixed point of $S_i:A_i\to A_i$  and that of $T^p:\left({\displaystyle {\cup }_{j\in \overline{p}}{A}_j}\right)|A_i\to \left({\displaystyle {\cup }_{j\in \overline{p}}{A}_j}\right)|A_i$; $\forall i\in \overline{p}$ , then $z_i\in A_i$ satisfies $d\left(z_i ,z_{i+1}\right)=D$ , and

$$z_i=T^p{z}_i=T^p{S}_i{z}_i=S_i{z}_i=S_i{T}^p{z}_i;\forall i\in \overline{p}$$

that is, $T^p$ commutes with $S_i$ at each of the fixed points so that

$$z_{i+1}=T^p{z}_{i+1}=T{z}_i=T^{p+1}{z}_i=S_i{z}_{i+1}=T^p{S}_i{z}_{i+1}=T{S}_i{z}_i=T^{p+1}{S}_i{z}_i;\forall i\in \overline{p}$$

In the case when ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\ne \varnothing$ then ${\displaystyle {\cap }_{i\in \overline{p}}{A}_{i0}}=\left\{z\right\}$ since $z_i=z$ ; $\forall i\in \overline{p}$.

The next proposed definition leading immediately to an “ad hoc” elementary result refers to the fact that it is only needed a contraction of $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ on one of the transitions from a set $A_i$ to its adjacent one $A_{i+1}$ to obtain a cyclic contraction under the whole disposal on the $p$ subsets. It is useful if the mapping $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ has different real constants $K_i$ for $i\in \overline{p}$. It is simply necessary for that purpose that one of contractive constants be small enough to deal with other transitions which can be non-contractive or even expansive.

Definition 1. 

The $p$-cyclic self-mapping $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ with $T\left(A_i\right)\subseteq A_{i+1}$; $\forall i\in \overline{p}$ is said to be an extended $p\left(\ge 2\right)$-cyclic contraction if it satisfies the following condition:

• $d\left(Tx , Ty\right)\le K_{i}d\left(x,y\right)+\left(1-K_i\right)D$  for any $x\in A_i$, $y\in A_{i+1}$ with real constants $K_i\in R_{0+}$, any $i\in \overline{p}$ and $K_T={\displaystyle {\Pi }_{i=1}^p\left[K_i\right]}\in \left[0 , 1\right)$.

The following result relies on the fact that the existence of a unique convex element in the intersection set ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$  of non-disjoint closed sets $A_i$;$\forall i\in \overline{p}$ suffices for the convexity of such an intersection.

Lemma 1. 

Assume that all the sets $A_i$for $i\in \overline{p}$ are nonempty, non-disjointed and closed and that there exists some $j\in \overline{p}$such that $A_j$is convex. Then, ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$is closed and strictly convex.

Proof. 

${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }$ is closed since $A_i$;$\forall i\in \overline{p}$ are closed. Also, ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\subseteq A_j}$ and $A_j$ is closed and convex then the non-empty closed set ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$ is also convex and strictly convex since all its boundary points are extreme since all points in any segment in ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$ are necessarily in its interior since it is closed. □

If the subsets $A_i$ intersect, then $S$ and $T$ have unique fixed points in the intersection set. Formally, one has the subsequent result:

Lemma 2. 

Assume that all the sets $A_i$; $\forall i\in \overline{p}$of a complete metric space $\left(X,d\right)$are nonempty, closed and non-disjointed. Then, $S$has a unique fixed point in ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$, and $T^p$and $T$have an identical unique fixed point in ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$.

Proof. 

Note that ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$ is nonempty and closed and then $D=0$. Also, one has for any given $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$:

$$d\left(S^{n+m}x , S^{n}x\right)\le {\displaystyle {\sum }_{k=n}^{n+m-1}d\left(S^{k+1}x, S^{k}x\right)}$$

$$\le \left({\displaystyle {\sum }_{j=0}^{m-1}{K}_S^j}\right)d\left(S^{n+1}x , S^{n}x\right)$$

$$\le K_S^n\left({\displaystyle {\sum }_{j=0}^{m-1}{K}_S^j}\right)d\left(Sx , x\right)$$

$$=K_S^n\left({\displaystyle {\sum }_{j=0}^{\infty }{K}_S^j}-{\displaystyle {\sum }_{j=m}^{\infty }{K}_S^j}\right)d\left(Sx , x\right)$$

$$=K_S^n\left({\displaystyle \frac{1}{1-K_S}}-{\displaystyle \frac{K_S^m}{1-K_S}}\right)d\left(Sx , x\right)$$

$$={\displaystyle \frac{K_S^n}{1-K_S}}\left(1-K_S^m\right)d\left(Sx , x\right); \forall m\in Z_{+}$$

(5)

Then, $d\left(S^{n+m}x , S^{n}x\right)\to 0$ as $n\to \infty$; $\forall m\in Z_{+}$ and for any given $\epsilon \in R_{+}$, there is some $N=N\left(\epsilon \right)\in Z_{0+}$ such that for any $n\left(\ge N\right)\in Z_{0+}$ and any $m\in Z_{+}$ such that $d\left(S^{n+m}x , S^{n}x\right)<\epsilon$, then, ${\left\{S^{n}x\right\}}_{n=0}^{\infty }\to \omega \left(\in {\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\right)$. In particular, since $S:A_i\to A_i$ is Lipschitz-continuous with Lipschitz constant $K_S<1$, one can interchange limit and distance to get:

$$\underset{n\to \infty }{lim}d\left(S^{n+1}x , S^{n}x\right)=\underset{n\to \infty }{lim}d\left(S{S}^{n}x , S^{n}x\right)=d\left(S\underset{n\to \infty }{lim}{S}^{n}x , \underset{n\to \infty }{lim}{S}^{n}x\right)=d\left(S\omega , \omega \right)=0$$

(6)

and $\omega \in Fix\left(S\right)$. Also, it is the unique fixed point, that is, $Fix\left(S\right)=\left\{\omega \right\}$. Otherwise, if ${\omega }_1\ne \omega \in Fix\left(S\right)$, then, we get the contradiction $d\left(\omega , {\omega }_1\right)=d\left(S\omega ,S{\omega }_1\right)\le K_{S}d\left(\omega , {\omega }_1\right)$ and $\left(1-K_S\right)d\left(\omega , {\omega }_1\right)\le 0$ so that $\omega ={\omega }_1$. All the sequences ${\left\{S^{n}y\right\}}_{n=0}^{\infty }$ converge to this point. Assume that there is some $y\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ such that ${\left\{S^{n}y\right\}}_{n=0}^{\infty }\to \omega ´\left(\ne \omega \right)$. Thus,

$$\underset{n\to \infty }{lim}d\left(S^{n}y , S^n\omega \right)\le d\left(y,\omega \right)\left(\underset{n\to \infty }{lim}{K}_S^n\right)=0$$

(7)

Then, $\omega ´=\omega$. A similar reasoning from (1) with $D=0$ concludes that ${\left\{T^{pn}x \right\}}_{n=0}^{\infty }$ is a Cauchy sequence, ${\left\{T^{pn}x \right\}}_{n=0}^{\infty }\to z\left(\in Fix\left(T^p\right){\displaystyle \cap \left({\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\right)}\right)$, with $Fix\left(T^p\right)=\left\{z\right\}$;$\forall x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$. It is proved that $z$ is also the unique fixed point of $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$. Otherwise, the following contradiction is obtained if $z$ is not a fixed point, that is, if $z\ne Tz$:

$$0\ne d\left(z,Tz\right)\le d\left(z, T^{p}z\right)+d\left(T^{p}z ,Tz\right)$$

$$=0+d\left(T^{p}z, Tz\right)\le K_T\left(T^{p-1}z, z\right)$$

$$=K_T\left(T^{p-1}z,T^p z\right)\le K_T{K}_T^{p-1}d\left(z , Tz\right)$$

$$=K_T^{p}d\left(z , Tz\right)<d\left(z,Tz\right)$$

(8)

Then, $z$ is a fixed point of $T$, which is unique under a similar reasoning that the one used before to prove the uniqueness of the fixed point of $S$ since $T$ is also a strict contraction if $D=0$. The proof is complete. The proof is complete. □

Equation (4) of Remark 2 is generalized to an extended $p\left(\ge 2\right)$-cyclic contraction as follows:

Theorem 1. 

Assume that all the sets $A_i$; $\forall i\in \overline{p}$of a metric space $\left(X,d\right)$are nonempty and closed. If $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$is an extended $p\left(\ge 2\right)$-cyclic contraction, then the following properties hold:

(i) There exists $\underset{n\to \infty }{lim}d\left(T^{pn}x ,T^{pn}y\right)=D$ for any $\left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ and any $i\in \overline{p}$; and $\underset{n\to \infty }{lim}d\left(T^{pn+j}x ,T^{pn+j}y\right)=D$; $\forall j\in \overline{p}$ and any given $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)}$.

(ii)  If ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\ne \varnothing$, then $\underset{n\to \infty }{lim}d\left(T^{pn+j}x ,T^{pn+j}y\right)=0$; $\forall j\in \overline{p}$and any given $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)}$ ; $d\left(T^{pn+j}x ,T^{p\left(m+n\right)+k}x\right)\to 0$ ; $\forall x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ ; $\forall j,k\in \overline{p-1} \cup \left\{0\right\}$ , and $\left\{T^{pn+j}x\right\}\to z\left(\in {\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\right)$ is a Cauchy sequence, then bounded for $x$ being finite, and $Fix\left(T^p\right)=Fix\left(T\right)=\left\{z\right\}$.

(iii)  The sequences ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$  are bounded for any finite $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$  irrespectively of the subsets of the cyclic disposal being disjoint or non-disjoint.

(iv)  The properties (i)–(iii) also hold if  $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$;$\forall i\in \overline{p}$  is an extended $p\left(\ge 2\right)$-cyclic contraction.

Proof. 

It follows for $\left(x ,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$;$\forall i\in \overline{p}$ that:

$$d\left(Tx ,Ty\right)\le K_{i}d\left(x,y\right)+\left(1-K_i\right)D$$

$$d\left(T^{2}x ,T^{2}y\right)\le K_{i+1}d\left(Tx,Ty\right)+\left(1-K_{i+1}\right)D\le K_{i+1}\left(K_{i}d\left(x,y\right)+\left(1-K_i\right)D\right)+\left(1-K_{i+1}\right)D$$

$$=K_i{K}_{i+1}d\left(x,y\right)+\left(1-K_i{K}_{i+1}\right)D$$

$$d\left(T^{p}x ,T^{p}y\right)=K_{T}d\left(x,y\right)+\left(1-K_T\right)D$$

so that

$$D\le \underset{n\to \infty }{lim sup}\left(d\left(T^{pn}x ,T^{pn}y\right)-K_T^{n}d\left(x,y\right)\right)=\underset{n\to \infty }{lim sup}d\left(T^{pn}x ,T^{pn}y\right)\le D$$

(9)

and then there exists $\underset{n\to \infty }{lim}d\left(T^{pn}x ,T^{pn}y\right)=D$ for any $\left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ and any $i\in \overline{p}$.

Also, note that

$$d\left(T^{pn+1}x ,T^{pn+1}y\right)\le K_{i}d\left(T^{pn}x ,T^{pn}y\right)+\left(1-K_i\right)D$$

$$d\left(T^{pn+2}x ,T^{pn+2}y\right)\le K_{i+1}{K}_{i}d\left(T^{pn}x ,T^{pn}y\right)+\left(1-K_{i+1}{K}_i\right)D$$

$$d\left(T^{\left(n+1\right)p-1}x ,T^{\left(n+1\right)p-1}y\right)\le \left({\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)d\left(T^{pn}x ,T^{pn}y\right)+\left(1-{\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)D$$

$$\le \left({\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)K_T^{n}d\left(T^{p}x ,T^{p}y\right)+K_T^n\left(1-{\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)D$$

then,

$$\underset{n\to \infty }{lim}d\left(T^{pn+j}x ,T^{pn+j}y\right)=D; \forall j\in \overline{p}, \forall \left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)}$$

(10)

Property (i) has been proved.

To prove Property (ii), first note that ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\ne \varnothing$ implies and it is implied by $D=0$. Thus, (10) holds with $D=0$. This implies for any $j=0$ and any $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)}$ that $d\left(T^{p\left(n+m\right)}x ,T^{p\left(n+m\right)}y\right)\to 0$ as $n\to \infty$; $\forall m\in Z_{0+}$, and

$$d\left(T^{pn+j}x ,T^{pn+j}y\right)\le \left({\displaystyle {\Pi }_{\mathcal{l}=i}^k\left[K_{\mathcal{l}}\right]}\right)d\left(T^{pn}x ,T^{pn}y\right)$$

(11)

for $j\in \overline{p}$ where $k=j-1$ if $j\ge i+1$ and $k=p+j-1$, otherwise, and then

$$d\left(T^{pn+j}x ,T^{pn+j}y\right)\le \left({\displaystyle {\Pi }_{\mathcal{l}=i}^k\left[K_{\mathcal{l}}\right]}\right)d\left(T^{pn}x ,T^{pn}y\right)$$

(12)

so that one has for any $m\in Z_{+}$:

$${\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{pn+j}y\right)}$$

$$\le \left(1+K_T+K_T^2+\dots +K_T^{m-1}\right)d\left(T^{pn}x ,T^{pn}y\right)={\displaystyle \frac{1-K_T^m}{1-K_T}}d\left(T^{pn}x ,T^{pn}y\right)<\infty$$

(13)

then, for any $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)}$, any $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ and any $j\in \overline{p}$, one gets by taking limits in the above expression as $n\to \infty$ since $D=0$:

$$\underset{n\to \infty }{lim}\left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{pn+j}y\right)}\right)=0$$

(14)

and by taking $y=T^{pm+1}x$; $\forall m\in Z_{+}$, $\forall k\in \overline{p}$ in the above expression:

$$\underset{n\to \infty }{lim}\left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{p\left(m+n\right)+j+1}x\right)}\right)=0$$

(15)

$$\underset{n\to \infty }{lim}\left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{p\left(m+n\right)+j}x\right)}\right)$$

$$\le \left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{p\left(m+n\right)+j+1}x\right)}\right)+\left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{p\left(m+n\right)+j+1}x ,T^{p\left(m+n\right)+j}x\right)}\right)$$

$$=0+0=0$$

(16)

$$\underset{n\to \infty }{lim}\left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{pn+k}x\right)}\right)=\underset{n\to \infty }{lim}\left({\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j}x ,T^{p\left(n+1\right)+k}x\right)}\right)$$

$$\le \underset{n\to \infty }{lim}\left({\displaystyle {\sum }_{\mathcal{l}=1}^k{\displaystyle {\sum }_{j=0}^{m}d\left(T^{pn+j+\mathcal{l}-1}x ,T^{p\left(n+1\right)+\mathcal{l}}x\right)}}\right)=k.0=0$$

(17)

Thus, there is no subsequence ${\left\{T^{p\left(n_k+j\right)}x\right\}}_{k=0}^{\infty }\subset {\left\{T^{p\left(n+j\right)}x\right\}}_{n=0}^{\infty }$ with ${\left\{n_k\right\}}_{k\in Z_{0+}}\subset Z_{+}$ being strictly sequence such that there is some $\epsilon \in R_{+}$ and some $N=N\left(\epsilon \right)\in Z_{0+}$ such that $T^{p\left(n_k+j\right)}x\ge \epsilon$ since the following will hold ${\displaystyle {\sum }_{j=0}^{m}d\left(T^{p\left(n+j\right)}x ,T^{p\left(n+j\right)+1}x\right)}\to \infty$, a contradiction. Therefore, ${\left\{T^{p\left(n+j\right)}x\right\}}_{n=0}^{\infty }$ is a Cauchy sequence which converges to some $z\left(\in {\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\right)$ for any given $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ and any $\forall j,k\in \overline{p-1} \cup \left\{0\right\}$. Since $\underset{n\to \infty }{lim}d\left(T^{pn+j}x ,T^{pn+j}y\right)=0$, take $y=T^{p}x$ to conclude, after permuting the order of the “limit” and “distance” functions, because of the Lipschitz continuity of $T^p$, that

$$\underset{n\to \infty }{lim}d\left(T^{pn+j}x ,T^{pn+j}\left(T^{p}x\right)\right)=\underset{n\to \infty }{lim}d\left(\left(T^{pn+j}x\right) ,T^p\left(T^{pn+j}x\right)\right)$$

$$=d\left(\underset{n\to \infty }{lim}\left(T^{pn+j}x\right) ,\underset{n\to \infty }{lim}{T}^p\left(T^{pn+j}x\right)\right)=d\left(z , T^{p}z\right)=0$$

(18)

Then, $z=T^{p}z$ so that $z\in Fix\left(T^p\right)$. Also, $z\in Fix\left(T^p\right)=\left\{z\right\}$. Assume, on the contrary, that $Fix\left(T^p\right)=\left\{z,z_1\right\}$. Then,$0\ne d\left(z_1,z\right)=d\left(T^p{z}_1 , T^{p}z\right)\le K_{T}d\left(z, z_1\right)<d\left(z, z_1\right)$, a contradiction so that $z_1=z$ and $Fix\left(T^p\right)=\left\{z \right\}$. On the other hand, if $z\notin Fix\left(T\right)$ then $0<d\left(z , Tz\right)=d\left(T^{p}z , T^{p+1}z\right)\le K_{T}d\left(z ,Tz\right)<d\left(z , Tz\right)$, a contradiction. Then, $z\in Fix\left(T\right)$ and it is unique. Assume that $Fix\left(T\right)=\left\{z , z_0\right\}$ then $0\ne d\left(z,z_0\right)=d\left(Tz,T{z}_0\right)=d\left(T^{p}z,T^p{z}_0\right)<d\left(z,z_0\right)$, a contradiction so that $Fix\left(T\right)=\left\{z \right\}$.

Property (iii) follows since by taking $y=Tx$ in the previous inequality to (10), which leads to:

$$d\left(T^{\left(n+1\right)p-1}x ,T^{\left(n+1\right)p}x\right)\le \left({\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)d\left(T^{pn}x ,T^{pn+1}x\right)+\left(1-{\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)D$$

$$\le \left({\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)K_T^{n}d\left(T^{p}x ,T^{p+1}x\right)+K_T^n\left(1-{\displaystyle {\Pi }_{j=1}^{p-1}\left[K_j\right] }\right)D$$

and then $d\left(T^{\left(n+1\right)p-1}x ,T^{\left(n+1\right)p}x\right)\to 0$ as $n\to \infty$ if $x$ is finite. For a fixed $z\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ lying in the same subset as $x$, one has

$$d\left(T^{\left(n+1\right)p}x ,z\right)\le d\left(T^{\left(n+1\right)p-1}x ,T^{\left(n+1\right)p}x\right)+d\left(T^{\left(n+1\right)p-1}x ,z\right)$$

If ${\left\{T^{\left(n+1\right)p}x\right\}}_{n=0}^{\infty }$ is unbounded for some $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ then there is some subsequence ${\left\{T^{\left(n_k+1\right)p}x\right\}}_{k=0}^{\infty }$ with $n_{k+1}>n_k$ such that $d\left(T^{\left(n_{k+1}+1\right)p}x ,z\right)>d\left(T^{n_{k}p}x ,z\right)$ and then

$$0<d\left(T^{\left(n_{k+1}+1\right)p}x , z\right)-d\left(T^{n_{k}p}x , z\right)\le d\left(T^{\left(n_{k+1}+1\right)p}x , T^{n_{k}p}x\right)+d\left(T^{n_{k}p}x , z\right)-d\left(T^{n_{k}p}x , z\right)$$

$$=d\left(T^{\left(n_{k+1}+1\right)p}x , T^{n_{k}p}x\right)$$

then, the subsequent contradiction is reached:

$$0<\underset{k\to \infty }{lim inf} d\left(T^{\left(n_k+1\right)p+1}x ,z\right)\le \underset{k\to \infty }{lim }d\left(T^{\left(n_{k+1}+1\right)p}x ,T^{n_{k}p}x\right)=0.$$

Thus, ${\left\{T^{\left(n+1\right)p}x\right\}}_{n=0}^{\infty }$ is bounded for all finite $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$. Property (iii) has been proved.

Property (iv) follows directly from the above properties since a $p\left(\ge 2\right)$-cyclic contraction is also an extended $p\left(\ge 2\right)$-cyclic contraction. □

Note that, as a consequence of Definition 1 of extended cyclic contractions, Theorem 1 only requires as a necessary condition for its fulfilment that $K_i<1$, for some $i\in \overline{p}$, provided that ${\displaystyle {\Pi }_{j\left(\ne i\right)=1}^p\left[K_j\right]}<K_i^{-1}$ if $K_i\in \left(0 , 1\right)$. Note also that trivially (1) a $p\left(\ge 2\right)$-cyclic contraction is also an extended $p$-cyclic contraction with all the contraction constants being identical and less than unity; (2) if $T$ is an extended $p$-cyclic contraction, then $T^p$ is a $p\left(\ge 2\right)$-cyclic contraction with contractive constant $K_T={\displaystyle {\Pi }_{i=1}^p\left[K_i\right]}$. In particular, it is seen that the constraint $d\left(T^{p}x ,T^{p}y\right)=K_{T}d\left(x,y\right)+\left(1-K_T\right)D$ in the previous inequality to (9) in the proof of Theorem 1 together with $K_T={\displaystyle {\Pi }_{i=1}^p\left[K_i\right]}$$<1$ ensures an extended cyclic contraction for any built subsequence of $p$ elements on the $p$ adjacent subsets irrespectively of the feature that some of the individual constants are not less than unity. This condition leads to (9) which is the basis to establish that, for positive integers $n$, sequences of $np$ elements built through compositions of the extended cyclic mapping T have similar properties as $n\to \infty$ as those of the sequences of $n$ elements of a cyclic self-mapping T. This point of view is also formally reflected in several of the subsequent results.

Lemma 3. 

Assume that $A_i$are nonempty and closed subsets of a metric space $\left(X , d\right)$with $D=d\left(A_i , A_{i+1}\right)\ge 0$; $\forall i\in \overline{p}$and that $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$is a $p\left(\ge 2\right)$-cyclic contraction self-mapping. Assume also that either $A_i$or $A_{i+1}$is boundedly compact; thus, if each ordered par of subsets in the set $\left\{\left(A_1,A_2\right) ,\left(A_3,A_4\right) ,\dots ,\left(A_{p-1},A_p\right)\right\}$has either its first element or its second element boundedly compact if $p$is even and, if $p$is odd, then the same rule applies to the set of pairs $\left\{\left(A_1,A_2\right) ,\left(A_3,A_4\right) ,\dots ,\left(A_{p-2},A_{p-1}\right)\right\}$and $A_p$is boundedly compact if $A_{p-1}$is not boundedly compact. Then, there is some finite $x_i\in A_i$such that $d\left(x_i, T{x}_i\right)=D$for each $i\in \overline{p}$.

The above property also holds if $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$is an extended $p\left(\ge 2\right)$-cyclic contraction self-mapping.

Proof. 

Since $D=d\left(A_i , A_{i+1}\right)\ge 0$, then $D=inf\left\{d\left(x , y\right) : x\in A_i , y\in A_{i+1}\right\}$ for any $i\in \overline{p}$ and $A_{i0}\left(\subset A_i\right)=\left\{x\in A_i :d\left(x , A_{i+1}\right)=D\right\}\left(\ne \varnothing \right)$; $\forall i\in \overline{p}$ so that $d\left(A_i ,A_{i+1}\right)=d\left(A_{i0} , A_{i+1,0}\right)$; $\forall i\in \overline{p}$. If $D=0$ then ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i}\ne \varnothing$ and $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is a (strict) contraction so that any sequence ${\left\{T^{n}x\right\}}_{n=0}^{\infty }\to z\left(=Tz\right)\in {\displaystyle {\cap }_{i\in \overrightarrow{p}}{A}_i}$, that is, $z=Tz$ is the unique fixed point of $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ which belongs to any $A_i$ for $i\in \overline{p}$. Since $d\left(z , Tz\right)=D=0$, the result follows. Now, consider the case $D>0$ for which the subsets of the cyclic disposal do not intersect. Take any $x\in A_i$ for any arbitrary $i\in \overline{p}$ so that ${\left\{T^{np+1}x\right\}}_{n=0}^{\infty }\subset A_{i+1}$ and ${\left\{T^{np}x\right\}}_{n=0}^{\infty }\subset A_i$. By the hypothesis that $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is a $p\left(\ge 2\right)$-cyclic contraction self-mapping, it follows that $d\left(T^{np}x , T^{np+1}x\right)\to D$ for any $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ as $n\to \infty$ (see Remark 2), so that for any given arbitrary constant $\epsilon \in R_{+}$ such that $d\left(T^{n_{k}p}x , A_{i0}\right)\le \epsilon /3$, for a subsequence ${\left\{T^{n_{k}p}\right\}}_{k=0}^{\infty }\subset {\left\{T^{np}x\right\}}_{n=0}^{\infty }$ for any $k\left(\ge N_1\right)\in Z_{0+}$, some strictly increasing ${\left\{n_k\right\}}_{k=0}^{\infty }\subset Z_{0+}$ and some $N_1=N_1\left(\epsilon \right)\in Z_{0+}$. In the same way, $d\left(T^{n_{k}p+1}x , A_{i+1,0}\right)\le \epsilon /3$ for a subsequence ${\left\{T^{n_{k}p+1}\right\}}_{k=0}^{\infty }\subset {\left\{T^{np+1}x\right\}}_{n=0}^{\infty }$, for all $k\left(\ge N_2\right)\in Z_{0+}$ and some $N_2=N_2\left(\epsilon \right)\in Z_{0+}$. Since either $A_i$ or $A_{i+1}$ is boundedly compact, the best proximity sets $A_{i0}\subset A_i$ and $A_{i+1,0}\subset A_{i+1}$ are nonempty. Now, assume that for any $z_i\in A_i$, there exists a real number $D_i>D$ such that $d\left(z_i , T{z}_i\right)=D_i$ so that

$$D<D_i=d\left(z_i , T{z}_i\right)\le d\left(z_i , T^{n_{k}p}x\right)+d\left(T^{n_{k}p}x , T^{n_{k}p+1}x\right)+d\left(T^{n_{k}p+1}x , T{z}_i\right).$$

Since $d\left(T^{np}x , T^{np+1}x\right)\to D$ as $n\to \infty$, then the subsequence of distances ${\left\{d\left(T^{n_{k}p}x , T^{n_{k}p+1}x\right)\right\}}_{k=0}^{\infty } \left(\subset {\left\{d\left(T^{np}x , T^{np+1}x\right)\right\}}_{n=0}^{\infty }\right)\to D$ as $k\to \infty$. Then, there is $N_3=N_3\left(\epsilon \right)$ such that $d\left(T^{n_{k}p}x , T^{n_{k}p+1}x\right)\le D+\epsilon /3$ for any $k\left(\ge N_3\right)\in Z_{0+}$. Thus, if $N=max\left(N_1 , N_2 , N_3\right)$ then, one has $D<D_i\le d\left(z_i , T{z}_i\right)\le D+\epsilon$, which implies that $\epsilon \ge D_i-D$. Since $\epsilon$ is arbitrary, it suffices to take $\epsilon \in \left(0 , D_i-D\right)$ to get a contradiction to $inf\left\{d\left(z_i , T{z}_i\right) : z_i\in A_i\right\}=D_i>D$ for any $i\in \overline{p}$ and then there is $x_i\in A_{i0}\left(\subset A_i\right)$ such that $d\left(x_i, T{x}_i\right)=D$; $\forall i\in \overline{p}$. The first part of the result has been proved. Now, if $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction, then for any integer $n\ge p$, there exists $m\in Z_{0+}$ such that $n=mp+j$ for some integer $j=j\left(n\right)\in \overline{p-1} \cup \left\{0\right\}$ and $m\to \infty$ as $n\to \infty$ for any $j\in \overline{p-1} \cup \left\{0\right\}$. Then, from the property $d\left(T^{np}x , T^{np+1}x\right)\to D$ as $n\to \infty$ for a $p\left(\ge 2\right)$-cyclic contraction and these considerations, it follows that $d\left(T^{mp+j}x , T^{mp+j+1}x\right)\to D$ as $m\to \infty$; $\forall j\in \overline{p-1} \cup \left\{0\right\}$ and any given $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ since:

$$D\leftarrow d\left(T^{mp}x , T^{mp+1}x\right)\le K_{T}d\left(T^{\left(m-1\right)p}x , T^{\left(m-1\right)p+1}x\right)+\left(1-K_T\right)D\to D\mathrm{as}m\to \infty$$

$$D\leftarrow d\left(T^{mp+1}x , T^{mp+2}x\right)\le K_{i+1}d\left(T^{mp}x , T^{mp+1}x\right)+\left(1-K_{i+1}\right)D\to D\mathrm{as}m\to \infty$$

$$D\leftarrow d\left(T^{mp+i}x , T^{mp+i+1}x\right)\le K_{i}d\left(T^{mp}x , T^{mp+1}x\right)+\left(1-K_i\right)D\to D\mathrm{as}m\to \infty$$

and the result follows in the same way as for the case of $p\left(\ge 2\right)$-cyclic contractions. The result also holds if $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction self-mapping by virtue of (10). □

It can be pointed out that the hypothesis of boundedly compactness of the alternate subsets in the cyclical disposal in Lemma 3 is invoked since, although the distances are finite, meaning $D=\underset{x\in A_i, y\in A_{i+1}}{inf}d\left(x,y\right)<+\infty$, it can happen that the set distance can correspond to points at infinity subject to finite mutual distances since the closed subsets are not assumed to be bounded.

The following result relies on the feature that if all the subsets of the cyclic disposal are nonempty and closed, it is only needed for one of the best-proximity points to be a singleton in order to achieve to convergence of the self-mapping $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ to best-proximity points even if the respective best proximity sets (the mentioned one excepted) are not singletons.

Lemma 4. 

The following properties hold:

(i)  Let $\left(X, d\right)$  be a metric space and let $A_i\subset X$for $i\in \overline{p}$be a set of $p$  nonempty closed subsets of $X$  such that $D=d\left(A_i ,A_{i+1}\right)>0$. Assume that $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$is a$p\left(\ge 2\right)$-cyclic contraction and assume also that, for some $j\in \overline{p}$,$A_j$  is boundedly compact and that $A_{j0}=\left\{z_j\right\}$  is a singleton. Then, the set $\left\{z_1,z_2,,\dots ,z_p\right\}$  of respective fixed points of $T^p$  from $A_i$  to $A_{i+1}$  is unique. This set is also the limit set of proximity points of $T$  from each subset to its adjacent one to which all the sequences generated by the cyclic contractive self-mapping converge within each subset.

(ii)  Property (i) holds if $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$  is an extended $p\left(\ge 2\right)$-cyclic contraction self-mapping.

Proof. 

Since $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is a $p\left(\ge 2\right)$-cyclic contraction, then $d\left(T^{pn}x , T^{pn}y\right)\to D$ as $n\to \infty$ for all pair $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}{\displaystyle \cup A_{i+1}\times A_i}\right)}$. Since all the subsets $A_i$ for $i\in \overline{p}$ are nonempty and closed and $D=inf\left\{d\left(x,y\right) :x\in A_i , y\in A_{i+1}\right\}>0$ for any $i\in \overline{p}$, then all the subsets $A_i$ have nonempty subsets of best-proximity points $A_{i0}\subset A_i$; $\forall i\in \overline{p}$. According to the hypothesis, it exists a subset $A_j\subset X$, for some $j\in \overline{p}$ such that $A_{j0}=\left\{z_j\right\}$ is a singleton. Also, $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is trivially nonexpansive and Lipschitz-continuous with unity Lipschitz constant since the constraint $d\left(x,y\right)\ge D$; $\forall \left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}{\displaystyle \cup A_{i+1}\times A_i}\right)}$ leads to:

$$d\left(Tx , Ty\right)\le Kd\left(x,y\right)+\left(1-K\right)D=\left(K+\left(1-K\right){\displaystyle \frac{D}{d\left(x ,y\right)}}\right)d\left(x,y\right)$$

$$\le \left(K+\left(1-K\right){\displaystyle \frac{D}{D}}\right)d\left(x,y\right)=d\left(x,y\right).$$

The continuity allows for the exchange of the orders of limit and distance functions so that $\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn}y\right)=\underset{n\to \infty }{lim}d\left(\underset{n\to \infty }{lim}{T}^{pn}x , T^{pn}y\right)=d\left(\underset{n\to \infty }{lim}{T}^{pn}x , \underset{n\to \infty }{lim}{T}^{pn}y\right)=D$

Take $x\in A_j$, $y=Tx\in A_{j+1}$; then, one has this result since $A_{j0}=\left\{z_j\right\}$,

$$\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn+1}x\right)=d\left(\underset{n\to \infty }{lim}{T}^{pn}x , \underset{n\to \infty }{lim}{T}^{pn}Tx\right)$$

$$=d\left(\underset{n\to \infty }{lim}{T}^{pn}x , T\left(\underset{n\to \infty }{lim}{T}^{pn}x\right)\right)$$

$$=d\left(z_j ,T{z}_j\right)=D$$

which follows from the analysis of the three subsequent possible cases for $x\in A_j$:

Case a: ${\left\{T^{pn}x\right\}}_{n=0}^{\infty }\to z_j\left(=T^p{z}_j\right)$. Since $T$ is Lipschiz, since it is nonexpansive, it follows that $\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn+1}x\right)=d\left(z_j , T\underset{n\to \infty }{lim}{T}^{pn}x\right)=d\left(z_j , T{z}_j\right)=D$.

Case b: ${\left\{ T^{pn}x\right\}}_{n=0}^{\infty }\to z_j^{´}\left(\ne z_j\right)\in A_j$ and since $A_{j0}=\left\{z_j\right\}$ then $d\left(z_j^{´} , T{z}_j^{´} \right)>D$ for all $z_j^{´}\ne z_j$. Then, $D=\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn+1}x\right)=d\left(z_j^{´} , T{z}_j^{´} \right)>D$ is a contradiction and Case b is impossible.

Case c: ${\left\{ T^{pn}x\right\}}_{n=0}^{\infty }$ does not converge in $A_j$. Since the set $A_j$ is boundedly compact, a subsequence ${\left\{T^{p{n}_k}\right\}}_{k=0}^{\subset }\infty \left({\left\{T^{pn}\right\}}_{n=0}^{\infty }\right)\subset A_j$ is convergent in the closed set $A_j$ to some ${\overline{z}}_j\left(\ne z_j\right)\in A_j$ for some strictly increasing sequence of nonnegative integers ${\left\{n_k\right\}}_{k=0}^{\infty }$. Then, the following contradiction follows:

$$D<\underset{k\to \infty }{lim}d\left({\overline{z}}_j , T\left(T^{p{n}_k}x\right)\right)=d\left({\overline{z}}_j , T{\overline{z}}_j\right)=\underset{k\to \infty }{lim}d\left(T^{p{n}_k}{\overline{z}}_j , T\left(T^{p{n}_k}x\right)\right)$$

$$=\underset{k\to \infty }{lim}d\left(T^{p{n}_k}{\overline{z}}_j , T^{p{n}_k}\left(Tx\right)\right)\le \left(1-\underset{k\to \infty }{lim}{K}_T^n\right)D+\underset{k\to \infty }{lim}{K}_T^{n}d\left({\overline{z}}_j, Tx\right)=D.$$

Thus, Case c is also impossible. As a result, ${\left\{ T^{pn}x\right\}}_{n=0}^{\infty }\to z_j$ for any $x\in A_j$. Now, consider, for any $x\in A_j$, the sequence ${\left\{T^{pn+k}x\right\}}_{n=0}^{\infty }\subset A_{j+k}$ if $0\le k\le p-j$ and ${\left\{T^{pn+k}x\right\}}_{n=0}^{\infty }\subset A_{j+k-p}$ if $p-j<k\le 2p-j$. Then, one has for $k\in \overline{p-1}$ since ${\left\{ T^{pn}x\right\}}_{n=0}^{\infty }\to z_j$,

$$D=\underset{n\to \infty }{lim}d\left(T^k\left(T^{pn}x\right) , T^{k+1}\left(T^{pn}x\right)\right)=d\left(T^k\underset{n\to \infty }{lim}\left(T^{pn}x\right) , T^{k+1}\underset{n\to \infty }{lim}\left(T^{pn}x\right)\right)$$

$$=d\left(T^k{z}_j , T^{k+1}{z}_j\right)=d\left(z_{\left(j+k\right)mod p} , z_{\left(j+k+1\right)mod p}\right)$$

where $z_{\left(j+k\right)mod p}=T^{np+k}{z}_j$ for any $n\in Z_{0+}$ is a fixed point of $T^p$ in $A_{\left(j+k\right)mod p}$

For sequences generated from initial points $\left(x,y\right)$ in adjacent subsets, one has that:

$$\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn}y\right)=\underset{n\to \infty }{lim}d\left(\underset{n\to \infty }{lim}{T}^{pn}x , T^{pn}y\right)$$

$$=\underset{n\to \infty }{lim}d\left(z_j , T^{pn}y\right)=d\left(z_j , \underset{n\to \infty }{lim}{T}^{pn}y\right)$$

$$=D$$

so that $T{z}_j\in \left\{A_{j+1,0}\right\}$ and $\underset{n\to \infty }{lim}{T}^{pn}y$ exists while it is in $A_{j+1,0}$. But, since $A_{j0}=\left\{z_j\right\}$, then $T^{p+1}\left(\underset{n\to \infty }{lim}{T}^{pn}y\right)=\underset{n\to \infty }{lim}{T}^{p\left(n+1\right)+1}y\in A_{j0}\left(=\left\{z_j\right\}\right)$ and $T^{p+2}\left(\underset{n\to \infty }{lim}{T}^{pn}y\right)=T{z}_j\in A_{j+1,0}$. Also, one has that

$$d\left(T^p\left(\underset{n\to \infty }{lim}{T}^{pn}x\right) , T^{p+1}\left(\underset{n\to \infty }{lim}{T}^{pn}x\right)\right)=d\left(T^p{z}_j,T^{p+1}{z}_j\right)=D=d\left(z_j, T{z}_j\right)$$

and one concludes for both the sequences initialized in $x$ and $Tx$ in $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ and the sequences initialized in $\left(x,y\right)\in {\displaystyle {\cup }_{i\in \overline{p}}\left(A_i\times A_{i+1}{\displaystyle \cup A_{i+1}\times A_i}\right)}$ that the unique finite best-proximity point $z_j$ of $A_j$ is the unique fixed point of $T^p:A_j\to A_{j+1}$ and $T{z}_j$ is a fixed point of $T^p:A_{j+1}\to A_{j+2}$ which is unique, since $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is single-valued, and finite since $z_j$ is finite and $T$ is continuous. Thus, the image of $T{z}_j$ is a singleton and it is possible to obtain similar conclusions for $T^i{z}_j$ for $i\in \overline{p-1} \cup \left\{0\right\}$. That is, although the only guaranteed singleton set of best-proximity points is $A_{j0}$, the set $\left\{z_1,z_2,,\dots ,z_p\right\}$ of fixed points of $T^p$ is unique, which is also the limit set of proximity points of $T$ to which all the sequences generated by the contractive self-mapping converge at each one of the subsets of the cyclic disposal. Property (i) has been proved.

Property (ii) follows directly from Property (i) since a $p\left(\ge 2\right)$-cyclic contraction is also an extended $p\left(\ge 2\right)$-cyclic contraction. □

Theorem 2. 

Let $\left(X,‖‖\right)$be a uniformly convex Banach space, let $A_i\subseteq X$; $\forall i\in \overline{p}$$\left(p\ge 2\right)$be nonempty closed strictly convex subsets of $X$with $d\left(A_i,A_{i+1}\right)=D>0$; $\forall i\in \overline{p}$, where $d:X\times X\to R_{0+}$is the norm-induced metric defined by $d\left(x,y\right)=‖x-y‖$; $\forall x,y\in X$. Let $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$be a $p\left(\ge 2\right)$-cyclic contraction with contractive constant $K\in \left[0 , 1\right)$and take $x\in A_{\mathcal{l}}$for some arbitrary $\mathcal{l}\in \overline{p}$. Then, the following properties hold:

(i) ${\left\{T^{\left(n+1\right)p+j+k-\mathcal{l}}x\right\}}_{n=0}^{\infty }\subset A_{\left(j+k\right)mod p}$; ${\left\{T^{\left(n+m+1\right)p+j+k-\mathcal{l}}x\right\}}_{n=0}^{\infty }\subset A_{\left(j+k\right)mod p}$; $\forall k\in \overline{p}$ are Cauchy sequences, and

$$d\left(T^{\left(n+m+1\right)p+j+k-\mathcal{l}}x , T^{\left(n+1\right)p+j+k-\mathcal{l}}x\right)\to 0 as n\to \infty ;$$

$$d\left(T^{\left(n+m+1\right)p+j+k+1-\mathcal{l}}x , T^{\left(n+1\right)p+k+j+1-\mathcal{l}}x\right)\to 0 as n\to \infty ;$$

$$d\left(T^{\left(n+m+1\right)p+j+k+1-\mathcal{l}}x , T^{\left(n+1\right)p+j+k-\mathcal{l}}x\right)\to D as n\to \infty ;$$

$$d\left(T^{\left(n+m+1\right)p+j+k-\mathcal{l}}x , T^{\left(n+1\right)p+j+k+1-\mathcal{l}}x\right)\to D as n\to \infty ,$$

for any $m\in Z_{0+}$ and any $k\in \overline{p-1} \cup \left\{0\right\}$.

(ii)${\left\{T^{\left(n+1\right)p+j+k-\mathcal{l}}x\right\}}_{n=0}^{\infty }\to z_{\left(j+k\right)mod p}\left(\subset A_{\left(j+k\right)mod p}\right)$ and $z_i\in A_i$ are unique best-proximity points in each $A_i$, that is, $d\left(z_{i }, T{z}_i\right)=d\left(A_i , A_{i+1}\right)=D$; $\forall i\in \overline{p}$.

(iii)  Propositions [(i)–(ii)] also hold if $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$  is an extended $p\left(\ge 2\right)$-cyclic contraction.

Proof. 

Note that $T^{j-\mathcal{l}}x\in A_j$ if $1\le \mathcal{l}\le j$ and $T^{j+p-\mathcal{l}}x\in A_j$ if $j<\mathcal{l}\le p$. Then, in both cases, $T^{j+p-\mathcal{l}}x\in A_j$. Property (i) follows from [1,2], (see Lemmas 3.7 and 3.8 in [1]), and Remark 2, for the norm-induced metric $d:X\times X\to R_{0+}$,$d\left(x,y\right)=‖x-y‖$; $\forall x,y\in X$. The fist part of Property (ii) is also direct from ([1], Lemmas 3.7 and 3.8). The convergence of the Cauchy sequences of Property (i) to unique best-proximity points in each subset of the cyclic disposal follows from ([1], Theorem 3.10). Property (iii) follows from Properties (i)–(ii) and Lemma 5(i). □

It is noticed that the part of Theorem 2(i) concerned with the convergence of distances to $D$ for sequences in adjacent subsets holds in any metric space $\left(X,d\right)$ without invoking the uniform convexity of $X$ and the involved sequences are bounded, see Equation (4). Also, the convexity of the sets of the cyclical disposal is invoked in the above theorem for the uniqueness of all the best-proximity points at each subset $A_i$. However, under the convexity of two adjacent subsets $A_j$ and $A_{j+1}$, which guarantees the uniqueness of the corresponding best proximity in $A_j$ point to its adjacent subset $A_{j+1}$ (see [1], Theorem 3.10 for $p=2$), it is guaranteed the convergence of the distances generated from any initial $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ as well as the convergence of sequences to one of the best-proximity points in the cyclic disposal through the cyclic contractive self-mapping $T$ on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$. This fact is addressed in the subsequent result valid for $p\ge 2$:

Corollary 1. 

In Theorem 2, assume that two adjacent subsets $A_j$and $A_{j+1}$are strictly convex for some $j\in \overline{p}$. Then, Property (i) of Theorem 2 holds. Furthermore,

(a) 

There are unique best-proximity points $z_j$ to $A_{j+1}$ in the strictly convex set $A_j$ and $z_{j+1}=T{z}_j$ in $A_{j+1}$ to $A_j$, that is, $A_{j0}=\left\{z_j\right\}$ and $A_{j+1,0}=\left\{z_{j+1}\right\}$ such that $d\left(z_{j }, T{z}_j\right)=d\left(A_j , A_{j+1}\right)=D$.

(b) 

The best proximity sets $\left\{A_{i0}\right\}$  for $i\in \overline{p}$  are nonempty and there exist $z_i\in A_{i0}$;$\forall i\left(\ne j\right)\in \overline{p}$   ($A_{i0}$  for $i\ne j,j+1$  are not non-necessarily singletons) such that $d\left(z_i ,T{z}_i\right)=D$;$\forall i\left(\ne j\right)\in \overline{p}$. The set $\left\{z_1 , z_2 ,\dots ,z_p\right\}$  is also the unique set of fixed points of  $T^p$  at each of the subsets $A_i$  for $i\in \overline{p}$. The above properties also hold if  $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$  is an extended $p\left(\ge 2\right)$-cyclic contraction.

Sketch of Proof. 

If $D=0$ then the mapping $T$ is a strict contraction and $z=z_j\in {\displaystyle {\cap }_{i\in \overline{p}}{A}_i}$ is the unique fixed point of $T$ and the set $\left\{z_1 , z_2 ,\dots ,z_p\right\}$ consists of the fixed point z and the proof is immediate. Proceed now with the case $D>0$. Since $A_j$ and are strictly convex ([1], Theorem 3.10) then $d\left(z_{j }, T{z}_j\right)=d\left(T^p{z}_j , T\left(T^p{z}_j\right)\right)=d\left(A_j , A_{j+1}\right)=D$ for a unique finite best-proximity point $z_j=T^p{z}_j$ as in the corresponding property of Theorem 2 so that $A_{j0}=\left\{z_j\right\}$ is a singleton. Similarly, $A_{j+1,}0$ is also a singleton. Since $A_i$ are all closed for $i\in \overline{p}$, all the sets of best-proximity points $A_{i0}$ are nonempty for any $i\in \overline{p}$ and since $T$ is non-expansive, then Lipschitz continuous, $T^i{z}_j$ are finite for $i\in \overline{p-1} \cup \left\{0\right\}$. Take any $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ which is necessary in some unique $A_{\mathcal{l}}$ for some $\mathcal{l}\in \overline{p}$ since $D>0$ so that $T^{j-\mathcal{l}}x\in A_j$ if $\mathcal{l}\le j\le p$ and $T^{p+j-\mathcal{l}}x\in A_j$ if $j<\mathcal{l}$. Then, from Theorem 2, one has for the given $j\in \overline{p}$ that ${\left\{T^{\left(n+1\right)p+j-\mathcal{l}}x\right\}}_{n=0}^{\infty }\to z_j\left(\subset A_j\right)$ and $A_{j0}=\left\{z_j\right\}$ is a singleton. Furthermore, from Property (i) of Theorem 2 and Lemma 4, one has the following convergence properties of distances at each subset of the cyclical disposal:${\left\{T^{\left(n+1\right)p+j+k-\mathcal{l}}x\right\}}_{n=0}^{\infty }\subset A_{\left(j+k\right)mod p}$; $d\left(T^{\left(n+1\right)p+j+k-\mathcal{l}}x , T^{\left(n+1\right)p+j+k+1-\mathcal{l}}x\right)\to D$ as $n\to \infty$; $\forall k\in \overline{p-1} \cup \left\{0\right\}$ and, since $z_j$ exists and it is the unique point in $A_j$ such that $d\left(z_{j }, T{z}_j\right)=d\left(A_j , A_{j+1}\right)=D$.

Then, since $z_j$ is unique and the above properties of convergence of distances of sequences irrespective of the initial conditions, it follows that $d\left(z_{i }, z_{i+1}\right)=d\left(T^{np+i-j}{z}_j , T^{np+i+1-j}\right)=d\left(A_i , A_{i+1}\right)=D$ for $n\in Z_{0+}$. Thus, $z_i\in A_{i0}$ are the unique best-proximity points in each $A_i$ which are reachable through the mapping $T$, that is, those fulfilling $d\left(z_{i }, T{z}_i\right)=d\left(A_i , A_{i+1}\right)=D$; $\forall i\in \overline{p}$ (irrespective of $A_{i0}$ being singletons or not if $i\ne j$. □

The next auxiliary result concerned with the uniqueness of the best-proximity point of a strictly convex subset to its adjacent one is later used. The result relies only on the (strict) convexity of the subset $A_j$ of the uniformly convex Banach space $\left(X , ‖‖\right)$ and it is not related to the cyclic contractive self-mapping.

Proposition 1. 

Assume that for some $j\in \overline{p}$, the nonempty and closed adjacent subsets $A_j$and $A_{j+1}$of the uniformly convex Banach space $\left(X , ‖‖\right)$satisfy, furthermore, that $A_j$is strictly convex. Then, the best proximity set to $A_{j+1}$in $A_j$is a singleton $A_{j0}=\left\{z_j\right\}$.

Proof. 

Define $x_{A_j}\left(\lambda \right)=\lambda x+\left(1-\lambda \right)y$ for any given $x,y\in A_j$ and $\lambda \in \left[0 , 1\right]$. Since $A_j$ is a nonempty, strictly convex and closed, the function $x_{A_j}\left(\lambda \right)$ for $\lambda \in \left[0 , 1\right]$ defines a segment $\left[x , y\right]$ of points in $A_j$ of given extreme points $x$ and $y$, such that $\left(x, y\right)\subset int A_j$ since $A_j$ is strictly convex. Note that $D=d\left(A_j ,A_{j+1}\right)=\underset{x_j\in A_j ,x_{j+1}\in A_{j+1}}{inf}d\left(x_j ,y_j\right)=\underset{x_j\in A_j ,x_{j+1}\in A_{j+1}}{inf}‖x_j-y_j‖$ and that the distance of the segment $\left[x , y\right]\subset A_j$, being itself a subset of the uniformly convex Banach space $\left(X , ‖‖\right)$, to $A_{j+1}$ is $d\left( \left[x , y \right], A_{j+1}\right)=\underset{\lambda \in \left[ 0 , 1\right] , x_{j+1}\in A_{j+1}}{inf}‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖\ge D$. Since the absolute value of the derivative of a norm of a differentiable function with respect to an argument is less than or equal to the norm of the derivative, one has that:

$$\underset{\lambda \in \left[ 0 , 1\right]}{min}\underset{\lambda \in \left[ 0 , 1\right] , x_{j+1}\in A_{j+1}}{inf}{‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖}^2$$

satisfies the set of inequalities:

$$\left|{\left.\underset{x_{j+1}\in A_{j+1}}{inf}{\displaystyle \frac{d}{d\lambda }}{‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖}^2\right]}_{\lambda \in \left[0 , 1\right]}\right|$$

$$\le 2\underset{\lambda \in \left[ 0 , 1\right] , x_{j+1}\in A_{j+1}}{inf}‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖ \underset{\lambda \in \left[ 0 , 1\right], x_{j+1}\in A_{j+1} }{max}{\displaystyle \frac{d}{d\lambda }}‖ \left(\lambda \left(x -y \right)+y- x_{j+1}\right)‖$$

$$\le 2\underset{\lambda \in \left[ 0 , 1\right] , x_{j+1}\in A_{j+1}}{inf}‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖ \underset{\lambda \in \left[ 0 , 1\right] ,x_{j+1}\in A_{j+1} }{max}‖ {\displaystyle \frac{d}{d\lambda }}\left(\lambda \left(x -y \right)+y- x_{j+1}\right)‖$$

$$\le 2\underset{\lambda \in \left[ 0 , 1\right] , x_{j+1}\in A_{j+1}}{inf}‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖\underset{x_{j+1}\in A_{j+1} }{max} ‖x-x_{j+1}‖$$

$$\le 2\underset{\lambda \in \left[ 0 , 1\right] , x_{j+1}\in A_{j+1}}{inf}‖\lambda x +\left(1-\lambda \right) y - x_{j+1}‖ ‖x-y‖$$

which is zero if $x=y$ which gives the minimum of the distance function from $\left[x , y\right]$ to $A_{j+1}$. That, is the minimum distance to $A_{j+1}$ from any segment $\left[x , y\right]$ of the nonempty, bounded and strictly convex set $A_j$ occurs when the segment is a point $x=y=z_j$. Now, assume that there is $z_j^{´}\left(\ne z_j\right)\in A_j$ such that $d\left(z_j^{´} , A_{j+1}\right)=d\left(z_j , A_{j+1}\right)=D$ and the best proximity set to $A_{j+1}$ in $A_j$ is a singleton $A_{j0}=\left\{z_j\right\}$. Take the segment $\left[z_j , z_j^{´} \right]\subset A_j$, with $\left(z_j , z_j^{´}\right)\subset int A_j$, the above previous discussion concludes that $d \left(\left[z_j , z_{j }^{´}\right], A_{j+1}\right)=d\left(A_j , A_{j+1}\right)=D$ occurs if and only if $z_j=z_j^{´}$. Thus, the best proximity set of $A_{j+1}$ in $A_j$ is a singleton $A_{j0}=\left\{z_j\right\}$. □

Corollary 1 is now re-addressed under Proposition 1, by invoking the strict convexity of just one of the adjacent subsets so that the best proximity set in such a subset to its adjacent one is a singleton $A_{j0}=\left\{z_j\right\}$. Then, Theorem 1(i) holds, the relevant sequences generated through $T^p$ in $A_j$ from any initial condition in ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ are Cauchy and converge to $\left\{z_j\right\}$. Then, their successive images in $A_{j+\mathcal{l}}$ are (non-necessarily unique) best-proximity points $z_{j+\mathcal{l}}=T^{\mathcal{l}}{z}_j$ in $A_{j+\mathcal{l}}$.

Corollary 2. 

In Theorem 2, assume that $A_j$is strictly convex for some $j\in \overline{p}$. Then, Property (i) of Theorem 2 holds and there is a unique best-proximity point $z_j$to $A_{j+1}$in $A_j$to $A_j$, that is, $A_{j0}=\left\{z_j\right\}$such that $d\left(z_{j }, T{z}_j\right)=d\left(A_j , A_{j+1}\right)=D$. Furthermore, the best proximity sets $\left\{A_{i0}\right\}$for $i\in \overline{p}$are nonempty and there exist $z_i\in A_{i0}$;$\forall i\left(\ne j\right)\in \overline{p}$($A_{i0}$for $i\ne j,j+1$are not non-necessarily singletons) such that $d\left(z_i ,T{z}_i\right)=D$;$\forall i\left(\ne j\right)\in \overline{p}$. The set $\left\{z_1 , z_2 ,\dots ,z_p\right\}$is also the unique set of fixed points of  $T^p$ at each of the subsets $A_i$ for $i\in \overline{p}$. These properties also hold if  $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction.

Sketch of Proof. 

It follows directly from Corollary 1 and Proposition 1 by using the fact that the $p\left(\ge 2\right)$ contractive self-mapping is single-valued. Then, $A_{j0}=\left\{z_j\right\}$ (Proposition 1), $z_{j+1}=T{z}_j\in A_{j+1, 0}$ and $z_{i+1}=T{z}_i$ for $i\in p$ with $z_{p+1}=z_p$ from Corollary 1. □

Corollary 2 is reformulated below leading to achieved similar properties while, instead of assuming that that the space is a uniformly Banach space, this suffices to assume that $\left(X, d\right)$ is a metric space, that one of the subsets is boundedly compact, instead of convex, and that its best-proximity point is a singleton. The proof is directly supported by Lemmas 3 and 4, Theorem 2 and Corollary 1.

Corollary 3. 

Assume that $A_i$are nonempty and closed subsets of a metric space $\left(X, d\right)$with $D=d\left(A_i , A_{i+1}\right)\ge 0$; $\forall i\in \overline{p}$and that $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$is a $p\left(\ge 2\right)$-cyclic contraction self-mapping. Assume also that the subset $A_j$is boundedly compact for some $j\in \overline{p}$and that $A_{j0}=\left\{z_j\right\}$(so that $d\left(z_{j }, T{z}_j\right)=d\left(A_j , A_{j+1}\right)=D$and $d\left(\omega , T\omega \right)>D$;$\forall \omega \left(\ne z_j\right)\in A_j$). Then, all the best proximity sets $\left\{A_{i0}\right\}$for $i\in \overline{p}$are nonempty and there exist $z_i\in A_{i0}$;$\forall i\left(\ne j\right)\in \overline{p}$($A_{i0}$for $i\ne j$are not non-necessarily singletons) such that $d\left(z_i ,z_{i+1}\right)=d\left(z_i ,T{z}_i\right)=D$;$\forall i\left(\ne j\right)\in \overline{p}$. The set $\left\{z_1 , z_2 ,\dots ,z_p\right\}$consists of the unique set of fixed points $z_i$of $T^p$at each one of the subsets $A_i$for $i\in \overline{p}$. The above properties also hold if  $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction.

Proof. 

For the case $D=0$, see the sketch of proof of Corollary 1. Consider the case $D>0$. From (4), note that $d\left(T^{np+1}x , T^{np}x\right)\to D$ as $n\to \infty$; $\forall x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$, that is, for any $x$ in any arbitrary subset $A_{\mathcal{l}}$; $\mathcal{l}\in \overline{p}$. Also, ${\left\{T^{n}x\right\}}_{n=0}^{\infty }\subset A_{\mathcal{l}}$ is bounded from Theorem 1(iii); for any given $x\in A_{\mathcal{l}}$ and any $\mathcal{l}\in \overline{p}$. For each given $\mathcal{l}\in \overline{p}$ such that $x\in A_{\mathcal{l}}$, it always exists $s=s\left(\mathcal{l}\right)\in \overline{p}$ such that ${\left\{T^{n+s}x\right\}}_{n=0}^{\infty }\subset A_j$ and such a sequence is bounded. Since $A_j$ is boundedly compact, there exists some subsequence ${\left\{T^{n_k+s}x\right\}}_{n=0}^{\infty }\subset A_j$, for a strictly increasing sequence ${\left\{n_k\right\}}_{k=1}^{\infty }\subset Z_{+}$, which is convergent ${\left\{T^{n_k+s}x\right\}}_{n=0}^{\infty }\to y=y\left(x\right)\left(\in A_j\right)$. Now, proceed by contradiction by assuming that all convergence points for any convergent subsequence of any sequence of the form ${\left\{T^{n_k+s}x\right\}}_{n=0}^{\infty }\subset A_j$ for any given $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ are not $z_j$, the unique best-proximity point to $A_{j+1}$ in $A_j$. Then, the sequence of distances ${\left\{d\left(T^{np+1}x , T^{np}x\right)\right\}}_{n=0}^{\infty }$ would not converge to $D$, a contradiction. Therefore, $d\left(z_j , T{z}_j\right)=D$. Note that $T^p{z}_j\in A_j$. If $T^p{z}_j\notin A_{j0}$ then $d\left(T^p{z}_j , T^{p+1}{z}_j\right)>D$ so that $T^p{z}_j\in A_{j0}=\left\{z_j\right\}$. As a result, $z_j$ is the unique fixed point of the composed self-mapping $T^p$ in $A_j$. Since ${\left\{d\left(T^{np+s}x , T^{np+s+1}x\right)\right\}}_{n=0}^{\infty }\to D$ for any $x\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ and any $s\in \overline{p-1}\cup \left\{0\right\}$, one gets in a similar way that $d\left(T^s{z}_j , T^{s+1}{z}_j\right)=d\left(z_j,T{z}_j\right)=D$ for $s\in \overline{p-1}\cup \left\{0\right\}$. Then, and since $T$ is single-valued, the above set $\left\{z_1 , z_2,\dots ,z_p\right\}$, subject to the property $z_{i+1}=T{z}_i=T^2{z}_{i-1}=T^i{z}_1$ for $i\in \overline{p-1} \cup \left\{0\right\}$, consists of the unique set of fixed points $z_i$ of $T^p$ at each of the subsets $A_i$ for $i\in \overline{p}$ which are also the best-proximity points reachable through $p\left(\ge 2\right)$ cyclic contractions on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$. Thus, they fulfil $z_i\in A_{i0}$ (but the $A_{i0}$ are not necessarily singletons for $i\left(\ne j\right)\in \overline{p}$.

The above properties also hold if $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction. □

It is possible to comment the following observations concerned with the feasibility of the best proximity set to be a singleton (see Lemma 4, Theorem 2, Corollary 1 and Corollary 3). The assumption that the distance from any point of the space to one of its nonempty closed subsets is reached at a unique point holds if the metric space is a uniformly convex Banach space and the considered subset is strictly convex, its set of best-proximity points to its adjacent subset $A_{j+1}$ is a singleton if such and adjacent subset is also strictly convex. Both proximity sets are singletons which fix the distance between such adjacent subsets while they are the unique points which set such a distance. However, in some cases, the convexity of the considered set may be overcome. For instance, assume that the focused on subset is imperfect, but still closed, that is, it has at least one isolated point which is obviously an adherent point $z_j\in A_j$ to the subset $A_j$ which is not either, also obviously, an accumulation point. If $z_j$ is located in the Banach space such that $d\left(z_j, A_{j+1}\right)<d\left(A_j\backslash \left\{z_j\right\}, A_{j+1}\right)$ then $D=d\left(A_j , A_{j+1}\right)=d\left(z , A_{j+1}\right)$ and then $A_{j0}=\left\{z_j\right\}$ while, obviously, $A_j$ is not convex.

The subsequent result is a slight extension of parallel results in [1] when the cyclic contraction involves more than two sunsets of a uniformly convex Banach space.

Lemma 5. 

Let $\left(X ,‖.‖\right)$be a uniformly convex Banach space with nonempty sets $A_i$for $i\in \overline{p}$with norm-induced metric $d\left(X,X\right)\to R_{0+}$defined by $d\left(x,y\right)=‖x-y‖$and then satisfying $d\left(A_i, A_{i+1}\right)=\underset{x\in A_i , y\in A_{i+1}}{inf}‖x-y‖=D$; $\forall i\in \overline{p}$. Then, the following properties hold:

(i)Assume that, for any given $j\in \overline{p}$, $A_j$ is closed and convex and $A_{j+1}$ is closed. Let ${\left\{x_{nj}\right\}}_{n=1}^{\infty }$and ${\left\{z_{nj}\right\}}_{n=1}^{\infty }$ be sequences in $A_j$ and ${\left\{y_{nj}\right\}}_{n=1}^{\infty }$ be a sequence in $A_{j+1}$ satisfying $d\left(x_{nj},y_{nj}\right)\to D$; $d\left(z_{nj},y_{nj}\right)\to D$ as $n\to \infty$. Then, $d\left(x_{nj},z_{nj}\right)\to 0$ as $n\to \infty$.

(ii)  Assume that $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$  is an extended $p\left(\ge 2\right)$-cyclic contraction and that, for any given $x\in A_i$, $z\in A_k$  and $y\in A_{\mathcal{l}}$  for some arbitrary $i,k,\mathcal{l}\in \overline{p}$, if$d\left(T^{j-i+pn}x , T^{j+1-\mathcal{l}+p\left(n+q\right)}y\right)\to D$; $d\left(T^{j-k+p\left(n+m\right)}z , T^{j+1-\mathcal{l}+p\left(n+q\right)}y\right)\to D$  as $n\to \infty$for any $m, q\in Z_{0+}$then

$$d\left(T^{j-i+pn}x , T^{j-k+p\left(n+m\right)}z\right)\to 0 as n\to \infty$$

for any $m\in Z_{0+}$.

Proof. 

If ${\left\{x_{np}\right\}}_{n=0}^{\infty }\subseteq A_j$, ${\left\{z_{np}\right\}}_{n=0}^{\infty }\subseteq A_j$${\left\{y_{np}\right\}}_{n=0}^{\infty }\subseteq A_{j+1}$, the proof of Property (i) follows directly from ([1], Lemma 2.5). The proof of Property (ii) follows from Property (i) as follows. Since $x\in A_i$ then $T^{j-i}x\in A_j$ if $p\ge j\ge i$ and $T^{p+j-i}x\in A_j$ if $1\le j<i$. Thus, ${\left\{T^{j-i+pn}x\right\}}_{n=1}^{\infty }\subset A_j$.

Under close reasoning, one concludes that ${\left\{T^{j-k+p\left(n+m\right)}x\right\}}_{n=1}^{\infty }\subset A_j$ and ${\left\{T^{j+1-\mathcal{l}+p\left(n+q\right)}x\right\}}_{n=1}^{\infty }\subset A_{j+1}$.

Then, Property (ii) follows directly from Property (i) since $T^{j-i+pn}x, T^{j-k+pn}z\in A_j$ and $T^{j+1-\mathcal{l}+pn}y\in A_{j+1}$; $\forall n\in Z_{0+}$. □

The next result relies on the fact that the basic results of Lemma 5 hold if just one of the subsets of the cyclic disposal is closed and boundedly compact but not necessarily convex.

Lemma 6. 

Let $\left(X ,‖.‖\right)$ be a uniformly convex Banach space with nonempty sets $A_i$ for $i\in \overline{p}$with norm- induced metric $d\left(X,X\right)\to R_{0+}$defined by $d\left(x,y\right)=‖x-y‖$and then satisfying $d\left(A_i, A_{i+1}\right)=\underset{x\in A_i , y\in A_{i+1}}{inf}‖x-y‖=D$; $\forall i\in \overline{p}$. Assume that $A_j$ is boundedly compact and that $A_{j0}=\left\{z_j\right\}$ for some $j\in \overline{p}$. Then, Property (ii) of Lemma 5 holds, $z_j=T^p{z}_j$ and, furthermore, $z_{i+1}\left(=T{z}_i\right)\in A_{i+1,0}$ (which are not necessarily singletons for $i\ne j$) for all $i\in \overline{p-1} \cup \left\{0\right\}$.

The result remains valid if $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction.

Proof. 

First, note that the convergence of distances of sequences allocated in adjacent subsets to $D$ is direct as a result of the first part of Lemma 5 (ii), proved in [1], which does not require the convexity of the subsets.

One proceeds now to prove the convergence to zero of distances of sequences in the same subset whose composition wit their images in adjacent subsets converge to $D$. First, note that, since $A_j$ is boundedly compact since $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is bounded, $A_j=\left\{z_j\right\}$ and there are subsequences ${\left\{T^{p{n}_k}x\right\}}_{k=0}^{\infty }\left(\subset {\left\{T^{pn}x\right\}}_{n=0}^{\infty }\subset A_j\right)$ and ${\left\{T^{p{m}_k}z\right\}}_{k=0}^{\infty }\left(\subset {\left\{T^{pn}z\right\}}_{n=0}^{\infty }\subset A_j\right)$ for some strictly increasing subsequences ${\left\{n_k\right\}}_{k=0}^{\infty }\subset Z_{0+}$ and ${\left\{m_k\right\}}_{k=0}^{\infty }\subset Z_{0+}$ which are convergent in $A_j$ to some ${\omega }_x\in A_j$ and, respectively, ${\omega }_z\in A_j$. Then, one has from Property (ii), since $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is non-expansive, then Lipschitz-continuous, that

$$\underset{k\to \infty }{lim}d\left(T^{p{n}_k}x , T^{p{m}_k}z\right)=\underset{k\to \infty }{lim}d\left({\omega }_x,T^{p{m}_k}z \right)=d\left({\omega }_x,{\omega }_z\right)=0$$

and

$$\underset{k\to \infty }{lim}d\left(T^{p{n}_k}x , T^{p{m}_k}z\right)=\underset{k\to \infty }{lim}d\left({\omega }_x,T^{p{m}_k}z \right)=d\left({\omega }_x,{\omega }_z\right)=0$$

and for $z_j$, being the unique best proximity to $A_{j+1}$ in $A_j$, that is, $A_{j0}=\left\{z_j\right\}$, note that ${\left\{T^{p{n}_k}{z}_j\right\}}_{k=0}^{\infty }\to {\overline{z}}_j$ (at this stage is not still proved that ${\overline{z}}_j=z_j$) since $A_j$ is boundedly compact and ${\left\{T^{p{n}_k}{z}_j\right\}}_{k=0}^{\infty }$ is a convergent subsequence of a bounded sequence, and one has that

$$\underset{k\to \infty }{lim sup }\left(d\left(T^{p{n}_k+1}{\omega }_x,T^{p{n}_k}{\overline{z}}_j\right)-K_T^{n_k}d\left({\omega }_x , {\overline{z}}_j\right)-\left(1-K_T^{n_k}\right)D\right)$$

$$=\underset{k\to \infty }{lim }\left(d\left(T^{p{n}_k+1}{\omega }_x,{\overline{z}}_j\right)-D\right)=d\left(\underset{k\to \infty }{lim}{T}^{p{n}_k}T{\omega }_x,{\overline{z}}_j\right)-D\le 0$$

since $K_T^{n_k}\to 0$ as $k\to \infty$ $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is single-valued then $d\left(T{\omega }_x ,{\overline{z}}_j\right)=D$. If ${\omega }_x\ne z_j$ then ${\omega }_x\notin A_{j0}=\left\{z_j\right\}$ while $T{\omega }_x , T{z}_j\in A_{j+1, 0}$ with $d\left(T{\omega }_x ,{\overline{z}}_j\right)=d\left(T{z}_j,z_j\right)=D$, a contradiction then ${\omega }_x=z_j={\overline{z}}_j$, that is, the limit of the convergent subsequence is the next proximity point. Proceeding in a similar way, one concludes that ${\omega }_z=z_j$. Thus,

$$\underset{k\to \infty }{lim}d\left(T^{p{n}_k}x , T^{p{m}_k}z\right)=\underset{k\to \infty }{lim}d\left(z_j,T^{p{m}_k}{z}_j \right)=d\left(z_j,z_j\right)=0.$$

And the above subsequences in $A_j$ fulfil the claimed property. Now, note that $n_k\to \infty$ as $n,k\to \infty$ with $\left(n-n_k\right)\to \infty$; then,

$$\underset{n\to \infty }{lim sup }d\left(T^{pn}x , T^{pn}{z}_j\right)\le \underset{n ,k\to \infty }{lim}\left(d\left(T^{pn}x , T^{p{n}_k}x\right)+d\left(T^{p{n}_k}x , T^{pn}{z}_j\right)\right)$$

$$=\underset{n ,k\to \infty }{lim}d\left(T^{pn}x , T^{p{n}_k}x\right)+\underset{k\to \infty }{lim}d\left(T^{p{n}_k}x , z_j\right)=\underset{n ,k\to \infty }{lim}d\left(T^{pn}x , T^{p{n}_k}x\right)+d\left(\underset{k\to \infty }{lim}{T}^{p{n}_k}x , z_j\right)$$

$$=\underset{n ,k\to \infty }{lim}d\left(T^{pn}x , T^{p{n}_k}x\right)+d\left(z_j , z_j\right)$$

$$=0+0=0$$

so that $\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn}{z}_j\right)=0$. In the same way, one proves that $\underset{n\to \infty }{lim}d\left(T^{pn}z , T^{pn}{z}_j\right)=0$ and combining this property with the above one and using the triangle inequality for distances yields:

$$\underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn}z\right)\le \underset{n\to \infty }{lim}d\left(T^{pn}x , T^{pn}{z}_j\right)+\underset{n\to \infty }{lim}d\left(T^{pn}z , T^{pn}{z}_j\right)=0+0=0.$$

Furthermore, $z_j=T^p{z}_j$, that is, the best-proximity point of $A_{j+1}$ in $A_j$ is the fixed point in $A_j$ of the composed self-mapping $T^p$. Otherwise, one would get the following contradiction $0<d\left(z_j , T^p{z}_j\right)=\underset{n\to \infty }{lim}d\left(T^{pn}{z}_j , T^{p\left(n+1\right)}{z}_j\right)=0$ using the asymptotic regularity, the non-expansivity, and the consequent continuity of the self-mappings $T$ on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ and $T^p$ on $A_j$. Thus, Property (iii) is proved if $x,z\in A_j$ and $A_{j0}=\left\{z_j\right\}$. Now, since $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ is single-valued, one has that

$$\underset{k\to \infty }{lim}d\left(T^{p{n}_k+1}x , T^{p{m}_k+1}z\right)=\underset{k\to \infty }{lim}d\left(T{z}_j,T\left(T^{p{m}_k}z\right) \right)=d\left(T{z}_j,T{z}_j\right)=d\left(z_{j+1} , z_{j+1}\right)=0$$

for any $x,z\in A_j$ with $z_{j+1}=T{z}_j\in A_{j+1,0}$ and, one has, in a similar way, that for any $i\in \overline{p-1}\cup \left\{0\right\}$,

$$\underset{k\to \infty }{lim}d\left(T^{i+p{n}_k}x , T^{i+p{m}_k}z\right)=\underset{k\to \infty }{lim}d\left(T^i{z}_j,T^i\left(T^{p{m}_k}z\right)\right)=d\left(T^i{z}_j,T^i{z}_j\right)=d\left(z_{j+i} , z_{j+i}\right)=0$$

if $0\le i\le p$ and $i\to i-p$ if $i>p$. □

3. Main Results on Combined Iterative Procedures Involving Mixed Compositions of the Self-Mappings T and S

This section considers the combined application of the self-mappings $T$ and $S$ in any order of composition and with any number of consecutive compositions of any of them before potential switching to the other one. Some of the main obtained results are supported by those given in the former section.

The composed iterations are built by defining recursively the sequence of composed mappings $F_k:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ for any $k\in Z_{0+}$ as follows:

$$F_{0}x=F_{0, 0}x\equiv F_{0}y=F_{0, 0}y=Tx=Sx=Ix=x, \mathrm{where} I \mathrm{is} \mathrm{the} \mathrm{identity} \mathrm{mapping} \mathrm{on} X;$$

$$F_{k}x\equiv F_{k,0}x=F_{k,k-1}{F}_{k-1,0}x=\left(Tx\vee Sx\right)\circ F_{k-1}x=\left(Tx\vee Sx\right)\circ F_{k-1 , 0} x$$

$$\mathrm{for} \mathrm{any}x\in A_i, \mathrm{any} i\in \overline{p}; k\in Z_{+}$$

(19)

The above notation reflects that $T$ and $S$ can be applied in any order and under any amount of consecutive applications of any of them along the construction process of the iterations $F_k$; $k\in Z_{0+}$.

Consider a $k$-th iteration $F_k$ such that:

$$k=t_k+s_k; t_k=n_{k}p+{\mathcal{l}}_k ; s_k=s_{ka}+s_{kb}$$

(20)

where all the above amounts are non-negative integers, and

$n_k\in Z_{0+}$ is the number of iterations happened for $T^p$ at the $k$-th iteration, that is, the number of total complete tours of $T$ on the whole cyclic disposal;

$n_{k}p+{\mathcal{l}}_k$ is the number of iterations occurred for $T$ at the $k$-th iteration $F_k$ with $0\le {\mathcal{l}}_k=t_k-n_{k}p<p$ so that ${\mathcal{l}}_k$ iterations of $T$ have not completed a whole tour around the complete cyclic disposal;

$s_k$ is the number of iterations happened for $S$ up until and included the $k$-th iteration $F_k$, with $s_{ka}$ being the number of iterations of $S$ fulfilling either the first condition of (2) or (3) up until the $k$-th iteration $F_k$ while $s_{kb}$ being that fulfilling the second condition of (2) up until the $k$-th iteration $F_k$.

Remark 3. 

It is found that $d\left(F_{k}x , F_{k}y\right)$denotes that the self-mappings $F_j\equiv F_{j,0}=\left(Tx\vee Sx\right)\circ F_{j-1}=\left(Tx\vee Sx\right)\circ F_{j-1 , 0}$for all $j\in \overline{k}$are identical for $x$and $y$. If there is, for instance a different mapping $S$for the sequence initialized at $x$and $T$for that initialized at $y$at the $k$-th iteration while the preceding ones are generated with the same previous iterative sequence of self-mappings, this is reflected with the notation $d\left(S{F}_{k}x , T{F}_{k}y\right)$.

Remark 4. 

Note from (20) that $k=n_{k}p+{\mathcal{l}}_k+s_{ka}+s_{kb}$. Thus, if $k=1$, then $n_k=0$ and ${\mathcal{l}}_k+s_{ka}+s_{kb}=1$ so that two of the amounts in this sum are null. Note also that if $k\to \infty$, then $max\left(n_k,s_k\right)\to \infty$; then, if $\underset{k\to \infty }{lim sup }{n}_k<\infty$, $\underset{k\to \infty }{lim}{s}_k=\infty$ but it can happen that $\underset{k\to \infty }{lim sup }{s}_{ka}<\infty$and $\underset{k\to \infty }{lim}{s}_{kb}=\infty$. In this last case, the number of contractive iterations of $S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$is finite as $k\to \infty$.

Remark 5. 

Since $S:A_j\to A_j$;$\forall j\in \overline{p}$, then if $x\in A_i$and $y\in A_{i+1}$then $F_{k}x\in A_{i+{\mathcal{l}}_k}$and $F_{k}y\in A_{i+1+{\mathcal{l}}_k}$so that, if $p\ge {\mathcal{l}}_k>i-p$, then $A_{i+{\mathcal{l}}_k}=A_{i+{\mathcal{l}}_k-p}$. In view of (2) and (19), (20), one has:

$$d\left(F_{k}x , F_{k}y\right)\left\{\begin{array}{c}\le \left(K_T^{n_{k}p+{\mathcal{l}}_k}d\left(x,y\right)+\left(1-K_T^{n_{k}p+{\mathcal{l}}_k}\right)D\right)K_S^{s_{ka}}\hspace{1em}if\hspace{1em}{s}_{ka}\le {\overline{s}}_{ka}\\ =D\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}otherwise\end{array}\right.;\forall k\in Z_{+}$$

(21)

for $\left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ , where

$${\overline{s}}_{ka}={\left| ln K_S\right|}^{-1}\left[ln \left(K_T^{n_{k}p+{\mathcal{l}}_k}d\left(x,y\right)+\left(1-K_T^{n_{k}p+{\mathcal{l}}_k}\right)D\right)-ln D\right]$$

$$={\left| ln K_S\right|}^{-1}ln \left({\displaystyle \frac{K_T^{n_{k}p+{\mathcal{l}}_k}\left(d\left(x,y\right)-D\right)}{D}}+1 \right);\forall k\in Z_{+}$$

(22)

Note from (22) that ${\overline{s}}_{ka}\to 0$ as $k\to \infty$.

The following result is given without explicit proofs since it is a direct consequence of the definitions of $T$ and $S$ and (19)–(20):

Lemma 7. 

Assume that $A_i\subset X$ for $i\in \overline{p}$ are non-empty closed subsets of a metric space $\left(X,d\right)$with $S\left(A_i\right)\subseteq A_i$, and $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ with $T\left(A_i\right)\subseteq A_{i+1}$ is a $p\left(\ge 2\right)$-cyclic self-mapping on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ such that $D=d\left(A_i ,A_{i+1}\right)=inf \left(d\left(x,y\right) : x\in A_i , y\in A_{i+1}\right)$;$\forall i\in \overline{p}$. Assume also that $x\in A_i$for some $i\in \overline{p}$.

Then, the following properties hold for all $k\in Z_{0+}$:

(1) 

$F_{k}x\in A_j$  for some $j=j\left(k\right)\in \overline{p}$.

(2) 

$S^m{T}^{np}{F}_{k}x\in A_j$  and $T^{np}{S}^m{F}_{k}x\in A_j$ for the same $j\in \overline{p}$ of Property 1 for any arbitrary $n, m\in Z_{0+}$.

(3) 

$S^{\theta }{T}^{\mathcal{l}p}{S}^m{T}^{np}{F}_{k}x\in A_j$  and $T^{np}{S}^m{T}^{\mathcal{l}p}{S}^{\theta }{F}_{k}x\in A_j$ for the same $j\in \overline{p}$ of Property 1 for any arbitrary $n, m, \mathcal{l} , \theta \in Z_{0+}$.

(4) 

$S^{\theta }{T}^{\mathcal{l}p+\omega }{S}^m{T}^{np+\delta }{F}_{k}x\in A_{j+\omega +\delta \left(mod p\right)}$  and $T^{np+\delta }{S}^m{T}^{\mathcal{l}p+\omega }{S}^{\theta }{F}_{k}x\in A_{j+\omega +\delta \left(mod p\right)}$ for the same $j\in \overline{p}$ of Property 1 for any arbitrary $n, m, \mathcal{l} , \theta \in Z_{0+}$  and any $\delta , \omega \in \overline{p-1}\cup \left\{0\right\}$ , where $j+\omega +\delta \left(mod p\right)=j+\omega +\delta$ if $\omega +\delta \le p-j$ ; $j+\omega +\delta \left(mod p\right)=j+\omega -p$ if $p-j<\omega +\delta \le 2p-j$ ; and $j+\omega +\delta \left(mod p\right)=j+\omega -2p$ if $2p-j<\omega +\delta \le 3p-j$.

(5) 

$F_{k+\sigma \left(k\right)}x=F_{k+\sigma \left(k\right),k}{F}_{k}x\in A_{j+\omega +\delta \left(mod p\right)}$

where $\sigma \left(k\right)=\theta \left(k\right)+m\left(k\right)+\left(\mathcal{l}\left(k\right)+n\left(k\right)\right)p+\omega \left(k\right)+\delta \left(k\right)$ and $n\left(k\right), m\left(k\right), \mathcal{l}\left(k\right) , \theta \left(k\right)\in Z_{0+}$and any $\delta \left(k\right) , \omega \left(k\right)\in \overline{p-1}\cup \left\{0\right\}$.

Remark 6. 

Note from Lemma 7 that:

-

Property 1 is a particular version of Property 2 if $n=m=0$.

-

Property 2 is a particular version of Property 3 if $\theta =\mathcal{l}=0$.

-

Property 3 is a particular version of Property 4 if $\omega =\delta =0$.

-

Property 5 trivially generalizes Property 4.

The following result establishes that, if either $x_k\left(=F_k{x}_0\right),y_k\left(=F_k{y}_0\right)\in A_i$ or $\left(x_k,y_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ for any $i\in \overline{p}$, with either $F_{k+1,k}{x}_k=T{x}_k$ or $F_{k+1,k}{y}_k=S{y}_k$, then for any $k\in Z_{0+}$,$d\left(F_{k+1}{x}_k , F_{k+1}{y}_k\right)<d\left(x_k , y_k\right)$ if $d\left(x_k,y_k\right)>D\ge 0$. This means that the distances which lie outside the interval $\left[ 0 , D\right]$ decrease strictly at any iteration performed under any of the two self-mappings irrespective if both distance arguments are either in the same set or in two adjacent subsets of the cyclic disposal. In that way, the distances which lie outside $\left[ 0 , D\right]$ behave locally as weak contractions. If the mapping $S$ is applied then the contraction is, furthermore, strict for distances of points in the same subset.

Lemma 8. 

Assume that $A_i\subset X$for $\forall i\in \overline{p}$ are non-empty closed subsets of a metric space $\left(X,d\right)$, such that $diam\left(A_i\right)\ge D=d\left(A_i ,A_{i+1}\right)$; $\forall i\in \overline{p}$, with $S\left(A_i\right)\subseteq A_i$, and $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$be self-mappings which satisfy (1)–(3), where $S\left(A_i\right)\subseteq A_i$is a contraction of contractive constant $K_S\in \left[0 , 1\right)$and $T\left(A_i\right)\subseteq A_{i+1}$is a $p\left(\ge 2\right)$-cyclic self-mapping on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$of contractive constant $K_T\in \left[0 , 1\right)$;$\forall i\in \overline{p}$.

The following properties hold:

(i) Assume that $x_k\left(=F_k{x}_0\right)\in A_i,y_k\left(=F_k{y}_0\right)\in A_i$ for any $i\in \overline{p}$ and that $F_{k+1,k}=S$ for some $k\in Z_{0+}$ . Then, $d\left(x_{k+1} ,y_{k+1}\right)=d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)<d\left(x_k,y_k\right)$ if $x_k\ne y_k$.

(ii) Assume that $\left(x_k,y_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ for any $i\in \overline{p}$ , and that either $F_{k+1,k}=S$ or $F_{k+1,k}=T$ for some $k\in Z_{0+}$, and assume also that $D=0$ . Then, $d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)<d\left(x_k,y_k\right)$ if $x_k\ne y_k$.

(iii) Assume that $\left(x_k,y_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$  for any $i\in \overline{p}$ , that $F_{k+1,k}=S$ and that $D>0$ . If $d\left(x_k,y_k\right)>K_S^{-1}D$ then $d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)<d\left(x_k,y_k\right)$ . If $d\left(x_k,y_k\right)\le K_S^{-1}D$ then $d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)=D\le d\left(x_k,y_k\right)$.

(iv) If $\left(x_k,y_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ for any $i\in \overline{p}$ and $F_{k+1,k}{x}_k=T{x}_k$ , $F_{k+1,k}{y}_k=T{y}_k$ and $d\left(x_k,y_k\right)>D>0$ for some $k\in Z_{0+}$  then $d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)<d\left(x_k,y_k\right)$ and if $d\left(x_k,y_k\right)=D$ then $d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)=D$.

Proof. 

Proof of Property (i): If $x_k,y_k\left(\ne x_k\right)\in A_i$ and $F_{k+1,k}=S$, then, from (3)

$$d\left(x_{k+1},y_{k+1}\right)=d\left(F_{k+1 ,k}{x}_k ,F_{k+1 ,k}{y}_k\right)\le K_{S}d\left(x_k,y_k\right)<d\left(x_k,y_k\right).$$

Proof of Property (ii): If $x_k\in A_i,y_k\left(\ne x_k\right)\in A_i$, note from the first condition of (2) with $D=0$ that the above inequality still holds for $F_{k+1,k}=S$ and $x_k,y_k\left(\ne x_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ since $D=0$. Also, if $F_{k+1,k}=T$, it follows from (1) for $D=0$ if $x_k,y_k\left(\ne x_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ for any $i\in \overline{p}$ that

$$d\left(F_{k+1 ,k}{x}_k ,F_{k+1 ,k}{y}_k\right)\le K_{T}d\left(x_k,y_k\right)<d\left(x_k,y_k\right)$$

Proof of Property (iii): Note that if $F_{k+1,k}=S$ and $d\left(x_k,y_k\right)>K_S^{-1}D$, one has under the first condition of (2) that there is some ${\epsilon }_k\left(>\left(K_S^{-1}-1\right)D\right)\in R_{+}$ such that:

$$d\left(x_k,y_k\right)=D+{\epsilon }_k>D>0\Rightarrow$$

$$d\left(F_{k+1 ,k}{x}_k ,F_{k+1 ,k}{y}_k\right)\le K_{S}d\left(x_k,y_k\right)=K_S\left(D+{\epsilon }_k\right)<D+{\epsilon }_k=d\left(x_k,y_k\right).$$

On the other hand, if $D\le d\left(x_k,y_k\right)\le K_S^{-1}D$, the first constraint arising from $\left(x_k,y_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ for any $i\in \overline{p}$ with $D>0$, then $d\left(x_k,y_k\right)\le K_S^{-1}D$$d\left(F_{k+1,k}{x}_k,F_{k+1,k}{y}_k\right)=D\le d\left(x_k,y_k\right)$ from the first condition of (2).

Proof of Property (iv): If $F_{k+1,k}=T$ and $\left(x_k,y_k\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ for any $i\in \overline{p}$ with $D>0$, one gets from (1) that if $D\le d\left(x_k,y_k\right)=D+{\epsilon }_k>D$; then,

$$D\le d\left(F_{k+1 ,k}{x}_k ,F_{k+1 ,k}{y}_k\right)\le K_{T}d\left(x_k,y_k\right)+\left(1-K_T\right)D$$

$$=K_{T}D+K_T{\epsilon }_k+D-K_{T}D$$

$$=K_T{\epsilon }_k+D<D+{\epsilon }_k$$

$$=d\left(x_k,y_k\right).$$

If $d\left(x_k,y_k\right)=D$, then one has from the above set of inequalities for ${\epsilon }_k=0$ that $d\left(F_{k+1 ,k}{x}_k ,F_{k+1 ,k}{y}_k\right)=D$. □

The following result is of interest concerning the convergence of distance sequences of iterations constructed with composite mappings of $T$ and $S$, (19), (20) and Lemma 8:

Lemma 9. 

Assume that $A_i\subset X$for $\forall i\in \overline{p}$ are non-empty closed subsets of a metric space $\left(X,d\right)$, such that $diam\left(A_i\right)\ge D=d\left(A_i ,A_{i+1}\right)$;$\forall i\in \overline{p}$, and $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$are self-mappings with $S\left(A_i\right)\subseteq A_i$being a (strict) contraction and $T\left(A_i\right)\subseteq A_{i+1}$being a $p\left(\ge 2\right)$-cyclic self-mapping on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$of contractive constant $K_T\in \left[0 , 1\right)$;$\forall i\in \overline{p}$. The following properties hold:

(i) If ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }$then $\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=0$; $\forall x,y\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$.

(ii) If ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i=\varnothing }$ and $F_{k}x , F_{k}y\in A_i$; $\forall k\in Z_{0+}$for some arbitrary $i\in \overline{p}$then $\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=0$.

(iii) If $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$commute, ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i=\varnothing }$, $F_{k}x \in A_i\left(=S^{k}x\right)$; $\forall k\in Z_{0+}$, $G_{k}x\left(=S^{k}x\right) \in A_i$; $\forall k\in {{\rm Z}}_{0+}\backslash \left\{j\right\}$  for some $j\in Z_{0+}$, $G_{j+1,j}=T$and $G_{k+1,k}=F_{k+1,k}=S$; $\forall k\left(>j\right)\in Z_{+}$. Then, $d\left(F_{j+\mathcal{l}}x , G_{m+1+j}y\right)=d\left(F_{j+\mathcal{l},j}{F}_{j}x,F_{m+j+1,j+1}T{F}_{j}y\right)=d\left(S^{\mathcal{l}}{F}_{j}x,F_{m+j+1,j+1}T{F}_{j}y\right)=d\left(S^{\mathcal{l}+j}x,S^{m}T{F}_{j}y\right)\ge D$; $\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k}x , G_{\mathcal{l}}y\right)=D$. If  $T$is replaced with  $T^{np}$in the above formulas, then the result still holds.

If ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }$, then the above limit becomes $\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k}x , G_{\mathcal{l}}y\right)=0$.

(iv) Assume ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i=\varnothing }$ and $x\in A_i$, $y\in A_{i+1}$ for some arbitrary $i\in \overline{p}$. If $k=n_{k}p+{\mathcal{l}}_k+s_{ka}+s_{kb}$ with ${\left\{{\mathcal{l}}_k\right\}}_{k=0}^{\infty }\subset \overline{p-1} \cup \left\{0\right\}$; $\forall n_k=n\left(k\right)\in Z_{0+}$, one has that  $F_{k}x\in A_{i+{\mathcal{l}}_k\left(mod p\right)}$, $F_{k}y\in A_{i+1+{\mathcal{l}}_k\left(mod p\right)}$; $\forall k\in Z_{0+}$, and $\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=D$. If ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }$, then the above limit becomes $\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=0$.

Also, if, $\mathcal{l}=m_{\mathcal{l}}p+{\mathcal{l}}_{\mathcal{l}}+s_{\mathcal{l}a}+s_{\mathcal{l}b}$; $\forall n_{\mathcal{l}}=n\left(\mathcal{l}\right)\in Z_{0+}$, one has $\underset{k.\mathcal{l}\to \infty }{lim}d\left(F_{k}x , F_{\mathcal{l}}y\right)=D$, in particular, $\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k}x , F_{\mathcal{l}}y\right)=0$ if ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }.$

(v)  Assume that $k=n_{k}p+{\mathcal{l}}_k+s_k$; $s_k=s_{ka}+s_{kb}$  and ${\mathcal{l}}_k\in \overline{p-1 } \cup \left\{0\right\}$; $\forall k\in Z_{0+}$, where $n_{k}p$is the total number of whole iterations of $T$on the whole cyclic disposal up until the $k$-th iteration in $F_k$, and $s_{ka}$  and $s_{kb}$  the number of iterations of $S$  fulfilling, respectively, the first condition of (2) or (3) and the second condition of (2) up until the $k$-th iteration. Thus, if $x\in A_i$  and $y\in A_{i+1}$for some $i\in \overline{p}$  then for any $k\in Z_{0+}$, one has $F_{k}x,F_{k+p}x\in A_{i+{\mathcal{l}}_k}\left(mod p\right)$; $F_{k}y,F_{k+p}y\in A_{i+{\mathcal{l}}_k+1}\left(mod p\right)$, and

(v.1) $d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}y\right)\to D$ for any $m\in Z_{0+}$as $k\to \infty$if $n_k\to \infty$.

(v.2) $d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}x\right)\to D$ for any $m\in Z_{0+}$as $k\to \infty$if $n_k\to \infty$.

(v.3) The limits (v.1)–(v.2) also hold as $k\to \infty$, $s_k\to \infty$for $n_k\left(=n\left(k\right)\in Z_{0+}\right)$ not necessarily diverging, any $m\in Z_{0+}$.

Proof. 

To prove Property (i), note that if ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i\ne \varnothing }$, then $D=0$. Thus, $s_{kb}=0$; $\forall k\in Z_{+}$, and, from (1) and the first condition of (2), or from (3), depending on to which set(s) $A_i$$\left(i\in \overline{p}\right)$ the points $F_{k}x$ and $F_{k}y$ belong to. Thus, one has from (21), (22):

$$d\left(F_{k}x , F_{k}y\right)\le K_S^{s_{ka}}{K}_T^{n_{k}p+{\mathcal{l}}_k}d\left(x,y\right) ; \forall x,y\in {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$$

(23)

Then, if $k\to \infty$, $max\left(n_k,s_k\right)\to \infty$(Remark 4). Since $s_{kb}=0$;$\forall k\in Z_{+}$, that is the number of iterations involving the mapping $S$ up until the $k$-th iteration of $F_k$ and, since $s_{ka} ,{\overline{s}}_{ka}\to 0$ as $k\to \infty$ from (7), then $K_S^{s_{ka}}\to 1$, $n_k\to \infty$ as $k\to \infty$ and $\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=D=0$. Property (i) has been proved.

To prove Property (ii), note that if ${\displaystyle {\cap }_{i\in \overline{p}}{A}_i=\varnothing }$, then $D>0$ and $x , y\in A_i$ for some arbitrary $i\in \overline{p}$, then $F_{k}x\left(=S{F}_{k-1}x \right), F_{k}y\left(=S{F}_{k-1}y\right)\in A_i$; $\forall k\in Z_{+}$ (that is, only the self-mapping $S$ is involved in building the whole set of sequences which are in $A_i$ for some $i\in \overline{p}$) so that one has from (3),

$$d\left(F_{k}x , F_{k}y\right)\le K_S^{s_k}d\left(x,y\right)=K_S^{s_{ka}}d\left(x,y\right)$$

(24)

If $k\to \infty$, then $s_k=s_{ka}\to \infty$ and, one gets from (24) that

$$\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k}x , F_{\mathcal{l}}y\right)=\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=0$$

(25)

and Property (ii) has been proved.

To prove Property (iii), note that the conditions of Property (ii) are kept from the initial conditions up until the $\left(j-1\right)$-th iteration, while there is an application of the self-mapping $T$ on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ at the $j\left(\in Z_{+}\right)$-th iteration for the sequence generated from the initial point $y\in A_i$. From the hypotheses, for $x,y\in A_i$, one determines that composite mappings $F_k$ and $G_k$ for all $k\in Z_{0+}$ only differ in $F_{j+1,j}=S$ and $G_{j+1,j}=T$ and $F_{k}x , F_{k}y\in A_i$; $\forall k\in Z_{0+}$, $G_{k}y\in A_{i+1}$ for all $k\left(>j\right)\in Z_{0+}$:

$$F_{k}x=S^{k}x\in A_i; \forall k\in Z_{0+}, G_{k}x=S^{k}y\in A_i; \forall k\left(<j\right)\in Z_{+};$$

(26)

$G_{k}y=G_{k,j+1}{G}_{j+1,j}{G}_{j}y=S^{k-j-1}T{S}^{j-1}y\in A_{i+1};$$\forall k\left(>j\right)\in Z_{0+}$ for some $j\in Z_{+}.$

Since for $k\ge j$, $S^{k}x$ and $S^{k-j-1}T{S}^{j}y$ are in disjointed adjacent sets $A_i$ and $A_{i+1}$, and since $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ commute, one has from (21), which derives from (1)–(3), that

$$D\le d\left(S^{k}x , S^{k-j-1}T{S}^{j}y\right)\le max\left(D ,K_S^{k-j-1}d\left(S^{j+1}x , T{S}^{j}y\right)\right)$$

which implies that $d\left(S^{k}x , S^{k-j-1}T{S}^{j}y\right)\to D$ as $k\to \infty$ with $S^{k}x\in A_i$; $\forall k\in Z_{0+}$, $Ty\in A_{i+1}$ and $S^{k-j-1}T{S}^{j}y$$\in A_{i+1}$ for sufficiently large $k$. From the first condition of (2), there is some finite non-negative integer $j{\left(j\right)}_1\ge j$ such that for $k\ge 1+j_1$, $K_S^{k-1}d\left(Sx , Ty\right)\le D$; then, for $k\ge 1+j_1$, one has $max \left( D ,D +K_S^{k-1}\left(1-K_T\right)D\right)=D$ and then $d\left(S^{k}x , S^{k-j-1}T{S}^{j}y\right)=D$ for $k\ge 1+j_1$ so that one gets as a result:

$$\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k,j+1}T{F}_{j}y\right)=\underset{k\to \infty }{lim}d\left(S^{k}x , S^{k-j-1}T{S}^{j}y\right)=\underset{k\to \infty }{lim}d\left(S^{k}x , S^{k-1}Ty\right)$$

$$=\underset{m,k\to \infty }{lim}d\left(F_{k}x , F_{m,j+1}T{F}_{j}y\right)=\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k}x , G_{\mathcal{l}}y\right)=D.$$

(27)

It is obvious that if $T$ is replaced with $T^{np}$ in the (27), the result still holds. Then, Property (iii) has been proved.

To prove Property (iv), note that $x\in A_i , y\in A_{i+1}$ for some $i\in \overline{p}$. Take $k=n_{k}p+{\mathcal{l}}_k+s_{ka}+s_{kb}$, $F_{k}x\in A_{i+{\mathcal{l}}_k\left(mod p\right)}$ and $F_{k}y\in A_{i+1+{\mathcal{l}}_k\left(mod p\right)}$; $\forall k\in Z_{0+}$. Then, it follows that

$$F_{k}x\in A_{i+{\mathcal{l}}_k}\mathrm{if} i\le p-{\mathcal{l}}_k\mathrm{and} F_{k}x\in A_{i+{\mathcal{l}}_k-p}\mathrm{if}i>p-{\mathcal{l}}_k; \mathrm{and}$$

$$F_{k}y\in A_{i+1+{\mathcal{l}}_k-p}\mathrm{if} i>p-1-{\mathcal{l}}_k\mathrm{and}{F}_{k}y\in A_{i+1+{\mathcal{l}}_k}\mathrm{if} i\le p-1-{\mathcal{l}}_k$$

and one gets from (6), (7) that $\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k+n_{\mathcal{l}}p}x , F_{k+m_{k}p}y\right)=\underset{k\to \infty }{lim}d\left(F_{k+n_{k}p}x , F_{k+m_{k}p}y\right)=D$; $\forall ,m_k=m\left(k\right)\in Z_{0+}$, since, if $k\to \infty$ for $k=n_{k}p+{\mathcal{l}}_k+s_{ka}+s_{kb}$ then either $n_k\to \infty$ or $\left(s_{ka}+s_{kb}\right)\to \infty$ for any ${\mathcal{l}}_k\in \overline{p-1} \cup \left\{0\right\}$, and note that

$$d\left(F_{k}x , F_{k}y\right)\le d\left(T^{n_{k}p+{\mathcal{l}}_k}x , T^{n_{k}p+{\mathcal{l}}_k}y\right)\le K_T^{n_{k}p}d\left(T^{{\mathcal{l}}_k}x , T^{{\mathcal{l}}_k}y\right)+\left(1-K_T^{n_{k}p}\right)D.$$

Thus, one has

(a)

if $n_k\to \infty$ then

$$\underset{k\to \infty }{lim sup}\left(d\left(F_{k}x , F_{k}y\right)-K_T^{n_{k}p}d\left(T^{{\mathcal{l}}_k}x , T^{{\mathcal{l}}_k}y\right)-\left(1-K_T^{n_{k}p}\right)D\right)=\underset{k\to \infty }{lim sup}\left(d\left(F_{km}x , F_{k}y\right)-D\right)\le 0$$

• so that there exists $\underset{k\to \infty }{lim}d\left(F_{k}x , F_{k}y\right)=D$;

(b)

if $s_k\to \infty$ then the above result follows as in Property (iii).

In the same way, it is obtained if, in addition, $\mathcal{l}=m_{\mathcal{l}}p+{\mathcal{l}}_{\mathcal{l}}+s_{\mathcal{l}a}+s_{\mathcal{l}b}$; $\forall n_{\mathcal{l}}=n\left(\mathcal{l}\right)\in Z_{0+}$, that $\underset{k,\mathcal{l}\to \infty }{lim}d\left(F_{k}x , F_{\mathcal{l}}y\right)=D$.

Property (iv) has been proved.

Property (v) is proved easily under a close reasoning as the one used in the proof of Property (iv). □

Remark 7. 

Note that it is relevant in Lemma 9 that:

(a) 

Any two iterative processes which only involve compositions of $S$  with itself in the same subset $A_i$  generate sequences which are contained in this subset.

(b) 

If two iterations only involve compositions of $S$ with itself but they are initialized in adjacent subsets, these generate sequences which always remain each in its corresponding subset but the maximum allowed distances at each iteration step is the distance $D$ between disjointed adjacent subsets. That means that the self-mapping $S$might be non-single valued from the second equation of (2), forced to be applied to this end if the distances exceed $D$.

(c) 

Any two iterations with initial points in the same subset generate sequences which lie asymptotically in adjacent subsets in the event that they mutually differ in the concourse of $np+1$  times the self-mapping $T$  for some given $n\in Z_{0+}$. The case when both of them just differ in a single contribution in the iterative composition of maps is sufficient for that property to hold. If the initial points are in adjacent subsets then both generated sequences lie asymptotically in adjacent subsets if they mutually differ in the concourse of $np+1$  times the self-mapping $T$  for some given $n\in Z_{0+}$. The number of contributions of $S$to the composed mappings in any of the mentioned iterations do not modify the above conclusions since the self-mapping $S$with domain in any subset has its image in the same subset.

The following assumptions will be invoked in the following as linked to results in the previous section and used to extend them to the iterative processes which involve compositions $F_k:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ for any $k\in Z_{0+}$ of the self-mappings $S, T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$, subject to (19) and (20), where all the subsets $A_i$ of the metric space $\left(X,d\right)$ are nonempty and closed.

Assumption 1. 

$A_i\subset X$for are non-empty closed subsets of a metric space $\left(X,d\right)$, where $d:X\times X\to R_{0+}$, with $d\left(A_i,A_{i+1}\right)=D>0$; $\forall i\in \overline{p}$; and $T,S:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$are self-mappings which satisfy (1)–(3), where $S\left(A_i\right)\subseteq A_i$is a contraction of contractive constant $K_S\in \left[0 , 1\right)$and $T\left(A_i\right)\subseteq A_{i+1}$is a $p\left(\ge 2\right)$-cyclic self-mapping on ${\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$of contractive constant $K_T\in \left[0 , 1\right)$;$\forall i\in \overline{p}$.

Assumption 2. 

The unique fixed points of the contractive self-mapping $S:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to \left({\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\right)\backslash A_{\mathcal{l}}\to A_{\mathcal{l}}$satisfy $z_{\mathcal{l}}\left(=S{z}_{\mathcal{l}}\right)\in A_{\mathcal{l}0}\left(\ne \varnothing \right)$; $\forall \mathcal{l}\in \overline{p}$.

Assumption 3. 

There is at least one $j\in \overline{p}$such that $A_j$is boundedly compact and $A_{j0}=\left\{z_j\right\}$is a singleton.

Assumption 4. 

At least one of the subsets $A_j$of $X$in the cyclic disposal is strictly convex for some $j\in \overline{p}$.

Assumption 2 states that the fixed points of $S$ in each subset are also best-proximity points of this subset to its adjacent subset. Also, from Assumption 3 (which holds directly if $A_j$ is convex, but it can also hold without convexity need in some cases if $A_j$ is not perfect) and Corollary 2, it follows that the best-proximity points $z_{\mathcal{l}}$ to $A_{\mathcal{l}+1}$ of $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ in $A_{\mathcal{l}}$ for $\mathcal{l}\in \overline{p}$ (also fixed points of $S:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to \left({\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\right)\backslash A_{\mathcal{l}}\to A_{\mathcal{l}}$;$\forall \mathcal{l}\in \overline{p}$ from Assumption 1) satisfy $z_{j+i}=T^i{z}_j\in A_{j+i,0}$ for $1\le i\le p-j$ and $z_{j-i}=T^{p-i}{z}_j\in A_{j-i ,0}$ for $1\le i\le j-1$. That is, if $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ contributes with infinitely many tours on the whole cyclic disposal in iterative processes, the sequences of distances of iteration in adjacent subsets converge to $D$ (Lemma 9(iv)). From Corollary 3, the sequences in adjacent subsets involved in such distances converge to the unique fixed points $z_{\mathcal{l}}$ of $T^p:\left({\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\right)\backslash A_{\mathcal{l}}\to \left({\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\right)\backslash A_{\mathcal{l}}$ for all $\mathcal{l}\in \overline{p}$ which are the, in general, non unique (except for $j\in \overline{p}$) best-proximity points of $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$ and the unique fixed points of $S:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to \left({\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\right)\backslash A_{\mathcal{l}}\to A_{\mathcal{l}}$;$\forall \mathcal{l}\in \overline{p}$ under Assumption 2. Assumption 4 guarantees the uniqueness of $z_j\in A_{j0}=\left\{z_j\right\}$ and then that of the fixed points of $T^p:\left({\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\right)\backslash A_{\mathcal{l}}\to \left({\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\right)\backslash A_{\mathcal{l}}$ for all $\mathcal{l}\in \overline{p}$.

The next result establishes the main result of this section which is supported by Theorem 2, Corollarys 1 and 3, Lemmas 4–6 and 9.

Theorem 3. 

Let $\left(X,‖‖\right)$be a uniformly convex Banach space under Assumptions 1–3, where $d:X\times X\to R_{0+}$is the norm-induced metric defined by $d\left(x,y\right)=‖x-y‖$; $\forall x,y\in X$, and consider an iterative process which involve compositions $F_k:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$for any $k\in Z_{0+}$of the self-mappings $S, T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$, subject to (19)–(20). Then, the following properties hold for any given $x\in A_i$and $y\in A_{i+1}$:

(i) The convergence of distances of Lemma 9(v) holds for $k=n_{k}p+{\mathcal{l}}_k+s_k$; $s_k=s_{ka}+s_{kb}$  and ${\mathcal{l}}_k\in \overline{p-1 } \cup \left\{0\right\}$; $\forall k\in Z_{0+}$, that is, irrespective of the order of the contributions of the self-mappings $S$  and $T$  to the iterative process ${\left\{F_k\left(.\right)=F_{k,k-1.}{F}_{k-1}\left(.\right)\right\}}_{k=0}^{\infty }$, the following holds for any $x\in A_i$, $y\in A_{i+1}$  for any $i\in \overline{p}$:$d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}y\right)\to D$ for any  $m\in Z_{0+}$as $k\to \infty$if $n_k\to \infty$ or if  $s_k\to \infty$. $d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}x\right)\to 0$ for any  $m\in Z_{0+}$as $k\to \infty$if $n_k\to \infty$ or if  $s_k\to \infty$provided that  ${\mathcal{l}}_k=j-i$if $i<j$ or ${\mathcal{l}}_k=j-i+p$, otherwise. $d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}x\right)\to 0$ for any  $m\in Z_{0+}$as $k\to \infty$if $n_k\to \infty$ or if  $s_k\to \infty$provided that  ${\mathcal{l}}_k=j-i$if $i<j$ or ${\mathcal{l}}_k=j-i+p$, otherwise.

Property (i) also holds if  $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction self-mapping.

(ii) The set $\left\{z_1,z_2,,\dots ,z_p\right\}$ of respective fixed points of $T^p$ from $A_i$ to $A_{i+1}$ is unique and it is also the set of proximity points of $T$ to which all the sequences generated by the contractive self-mapping converge at each one of the subset if $n_k\in Z_{+}$ (that is, if there is at lest a complete tour of $T$ in the iterative process ${\left\{F_k\left(.\right)=F_{k,k-1}{F}_{k-1}\left(.\right)\right\}}_{k=0}^{\infty }$as $k\to \infty$.

Property (ii) also holds if $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is an extended  $p\left(\ge 2\right)$-cyclic contraction.

Proof. 

Property (i) follows directly from Lemma 9(v) since a uniformly convex Banach space is a complete metric space and from Lemma 5(ii). Note that if $x\in A_i$, $y\in A_{i+1}$ for any $i\in \overline{p}$:

$d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}y\right)\to D$ for any $m\in Z_{0+}$ as $k\to \infty$ if $n_k\to \infty$ or if $s_k\to \infty$, and

$d\left(F_{n_{k}p+{\mathcal{l}}_k+s_k}x , F_{\left(n_k+m\right)p+{\mathcal{l}}_k+s_k}x\right)\to 0$ for any $m\in Z_{0+}$ as $k\to \infty$ if $n_k\to \infty$. If $n_k\equiv 0$ then

$d\left(F_{{\mathcal{l}}_k+s_k}x , F_{mp+{\mathcal{l}}_k+s_k}x\right)\to 0$ for any $m\in Z_{0+}$ as $k\to \infty$ as $s_k\to \infty$ from (3) since ${\left\{F_{mp+{\mathcal{l}}_k+s_k}x\right\}}_{k=1}^{\infty }\left(\subset A_i\right)\to z_i\left(=S{z}_i\right)$ since $F$ is a strict contraction. Finally, if $n_k\in {{\rm Z}}_{+}$ for (at least) a $k\in Z_{0+}$ while $\underset{k\to \infty }{lim sup} n_k<\infty$, then ${\left\{F_{mp+{\mathcal{l}}_k+s_k}x\right\}}_{k=1}^{\infty }\left(\subset A_{\mathcal{l}}\right)\to z_{\mathcal{l}}\left(=S{z}_{\mathcal{l}}\right)$ for some $\mathcal{l}\in \overline{p}$ from (2), (3) and Assumption 2. The result also holds from Lemma 4(ii) if $T:{\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}\to {\displaystyle {\cup }_{\mathrm{i}\in \overline{\mathrm{p}}}{A}_i}$ is an extended $p\left(\ge 2\right)$-cyclic contraction.

Property (ii) follows from Assumption 3, Lemma 4(i) and Lemma 6, and from Assumption 3, Lemma 4(ii) and Lemma 6 if $T$ is an extended $p\left(\ge 2\right)$-cyclic contraction with at least a contribution of one complete tour of $T$ as $k\to \infty$. □

The next result establishes the second main result of this section. The proof is similar by replacing Assumption 3 by Assumption 4, related to the strict convexity of one of the subsets of the cyclic disposal, and by taking into account Lemma 5 and Proposition 1.

Theorem 4. 

Let $\left(X,‖‖\right)$be a uniformly convex Banach space under Assumptions 1, 2 and 4. Assume that $A_j$is strictly convex for some $j\in \overline{p}$. Also, let $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$be a $p\left(\ge 2\right)$-cyclic contraction with contractive constant $K\in \left[0 , 1\right)$and take $x\in A_{\mathcal{l}}$for some arbitrary $\mathcal{l}\in \overline{p}$. Then, the following properties hold:

(i) Property (i) of Theorem 3 holds.

(ii) There is a unique best-proximity point $z_j$ to $A_{j+1}$ in the strictly convex set $A_j$ (Assumption 4), that is, $A_{j0}=\left\{z_j\right\}$, such that $d\left(z_{j }, T{z}_j\right)=d\left(A_j , A_{j+1}\right)=D$; and there exists $z_i\in A_{i0}$; $\forall i\left(\ne j\right)\in \overline{p}$ (such that $A_{i0}$ for $i\ne j$ are not empty and non-necessarily singletons) such that $d\left(z_i ,T{z}_i\right)=D$; $\forall i\left(\ne j\right)\in \overline{p}$.

(iii)  Propositions [(i)–(ii)] also hold if $T:{\displaystyle {\cup }_{i\in \overline{p}}{A}_i}\to {\displaystyle {\cup }_{i\in \overline{p}}{A}_i}$  is an extended $p\left(\ge 2\right)$-cyclic contraction.

4. Examples

Example 1. 

Consider the metric space $\left(R , d\right)$, where $d:R\times R\to R_{0+}$is the Euclidean metric, and consider the real intervals $A_1=\left(-\infty ,-1\right]$and $A_2=\left[1 , +\infty \right)$such that $D=d\left(A_1 , A_2\right)=2$. Note that the Euclidean metric in this case coincides with the taxi-cab metric and that the metric space is a uniformly convex metric space and then complete. Define the self-mapping $T$on $A_1\cup A_2$by $Tx=-x$which is a non-expansive 2-cyclic self-mapping, since $\left|T{x}_1-T{x}_2\right|=\left|x_2-x_1\right|=\left|x_1-x_2\right|$for $x_1, x_2\in A_1\cup A_2\times A_2\cup A_1$, with $T\left(A_1\right)=A_2$and $T\left(A_2\right)=A_1$, and $z_1=-1$and $z_2=1$are the best-proximity points with $T{z}_1=z_2$in $A_1$to $A_2$and $T{z}_2=z_1$in $A_2$to $A_1$.

Consider a contractive self-mapping $S$  on $A_1\cup A_2$  satisfying $S\left(A_1\right)\subset A_2$  and $S\left(A_2\right)\subset A_1$  such that for some $K_S\in \left(0 , 1\right)$,$\left|S{x}_1-S{x}_2\right|\le K_S\left|x_1-x_2\right|$  if $x_1 , x_2\in A_i$  or if $x_1, x_2\in A_1\cup A_2\times A_2\cup A_1$  with $d\left(x_1 ,x_2\right)=\left|x_1-x_2\right|>2K_S^{-1}$  and $d\left(x_1 ,x_2\right)=\left|z_1-z_2\right|=2$, otherwise, with $S{z}_i=z_i \mathrm{for} i=1,2.$ Note that the unique best-proximity points Z_i in A_i to $A_{i+1}$$\left(3\equiv 1\left(mod 2\right)\right)$ are the respective unique fixed points of in S in $A_i$  for $i=1, 2$. Note that:

(1) 

If only composed mappings of $S$ contribute in an iteration process ${\left\{F_{k}x\right\}}_{k=0}^{\infty }$, then ${\left\{F_{k}x\right\}}_{k=0}^{\infty }\to z_i$ and $d\left(F_{k+1}x , F_{k}x\right)\to 0$ as $k\to \infty$ if $x\in A_i$ for $i=1,2.$

(2) 

If only composed mappings of $T$ contribute in an iteration process ${\left\{F_{k}x\right\}}_{k=0}^{\infty }$, then ${\left\{F_{k}x\right\}}_{k=0}^{\infty }\in \left\{-x , x\right\}$, with $F_{k+1}x=-F_{k}x$, for all $k\in Z_{0+}$ any $x\in A_1\cup A_2$ but the iterative process does not converge and $d\left(F_{k+1}x , F_{k}x\right)\to 2$ as $k\to \infty$ while $d\left(F_{2k}x , F_{2k+2}x\right)\to 0$ and $d\left(F_{2k+1}x , F_{2k+3}x\right)\to 0$ as $k\to \infty$. In particular, if $x\left(=F_{0}x\right)\in \left\{z_1 , z_2\right\}$, then $F_{2k}{z}_i=z_i$ and $F_{2k+1}{z}_i=z_{i+1}$ for $k\in Z_{0+}$; $i=1, 2$.

(3) 

If a finite even number of compositions of $T$ contribute to an iterative process ${\left\{F_{k}x\right\}}_{k=0}^{\infty }$ which has infinitely many compositions of $S$, and irrespective of the order of occurrence of the whole set of compositions, ${\left\{F_{k}x\right\}}_{k=0}^{\infty }\to z_i$ if $x\in A_i$; $i=1,2$. If a finite odd number of compositions of $T$ contribute to the iterative process, then ${\left\{F_{2k}x\right\}}_{k=0}^{\infty }\to z_{i+1}$ if $x\in A_i$; $i=1,2$. In both cases, $d\left(F_{k+1}x , F_{k}x\right)\to 0$ as $k\to \infty$.

(4) 

If infinitely many compositions of $T$contribute to an iterative process ${\left\{F_{k}x\right\}}_{k=0}^{\infty }$ which has either a finite number or infinitely many compositions, of $S$, and irrespective of the order of occurrence of the whole set of compositions, ${\left\{F_{2k}x\right\}}_{k=0}^{\infty }\to z_i$ and ${\left\{F_{2k+1}x\right\}}_{k=0}^{\infty }\to z_{i+1}$if $x\in A_i$; $i=1,2$; and the sequence ${\left\{F_{k}x\right\}}_{k=0}^{\infty }\in \left\{z_1 , z_2\right\}$ does not converge while $d\left(F_{2k}x , F_{2k+2}x\right)\to 0$ and $d\left(F_{2k+1}x , F_{2k+3}x\right)\to 0$ as $k\to \infty$.

Example 2. 

Modify Example 1 with $Tx=-x+\sigma \lambda \left(x\right)$, where $\sigma \in R_{+}$, and $\lambda :R\to R$is continuous, with $\lambda \left(x\right)=0$for $x\in \left[-1 , 1\right]$. Thus, $T\left(\pm 1\right)=\mp 1$, and for $x\in A_1$and $y\in A_2$, one has that $T$is $2$-cyclic contractive if

$$\left|Tx -Ty\right|=\left|y -x+\sigma \lambda \left(x\right)-\sigma \lambda \left(y\right)\right|=\left|y -x-\sigma \left(\lambda \left(y\right)-\lambda \left(x\right)\right)\right|=\left|y +\left|x\right|-\sigma \left(\lambda \left(y\right)-\lambda \left(x\right)\right)\right|$$

$$=\left|y -x-\sigma \left(\lambda \left(y\right)-\lambda \left(x\right)\right)\right|-\sigma \left| \lambda \left(y\right)-\lambda \left(x\right)\right|$$

Assume that $-x+\sigma \lambda \left(x\right)\ge 1$ and $-y+\sigma \lambda \left(y\right)\le -1$ , then $\sigma \lambda \left(x\right)\ge 1+x=1-\left|x\right|$ and $\sigma \lambda \left(y\right)\le y-1=\left|y\right|-1$ , then $\left|y\right|-1\ge 1-\left|x\right|$ ; equivalently, the necessary constraint $\left|y\right|+\left|x\right|=y-x\ge 2$ , is guaranteed if $\left[\left(y\in A_2\right)\wedge \left(y\in A_1\right)\right]\Rightarrow$$\lambda \left(y\right)>\lambda \left(x\right)$. Thus, for some real constant $K_T\in \left[0 , 1\right)$, one has that $T$ is $2$ -cyclic contractive if

$$\left|Tx -Ty\right|=\left|y -x\right|-\sigma \left|\lambda \left(y\right)-\lambda \left(x\right)\right|\le K_T\left|y-x\right|+2\left(1-K_T\right)$$

which holds if

$$\left(1-K_T\right)\left( \left|x -y\right|-2\right)\le \sigma \left|\lambda \left(y\right)-\lambda \left(x\right)\right|\le \sigma {\sigma }^{-1} \left(y+\left|x\right|-2\right)=y+\left|x\right|-2$$

Whose second inequality follows since $\lambda \left(y\right)-\lambda \left(x\right)\le {\sigma }^{-1}\left(y+\left|x\right|-2\right)$  obtained from the former constraints $\sigma \lambda \left(x\right)\ge 1-\left|x\right|$  and $\sigma \lambda \left(y\right)\le y-1$. As a result, $T$ is $2$ -cyclic contractive if for $\sigma \in R_{+}$ , ${\sigma }^{-1} \left(1-\left|x\right|\right)\le \lambda \left(x\right)\le {\sigma }^{-1} \left(\left|x\right|-1\right)$ for $x\in A_1\cup A_2$ . The properties of Example 1 hold for iterative processes of the above cyclic contractive self-mapping and contractive self-mapping $S$, whose fixed points at each subset is the best-proximity point to its adjacent subset as in Example 1.

Examples 1 and 2 are easily extendable to the real vector case as follows:

Example 3. 

Consider the complete metric space $\left(R^n , d\right)$, where $d:R\times R\to R_{0+}$is the supremum metric of all the $n$-vector components obtained from the component-wise taxi-cab metric, and consider the subsets $A_i=A_i\times A_{i }\times \stackrel{n}{\dots }\times A_i$for $i=1, 2$of $\left(R^n , d\right)$, where $A_1=\left(-\infty ,-1\right]$and $A_2=\left[1 , +\infty \right)$such that $D=d\left(A_1 , A_2\right)=d\left(A_1 , A_2\right)=2$. The best-proximity points are real $n$-vectors $z_i$; $i=1,2$with all their respective components being, respectively, $-1$and $+1$. The contractive self-mapping $S$on $A_i$are anyone with fixed points being the best-proximity points between adjacent subsets. Then, the derivations and conclusions of Example 2, inherited from Example 1, hold.

There are dynamic systems present in the real word which have several potential configurations which can be activated at different time instants as long as the process under study evolves in different production phases or because of the concourse of certain events which activate switching. Switching actions can cause instability if they occur repeatedly in very short time intervals even if the various alternative configurations are stable. Also, switching actions can translate into stability of the switched configuration if the switching map runs properly through time or is governed by events, for instance, if signals exceed a certain threshold, the switching might be activate to prevent against instability, even if the individual configurations are instable, [20,21,22,23]. The following two examples link those problems with the combined iterations involving cyclic self-maps and strict contractions. Two sets of the metric space are defined for the state evolution of a time-varying linear dynamic system of $n$-th order with constant distinct parameterizations in each of both subsets. The two individual parameterizations are assumed to be stable and governed by a strict contraction, while they keep either non-negative solutions or non-positive ones depending on which set has been activated.

Example 4. 

Let $A_i={\left(A_i\right)}_{jk}\in R^{n\times n}$; $i=1,2$; $j,k\in \overline{m}$be real Metzler stability matrices, that is, ${\left(A_i\right)}_{jk}\ge 0$if $j\ne k$and $A_i$have all their eigenvalues with negative real parts. Since they are Metzler, they are stable iff ${\left(-A_i\right)}^{-1}\succ 0$, that is, they are non-singular with non-negative entries, [24,25,26]. Define the $n$-th order differential systems ${\dot{x}}_i\left(t\right)=A_i{x}_i\left(t\right)$with initial conditions $x_i\left(0\right)=x_{i0}\in R^n$;$i=1,2$. Since the matrices of dynamics are Metzler, then the fundamental matrices $e^{A_{i}t}$of the corresponding differential systems are non-singular with non-negative entries for all time and the solutions satisfy $x_i\left(t\right)\succ 0$for $t\in R_{0+}$if $x_i\left(0\right)\succ 0$. In the same way, $-x_i\left(t\right)\succ 0$for $t\in R_{0+}$if $-x_i\left(0\right)\succ 0$. This feature is equivalently denoted by $x_i\left(t\right)\prec 0$for $t\in R_{0+}$if $x_i\left(0\right)\prec 0$.

Consider the Banach space $\left(R^n , ‖.‖\right)$  and consider also consider the first hyperorthant $R_{0+}^n$  of $R^n$ and its opposed one $\left(-R_{0+}^n\right)$  as subsets of$R^n$. Clearly $R_{0+}^n\cap \left(-R_{0+}^n\right)=0\left(\in R^n\right)$ and both subsets are unbounded, closed and convex if $n\ge 2$. Rename $A_1=-R_{0+}^n$ and $A_2=R_{0+}^n$ so that all the solutions $x_1\left(t\right)$ of ${\dot{x}}_1\left(t\right)=A_1{x}_1\left(t\right)$ are in $A_1$ for all time under any given non-positive initial conditions and all those of ${\dot{x}}_2\left(t\right)=A_2{x}_2\left(t\right)$ are in $A_2$ for all time under any given non-negative initial conditions as a result of the properties of Metzler matrices.

Furthermore, as a consequence of the properties of fundamental matrix functions of stability matrices, one has that there exist positive constants $k_i\ge 1$ and ${\rho }_i>0$ for $i=1, 2$ such that $‖x_i\left(t\right)‖\le k_i{e}^{-{\rho }_i \left(t-t_0\right)}‖x_i\left(t_0\right)‖$ for all $t\ge t_0$ and any  $t_0\in R_{0+}$, where $k_i\ge 1$  for $i=1, 2$ is norm-dependent and $\left(-{\rho }_i\right)<0$ is the maximum real part of the dominant eigenvalue of the stability Metzler matrix $A_i$, for $i=1, 2$, if it has multiplicity equal to one, and any greater negative real number, otherwise.

Thus, in terminology of systems theory, note that both differential systems are globally exponentially stable. Note also that $\left(R^n , ‖.‖\right)$ is a complete metric space  $\left(R^n , d\right)$ if $d:R^n\times R^n\to R_{0+}$ is the norm-induced metric and that  $d\left(A_1 , A_2\right)=0$. Fix a sampling period $T_s>0$ for picking-up the discrete samples of the solutions  $x_i\left(k{T}_s\right)$for $i=1,2$ and $k\in Z_{0+}$.

Assume that the initial condition of the differential system  $x_0=x\left(0\right)$ is either in $A_1$ or in $A_2$. Now, define strictly increasing sequences of non-negative integer numbers  ${\left\{n_k\left(i\right)\right\}}_{k\in Z_{0+}}$ for $i=1,2$ such that a self-mapping  $T$ from $A_1\cup A_2\times Z_{0+}$ to $A_1\cup A_2$ is defined by:

$$Tx\left(j_k\right)=x\left(j_k+1\right)$$

(28)

if $j_k\in \left(n_k\left(i\right) , n_{k+1}\left(i\right)\right)\cap Z_{0+}$ and $x\left(n_k\left(i\right)\right)\in A_i$ for $i=1, 2$ ;

$$Tx\left(n_{k+1}\right)=-\lambda \left(n_{k+1}\right) x\left(n_{k+1}-1\right)$$

(29)

If $j_k\in \left(n_k , n_{k+1}\right)\cap Z_{0+}$ , for some $\lambda \left(n_{k+1}\right)\in \left(0 , K_T\right)$ , all $k\in Z_{0+}$ , and some $K_T<1$ . That is, $T$ is a non-expansive self-mapping on $R^n$  which is also a $2$-cyclic contraction in $A_1\cup A_2$ at the sampling instants ${\left\{n_k{T}_s\right\}}_{k=0}^{\infty }$ and it governs switching in-between $A_1$  and $A_2$.

A self-mapping $S$  on $R^n$  which defines the solution trajectory in $A_1\cup A_2$  is contractive in both $A_1$  and $A_2$  if the sequences ${\left\{n_k\left(i\right)\right\}}_{k\in Z_{0+}}$  for $i=1,2$  satisfy $k_i{e}^{-{\rho }_i\left(n_{k+1}\left(i\right)-n_k\left(i\right)\right)T_s}\le K_S<1$  for $i=1,2$, equivalently, the sampling interval between consecutive values of the sequence ${\left\{n_k\left(i\right)\right\}}_{k\in Z_{0+}}$  is large enough to satisfy the constraints:

$$n_{k+1}\left(i\right)\ge n_k\left(i\right)+{\rho }_i^{-1}{T}_s^{-1}\left(ln k_i+\left|ln K_S\right|\right)$$

(30)

for $i=1,2$ and some $K_S\in \left[0 , 1\right)$. That is, the restriction of $S$  to any of the subsets $A_1$  and $A_2$ is contractive if there is a sufficiently large number of samples at each of the parameterizations $A_1$  and $A_2$ before switching (activated by the mapping $T$), respectively to $A_2$ and $A_1$ . The reason for those constraints is that the constants$k_i$might exceed unity so that small residence times at a parameterization before switching in-between $A_1$  and $A_2$ can translate into either expansiveness of the solutions or into the achievement of non-expansive (although non-contractive) properties.

Note by direct inspection that $0\in R^n$ is the unique fixed point of $S$and the unique best-proximity point of to $A_1$ in $A_2$ and vice versa. Note also that because of the definition of the mapping $T$ the solution trajectory never switches in-between $A_1$ in $A_2$ at time instants in an open interval defined by consecutive sampling instants $k{T}_s$ and $\left(k+1\right)T_s$ so an identical sign for all the components of $x\left(t\right)$ is always kept except at switching time instants which take place at sampling instants accordingly to (28)–(30).

In summary, we have proved the following result:

Proposition 2 (for Example 4). 

Consider a switched linear time-varying dynamic system described by ${\dot{x}}_1\left(t\right)=A_1{x}_1\left(t\right)$when the solution trajectory is in $A_1$and by ${\dot{x}}_2\left(t\right)=A_2{x}_2\left(t\right)$when the solution trajectory is in $A_2$, where both matrices of dynamics are stability Metzler matrices and that $x_0=x\left(0\right)$is either in $A_1$or in $A_2$. Assume that the self-mappings $T$and $S$for the discretization of the solution trajectory satisfy (28)–(30) for a given sampling period $T_s>0$. Then, the solution trajectory is bounded for all time for any given finite initial conditions and $x\left(k{T}_s\right)\to 0$as $k\to \infty$and $x\left(t\right)\to 0$as $t\to +\infty$.

Example 5. 

In Example 4, the stabilization of the cyclic configuration without requiring convergence can be governed just by monitoring the self-mapping $T$without requiring the contractiveness of $S$. That is, even if one (or both) differential system(s) is not asymptotically stable (and even if both of them are unstable), that is, if the Metzler matrices of dynamics fail to fulfil ${\left(-A_i\right)}^{-1}\succ 0$; $i=1,2$, or if they are still stability matrices while the residence time of the solution trajectory within each of the subsets violates the constraint (30).

The following concerned immediate result is pertinent:

Proposition 3 (for Example 5). 

Assume that at least one of the Metzler matrices $A_i$(i=1,2) is not a stability matrix and that, for some prefixed arbitrary real constant $M>0$, the self-mapping $T$on $A_1\cup A_2$is defined such as to satisfy:$x\left[ \left(k+1\right)T_s\right]=-{\theta }_{k}x\left(k{T}_s\right)$ if $‖x\left(k{T}_s\right)‖\ge M$ with ${\left\{{\theta }_k\right\}}_{k=0}^{\infty }\subset {\left\{{\displaystyle \frac{\lambda M}{‖x\left(k{T}_s\right)‖}}\right\}}_{k=0}^{\infty }\subset \left[0 , \lambda \right]\subset \left[0 , 1\right)$ for any given prescribed minimum sample threshold $z_m\in Z_{+}$ for any given real constant $\lambda \in \left[ 0 , 1\right)$. Then, there is some real constant $M´=M´\left(M, \lambda \right)>0$ such that if $x_0\in A_1$, or if $x_0\in A_2$ , and $‖x_0‖\le \lambda M$ , then $\underset{t\ge 0}{sup} ‖x\left(t\right)‖\le M+M´$.

Proof. 

${\left\{‖x\left(k{T}_s\right)‖\right\}}_{k=0}^{\infty }$ is bounded if $‖x\left(0\right)‖\le \lambda M$ since, if $‖x\left(k{T}_s\right)‖\ge M$ for some $k\in Z_{+}$, then $‖x\left[k+1\right]T_s‖\le \lambda M<M$ and $‖x\left[k+1\right]T_s‖<min\left(M , ‖x\left(k{T}_s\right)‖\right)$. Otherwise, if $‖x\left[k+1\right]T_s‖\ge ‖x\left(k{T}_s\right)‖$, then $M\le ‖x\left(k{T}_s\right)‖\le ‖x\left[\left(k+1\right) T_s\right]‖\le \lambda M<M$, a contradiction. Since the solution trajectory at sampling instants is uniformly bounded, the solution trajectory in-between consecutive sampling instants is also bounded according to:

$$‖x\left(t\right)‖\le max\left(K_1,K_2\right)e^{max\left(\left|{\rho }_1\right| , , \left|{\rho }_2\right|\right)T_s}‖x\left(k{T}_s\right)‖\le max\left(K_1,K_2\right)e^{max\left(\left|{\rho }_1\right| , , \left|{\rho }_2\right|\right)T_s}M$$

for all $t\in \left[kT , \left(k+1\right)T\right)$ and any $k\in Z_{0+}$. Thus, the solution trajectory is uniformly bounded for all time since $\underset{t\ge 0}{sup}‖x\left(t\right)‖\le max\left(K_1,K_2\right)e^{max\left(\left|{\rho }_1\right| , , \left|{\rho }_2\right|\right)T_s}M$. □

Example 6. 

Consider the infinite-dimensional Banach space $\left(L^p\left[-a , a\right] , ‖‖\right)$of real $n$-vector functions $f\in L^p \left[-a , a\right]$for any given $p\in Z_{+}$ and any given real $a$ with $+\infty >a>0$, and consider a set $B=B_1=B_2\subset R^n$. Note that, since $B_1=B_2$, the distance $D=d\left(B_1 , B_2\right)=\underset{x\in B_1 ,y\in B_2}{inf}‖x-y‖$ is zero. Define the two-cyclic self-mapping $T\left(B\right)\subset B$ by $\left(Tf\right)\left(x\right)=\left(-\Lambda f\right)\left( x\right)$, for any $x\in \left[-a , a\right]$, where $\Lambda \in R^{n\times n}$ is a diagonal matrix with diagonal entries in the interval $\left(-\beta , 0\right]\subset \left(-1 , 0\right)$. Consider the sequence generated by$f_{n+1}\left(x\right)=\left(T{f}_n\right)\left(x\right)=\left(-\Lambda f_n\right)\left( x\right)$; $\forall x\in \left[-a , a\right]$. Then, clearly ${\left\{ {\left(Tf\right)}_n\left(x\right)\right\}}_{n=0}^{\infty }\to 0$, for initial value$f_0\left(x\right)=f\left(x\right)$; $\forall x\in \left[-a , a\right]$ and any $f\in L^p\left[-a , a\right]$, so that $f_n$converges to the null real $p$-vector function in  $\left(L^p\left[-a , a\right] , ‖‖\right)$  as $n\to \infty$.

5. Conclusions

This paper has dealt with the properties of the convergence of distances between sequences and with the convergence of sequences of points in iterative processes which jointly involve cyclic contractive self-mappings and strict contractions. The cyclic contractions can involve more than two nonempty closed subsets in a metric space. It is assumed, in general, that the compositions of a cyclic contraction are combined with compositions of a strict contraction, which operates in each of the individual subsets. The compositions of both self-mappings can take place in any order and they can involve any number of mutual compositions of each of both contractive self-mappings with themselves and with the other self-mapping. It is proved that if one of the best-proximity points in the cyclic disposal is unique in a boundedly compact subset of the metric space, this suffices to achieve unique asymptotic limit cycles formed by one best-proximity point per each adjacent subset. The same conclusion is reached if one of the subsets is strictly convex and the metric space is a uniformly convex Banach space. That property holds for all the sequences with arbitrary initial points in the union of all the subsets of the cyclic disposal which converge to such a limit cycle.