Cyclic Large Contractions in Metric and Normed Spaces Under Eventual Perturbations

Abstract

Some properties on large contractions in metric spaces are proven. In particular, such contractions are proven to be asymptotically regular. In addition, if the metric space is complete, then the sequences that they generate are bounded, Cauchy, and convergent to a unique fixed point. Also, cyclic large contractions are an area of focus. It is proven that, if subsets of the cyclic disposal are nonempty closed and they intersect, all the sequences are bounded and Cauchy, and they converge to a unique fixed point located in the intersection of such subsets if the metric space is complete. If the subsets have a pair-wise empty intersection, then the boundedness of such sequences is proven without the need to assume the boundedness of the subsets in the cyclic disposal. The convergence of the sequences to a unique limit cycle of best proximity points, with one per subset in the cyclic disposal, is proven provided that the metric space is complete and that one of such subsets is boundedly compact with a singleton best proximity set. For that property to hold, it is not assumed that the remaining best proximity points are necessarily singletons. It has also been proven that all the subsequences contained within each of the subsets are Cauchy and they converge to a unique best proximity point, even if the corresponding best proximity sets is not a singleton. Furthermore, the hypothesis that one of the best proximity sets between adjacent subsets is a singleton can be weakened for any particular cyclic large contraction. Later on, eventual perturbations of the cyclic large self-mappings in normed spaces are discussed. If the norm of the perturbation additive operator is small enough, it is proven that the perturbed cyclic self-mapping maintains the property of being a cyclic large contraction associated with the unperturbed nominal cyclic large contraction. The maximum upper-bound of the perturbed operator ensures that such a property is given in an explicit manner.

1. Introduction

Cyclic self-mappings have been exhaustively studied in general metric spaces and in Banach spaces. See, for instance, [1,2,3,4,5,6,7,8,9] and some references therein. Also, cyclic ${\varphi }_A$-cyclic contractions and cyclic mappings in multiplicative metric spaces, in partially ordered and in orbitally complete metric spaces, have been addressed in [10,11,12,13] and in some of the references therein. Investigations have also been extended to cover b-metric and b-metric-like spaces and Hardy–Rogers contractions [14]. On the other hand, some proximity properties related to BSS cyclic mappings, S-cyclic mappings, Kannan S-coupled cyclic mappings, and Busemann convex metric spaces have been reported in [15,16,17,18]. It can be pointed out that the approximative compactness and the bounded compactness properties ensure the non-emptiness of the sets of best proximity points [2,19,20]. The framework of cyclic contractions is also a useful tool for the stabilization of switched dynamic systems where switches activate different model parameterizations or configurations [21,22,23,24,25,26,27,28,29]. On the other hand, it is well-known that enriched contractions involve some extra parameters apart from the usual contractive constants. In this way, the properties of boundedness and convergence of distances and sequences might be achieved under weaker conditions [30,31] than in the case of strict contractions. In particular, Chatterjea-enriched contractions have been investigated in [32] and some background studies in the literature. Enriched Ćirić–Reich–Rus contractions and quasi-contractions have been dealt with in [33,34] in Banach spaces and in convex metric spaces and in [35] for Kannan-type enriched contractions. In [36], enriched contractive and non-expansive mappings were dealt with in such a way that the introduction of the Mann iteration scheme with Kasnoselskij-type iterations was allowed. On the other hand, fixed point results for enriched Kannan mappings in CAT(0) spaces and enriched rational-type contractions in both quasi-Banach spaces and in generalized convex b-metric spaces were obtained in [37,38]. In [39], convergence results for Krasnoselskij-type iterations-based were obtained for cyclic contractions. In [40], generalized cyclic contractions in Banach spaces and a concerned fixed point theorem were investigated. Also, non-expansive mappings have been studied in [41] in ordered CAT(0) spaces for the approximation of fixed points. Furthermore, some iterative schemes concerned with enriched contractive and asymptotically non-expansive maps were investigated in Banach spaces in [42]. More recently, enriched cyclic contractive self-mappings in metric spaces and in uniformly convex Banach spaces were proposed and investigated in [43] for cyclical disposals composed of subsets of metric spaces.

It can be pointed out that so-called large contractions [1,44,45,46,47,48,49,50,51,52,53,54] are a class of weak contractions which maintain many of the relevant properties of strict contractions when distances between sequences are close to zero. In particular, large contractions of the Burton type are described in [1], and they are invoked as a main basic supporting tool for the results of this research. A fixed point theorem was obtained in [44] for large contractions via a c-comparison function in complete metric spaces. In [45], necessary and sufficient conditions are given in complete metric spaces of continuous scalar functions, $f\left(x\right)$ for $\left(x-f\left(x\right)\right)$, to be a large contraction. In [46], the fixed points of large enriched contractions are investigated in convex metric spaces and in G-convex metric spaces. Sufficient conditions are derived for the existence and uniqueness of fixed points. It is demonstrated that the Krasnoselskij iteration converges to the unique fixed point for large enriched contractions. The concepts of large s-simulation functions and large $Z_s$-contractions are developed in [47] with derivations of fixed point results for large $Z_s$-contractions. A fixed point theorem is proved in [48] for large contraction mappings with applications to fractional delay-differential equations formulated via the Caputo derivative. In [49], the existence of periodic solutions of a third-order iterative differential equation is investigated. The main tool used in the study is the Krasnoselskij–Burton’s theorem for a mapping which is the sum of a compact mapping and a large contraction. Such theorem is also used in [50] to establish sufficient conditions for the existence of periodic solutions of a class of discrete nonlinear Volterra-type equations subject to infinite delay. On the other hand, Krasnoselskij-type theorems are also applied in [51] to investigate the periodic solutions of a class of nonlinear neutral integro-differential equations. Also, the stability around the zero solution of a class of neutral nonlinear differential equations with a variable delay is investigated in [52] via Krasnoselskij–Burton’s theorem, involving the sum of a large contraction and a compact mapping in the mapping which defines the solution. On the other hand, the periodic solutions of a nonlinear totally iterative differential equation of arbitrary finite order via the above theorem are investigated in [53]. Also, the existence of a non-null periodic solution and that of a periodic positive solution in a highly nonlinear differential equation is addressed in [54] via a Krasnoselskij’s fixed point theorem which involves a large contraction. The objective of this article is to study the properties of cyclic large contractions in metric spaces. In particular, we investigate the boundedness, Cauchyness, and convergence of distances of sequences running in adjacent subsets to the distance between adjacent subsets, which coincide with the distance between the corresponding best proximity points. We also investigate the maintenance of the cyclic large contraction property under reasonably small perturbations of cyclic self-mappings in normed spaces. If the metric space is complete, then the subsets of the cyclic disposal are pair-wise disjoint, and if one of them is boundedly compact, with its best proximity set to its adjacent subset being furthermore a singleton, then any sequence generated on the union of the subsets converges to a unique limit cycle of best proximity points, with one per subset, even in the event that all the best proximity sets are not singletons.

In so-called Rakotch’s contractions, the contraction constant inherent to strict contractions is replaced by a monotonically decreasing function of the distance, while a large contraction is weak for any points, which then becomes a strict contraction for distances exceeding any given threshold which can be arbitrarily fixed. Also, it should be noted that a Rakotch contraction [55] is a Burton large contraction [1] and that a Burton large contraction lies in the class of Rakotch contractions for a particular constant contractive function.

On the other hand, it should also be noted that approximate fixed points of non-expansive mappings have been studied in [56] in the union of two sets in a complete hyperbolic space. The Baire category approach is used to prove that the iterates converge to a unique fixed point located in the intersection of both sets provided that they are not disjoint. If both sets are disjoint, for most mappings, the distance between iterates is proven to converge to that in between those two sets by still using the Baire category approach.

The rest of this paper is organized as follows: Section 2 recalls the definition of large contractions in metric spaces and shows that strict contractions are also large contractions. Some properties of large contractions are proven, particularly the fact that they are asymptotically regular and the fact that the sequences that they generate are bounded, Cauchy, and convergent to a unique fixed point if the metric space is complete. Section 3 gives the definition of cyclic contractions in metric spaces and provides and proves further results; in particular, it is shown that strict contractions are also large contractions, and the convergence of distances of sequences to the distance between the adjacent sets of a cyclic disposal is proven. Those sets are nonempty closed subsets of the metric space. In the event that the subsets are nonempty and closed and also intersect, all the sequences are bounded, even if the subsets are not bounded and Cauchy; additionally, if the metric space is complete, such sequences converge to a unique fixed point located in the intersection of such subsets. In the event that the subsets of the cyclic disposal have a pair-wise empty intersection, the boundedness of such sequence is proven without the need to assume the boundedness of the subsets. It is also proven that the sequences converge to a unique limit cycle of best proximity points, with one per subset in the cyclic disposal, provided that the metric space is complete and that one of these subsets is boundedly compact with a singleton best proximity set. However, it is not assumed that the remaining best proximity points are necessarily singletons. In parallel, it is seen that all the subsequences running within each of the subsets of the cyclic disposal converge to a unique best proximity point, even if the corresponding best proximity sets are not singletons, and they are also Cauchy sequences. We also prove that the former hypothesis that one of the best proximity sets between adjacent subsets is a singleton can be weakened, for a given particular cyclic large contraction, if there is just a single best proximity point in one of the subsets whose distance with its image in the adjacent subset equals the distance between adjacent subsets. This property is supported by the fact that even if only a best proximity point is reached by the subsequences of a given cyclic self-mapping in one of the subsets of the cyclic disposal, this does not necessarily imply that the best proximity set to which it belongs is necessarily a singleton. Section 4 considers the eventual perturbations of the cyclic large self-mappings in normed spaces. In the event that the norm of the perturbation additive operator is small enough, it is proven that the perturbed cyclic self-mapping maintains the property of being a cyclic large contraction associated with the nominal, i.e., an unperturbed, cyclic large contraction. It is assumed that both the unperturbed and perturbed mappings are cyclic on the same disposal of nonempty closed subsets of the metric space, but this fact does not require the perturbed operator to be cyclic. The maximum upper-bound of the perturbed operator which ensures that such a property is kept is given. Finally, conclusions are provided at the end of the paper.

2. Some Preliminary Features and Results on Large Non-Cyclic Contractions

Some properties of large contractions in the non-cyclic case in metric spaces are proven in this section, like the fact that they are asymptotically regular, while the generated iterates are bounded and Cauchy. If the metric space is complete, then such sequences converge to a unique fixed point.

The following notation is used:

$${\mathit{Z}}_{0+}={\mathit{Z}}_{+}\cup \left\{0\right\}; {\mathit{Z}}_{+}=\left\{z\in \mathit{Z}\hspace{0.33em}:z>0\right\}; {\mathit{Z}}_{0-}={\mathit{Z}}_{-}\cup \left\{0\right\}; {\mathit{Z}}_{-}=\left\{z\in \mathit{Z}\hspace{0.33em}:z<0\right\};$$

$${\mathit{R}}_{0+}={\mathit{R}}_{+}\cup \left\{0\right\}; {\mathit{R}}_{+}=\left\{z\in \mathit{R}\hspace{0.33em}:z>0\right\}; {\mathit{R}}_{0-}={\mathit{R}}_{-}\cup \left\{0\right\}; {\mathit{R}}_{-}=\left\{z\in \mathit{R}\hspace{0.33em}:z<0\right\};$$

$\mathit{Z},\mathit{R}, \mathrm{and}\mathbf{C}$ are, respectively, the sets of integer, real, and complex numbers;

$\bar{n}=\left\{1,2,\cdots ,n\right\}$;

$I_d$ is the identity matrix of order $d$;

The symbol $\vee$ stands for the logic disjunction “or”; the symbol $\wedge$ stands for the logic conjunction “and”;

$clX$ is the closure of the set $X$;

$Fix\left(T\right)$ denotes the set of fixed points of a self-mapping $T$ on $X$, where $\left(X,d\right)$ is a metric space.

The following definition borrowed from [1] is pertinent:

Definition 1. 

Let $\left(M,\hspace{0.33em}\rho \right)$ be a metric space. Then, $B:M\to M$ is said to be a large contraction if for each pair $\phi ,\hspace{0.33em}\psi \in M$ with $\phi \ne \psi$ then $\rho \left(B\phi \hspace{0.33em},\hspace{0.33em}B\psi \right)<\rho \left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ and if for each $\mathsf{ϵ} > 0$ there exists $\delta <1$ such that $\left[\phi ,\psi \in M,\hspace{0.33em}\rho \left(\phi ,\hspace{0.33em}\psi \right)\ge \epsilon \right]\Rightarrow \rho \left(B\phi \hspace{0.33em},\hspace{0.33em}B\psi \right)\le \delta \rho \left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right).$

It can be seen that a Rakotch contraction [55] is a Burton large contraction, [1]. Let $\left(X,d\right)$ be a metric space. A Rakotch contraction $T:X\to X$ satisfies a contractive condition of the form:

$$d\left(Tx\hspace{0.33em},Ty\right)\le \Phi \left(d\left(x,y\right)\right)d\left(x,y\right); \forall x,y\in X$$

where $\Phi :{\mathit{R}}_{0+}\to \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ is a decreasing function. As shown in one of the references, we can take $\delta =\delta \left(t\right)=\left\{\mathit{sup}\Phi \left(d\left(x,y\right)\right)\hspace{0.33em}:\hspace{0.33em}x,y\in X\hspace{0.33em}\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{h}\hspace{0.33em}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\hspace{0.33em}d\left(x,y\right)\ge t\right\}<1$ for any given $t>0$, which means that $d\left(Tx\hspace{0.33em},Ty\right)\le \delta d\left(x,y\right)$ for each $t>0$ such that $d\left(x,y\right)\ge t$. Furthermore, $d\left(Tx\hspace{0.33em},\hspace{0.33em}Ty\right)<d\left(x,y\right)$ if $x\ne y$. Therefore, a Rakotch contraction is a Burton large contraction. Also, a Burton large contraction $T:X\to X$ lies in the class of Rakotch contractions. Take any $t>0$ such that $d\left(x,y\right)\ge t$, then one has that $d\left(Tx\hspace{0.33em},Ty\right)\le \delta d\left(x,y\right)$ for some $\delta \in \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ and, furthermore, $d\left(Tx\hspace{0.33em},Ty\right)<d\left(x,y\right)$ if $x\ne y$. Now, take a constant function $\Phi \left(d\left(x,y\right)\right)=\delta$ for all $x,y\in X$ such that $d\left(x,y\right)\ge t$ so that $d\left(Tx\hspace{0.33em},Ty\right)\le \delta d\left(x,y\right)=\Phi \left(d\left(x,y\right)\right)d\left(x,y\right)$ if $d\left(x,y\right)\ge t$ for any given $t>0$. Note that the function $\Phi :{\mathit{R}}_{0+}\to \left[0\hspace{0.33em},\hspace{0.33em}\delta \right]\hspace{0.33em}\left(\subset \left[0\hspace{0.33em},\hspace{0.33em}1\right)\right)$, since constant, is also decreasing and thus it lies in the class of Rakotch contractions.

The following result is immediate:

Proposition 1. 

Let $\left(X,d\right)$ be a metric space. If $T:X\to X$ is a (strict) contraction, then it is a large contraction.

Proof. 

Since $T:X\to X$ is a (strict) contraction, then there exists a real constant $K\in \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ such that $d\left(Tx\hspace{0.33em},\hspace{0.33em}Ty\right)\le Kd\left(x,y\right)<d\left(x,y\right)$; $\forall x,y\left(\ne x\right)\in X$. Furthermore, for any given $\epsilon \in {\mathit{R}}_{+}$, if $d\left(x,y\right)\ge \epsilon$, then it still holds that $d\left(Tx\hspace{0.33em},\hspace{0.33em}Ty\right)\le Kd\left(x,y\right)$ so that it holds trivially that $\left[\forall \epsilon \in {\mathit{R}}_{+}\hspace{0.33em}:x,y\in X,\hspace{0.33em}d\left(x,\hspace{0.33em}y\right)\ge \epsilon \right]\Rightarrow d\left(Tx\hspace{0.33em},\hspace{0.33em}Ty\right)\le Kd\left(x\hspace{0.33em},\hspace{0.33em}y\right)$. Then, $T:X\to X$ is a large contraction. □

The converse of Proposition 1 is not true since a large contraction can be non-strict. Therefore, the set of large contractions of a metric space contains the set of strict contractions, but the sets are not identical. It turns out that a large contraction is asymptotically regular since it is a weak contraction. An alternative “ad hoc” proof based on the properties of large contractions is addressed in the following result, which also proves the boundedness, the Cauchyness, and the existence and uniqueness of a fixed point of the sequences generated through a large contraction, $T:X\to X$. It can be pointed out that since all sequences are proven to be bounded, the need for the existence of a bounded sequence, as assumed in [1], to guarantee the existence and uniqueness of a fixed point is not invoked.

Theorem 1. 

Let $\left(X,d\right)$ be a metric space. If $T:X\to X$ is a large contraction, then the following properties hold:

(i) $T:X\to X$ is asymptotically regular, that is, $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=0$; $\forall x\in X$.

(ii) ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}x\right)\right\}}_{n=0}^{\infty }\subset \left[0,\hspace{0.33em}+\infty \right)$, $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}x\right)=0$; $\forall m\in {\mathit{Z}}_{0+}$, $\forall x\in X$, and $\underset{n,m\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}x\right)=0$; $\forall x\in X$.

(iii) ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}y\right)\right\}}_{n=0}^{\infty }\subset \left[0,\hspace{0.33em}+\infty \right)$, $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}y\right)=0$; $\forall m\in {\mathit{Z}}_{0+}$, $\forall x\in X$, and $\underset{n,m\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}y\right)=0$; $\forall x,y\in X$.

(iv) Any sequence ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is bounded and it is a Cauchy sequence; $\forall x\in X$.

(v) If $\left(X,d\right)$ is, in addition, complete, then all the Cauchy sequences ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ are convergent; $\forall x\in X$ and they converge to a fixed point of $T:X\to X$ which is unique.

Proof. 

To prove Property (i), first note that since $T:X\to X$ is a large contraction, then it is also a weak contraction and then is continuous everywhere in $X$ since, for any given $\vartheta \in {\mathit{R}}^{+}$, there exist $\upsilon =\upsilon \left(\vartheta \right)\in {\mathit{R}}^{+}$ such that $\forall x\in X$, $\left[d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)<\upsilon \right]\Rightarrow \left[d\left(Tx,T^{2}x\right)<d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)<\vartheta =\upsilon \right]$ so that it suffices to choose a $\upsilon$ value equal to $\vartheta$. Consider the two subsequent cases which imply that $T:X\to X$ is not asymptotically regular:

Case a: For any given $x\in X$, $d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)\to \theta \left(>0\right)$ as $n\to \infty$. Then, since $T:X\to X$ is continuous for all $x\in X$, there exists some $N=N\left(x\right)\in {\mathit{Z}}_{0+}$ such that $d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)\ge \theta /2$ for all $n\ge N$. Since $T:X\to X$ is a large contraction, then there exists some $\delta =\delta \left(\theta \right)\in \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ such that $\theta /2\le d\left(T^{n+1}x\hspace{0.33em},\hspace{0.33em}{T}^{n+2}x\right)\le \delta \theta /2<\theta /2$, a contradiction, unless $\theta =0$, which implies that $T:X\to X$ is asymptotically regular.

Case b: $d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)$ does not have a limit as $n\to \infty$ for any given $x\in X$. Then, there is no $n\in {\mathit{Z}}_{0+}$ such that $d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=0$ since, otherwise, $d\left(T^{n+m}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m+1}x\right)=0$; $\forall m\in {\mathit{Z}}_{0+}$, since $T:X\to X$ is a weak contraction. But then $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=0$, which means that it has a limit, which contradicts the claim that it does not have a limit. Property (i) is thus proven.

To prove Property (ii), note that since $T:X\to X$ is a large contraction, then for any given $\epsilon \in {\mathit{R}}^{+}$, there exists some real constant $\delta =\delta \left(\epsilon \right)\in \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ such that

$$\begin{gathered} \left[\left(d\left(T^{2}x\hspace{0.33em},\hspace{0.33em}Tx\right)<d\left(x,Tx\right)\right)\iff x\ne Tx\hspace{0.33em}\right]\hspace{0.33em}\wedge \left[\left(d\left(T^{2}x\hspace{0.33em},\hspace{0.33em}Tx\right)=d\left(x,Tx\right)=0\right)\iff x=Tx\hspace{0.33em}\right] \\ \wedge \left[d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)\ge \epsilon \right]\Rightarrow \left[d\left(Tx,T^{2}x\right)\le \delta d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)\right]; \forall x\in X \end{gathered}$$

(1)

Then, for any $x\in X$, there exists $\widehat{\lambda }=\widehat{\lambda }\left(x,Tx,T^{2}x\right)\in \left[0\hspace{0.33em},\hspace{0.33em}\delta \right)\cap {\mathit{R}}_{0+}$ with $\widehat{\lambda }=\left\{\begin{array}{c}d\left(Tx,T^{2}x\right)/d\left(x,Tx\right) if Tx\ne x\\ 0 if Tx=x\end{array}\right.$, such that, for any $\lambda \in \left[0\hspace{0.33em},\hspace{0.33em}\widehat{\lambda }\right]\cap {\mathit{R}}_{0+}$, one has

$$\lambda d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)\le \widehat{\lambda }d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)=d\left(Tx,T^{2}x\right)\le \delta d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right); \forall x\in X.$$

(2)

As a result, there exist, in general, non-unique real sequences ${\left\{{\delta }_n\right\}}_{n=1}^{\infty }\subset \left[0\hspace{0.33em},\hspace{0.33em}1\right)$, ${\left\{{\lambda }_n\right\}}_{n=1}^{\infty }\left(\subset {\left\{{\delta }_n\right\}}_{n=1}^{\infty }\right)\subset \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ with ${\delta }_n={\delta }_n\left(T^{n-1}x\hspace{0.33em},T^{n}x,{\lambda }_n\right)$, ${\lambda }_n={\lambda }_n\left(T^{n-1}x,T^{n}x,T^{n+1}x\right)$; $\forall n\in {\mathit{Z}}_{0+}$ with ${\delta }_1=\delta$ and ${\lambda }_n=0$ if $d\left(T^{n}x,T^{n+1}x\right)=0$ such that

$$\begin{gathered} d\left(T^{n-1}x\hspace{0.33em},\hspace{0.33em}{T}^{n}x\right)\ge {\epsilon }_{n-1}\ge {\lambda }_{n-1}d\left(T^{n-2}x\hspace{0.33em},\hspace{0.33em}{T}^{n-1}x\right)\Rightarrow \\ {\lambda }_{n}d\left(T^{n-1}x\hspace{0.33em},\hspace{0.33em}{T}^{n}x\right)\le d\left(T^{n}x,T^{n+1}x\right)\le {\delta }_{n}d\left(T^{n-1}x\hspace{0.33em},\hspace{0.33em}{T}^{n}x\right)\le \left({\prod }_{j=1}^n\left[{\delta }_j\right]\right)d\left(x,Tx\right)\le {\bar{\delta }}^n\left(1,n\right)d\left(x,Tx\right); \forall x\in X \end{gathered}$$

(3)

where $\bar{\delta }\left(1,n\right)={\left({\prod }_{j=1}^n\left[{\delta }_j\right]\right)}^{1/n}\left(\in \left[0\hspace{0.33em},\hspace{0.33em}1\right)\right)$ is the geometric mean of the set $\left\{{\delta }_1,{\delta }_2,\dots ,{\delta }_n\right\}$. Thus,

$$\begin{gathered} d\left(T^{n+m}x,T^{n}x\right)\le \sum _{j=0}^{m-1}d\left(T^{n+j}x,T^{n+j+1}x\right)\le \left(\sum _{j=0}^{m-1}{\bar{\delta }}^{n+j}\left(1,n+j\right)\right)d\left(x,Tx\right) \\ =\left(\prod _{j=1}^{m-1}\left[{\delta }_{n+j}\right]+...+{\delta }_{n+2}{\delta }_{n+1}+{\delta }_{n+1}+1\right){\bar{\delta }}^n\left(1,n\right)d\left(x,Tx\right) \\ \le {\displaystyle \frac{1-{\tilde{\delta }}^{n+m}\left(n+1\hspace{0.33em},\hspace{0.33em}n+m-1\right)}{1-\widehat{\delta }\left(n+1\hspace{0.33em},\hspace{0.33em}n+m-1\right)}}{\bar{\delta }}^n\left(1,n\right)d\left(x,Tx\right) \\ \le {\displaystyle \frac{1-{\tilde{\delta }}^m\left(n+1\hspace{0.33em},\hspace{0.33em}n+m-1\right)}{1-\widehat{\delta }\left(n+1\hspace{0.33em},\hspace{0.33em}n+m-1\right)}}{\widehat{\delta }}^n\left(1,n\right)d\left(x,Tx\right)<+\infty ; \forall m\in {\mathit{Z}}_{0+}, \forall x\in X \end{gathered}$$

(4)

where $\widehat{\delta }\left(n+1\hspace{0.33em},\hspace{0.33em}n+m-1\right)=\underset{1\le j\le m-1}{max}{\delta }_{n+j}$, so that $\widehat{\delta }\left(1\hspace{0.33em},\hspace{0.33em}n\right)=\underset{1\le j\le n}{max}{\delta }_j$, and $\tilde{\delta }\left(n+1\hspace{0.33em},\hspace{0.33em}n+m-1\right)=\underset{1\le j\le m-1}{min}{\delta }_{n+j}$. Since ${\widehat{\delta }}^n\left(1,n\right)\to 0$ as $n\to \infty$, one can see from (4) that $d\left(T^{n+m}x,T^{n}x\right)\to 0$ as $n\to \infty$; $\forall m\in {\mathit{Z}}_{0+}$ and $d\left(T^{n+m}x,T^{n}x\right)\to 0$ as $n,m\to \infty$; $\forall x\in X$. Since the sequences of distances converge to zero, then they have to be bounded, too, that is, ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}x\right)\right\}}_{n=0}^{\infty }\subset \left[0,\hspace{0.33em}+\infty \right)$. Property (ii) is thus proven.

Property (iii) is proven by replacing Equation (3) with Equation (4) for any $x,y\in X$ and following the same line of reasoning as in the proof of Property (ii) by noting that

$${\lambda }_{n}d\left(T^{n-1}y\hspace{0.33em},\hspace{0.33em}{T}^{n-1}x\right)\le d\left(T^{n}y,T^{n}x\right)\le {\delta }_{n}d\left(T^{n-1}y\hspace{0.33em},\hspace{0.33em}{T}^{n-1}x\right)\le \left({\prod }_{j=1}^n\left[{\delta }_j\right]\right)d\left(y,x\right)\le {\bar{\delta }}^n\left(1,n\right)d\left(y,x\right); \forall x,y\in X.$$

To prove Property (iv), proceed by contradiction by assuming that ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is unbounded for some $x\in X$. Then, there exists a strictly increasing sequence ${\left\{n_k\right\}}_{k=0}^{\infty }\subset {\mathit{Z}}_{0+}$ with $\mathit{sup}\left(n_{k+1}-n_k\right)\le \widehat{n}<+\infty$ such that ${\left\{\left|T^{n_k}x\right|\right\}}_{k=0}^{\infty }$ is strictly increasing (then diverging to $+\infty$) since ${\left\{d\left(x\hspace{0.33em},T^{n_k}x\right)\right\}}_{k=0}^{\infty }$ is strictly increasing, then diverging. Then, it follows by taking limits as $k\to \infty$ in $d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}x\right)\le d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k}x\right)+d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}x\right)$ and invoking the fact that if ${\left\{\left|T^{n_k}x\right|\right\}}_{k=0}^{\infty }$ is strictly increasing, then the subsequent contradiction holds:

$$0<\underset{k\to \infty }{\mathit{lim}\mathit{sup}}\left(d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}x\right)-d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}x\right)\right)\le \underset{k\to \infty }{\mathit{lim}}d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k}x\right)=0$$

As a result, ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is bounded; $\forall x\in X$ and the first part of Property (iv) are thus proven. Now, proceed again by contradiction to prove that ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is Cauchy. Assume that there is a sequence, ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$, which is not Cauchy for some $x\in X$. Then, for any given $\epsilon \in {\mathit{R}}_{+}$, there is some $m=m\left(x,\epsilon \right)\in {\mathit{Z}}_{0+}$ such that, for some $N=N\left(x,\epsilon \right)\in {\mathit{Z}}_{0+}$, it follows that $d\left(T^{n+j}x\hspace{0.33em},\hspace{0.33em}{T}^{n+k}x\right)\ge \epsilon$ for some $j,k\in {\mathit{Z}}_{0+}\cap \left(\overline{m}\cup \left\{0\right\}\right)$ and $d\left(T^{n+i}x\hspace{0.33em},\hspace{0.33em}{T}^{n+\mathcal{l}}x\right)<\epsilon$; $\forall i,\hspace{0.33em}\mathcal{l}\left(\ne j,k\right)\in {\mathit{Z}}_{0+}\cap \left(\overline{m}\cup \left\{0\right\}\right)$. Thus, one can see that there is no zero limit $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+m}x\right)$ for some $m\in {\mathit{Z}}_{0+}$ and some $x\in X$ contradicting Property (ii). Therefore, ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is Cauchy; $\forall x\in X$. Property (iv) is thus proven.

The first part of Property (v) follows directly from Property (iv) since all Cauchy sequences are convergent in a complete metric space. Thus, ${\left\{T^{n}x\right\}}_{n=0}^{\infty }\to z\left(\in X\right)$ Assume that $z\ne Tz$ and proceed by contradiction, leading to

$$0<d\left(z,\hspace{0.33em}Tz\right)\le d\left(z,\hspace{0.33em}{T}^{n}x\right)+d\left(T\left(T^{n-1}x\right),Tz\right); \forall x\in X.$$

Now, since $T:X\to X$ is continuous, then the order of the distance and the limit may be interchanged, representing a weak contraction; the above relation implies the subsequent contradiction:

$$\begin{gathered} 0<d\left(z,\hspace{0.33em}Tz\right)\le \underset{n\to \infty }{lim}d\left(z,\hspace{0.33em}{T}^{n}x\right)+\underset{n\to \infty }{lim}d\left(T\left(T^{n-1}x\right),Tz\right) \\ =d\left(z,\hspace{0.33em}\underset{n\to \infty }{lim}{T}^{n}x\right)+d\left(T\left(\underset{n\to \infty }{lim}{T}^{n-1}x\right),Tz\right)=d\left(z,z\right)+d\left(Tz,Tz\right)=0. \end{gathered}$$

Then, $z=Tz$ so that $z\in Fix\left(T\right)$ and all the Cauchy sequences converge to a fixed point of $T:X\to X$. It remains to be proven that $Fix\left(T\right)=\left\{z\right\}$. Assume that $Fix\left(T\right)=\left\{z,z_1\right\}$. Since $T:X\to X$ is a large contraction and $z\ne z_1$, then $0<d\left(z,z_1\right)=d\left(Tz,T{z}_1\right)<d\left(z,z_1\right)$ and hence a contradiction. Thus, $z=z_1$ and $Fix\left(T\right)=\left\{z\right\}$. Property (v) is thus proven. □

3. Results of Cyclic Large Contractions

Large contractions in metric spaces $\left(X,d\right)$ in the cyclic case are focused on for the case of cyclical disposals of $p\left(\ge 2\right)$ nonempty closed subsets of $X$ for large contractive self-mappings defined on the union of such subsets. In particular, the sequences of distances between adjacent subsets converge to the distances between such subsets, and the iterates are bounded for any initial condition. In the event that the subsets have a pair-wise intersection, then the self-mappings are asymptotically regular and, if the metric space is complete, then such sequences converge to a unique fixed point. If $\left(X,d\right)$ is complete, the subsets of the cyclic disposal are pair-wise disjoint, and one of them is boundedly compact, with its best proximity set to its adjacent subset, which is a singleton, and then any sequence generated on the union of the subsets converges to a unique limit cycle of best proximity points, with one per subset. In this case, each subsequence of the above one contained in one of the subsets converges to a unique best proximity point to the next adjacent subset, even if some of the corresponding best proximity sets are not singletons.

The following definition generalizes Definition 1 and refers to a large contraction for the case of cyclic mappings on a set of subsets of a metric space.

Definition 2. 

Let $\left(X,\hspace{0.33em}d\right)$ be a metric space with a set of $p$ of nonempty subsets $A_i$ of $X$; $\forall i\in \bar{p}$ such that $dist\left(A_i,A_{i+1}\right)=d\left(A_i,A_{i+1}\right)=D$. Then, the self-mapping $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is said to be a cyclic large contraction if the following conditions hold:

1. $T\left(A_i\right)\subset A_{i+1}$; $\forall i\in \bar{p}$.

2. $d\left(T\phi ,T\psi \right)<d\left(\phi ,\psi \right)$ for each given pair $\left(\phi ,\psi \right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$; $\forall i\in \bar{p}$, such that $d\left(\varphi ,\psi \right)>D$.

3. For each $\epsilon >0$ there exists $\delta <1$ such that

$$\left[\left(\phi ,⥂\psi \right)\in \left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)\hspace{0.33em}for\hspace{0.33em}any\hspace{0.33em}i\in \bar{p},\hspace{0.33em}\hspace{0.33em}d\left(\phi ,\hspace{0.33em}\psi \right)\ge D+\epsilon \right]\Rightarrow d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)\le \delta \left(d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)-D\right)+D.$$

Now, recall that the sets of best proximity points $A_{i0}\subset A_i$ of each subset $A_i$ to its adjacent one $A_{i+1}$ in the cyclic disposal are $A_{i0}=\left\{z_i\in A_i:d\left(z_i,A_{i+1}\right)=D\right\}$; $\forall i\in \bar{p}$.

Remarks 1. 

The following discussion is pertinently related to Definition 2. Note that condition 2 of Definition 2 for $D=0$ is identical to its parallel condition of the non-cyclic case $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)<d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ if $\phi \ne \psi$ of Definition 1. If $D>0$, one can see from the third condition of Definition 2 that

(1) $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)=d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\le \delta d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)+\left(1-\delta \right)D\Rightarrow d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=D\Rightarrow d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)=d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=D$ and, in this case, the second condition does not apply.

(2) From the third condition, one sees that if $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)>d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ that $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)<d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)\le \delta d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)+\left(1-\delta \right)D\Rightarrow \left(1-\delta \right)d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)<\left(1-\delta \right)D\Rightarrow d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)<D$, which contradicts $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\ge D$ since $\left(\phi ,\hspace{0.33em}\psi \right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$. Therefore, $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)\le d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ so that $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is non-expansive.

(3) Again, from the third condition, note that if $\epsilon +D\le d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)<d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)$, then this would imply that $\epsilon +D\le d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)<d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)\le \delta d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)+\left(1-\delta \right)D\Rightarrow \epsilon +\delta D<\delta d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\Rightarrow d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)-D>\epsilon /\delta >\epsilon$, which fails to hold if $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\in \left(D+\epsilon \hspace{0.33em},\hspace{0.33em}D+\epsilon /\delta \right]$; hence, it represents a contradiction to the joint constraints $\epsilon +D\le d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ and $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\le d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)$. As a result, $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)\le d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ for any $\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$; $\forall i\in \bar{p}$. Also, if $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)\le \delta \left(d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)-D\right)+D$, then $\left(1-\delta \right)d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=\left(1-\delta \right)D\Rightarrow d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=D$ so that $\phi$ and $\psi$ are the best proximity points in adjacent subsets, and then one can conclude that, for any points in adjacent subsets which are not best proximity points, $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)<d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ if $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)\ge D+\epsilon$ as expected. Then $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)<d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$; $\forall \left(\phi \hspace{0.33em},\hspace{0.33em}\psi \ne T\phi \right)\in A_i\times A_{i+1}$; $\forall i\in \bar{p}$ and $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)=d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ $\iff d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=D$ so that $\phi$ and $\psi$ are the best proximity points in adjacent subsets.

As a result, concerning Definition 2, one concludes that

(a) The second condition always holds except at the best proximity points between adjacent subsets of the cyclic disposal;

(b) The third condition implies that $d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)=D\iff d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)=D$;

(c) If $\epsilon +D\le d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ then $d\left(T\phi \hspace{0.33em},\hspace{0.33em}T\psi \right)<d\left(\phi \hspace{0.33em},\hspace{0.33em}\psi \right)$ as expected from the extension of Definition 1 to the cyclic case.

The following result is a direct consequence of Theorem 1 when the subsets $A_i$ of the cyclic disposal are nonempty, closed, and have a nonempty intersection:

Corollary 1. 

Let $\left(X,d\right)$ be a complete metric space with a set of $p$ nonempty closed subsets $A_i\subset X$; $\forall i\in \bar{p}$ which have a nonempty intersection. Assume that $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction. Then, $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ has a unique fixed point in ${\bigcap }_{i\in \bar{p}}{A}_i$ to which all sequences ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$, which are also Cauchy sequences and then bounded, converge for any given initial point $x\in {\bigcup }_{i\in \bar{p}}{A}_i$.

Proof. 

Since ${\bigcap }_{i\in \bar{p}}{A}_i\ne \varnothing$, then $A_i\cap A_j\ne \varnothing$, $\forall i,j\left(\ne i\right)\in \bar{p}$ and $d\left(A_i,A_{i+1}\right)=D=0$; $\forall i\in \bar{p}$, and the third condition of Definition 2 is identical to the parallel condition of Definition 1 by replacing the metric space with ${\bigcup }_{i\in \bar{p}}{A}_i$. Also, since $T\left(A_i\right)\subset A_{i+1}$ and $A_i$ is nonempty and closed, $\forall i\in \bar{p}$, one can see from Theorem 1[(iv)–(v)] that ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$; $\forall x\in {\bigcup }_{i\in \bar{p}}{A}_i$ is a bounded Cauchy sequence convergent to some $z=z\left(x\right)$, which is in ${\bigcap }_{i\in \bar{p}}{A}_i$, which is nonempty and closed, since $\left(X,d\right)$ is a complete metric space. From Theorem 1(i), the continuity of $T$ on ${\bigcup }_{i\in \bar{p}}{A}_i$ if $z\notin Fix\left(T\right)$, and we reach the following contradiction:

$$0<\underset{n\to \infty }{lim}d\left(Tz,\hspace{0.33em}z\right)=\underset{n\to \infty }{lim}d\left(z,T\left(T^{n}x\right)\hspace{0.33em}\right)\le \underset{n\to \infty }{lim}d\left(z,T^{n}x\hspace{0.33em}\right)+\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=d\left(z,\underset{n\to \infty }{lim}{T}^{n}x\hspace{0.33em}\right)+0=d\left(z,z\right)=0.$$

Thus, $z\in Fix\left(T\right)$. Assume that there are two distinct fixed points, $z_1$ and $z$, of $T$ on ${\bigcup }_{i\in \bar{p}}{A}_i$; then, since a large contraction is also a weak contraction for any $x,y\left(\ne x\right)\in {\bigcup }_{i\in \bar{p}}{A}_i$, one gets the contradiction $0<d\left(z,z_1\right)=d\left(Tz,T{z}_1\right)<d\left(z,z_1\right)$ if $FixT=\left\{z,z_1\right\}$. Thus, $FixT=\left\{z\right\}$. □

In the event that $d\left(A_i,A_{i+1}\right)=D>0$; $\forall i\in \bar{p}$, then $A_i\cap A_j=\varnothing$; $\forall i,j\left(\ne i\right)\in \bar{p}$ and ${\bigcap }_{i\in \bar{p}}{A}_i=\varnothing$, and it turns out that a cyclic large contraction $T$ on ${\bigcup }_{i\in \bar{p}}{A}_i$ cannot be asymptotically regular, and the sequences that it generates cannot converge and they cannot be Cauchy. Thus, Corollary 1 no longer holds.

In the event that the distance between adjacent subsets is eventually larger than zero, we have the following result for cyclic large contractions:

Theorem 2. 

Let $\left(X,d\right)$ be a metric space with a set of $p$ of nonempty closed subsets $A_i\subset X$; $\forall i\in \bar{p}$ such that $d\left(A_i,A_{i+1}\right)=D>0$. If $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction, then the following properties hold:

(i) $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=D$; $\forall x\in {\bigcup }_{i\in \bar{p}}{A}_i$

$\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n}y\right)=D$; $\forall \left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$; $\forall i\in \bar{p}$.

(ii) ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+mp+1}x\right)\right\}}_{n=0}^{\infty }\subset \left[0,\hspace{0.33em}+\infty \right)$; $\forall x\in {\bigcup }_{i\in \bar{p}}{A}_i$, $\forall m\in {\mathit{Z}}_{0+}$.

${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+mp}y\right)\right\}}_{n=0}^{\infty }\subset \left[0,\hspace{0.33em}+\infty \right)$; $\forall \left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$, $\forall m\in {\mathit{Z}}_{0+}$.

(iii) ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is bounded for all $x\in {\bigcup }_{i\in \bar{p}}{A}_i$.

(iv) If $\left(X,d\right)$ is, in addition, complete and $A_i$ is boundedly compact for some $i\in \bar{p}$ and $A_{i0}=\left\{z\in A_i:\hspace{0.33em}d\left(z,\hspace{0.33em}Tz\right)=D\right\}\hspace{0.33em}=\left\{z_i\right\}$, that is, the best proximity set of $A_i$ to $A_{i+1}$ is a singleton, then

(1) 

Any sequence ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$; $\forall x\in {\bigcup }_{i\in \bar{p}}{A}_i$ converges to a unique limit cycle formed with best proximity points, with one per adjacent subset, $\left(z_1,z_2,\dots ,z_p\right)\in A_{10}\times A_{20}\times ...\times A_{p0}$.

(2) 

If $x\in A_j$ for some $j\in \bar{p}$, then ${\left\{T^{np+\mathcal{l}}\right\}}_{n=0}^{\infty }\to z_{j+\mathcal{l}}$ and it is a Cauchy sequence.

Proof. 

From Definition 2, one has for any given $x\in {\bigcup }_{i\in \bar{p}}{A}_i$ that $D\le d\left(Tx\hspace{0.33em},\hspace{0.33em}{T}^{2}x\right)<\mathit{min}\left(D+\epsilon \hspace{0.33em},d\left(x,Tx\right)\right)$, and the following two cases can arise:

Case 1: $\left[\left(x,Tx\right)\in \left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)\hspace{0.33em}for\hspace{0.33em}any\hspace{0.33em}i\in \bar{p},\hspace{0.33em}\hspace{0.33em}d\left(x,\hspace{0.33em}Tx\right)\ge D+\epsilon \right]\Rightarrow$

$d\left(Tx\hspace{0.33em},\hspace{0.33em}{T}^{2}x\right)\le \delta \left(d\left(x\hspace{0.33em},\hspace{0.33em}Tx\right)-D\right)+D$.

Then, there exist integers $n_0,n_{01}\left(>n_0\right)\in {\mathit{Z}}_{0+}$ such that

$$D+\epsilon \le d\left(T^{n+1}x\hspace{0.33em},\hspace{0.33em}{T}^{n}x\right)\le \delta \left(d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n-1}x\right)-D\right)+D\le {\delta }^n\left(d\left(x,Tx\right)-D\right)+D; \forall n\in \left[1,\hspace{0.33em}{n}_0\right]\cap {\mathit{Z}}_{+}$$

(5)

$$D\le d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)<\mathit{min}\left(D+\epsilon \hspace{0.33em},d\left(T^{n-1}x,T^{n}x\right)\right)=D+\epsilon \le d\left(T^{n-1}x,T^{n}x\right); \forall n\in \left[n_0+1,\hspace{0.33em}{n}_{01}\right]\cap {\mathit{Z}}_{+}$$

(6)

$$D\le d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)<d\left(T^{n-1}x,T^{n}x\right)<D+\epsilon ; \forall n\left(>n_{01}\right)\in {\mathit{Z}}_{+}$$

(7)

Case 2: $D\le d\left(x,\hspace{0.33em}Tx\right)=d\left(T^{n_0-1}x,\hspace{0.33em}{T}^{n_0}x\right)<D+\epsilon$.

Case 2 follows from (7) with $n_0=1$. Cases 1–2 both lead to $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=D$; $\forall x\in {\bigcup }_{i\in \bar{p}}{A}_i$. On the other hand, if $\left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$; $\forall i\in \bar{p}$ then $D\le d\left(Tx\hspace{0.33em},\hspace{0.33em}Ty\right)<d\left(x,y\right)$ if $d\left(x,y\right)>D$, which implies that $\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n}y\right)=D$. Property (i) is thus proven.

To prove Property (ii), use contradiction arguments. If ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+mp+1}x\right)\right\}}_{n=0}^{\infty }$ is unbounded for some $x\in {\bigcup }_{i\in \bar{p}}{A}_i$, then there is some distance subsequence ${\left\{d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)\right\}}_{k=0}^{\infty }\to +\infty$ for a strictly increasing sequence of integer numbers ${\left\{n_k\right\}}_{k=0}^{\infty }\left(\subset {\mathit{Z}}_{+}\right)$, that is, $d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)<d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}{T}^{n_{k+1}+mp+1}x\right)$. But then, there is some finite $N\in {\mathit{Z}}_{0+}$ such that

$$\begin{gathered} d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)<d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}{T}^{n_{k+1}+mp+1}x\right) \\ \le {\delta }^{n_{k+1}-n_k}\left(d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)-D\right)+D\le \delta \left(d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)-D\right)+D \end{gathered}$$

$\left(1-\delta \right)d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)<\left(1-\delta \right)D$, hence a contradiction to the property $d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp+1}x\right)\ge D$. Therefore, such a diverging subsequence of distances cannot exist so that ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+mp+1}x\right)\right\}}_{n=0}^{\infty }$ is bounded for all $x\in {\bigcup }_{i\in \bar{p}}{A}_i$. In the same way, it is proven under a similar contradiction argument that some distance subsequence ${\left\{d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp}y\right)\right\}}_{k=0}^{\infty }\to +\infty$ cannot exist for some $\left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$ and some $m\in {\mathit{Z}}_{0+}$, for some strictly increasing sequences of integer numbers ${\left\{n_k\right\}}_{k=0}^{\infty }\left(\subset {\mathit{Z}}_{+}\right)$, that is, $d\left(T^{n_k}x\hspace{0.33em},\hspace{0.33em}{T}^{n_k+mp}y\right)<d\left(T^{n_{k+1}}x\hspace{0.33em},\hspace{0.33em}{T}^{n_{k+1}+mp}y\right)$. This fact shows that ${\left\{d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+mp}y\right)\right\}}_{n=0}^{\infty }\subset \left[0,\hspace{0.33em}+\infty \right)$; $\forall \left(x,y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$, $\forall m\in {\mathit{Z}}_{0+}$. Property (ii) is thus proven.

To prove Property (iii), one can use contradiction arguments again. Assume that ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is unbounded for some $x\in {\bigcup }_{i\in \bar{p}}{A}_i$ so that $d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}x\right)\to +\infty$ as $n\to \infty$. Then, such a sequence of distances has a diverging subsequence of strictly increasing members:

$$+\infty \leftarrow d\left(T^{n_{k+j_k}}x\hspace{0.33em},\hspace{0.33em}x\right)>d\left(T^{n_{k+j_{k-1}}}x\hspace{0.33em},\hspace{0.33em}x\right)>..>d\left(Tx,x\right)$$

where $n_{k+1}>n_k$; $\forall k\in {\mathit{Z}}_{0+}$. By choosing $\left(n_{k+j_k}-n_k\right)\to +\infty$ as $k\to \infty$, one reaches the subsequent contradiction as $k\to \infty$:

$$\begin{gathered} \left(+\infty \leftarrow \right)\left|d\left(T^{p{n}_{k+j_k}}x\hspace{0.33em},\hspace{0.33em}x\right)-d\left(T^{p{n}_k}x\hspace{0.33em},\hspace{0.33em}x\right)\right| \\ \le d\left(T^{p{n}_{k+j_k}}x\hspace{0.33em},\hspace{0.33em}{T}^{p{n}_k+1}x\hspace{0.33em}\right)+d\left(x\hspace{0.33em},\hspace{0.33em}{T}^{p{n}_k+1}x\hspace{0.33em}\right)\left(\to D+D=2D<+\infty \right). \end{gathered}$$

As a result, no subsequence ${\left\{T^{n_k}x\right\}}_{k=0}^{\infty }$ of ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is unbounded for some $x\in {\bigcup }_{i\in \bar{p}}{A}_i$ so that ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ is bounded for all $x\in {\bigcup }_{i\in \bar{p}}{A}_i$. Property (iii) is thus proven.

To prove Property (iv), note that if $\left(X,d\right)$ is complete and $A_i$ is boundedly compact for some $i\in \bar{p}$ and $A_{i0}=\left\{z\in A_i:\hspace{0.33em}d\left(z,\hspace{0.33em}Tz\right)=D\right\}\hspace{0.33em}=\left\{z_i\right\}$, then any $x\in {\bigcup }_{i\in \bar{p}}{A}_i$ is in a subset $A_j$ of $X$ for some $j\in \bar{p}$ and then $T^{i-j+\mathcal{l}p}x\in A_i$ for all $\mathcal{l}\in {\mathit{Z}}_{0+}$ if $j\le i$ and for all $\mathcal{l}\in {\mathit{Z}}_{+}$ if $i<j\le p$. Thus, ${\left\{T^{i-j+np}x\right\}}_{n=1}^{\infty }\subset A_i$. Since $A_i$ is boundedly compact and closed, there is a subsequence ${\left\{T^{i-j+n_{k}p}x\right\}}_{k=1}^{\infty }$ of the bounded sequence ${\left\{T^{i-j+np}x\right\}}_{n=1}^{\infty }\subset A_i$ (Property (i)), with ${\left\{n_k\right\}}_{k=1}^{\infty }\subset \left({\mathit{Z}}_{0+}\right)$ strictly increasing, which is convergent in $A_i$, that is, ${\left\{T^{i-j+n_{k}p}x\right\}}_{k=1}^{\infty }\to {\omega }_i\left(\in A_i\right)$. Furthermore, from Property (i),$\underset{n\to \infty }{lim}d\left(T^{n}x\hspace{0.33em},\hspace{0.33em}{T}^{n+1}x\right)=D$; $\forall x\in {\bigcup }_{i\in \bar{p}}{A}_i$ implies that all the distance subsequences are also convergent so that, by virtue of the continuity of large contraction mapping, we can interchange the orders of the limit and distance operations to conclude that

$$\begin{array}{l}\underset{k\to \infty }{lim}d\left(T^{i-j+n_{k}p}x\hspace{0.33em},\hspace{0.33em}{T}^{i-j+n_{k}p+1}x\right)=\underset{k\to \infty }{lim}d\left(T^{i-j+n_{k}p}x\hspace{0.33em},\hspace{0.33em}{T}^{i-j+n_{k}p}\left(Tx\right)\right)\\ =\underset{k\to \infty }{lim}d\left(\underset{k\to \infty }{lim}{T}^{i-j+n_{k}p}x\hspace{0.33em},\hspace{0.33em}{T}^{i-j+n_{k}p}\left(Tx\right)\right)=d\left({\omega }_i\hspace{0.33em},\hspace{0.33em}T\underset{k\to \infty }{lim}{T}^{i-j+n_{k}p}x\right)=d\left({\omega }_i\hspace{0.33em},\hspace{0.33em}T{\omega }_i\right)=D.\end{array}$$

(8)

Therefore, ${\omega }_i=z_i\left(\in A_{i0}=\left\{z_i\right\}\right)$ and the best proximity set $A_{i+1,0}\subset A_{i+1}$ is nonempty (although not necessarily a singleton) and ${\left\{T^{i-j+n_{k}p}x\right\}}_{k=1}^{\infty }\to T{\omega }_i\left(\in A_{i+1,0}\right)$. Furthermore, no other subsequence of ${\left\{T^{i-j+np}x\right\}}_{n=1}^{\infty }\subset A_i$ for any $x\in A_j$; $\forall j\in \bar{p}$ can converge to another best proximity point of $A_i$ since $A_{i0}$ is a singleton or to another point which is not a best proximity point since Property (i) would then be violated and (8) would not hold. Also, if there is some subsequence of ${\left\{T^{i-j+np}x\right\}}_{n=1}^{\infty }\subset A_i$ for any $x\in A_j$; $\forall j\in \bar{p}$ which does not converge to $z_i$, again (8) and then Property (i) would fail to hold. Thus, all the sequences ${\left\{T^{i-j+np}x\right\}}_{n=1}^{\infty }\subset A_i$ for any given $x\in A_j$; $\forall j\in \bar{p}$ converge to $z_i$. Since $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is single-valued, ${\left\{T^{i-j+np+1}x\right\}}_{n=1}^{\infty }\left(\subset A_{i+1}\right)\to T{z}_i\left(\in A_{i+1,0}\right)$ and ${\left\{T^{i-j+np+\mathcal{l}}x\right\}}_{n=1}^{\infty }\left(\subset A_{i+\mathcal{l}}\right)\to T^{\mathcal{l}}{z}_i\left(\in A_{i+\mathcal{l},0}\right)$ for $\mathcal{l}\in \overline{p-1}\cup \left\{0\right\}$. Any sequence ${\left\{T^{n}x\right\}}_{n=0}^{\infty }$ converges to a unique limit cycle formed with best proximity points, with one per adjacent subset, $\left(z_1,z_2,\dots ,z_p\right)\in A_{10}\times A_{20}\times ...\times A_{p0}$; $x\in {\bigcup }_{i\in \bar{p}}{A}_i$, and if $x\in A_j$ for some $j\in \bar{p}$, then ${\left\{T^{np+\mathcal{l}}x\right\}}_{n=0}^{\infty }\left(\subset {\left\{T^{np}x\right\}}_{n=0}^{\infty }\subset A_{j+\mathcal{l}}\right)\to z_{j+\mathcal{l}}$; $\forall \mathcal{l}\in \overline{p-1}\cup \left\{0\right\}$. Also, each sequence ${\left\{T^{np+\mathcal{l}}x\right\}}_{n=0}^{\infty }$ is Cauchy; $x\in {\bigcup }_{i\in \bar{p}}{A}_i$. Assume that the sequence is not a Cauchy sequence. Then, for any real constant $\epsilon >0$, some $\mathcal{l}\in \overline{p-1}\cup \left\{0\right\}$, some $N\in {\mathit{Z}}_{0+}$, and any finite integer $m=m\left(\epsilon ,\hspace{0.33em}n\right)$ for $n\left(\ge n_0\right)\in {\mathit{Z}}_{0+}$, there is $m_1\in \left[1\hspace{0.33em},\hspace{0.33em}m\right]\cap {\mathit{Z}}_{+}$ such that if $d\left(T^{np+j-1}x, T^{np+j1}x\right)<\epsilon$ for $j\left(\ne m_1\right)\in \left[1\hspace{0.33em},\hspace{0.33em}m\right]\cap {\mathit{Z}}_{+}$ then $d\left(T^{np+m_1-1}x, T^{np+\mathcal{l}+m_1}x\right)\ge \epsilon$. Thus, $d\left(T^{np+\mathcal{l}}x, T^{\left(n+m\right)p+\mathcal{l}}x\right)$ does not converge to zero as $n\to \infty$ so that ${\left\{T^{np+\mathcal{l}}x\right\}}_{n=0}^{\infty }$ is not convergent. Thus, each sequence ${\left\{T^{np+\mathcal{l}}x\right\}}_{n=0}^{\infty }$ is Cauchy; $x\in {\bigcup }_{i\in \bar{p}}{A}_i$. □

The following result is obvious:

Corollary 2. 

Theorem 2 holds “mutatis-mutandis” if $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic contraction.

The condition of Theorem 2(iv) may be weakened by replacing the condition that $A_{i0}$ is a singleton with the one in which there is a unique point $z_i$ in $A_{i0}$ such that $d\left(z_i,\hspace{0.33em}T{z}_i\right)=D$. This means that $z_i$ and $T{z}_i$ are the unique best proximity points in the adjacent subsets $A_i$ and $A_{i+1}$ for some $i\in \bar{p}$, which belong to sequences generated by the self-mapping $T$, but it can eventually exist $z_i^{\prime }\left(\ne z_i\right)\in A_{i0}$ and $T^{\prime }\left(\ne T\right):{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ such that $d\left(z_i^{\prime },\hspace{0.33em}{T}^{\prime }{z}_i^{\prime }\right)=D$.

Corollary 3. 

Theorem 2(iv) also holds if $\left(X,d\right)$ is complete under the weaker condition that a subset $A_i$ of $X$ is boundedly compact for some $i\in \bar{p}$ and $d\left(z_i,\hspace{0.33em}T{z}_i\right)=D$ for a unique $z_i\in A_{i0}$.

Proof. 

All the sequences generated through the cyclic self-mapping $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ from any initial point $x\in {\bigcup }_{i\in \bar{p}}{A}_i$, which are bounded from Theorem 2(iii), have a convergent subsequence in $A_i$ since $\left(X,d\right)$ is complete and $A_i$ is nonempty, closed, and boundedly compact. From Theorem 2(i), such a convergent subsequence has to converge to $z_i\in A_{i0}$, which is unique by hypothesis and by the continuity of the self-mapping (see the proof of Property (iv)), and because it is single-valued, the whole sequences in $A_i$ also have to converge to such a unique $z_i\in A_{i0}$. The rest of the proof follows as in Theorem 2(iv). □

Example 1. 

The above corollary might be determined through the following simple geometric example. Consider two subsets $A_{1\rho }=-A_{2\rho }$ of ${\mathit{R}}^2$ where $A_{2\rho }=\left\{\left(x,y\right)\in {\mathit{R}}^2:\hspace{0.33em}x\ge \rho \right\}$ for some real $\rho >0$. It turns out that, for the Euclidean distance, $d\left(A_{1\rho }\hspace{0.33em},\hspace{0.33em}{A}_{2\rho }\right)=2\rho$, and note that $A_{1\rho }$ and $A_{2\rho }$ are closed and non-strictly convex. The straight lines $\left(x,y\right)$ in ${\mathit{R}}^2$ defined by $x=-\rho$ and $x=\rho$ of ${\mathit{R}}^2$ are the best proximity sets $A_{1\rho 0}$ and $A_{2\rho 0}$ of $A_{1\rho }$ and $A_{2\rho }$. There are infinitely many corresponding pairs of best proximity points in both sets. For a mapping $T_y:A_{1\rho }\cup A_{2\rho }\to A_{1\rho }\cup A_{2\rho }$ defined by $T_y\left(x,y\right)=\left(-\alpha x,y\right)$ if $\left|x\right|>\rho$; $T_y\left(\mp \rho ,y\right)=\left(\pm \rho ,y\right)$ for $\left(x,y\right)$ in ${\mathit{R}}^2$ for fixed $y\in \mathit{R}$, it follows that $z_{1\rho }\left(=\left(-\rho ,y\right)\right)\in A_{1\rho 0}$ is the unique best proximity point in $A_{1\rho }$ such that $T{z}_{1\rho }\left(=z_{2\rho }=\left(\rho ,y\right)\right)\in A_{2\rho 0}$. Define in the same way a self-mapping $T_{y^{\prime }}:A_{1\rho }\cup A_{2\rho }\to A_{1\rho }\cup A_{2\rho }$ for some given $y^{\prime }\ne y$. The corresponding best proximity points for such mapping are now $\left(-\rho ,y^{\prime }\right)$ and $\left(\rho ,y^{\prime }\right)$. The sequences generated through $T$ and those generated through $T^{\prime }$ each have unique and distinct corresponding best proximity points in $A_{1\rho }$ and $A_{2\rho }$.

4. Keeping Cyclic Large Contractions Under Perturbations of a Cyclic Large Contractive Mapping

The above results are extended to the case of cyclic disposal for sums of operators in normed spaces in such a way that one of the operators can be interpreted as a nominal one, while the other one may be interpreted as its perturbation. We investigate the smallness in terms of the norm of the perturbation operator such that the perturbed operator is guaranteed to be a cyclic large contraction provided that the nominal one is a cyclic large contraction.

The following definition extends the cyclic disposal of sums of operators:

Definition 3. 

Let $\left(X,\hspace{0.33em}‖.‖\right)$ be a normed space with a set $p$ of nonempty subsets $A_i$ of $X$; $\forall i\in \bar{p}$ such that $dist\left(A_i,A_{i+1}\right)=d\left(A_i,A_{i+1}\right)=D$, where $d:X\times X\to {\mathit{R}}_{0+}$ is a norm-induced metric that is homogeneous and translation-invariant. Then, $\bar{T}\left(=T+\tilde{T}\right):{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is said to be a cyclic operator, where $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is also cyclic, that is, $\bar{T}\left(A_i\right)\subset A_{i+1}$, $T\left(A_i\right)\subset A_{i+1}$; $\forall i\in \bar{p}$.

The above definition can determine the situation when $T$ is a nominal (or reference) version of $T$ and $\tilde{T}$ is a perturbation operator of $T$.

Remark 2. 

Note that a non-expansive $\bar{T}=T+\tilde{T}$ does not imply that $T$ and $\tilde{T}$ are non-expansive since if $T$ is non-expansive then $\bar{T}$ may be expansive. Assume, for instance, that $T$ is non-expansive with $A_i=A=X$; $\forall i\in \bar{p}$ and $0\in A$. Then, by taking into account the homogeneity and translation invariance of the distance, one obtains the following for any $x,y\in X$:

$$\begin{gathered} d\left(\bar{T}x,\bar{T}y\right)=d\left(Tx+\tilde{T}x,Ty+\tilde{T}y\right)=d\left(Tx,Ty+\tilde{T}y-\tilde{T}x\right)\le ‖Tx-Ty‖+‖\tilde{T}y-\tilde{T}x‖ \\ =d\left(x,y\right)+‖\tilde{T}y-\tilde{T}x‖=d\left(x,y\right)+d\left(\tilde{T}y,\tilde{T}x\right)=d\left(x,y\right)+d\left(\tilde{T}y-\tilde{T}x,0\right) \end{gathered}$$

where $‖\tilde{T}‖=\underset{‖x‖\le 1}{sup}‖Tx‖$, which can imply that $d\left(Tx,\hspace{0.33em}Ty\right)>d\left(x,y\right)$. Also, if $T$ is contractive with constant $K\in \left(0\hspace{0.33em},\hspace{0.33em}1\right)$, then

$$d\left(\bar{T}x,\bar{T}y\right)=d\left(Tx+\tilde{T}x,Ty+\tilde{T}y\right)=d\left(Tx,Ty+\tilde{T}y-\tilde{T}x\right)\le Kd\left(x,y\right)+‖\tilde{T}y-\tilde{T}x‖\le \left(K+‖\tilde{T}‖\right)d\left(x,y\right)$$

so that $\bar{T}$ is guaranteed non-expansive under the condition $‖\tilde{T}‖<1-K$.

Now, assume that $\tilde{T}=-\lambda T$ for some real constant $\lambda \in \left(0\hspace{0.33em},\hspace{0.33em}1\right]$, and then note that $‖\bar{T}‖=\left(1-\lambda \right)‖T‖\le 1$ and $\bar{T}$ is non-expansive if $‖T‖\le {\left(1-\lambda \right)}^{-1}$, while $T$ is expansive if $‖T‖>1$.

The following auxiliary results will be used later on:

Lemma 1. 

Let $\left(X,\hspace{0.33em}‖‖\right)$ be a normed space. Then, the following properties hold:

(i) Consider the operator ${\bar{T}}_n:X\to X$ and ${\bar{T}}_n={\prod }_{i=1}^n\left[T+{\tilde{T}}_i\right]$, such that $\bar{T},{\tilde{T}}_n:X\to X$ with $‖T‖\le K$, $‖{\tilde{T}}_n‖\le {\tilde{K}}_n$; $\forall n\in {\mathit{Z}}_{+}$. Then, $‖{\bar{T}}_n‖\le K^n+{\widehat{\tilde{K}}}_n\le {\bar{K}}^n$; $\forall n\in {\mathit{Z}}_{+}$, if

$${\tilde{K}}_{n+1}\le {\displaystyle \frac{{\bar{K}}^{n+1}-K^{n+1}-K^{n-1}{\widehat{\tilde{K}}}_n}{K^n+{\widehat{\tilde{K}}}_n}}; {\widehat{\tilde{K}}}_n=K^{n-1}\left({\widehat{\tilde{K}}}_{n-1}+{\tilde{K}}_n\right)+{\widehat{\tilde{K}}}_{n-1}{\tilde{K}}_n; \forall n\in {\mathit{Z}}_{+}$$

with ${\tilde{K}}_0={\widehat{\tilde{K}}}_0=0$, ${\widehat{\tilde{K}}}_1={\tilde{K}}_1$.

(ii) Assume that $\bar{T}:X\to X$ is defined by $\tilde{T}=\lambda T$ for some $\lambda \in {\mathit{R}}_0^{+}$ so that $\bar{T}=T+\tilde{T}=\left(1+\lambda \right)T$. If $‖T‖\le K$ then $‖\bar{T}‖\le \bar{K}$ if $\lambda \in \left[-1,\hspace{0.33em}\left(\bar{K}-K\right)/K\right)$ for some real constants $K,\hspace{0.33em}\bar{K}\left(\ge K\right)$.

Proof. 

If ${\bar{T}}_n={\prod }_{i=1}^n\left[T+{\tilde{T}}_i\right]$ and ${\bar{T}}_{n+1}={\prod }_{i=1}^{n+1}\left[T+{\tilde{T}}_i\right]={\bar{T}}_n\left(T+{\tilde{T}}_{n+1}\right)$; $\forall i\in {\mathit{Z}}_{+}$, then,

$$‖{\bar{T}}_1‖\le ‖T‖+‖{\tilde{T}}_1‖\le K+{\tilde{K}}_1\le \bar{K}$$

if ${\tilde{K}}_1\le \bar{K}-K={\displaystyle \frac{\bar{K}-K-{\widehat{\tilde{K}}}_0}{K^0+{\widehat{\tilde{K}}}_0}}$ with ${\widehat{\tilde{K}}}_0=0$;

$$‖{\bar{T}}_2‖\le \left(K+{\tilde{K}}_1\right)\left(K+{\tilde{K}}_2\right)\le K^2+K\left({\tilde{K}}_1+{\tilde{K}}_2\right)+{\tilde{K}}_1{\tilde{K}}_2=K^2+{\widehat{\tilde{K}}}_2\le {\bar{K}}^2$$

if ${\tilde{K}}_2\le {\displaystyle \frac{{\bar{K}}^2-K^2-{\widehat{\tilde{K}}}_1}{K+{\widehat{\tilde{K}}}_1}}$ with ${\widehat{\tilde{K}}}_1={\tilde{K}}_1$;

$$\begin{gathered} ‖{\bar{T}}_3‖\le ‖{\bar{T}}_2‖\left(K+{\tilde{K}}_2\right)\le \left(K+{\tilde{K}}_1\right)\left(K+{\tilde{K}}_2\right)\left(K+{\tilde{K}}_3\right) \\ \le \left(K^2+{\widehat{\tilde{K}}}_2\right)\left(K+{\tilde{K}}_3\right)\le K^3+K\left({\widehat{\tilde{K}}}_2+K{\tilde{K}}_3\right)+{\widehat{\tilde{K}}}_2{\tilde{K}}_3=K^3+{\widehat{\tilde{K}}}_3\le {\bar{K}}^3 \end{gathered}$$

if ${\tilde{K}}_3\le {\displaystyle \frac{{\bar{K}}^3-K^3-K{\widehat{\tilde{K}}}_2}{K^2+{\widehat{\tilde{K}}}_2}}$ with ${\widehat{\tilde{K}}}_2=K\left({\widehat{\tilde{K}}}_1+{\tilde{K}}_2\right)+{\widehat{\tilde{K}}}_1{\tilde{K}}_2=K\left({\tilde{K}}_1+{\tilde{K}}_2\right)+{\tilde{K}}_1{\tilde{K}}_2$

$$\begin{gathered} ‖{\bar{T}}_n‖\le ‖{\bar{T}}_{n-1}‖\left(K+{\tilde{K}}_n\right)\le \left(K^{n-1}+{\widehat{\tilde{K}}}_{n-1}\right)\left(K+{\tilde{K}}_n\right) \\ \le K^n+K^{n-2}\left({\widehat{\tilde{K}}}_{n-1}+K{\tilde{K}}_n\right)+{\widehat{\tilde{K}}}_{n-1}{\tilde{K}}_n=K^n+{\widehat{\tilde{K}}}_n\le {\bar{K}}^n \\ ‖{\bar{T}}_{n+1}‖\le ‖{\bar{T}}_n‖\left(K+{\tilde{K}}_{n+1}\right)\le \left(K^n+{\widehat{\tilde{K}}}_n\right)\left(K+{\tilde{K}}_{n+1}\right) \\ \le K^{n+1}+K^{n-1}\left({\widehat{\tilde{K}}}_n+K{\tilde{K}}_{n+1}\right)+{\widehat{\tilde{K}}}_n{\tilde{K}}_{n+1}=K^{n+1}+{\widehat{\tilde{K}}}_{n+1}\le {\bar{K}}^{n+1} \end{gathered}$$

if ${\tilde{K}}_{n+1}\le {\displaystyle \frac{{\bar{K}}^{n+1}-K^{n+1}-K^{n-1}{\widehat{\tilde{K}}}_n}{K^n+{\widehat{\tilde{K}}}_n}}$ with ${\widehat{\tilde{K}}}_n=K^{n-1}\left({\widehat{\tilde{K}}}_{n-1}+{\tilde{K}}_n\right)+{\widehat{\tilde{K}}}_{n-1}{\tilde{K}}_n$; $\forall n\in {\mathit{Z}}_{0+}$.

Property (i) is thus proven. Property (ii) follows directly from $‖\bar{T}‖\le \left(1+\lambda \right)‖T‖\le \left(1+\lambda \right)K\le \bar{K}$. □

Theorem 3. 

Let $\left(X,‖.‖\right)$ be a normed space with a set $p$ of nonempty closed subsets $A_i\subset X$; $\forall i\in \bar{p}$ such that $d\left(A_i,A_{i+1}\right)=D>0$ for the norm-induced metric. Assume that $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic contraction with $‖T‖\le K<1$ and that $\tilde{T}:X\to X$ such that $\bar{T}\left(=T+\tilde{T}\right):{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$. Then, the following properties hold:

(i) The cyclic contraction $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is also a cyclic large contraction with the norm-induced metric and contraction constant $K$ and satisfies the properties of Theorem 2[(i)–(iii)] and Theorem 2(iv) if, in addition, $\left(X,‖.‖\right)$ is a Banach space.

(ii) Assume that ${\tilde{T}}_n:X\to X$ such that $‖{\tilde{T}}_n‖\le {\tilde{K}}_n$; $n\in Z_{0+}$ and that the sequence ${\left\{{\tilde{K}}_n\right\}}_{n=0}^{\infty }$ of norm upper-bounds is small enough according to the conditions of Lemma 1(i) for some real constant $\bar{K}\in \left[K,\hspace{0.33em}1\right)$. Then, $\bar{T}:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic contraction of constant $\bar{K}$.

(iii) Assume that $\tilde{T}\left(=\lambda T\right):X\to X$ for some $\lambda \in {\mathit{R}}_0^{+}$ so that $\bar{T}=T+\tilde{T}=\left(1+\lambda \right)T$. If $\lambda \in \left[-1,\hspace{0.33em}\left(\bar{K}-K\right)/K\right)\subset \left[-1,\hspace{0.33em}\left(1-K\right)/K\right)$, then $‖\bar{T}‖\le \bar{K}\left(\in \left[K\hspace{0.33em},\hspace{0.33em}1\right)\right)$ so that $\bar{T}:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic contraction of constant $\bar{K}$.

Proof. 

Property (i) follows directly from Theorem 2 and Corollary 2. Property (ii) follows directly from Lemma 1(i), Theorem 2, and Corollary 2. Property (iii) follows directly from Lemma 1(ii), Theorem 2, and Corollary 2 since a cyclic contraction is a cyclic large contraction. □

Corollary 4. 

The conditions of Theorem 3[(i)–(ii)] imply that $\bar{T}:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is also a cyclic large contraction.

Remark 3. 

Note that it is not necessary that the perturbation mapping $\tilde{T}$ be defined on the restriction ${\bigcup }_{i\in \bar{p}}{A}_i$ of $X$ in order for both $T$ and $\bar{T}$ to be cyclic on ${\bigcup }_{i\in \bar{p}}{A}_i$. In other words, $\bar{T},T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ with the constraints $T\left(A_i\right)\subset A_{i+1}$, $\bar{T}\left(A_i\right)\subset A_{i+1}$; $\forall i\in \bar{p}$ does not require that $\tilde{T}:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$, $\tilde{T}\left(A_i\right)\subset A_{i+1}$; $\forall i\in \bar{p}$.

Example 2. 

Suppose that $0\in X$, $d\left(A_i\hspace{0.33em},\hspace{0.33em}{A}_{i+1}\right)=D>0$ and $\tilde{T}\left(z_i\right)=0$ if $z_i\in A_{i0}$; $\forall i\in \bar{p}$. Then, $0\notin {\bigcup }_{i\in \bar{p}}{A}_i$ and $\tilde{T}:X\to X$ is not restricted to be a mapping on ${\bigcup }_{i\in \bar{p}}{A}_i$. And non-trivial mappings $\tilde{T}:X\to X$, non-necessarily restricted to be defined on ${\bigcup }_{i\in \bar{p}}{A}_i$, can be defined by construction satisfying $\tilde{T}=\bar{T}-T$ such that both $T,\bar{T}$ are cyclic mappings on ${\bigcup }_{i\in \bar{p}}{A}_i$. For instance, assume that $X=\mathit{R}$, $d$ is the Euclidean distance, $A_1=\left(-\infty \hspace{0.33em},\hspace{0.33em}-a\right]$, $A_2=\left[a\hspace{0.33em},\hspace{0.33em}+\infty \right)$, $T:A_1\cup A_2\to A_1\cup A_2$ is defined by $Tx=-\alpha x$ with $\alpha \in \left(0\hspace{0.33em},\hspace{0.33em}1\right)$ if $\left|\alpha x\right|\ge a>0$ and $Tx=-a\mathit{sgn}\left(x\right)\left(=\pm a\right)$; otherwise, for all $x\in A_1\cup A_2$, $\tilde{T}:X\to X$ is defined by $\tilde{T}x=\lambda \alpha x$ if $\left|\alpha x\right|\ge a>0$ and $\tilde{T}x=0$, and otherwise with $1>\lambda >1-{\alpha }^{-1}$. Then, $\bar{T}x=\left(T+\tilde{T}\right)x=-\alpha \left(1-\lambda \right)x$ and $T,\bar{T}:A_1\cup A_2\to A_1\cup A_2$, $T\left(A_1\right)\subset A_2,\bar{T}\left(A_1\right)\subset A_2$, $T\left(A_2\right)\subset A_1,\bar{T}\left(A_2\right)\subset A_1$ so that $T$, $\bar{T}$ are cyclic on $A_1\cup A_2$ while $\tilde{T}:X\to X$ is not cyclic on $A_1\cup A_2$.

Theorem 4. 

Let $\left(X,‖.‖\right)$ be a normed space with a set of $p$ nonempty closed subsets $A_i\subset X$; $\forall i\in \bar{p}$ such that $d\left(A_i,A_{i+1}\right)=D>0$ for the norm-induced metric. Assume that:

(1) $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction such that for any real constants $\epsilon \in {\mathit{R}}_{+}$ and $\delta \in \left[0\hspace{0.33em},1\right)$:

$$\begin{gathered} \left[\left(x,⥂y\right)\in \left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)for\hspace{0.33em}any\hspace{0.33em}i\in \bar{p},\hspace{0.33em}\hspace{0.33em}‖x-y‖\ge D+\epsilon \right]\Rightarrow \\ ‖Tx-Ty‖\le \delta \left(‖x-y‖-D\right)+D; \end{gathered}$$

(2) $\tilde{T}:X\to X$, with $‖\tilde{T}‖=\mathit{sup}\hspace{0.33em}\left(‖\tilde{T}x‖\hspace{0.33em}:\hspace{0.33em}x\in X,\hspace{0.33em}‖x‖\le 1\right)\le \tilde{K}$, is an additive perturbation operator of $T$ such that $\bar{T}\left(=T+\tilde{T}\right):{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ and $\tilde{K}\in \left[0\hspace{0.33em},\hspace{0.33em}\bar{\delta }-\delta \right)\left(\subset \left[0\hspace{0.33em},\hspace{0.33em}1\right)\right)$ for some real constant $\bar{\delta }\in \left[\delta \hspace{0.33em},\hspace{0.33em}1\right)$.

Then, $\bar{T}\left(=T+\tilde{T}\right):{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction, that is, $‖\bar{T}x-\bar{T}y‖<‖x-y‖$; $\forall \left(x,⥂y\right)\in \left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right) \mathrm{for}\hspace{0.33em}\mathrm{any}\hspace{0.33em}i\in \bar{p}$, such that $‖x-y‖>D$, and

$$\begin{gathered} \left[\left(x,⥂y\right)\in \left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right) for\hspace{0.33em}any\hspace{0.33em}i\in \bar{p},\hspace{0.33em}\hspace{0.33em}‖x-y‖\ge D+\epsilon \right]\Rightarrow \\ ‖\bar{T}x-\bar{T}y‖\le \bar{\delta }\left(‖x-y‖-D\right)+D. \end{gathered}$$

Proof. 

Since $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction, then $T\left(A_i\right)\subset A_{i+1}$; $\forall i\in \bar{p}$, $d\left(Tx\hspace{0.33em},\hspace{0.33em}Ty\right)<d\left(x\hspace{0.33em},\hspace{0.33em}y\right)$ for each given pair $\left(x,\hspace{0.33em}y\right)\in A_i\times A_{i+1}\cup A_{i+1}\times A_i$; $\forall i\in \bar{p}$ if $‖x-y‖=d\left(x,\hspace{0.33em}y\right)>D$ and, for each positive real constant $\epsilon$, there exists $\delta <1$ such that

$$\begin{gathered} \left[\left(x,⥂y\right)\in \left(A_i\times A_{i+1}\cup A_{i+1}\times A_i\right)for\hspace{0.33em}any\hspace{0.33em}i\in \bar{p},\hspace{0.33em}\hspace{0.33em}‖x-y‖\ge D+\epsilon \right]\Rightarrow \\ ‖Tx-Ty‖\le \delta ⥂\left(‖x-y‖-D\right)+D. \end{gathered}$$

On the other hand, if $‖x-y‖\ge D+\epsilon$ for a positive real constant $\epsilon$, then one can observe for such real constant $\delta \in \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ that

$$\begin{gathered} ‖Tx-Ty‖\le \delta ‖x-y‖+\left(1-\delta \right)D \\ ‖\bar{T}x-\bar{T}y‖=d\left(Tx,Ty+\tilde{T}y-\tilde{T}x\right)\le d\left(Tx,Ty\right)+‖\tilde{T}y-\tilde{T}x‖ \\ =\left(\delta +\tilde{K}\right)‖x-y‖+\left(1-\delta \right)D\le \left(\delta +\tilde{K}\right)‖x-y‖+\left(1-\delta -\tilde{K}\right)D \\ =\bar{\delta }‖x-y‖+\left(1-\bar{\delta }\right)D \end{gathered}$$

where $\bar{\delta }=\delta +\tilde{K}$. Assume that $\bar{\delta }<1$, which implies that $\tilde{K}\in \left[0\hspace{0.33em},\hspace{0.33em}\bar{\delta }-\delta \right)\left(\subset \left[0\hspace{0.33em},\hspace{0.33em}1\right)\right)$ and $\bar{\delta }\in \left[\delta \hspace{0.33em},\hspace{0.33em}1\right)$ since $\delta \in \left[0\hspace{0.33em},\hspace{0.33em}1\right)$ and since $T:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction. Furthermore, $‖\bar{T}x-\bar{T}y‖<‖x-y‖$ if $‖x-y‖>D$, which is now proven by contradiction arguments. Assume that this is not true so that $‖\bar{T}x-\bar{T}y‖\ge ‖x-y‖=D+\mu$ for some $\mu \in {\mathit{R}}_{0+}$. Then, either $‖x-y‖=D$ and $\mu =0$, or $‖x-y‖>D$ an $\mu \in {\mathit{R}}^{+}$.Then $‖x-y‖<{\bar{\delta }}_{\mu }‖x-y‖+\left(1-{\bar{\delta }}_{\mu }\right)D$ for some ${\bar{\delta }}_{\mu }\in \left[\delta \hspace{0.33em},\hspace{0.33em}1\right)$ so that $\left(1-{\bar{\delta }}_{\mu }\right)‖x-y‖<\left(1-{\bar{\delta }}_{\mu }\right)D$ which is equivalent to $‖x-y‖<D$, a contradiction. As a result, $‖\bar{T}x-\bar{T}y‖<‖x-y‖$ if $‖x-y‖>D$. It has been proven that if $\tilde{K}\in \left[0\hspace{0.33em},\hspace{0.33em}\bar{\delta }-\delta \right)\left(\subset \left[0\hspace{0.33em},\hspace{0.33em}1\right)\right)$ and $\bar{\delta }\in \left[\delta \hspace{0.33em},\hspace{0.33em}1\right)$,

(1) $‖\bar{T}x-\bar{T}y‖<‖x-y‖$ if $‖x-y‖>D$;

(2) $‖\bar{T}x-\bar{T}y‖\le \bar{\delta }‖x-y‖+\left(1-\bar{\delta }\right)D$ if $‖x-y‖\ge D+\epsilon$ for any given $\epsilon \in {\mathit{R}}_{+}$.

Thus, $\bar{T}:{\bigcup }_{i\in \bar{p}}{A}_i\to {\bigcup }_{i\in \bar{p}}{A}_i$ is a cyclic large contraction. □

It turns out that the cyclic mapping $\bar{T}=T+\tilde{T}$ on ${\bigcup }_{i\in \bar{p}}{A}_i$, which satisfies Theorem 4, also satisfies Theorem 2 since it is a large contraction, under the assumption that $T$ is also a large contraction on ${\bigcup }_{i\in \bar{p}}{A}_i$. Thus, under the conditions of Theorem 4, both self-mappings $T$ and $\bar{T}$ on ${\bigcup }_{i\in \bar{p}}{A}_i$ are large contractions so that they satisfy the properties of Theorem 2. It can be pointed out that the perturbed operator $\tilde{T}$ of $T$ is not necessarily structured since it suffices that it has an upper-bounded norm with an upper-bound in the range $\left[0\hspace{0.33em},\hspace{0.33em}\bar{\delta }-\delta \right)$.

5. Conclusions

This paper has proven some basic properties of large contractions in metric spaces like, for instance, the fact that they are asymptotically regular and that the generated sequences are bounded and Cauchy and, in addition, convergent to a unique fixed point if the metric space is complete. Cyclic large contractions are focused on in metric spaces, and it is proven that strict contractions are also large contractions and that the distances of sequences converge to that between the adjacent subsets of the cyclic disposal of nonempty closed subsets of the metric space. In the event that such subsets intersect, all the sequences are proven to be bounded (even if the subsets are not bounded) and Cauchy and that they converge to a unique fixed point located in the intersection of such subsets. In the event that the subsets have a pair-wise empty intersection, the boundedness of such sequences is proven without the need to assume the boundedness of the subsets of the cyclic disposal.

We also prove the convergence of the sequences to a unique limit cycle of best proximity points, with one per subset in the cyclic disposal, provided that the metric space is complete and that one of such subsets is boundedly compact with a best proximity set, which is a singleton. It is not assumed that the remaining best proximity points are necessarily singletons. In parallel, all their subsequences running within each of the subsets of the cyclic disposal converge to a unique best proximity point, even if the corresponding best proximity sets are not singletons, and they are also Cauchy sequences. It is also proven that the former hypothesis that one of the best proximity sets between adjacent subsets is a singleton can be weakened, for a particular cyclic large contraction, if there is just a single best proximity point in one of the subsets mapped to its image by the cyclic self-mapping. This property is based on the simple idea that only a best proximity point is reached by the subsequences of a given cyclic self-mapping in one of the subsets of the cyclic disposal, though it does not imply that the best proximity set to which it belongs is necessarily a singleton.

Later on, eventual perturbations of the cyclic large self-mappings in normed spaces are admitted. In the event that the norm of the perturbation additive operator is small enough, it is proven that the perturbed cyclic self-mapping keeps the cyclic large contraction property associated with the unperturbed cyclic large contraction. It is assumed that both the unperturbed and the perturbed mappings are cyclic on the same union of nonempty closed subsets of the metric space while it is not required for the perturbed operator to be cyclic. The maximum upper-bound of the perturbed operator, which ensures that such a property holds, is explicitly characterized.

Extensions of this work may rely on the use of orthogonal metric spaces to define orthogonal Cauchy sequences or subsequences in large contractions and on the study of their relevant properties. Further extensions of this intended work regarding the case of a cyclical disposal on a set of subsets of metric spaces, may also be carried out.