On Ω Class of Mappings in a p-Cyclic Complete Metric Space

Abstract

In this manuscript, we introduce the concept of $\mathsf{\Omega }$-class of self mappings on a metric space and a notion of p-cyclic complete metric space for a natural number $(p\ge 2)$. We not only give sufficient conditions for the existence of best proximity points for the $\mathsf{\Omega }$-class self-mappings that are defined on p-cyclic complete metric space, but also discuss the convergence of best proximity points for those mappings.

1. Introduction

In the classical Banach fixed point theorem, the undertaking operator is necessarily continuous due to contraction inequality. This simple observation brings a natural question: Does a discontinuous contraction mapping possess a fixed point? The answer to this question is affirmative. Indeed, there are various approaches to overcome weakness of the discontinuous mapping for guaranteeing a fixed point. One of the significant results was constructed by Bryant [1] who proved the following result: In a complete metric space, if, for some positive integer $n\ge 2$, the nth iteration of the given mapping forms a contraction, then it possess a unique fixed point. Another outstanding approach was proposed by Kirk, Srinivasan and Veeramani [2] by introducing the notion of cyclic contraction. More precisely, every cyclic contraction in a complete metric space possess a unique fixed point. This statement is plain but significant when we compare with the results of Bryant. Attendantly, the concept of the cyclic contractions has been investigated densely by a considerable number of authors who bring several variants of the notion and derive a number of interesting results (see, e.g., [3,4,5,6,7,8,9,10,11,12,13,14,15,16] and the references therein).

Let there be a self-mapping on a metric space $(X,d)$. Suppose that A and B are non-empty subsets of X such that $X=A\cup B$. A self-mapping T on $A\cup B$ is called cyclic [2]

$$T\left(A\right)\subset B\mathrm{and}T\left(B\right)\subset A.$$

Further, a mapping T is called cyclic contraction [2] if there is a $k\in [0,1)$ such that the following inequality is satisfied:

$$d(Tx,Ty)\le kd(x,y),\mathrm{for}\mathrm{all}x\in X_i,y\in X_{i+1},i\in \{1,2,\dots ,m\}.$$

After this initial construction, several extension of cyclic mappings and cyclic contractions have been introduced. In this paper, we mainly follow the notations defined in [9].

2. Motivation

In [9], a notion of p-cyclic map is introduced. Let $B_1,B_2,\dots ,B_p(p\ge 2)$ be non-empty sets. A p-cyclic map $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is defined such that $T\left(B_i\right)\subseteq B_{i+1}$, $\forall i\in \{1,2,\dots ,p\}$. $x=x_0\in B_i$ defines a sequence $\left\{x_n\right\}\subseteq {\cup }_{i=1}^p{B}_i$ as $x_n=T\left(x_{n-1}\right)$. Then, $\left\{x_{pn}\right\}$ is a subsequence in $B_i$, $\left\{x_{pn+1}\right\}$ is a subsequence in $B_{i+1}$ and so on. The arrangement of such a sequence formed by a p-cyclic map motivated us to introduce a notion of p-cyclic sequence (Definition 6(1)). If $B_i$s are subsets of a metric space $(M,\rho )$, then we observe that, to obtain a best proximity point of T under various contractive conditions (some of them given in the literature), it is enough to prove that: given $ϵ>0$, there exists $N_0\in \mathbb{N}$ such that

$$\rho (x_{pn},x_{pm+1})<dist(B_i,B_{i+1})+ϵ,\forall n,m\ge N_0.$$

This observation motivated us to introduce a concept of p-cyclic Cauchy sequence and p-cyclic complete metric space (Definition 6). In addition, while investigating the behavior of such p-cyclic maps, it is often the case that, if $\rho (x,y)>dist(B_i,B_{i+1})$, then $\rho (Tx,Ty)<\rho (x,y)$ and, if $\rho (x,y)=dist(B_i,B_{i+1})$, then $\rho (Tx,Ty)=\rho (x,y),x\in B_i,y\in B_{i+1}$. This motivated us to call a p-cyclic map with this property as p-cyclic strict contraction map (Definition 7). Note that, if the distances between the adjacent sets are zero, then a p-cyclic strict contraction map is a strict contraction map in the usual sense. All such maps invariably satisfy the condition: $x,y\in B_i,\rho (T^{pn}x,T^{pn+1}y)⟶dist(B_i,B_{i+1})$ as $n⟶\infty$. In this paper, all p-cyclic maps which satisfy the above two properties are said to belong to class $\mathsf{\Omega }$ (Definition 8). Finally, we prove the existence and convergence of best proximity points of $\mathsf{\Omega }$ class of mappings in a p-cyclic complete metric space.

3. Preliminaries

In what follows, we give some definitions and fundamental results that are essential to understand and prove the main results.

Definition 1

([9], Definitions 3.1). For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Define $B_{p+i}:=B_i$, for all $i\in \{1,2,\dots ,p\}$. A map $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is called a $\mathit{p}$-cyclic map if $T\left(B_i\right)\subseteq B_{i+1}$, $\forall i\in \{1,2,\dots ,p\}$. If $p=2$, the map T is called cyclic. A point $x\in B_i$ is said to be a best proximity point of T in $B_i$, if $\rho (x,Tx)=dist(B_i,B_{i+1})$, where $dist(B_i,B_{i+1}):=inf\{\rho (x,y):x\in B_i,y\in B_{i+1}\}.$

In this paper, we give the conditions for the underlying space and for the subsets of the space, to have a unique best proximity point under a p-cyclic map, if it exists, irrespective of the contraction condition imposed on the map.

Proposition 1.

Let $B_1,B_2,\dots ,B_p$, ($p\ge 2$) be non-empty convex subsets of a strictly convex norm linear space M such that $dist(B_i,B_{i+1})>0,i\in \{1,2,\dots ,p\}$. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic map. Then, T has at most one best proximity point in $B_i$, $1\le i\le p$.

Proof. 

Let $x,y\in B_i$ be such that $\parallel x-Tx\parallel =dist(B_i,B_{i+1})=\parallel y-Ty\parallel ,$$1\le i\le p$. If $x\ne y,$ then $\parallel \frac{x-Tx}{dist(B_i,B_{i+1})}\parallel =1$ and $\parallel \frac{y-Ty}{dist(B_i,B_{i+1})}\parallel =1$. Since M is strictly convex, $\parallel \frac{(x-Tx)+(y-Ty)}{2dist(B_i,B_i+1)}\parallel <1.$ Thus, we get $\parallel \left(\frac{x+y}{2}\right)-\left(\frac{Tx+Ty}{2}\right)\parallel <dist(B_i,B_{i+1})$, which is a contradiction. Hence, $x=y.$ □

Let T be a p-cyclic map as given in Definition 1.

T is said to be p–cyclic non expansive map if for all $x\in B_i,y\in B_{i+1},$ the following holds:

$$\rho (Tx,Ty)\le \rho (x,y),\forall i\in \{1,2,\dots ,p\}.$$

The Lemma given below naturally follows for a p-cyclic non expansive map.

Lemma 1

([9], Lemma 3.3). For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. If $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is a p-cyclic non-expansive map, then

$$dist(B_i,B_{i+1})=dist(B_{i+1},B_{i+2})=dist(B_1,B_2),\forall i\in \{1,2,\dots ,p\}.$$

(1)

In addition, if $\xi \in B_i\cap \mathcal{B}{\left(T\right)}_i\ne \varnothing$, then $T^j\xi \in B_{i+j}\cap \mathcal{B}{\left(T\right)}_{i+j}\ne \varnothing$, for all $j=1,2,\dots ,(p-1)$, where $\mathcal{B}{\left(T\right)}_k$ is the set of best proximity point of the mapping T in $B_k$.

In [5], the following lemma is proved, which is again proved here. This lemma is crucial to prove that a given sequence is Cauchy.

Lemma 2

([5], Lemma 3.7). For a uniformly convex Banach space $(X,\parallel ·\parallel )$, we suppose that $C,D$ are non-empty closed subsets of X and $\left\{a_n\right\},\left\{b_n\right\}\subset C$ and $\left\{d_n\right\}\subset D$. If C is convex such that

(i) 

$\parallel a_n-d_n\parallel ⟶dist(C,D)$; and

(ii) 

for every $ϵ>0$ there exists $N\in \mathbb{N}$ such that for all $m>n\ge N$, $\parallel a_m-d_n\parallel \le dist(C,D)+ϵ$,

then for all $ϵ>0$, there exists $N_1\in \mathbb{N}$ such that for all $m>n\ge N_1$, $\parallel a_m-b_n\parallel \le ϵ$.

Proof. 

If $dist(C,D)=0$, then, for a given $ϵ>0$, we can find positive integers $N_2$ and $N_0$ such that $\parallel b_n-d_n\parallel \le \frac{ϵ}{2}$ for all $n\ge N_2$ and $\parallel a_m-d_n\parallel \le \frac{ϵ}{2}$ for all $m>n>N_0$. Now, choosing $N_1=max(N_0,N_2)$, then

$$\parallel a_m-d_n\parallel \le \parallel a_m-d_n\parallel +\parallel b_n-d_n\parallel \le ϵ,forallm>n>N_1.$$

Suppose $dist(C,D)>0$; assuming the contrary, there exists ${ϵ}_0>0$ such that, for every $k\in \mathbb{N}$, there exists $m_k>n_k\ge k$, for which $\parallel a_{m_k}-d_{n_k}\parallel \ge {ϵ}_0$. Choose $0<\gamma <1$ such that ${ϵ}_0/\gamma >dist(C,D)$ and choose $ϵ$ such that $0<ϵmin\left(\frac{{ϵ}_0}{\gamma }-dist(C,D),\frac{dist(C,D)\delta \left(\gamma \right)}{1-\delta \left(\gamma \right)}\right)$, where $\delta \left(\gamma \right)$ is the modulus of convexity. For this $ϵ>0$, there exists $N_0$ such that for all $m_k>n_k\ge N_0$, $\parallel a_{m_k}-d_{n_k}\parallel <dist(C,D)+ϵ$. In addition, there exists $N_2$ such that $\parallel b_{n_k}-d_{n_k}\parallel <dist(C,D)+ϵ$ for all $n_k\ge N_2$. Choose $N_1=max(N_0,N_2)$. By uniform convexity, for all $m_k>n_k\ge N_1$,

$$\parallel \frac{a_{m_k}+d_{n_k}}{2}-d_{n_k}\parallel \le \left(1-\delta \left(\frac{{ϵ}_0}{dist(C,D)+ϵ}\right)\right)(dist(C,D)+ϵ).$$

Since

$$ϵ<\frac{{ϵ}_0}{\gamma }-dist(C,D)\Rightarrow \frac{{ϵ}_0}{dist(C,D)+ϵ}>\gamma ,$$

which implies $1-\delta \left(\frac{{ϵ}_0}{dist(C,D)+ϵ}\right)<1-\delta \left(\gamma \right)$,

$$\begin{array}{ccc}\hfill lll\parallel \frac{a_{m_k}+d_{n_k}}{2}-d_{n_k}\parallel & \le & (1-\delta (\gamma \left)\right)\left(dist\right(C,D\left)\right)+ϵ\hfill \\ & \le & (1-\delta \left(\gamma \right))\left(dist(C,D)+\frac{dist(C,D)}{\delta }\left(\gamma \right)1-\delta \left(\gamma \right)\right)\hfill \\ & <& dist(C,D),\hfill \end{array}$$

which is a contradiction. Hence, the lemma holds. □

Next, we recall few p-cyclic maps with some contraction conditions imposed on them, which are defined in [3,9,10].

Definition 2

([10], Definition 3.1). For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic map, T is said to be p-cyclic contraction, if there exists $k\in (0,1)$ such that for all $x\in B_i$ and $y\in B_{i+1}$, we have

$$\rho (Tx,Ty)\le k\rho (x,y)+(1-k)dist(B_i,B_{i+1}),\forall i\in \{1,2,\dots ,p\}.$$

Definition 3

([9], Definition 3.5). Let $B_1,B_2,\dots ,B_p(p\ge 2)$ be non-empty subsets of a metric space $(M,\rho )$. A p-cyclic map $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is said to be p-cyclic MK-contraction, if for all $ϵ>0$ there exists $\delta >0$ such that whenever $\rho (x,y)<dist(B_i,B_{i+1})+ϵ+\delta ,$ we have

$$\rho (Tx,Ty)<dist(B_i,B_{i+1})+ϵ,$$

where $x\in B_i,y\in B_{i+1}$ and $1\le i\le p.$

Definition 4

([17], Definition 2). Let $\psi :[0,\infty )⟶[0,\infty )$ be a map such that $\psi \left(0\right)=0$ and $\psi \left(\vartheta \right)>0$ if $\vartheta >0.$ We say that ψ is an L–function if for all $\vartheta >0$ there exists $\delta >0$ such that $\psi \left(t\right)\le \vartheta$ for all $t\in [\vartheta ,\vartheta +\delta ]$.

Definition 5

([3], Definition 2.1). For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and C and D are non-empty subsets of M. A cyclic map $T:C\cup D⟶C\cup D$ is said to be cyclic φ –contraction if

$$\rho (Tx,Ty)\le \rho (x,y)-\phi \left(\rho \right(x,y\left)\right)+\phi \left(dist\right(C,D\left)\right)forallx\in C,y\in D,$$

where $\phi :[0,\infty )⟶[0,\infty )$ is a strictly increasing map.

4. p-Cyclic Sequence and p-Cyclic Complete Metric Space

In this article, ${\mathbb{N}}_0$ refers to $\mathbb{N}\cup \left\{0\right\}$. The notion of p-cyclic sequence is given as follows:

Definition 6.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M.

1. 

A sequence ${\left\{x_n\right\}}_{n=1}^{\infty }\subseteq {\cup }_{i=1}^p{B}_i$ is called a p–cyclic sequence if $x_{pn+i}\in B_i$, for all $n\in {\mathbb{N}}_0$ and $i=1,2,\dots ,p.$

2. 

We say that ${\left\{x_n\right\}}_{n=1}^{\infty }$ is a p–cyclic Cauchy sequence, if for given $ϵ>0$ there exists an $N_0\in \mathbb{N}$ such that $forsomei\in \{1,2,\dots ,p\},$ we have

$$\rho (x_{pn+i},x_{pm+i+1})<dist(B_i,B_{i+1})+ϵ,\forall m,n\ge N_0.$$

(2)

3. 

A p-cyclic sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ in ${\cup }_{i=1}^p{B}_i$ is said to be p–cyclic bounded, if ${\left\{x_{pn+i}\right\}}_{n=0}^{\infty }$ is bounded in $B_i$ for some $i\in \{1,2,\dots ,p\}.$

4. 

Let ${\left\{x_n\right\}}_{n=1}^{\infty }$ be a p-cyclic sequence in ${\cup }_{i=1}^p{B}_i$. If for some $j\in \{1,2,\dots ,p\}$ the subsequence $\left\{x_{pn+j}\right\}$ of ${\left\{x_n\right\}}_{n=1}^{\infty }$ converges in $B_j$, then we say that ${\left\{x_n\right\}}_{n=1}^{\infty }$ is p–cyclic convergent.

5. 

Under the assumption that $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty closed subsets of a metric space $(M,\rho )$, we say that ${\cup }_{i=1}^p{B}_i$ is p-cyclic complete if every p-cyclic Cauchy sequence in ${\cup }_{i=1}^p{B}_i$ is p-cyclic convergent.

6. 

If there are subsets $B_1,B_2,\dots ,B_p(p\ge 2)$ of $(M,\rho )$ such that $M={\cup }_{i=1}^p{B}_i$ and ${\cup }_{i=1}^p{B}_i$ is p-cyclic complete, then we call $(M,\rho )$ is p-cyclic complete.

Remark 1.

Note that a p-cyclic sequence that is a Cauchy sequence in the usual sense is a p-cyclic Cauchy sequence. On the other hand, p-cyclic Cauchy sequences need not be Cauchy sequences in the usual sense, even if $dist(B_i,B_{i+1})=0\forall i\in \{1,2,\dots ,p\}$.

The following examples illustrate the notion of p-cyclic sequence and p-cyclic Cauchy sequence.

Example 1.

Consider $\mathbb{R}$ with the usual metric. Let $I_1=[0,1]$, $I_2=[1,2]$ and $I_3=[2,3]$. The sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ defined by $x_n=n-1\left(mod3\right)+\frac{1}{n}$, $n=1,2,3,\dots .$ is a three-cyclic sequence in ${\cup }_{i=1}^3{I}_i$ but not a three-cyclic Cauchy sequence.

Example 2.

Let $X={\mathbb{R}}^2$ be a Euclidean space. Let the subsets $B_i$, $i=1,2,3,4$ be as follows:

$$B_1=\{(0,1+x):0\le x\le 1\},B_2=\{(1+x,0):0\le x\le 1\},$$

$$B_3=\{(0,-(1+x)):0\le x\le 1\}\mathrm{and}{B}_4=\{(-(1+x),0):0\le x\le 1\}.$$

Then, $dist(B_i,B_{i+1})=\sqrt{2}$, for $i=1,2,3,4$, where $B_5=B_1$.

Let us define a sequence ${\left\{(x_n,y_n)\right\}}_{n=1}^{\infty }$ in ${\cup }_{i=1}^4{B}_i$ as follows:

$$(x_n,y_n)=\left\{\begin{array}{ll}(0,1+\frac{1}{t^n}),& n=4k+1,k=0,1,2,3,\dots ;\\ (1+\frac{1}{t^n},0),& n=4k+2,k=0,1,2,3,\dots ;\\ (0,-(1+\frac{1}{t^n})),& n=4k+3,k=0,1,2,3,\dots ;\\ (-(1+\frac{1}{t^n}),0),& n=4k+4,k=0,1,2,3,\dots .\end{array}\right.$$

where $t\in \mathbb{N}andt>1$.

Then, ${\left\{(x_n,y_n)\right\}}_{n=1}^{\infty }$ is a four-cyclic Cauchy sequence in ${\cup }_{i=1}^4{B}_i$.

The following Proposition shows that a p-cyclic Cauchy sequence is p-cyclic bounded.

Proposition 2.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Then, every p-cyclic Cauchy sequence in ${\cup }_{i=1}^p{B}_i$ is p-cyclic bounded.

Proof. 

Let ${\left\{x_n\right\}}_{n=1}^{\infty }$ be a p-cyclic Cauchy sequence in ${\cup }_{i=1}^p{B}_i$. Then, for some $i\in \{1,2,\dots ,p\}$, there exists an $N\in \mathbb{N}$ such that

$$\rho (x_{pn+i},x_{pN+i+1})<1+dist(B_i,B_{i+1})foralln\ge N.$$

Therefore, for all $n\in \mathbb{N}$, $x_{pn+i}\in B(x_{pN+i+1},r)$ where

$$r=max\{1+dist(B_i,B_{i+1}),\rho (x_{pn+1},x_{pN+i+1}):n=1,2,3,\dots ,N\}.$$

Thus, ${\left\{x_{pn+i}\right\}}_{n=0}^{\infty }$ is bounded for some $i\in \{1,2,\dots ,p\}$. Hence, ${\left\{x_n\right\}}_{n=1}^{\infty }$ is p-cyclic bounded. □

Remark 2.

A complete metric space need not be p-cyclic complete. For example, let us consider $(l^{\infty },\parallel .{\parallel }_{sup})$ and let $X={\cup }_{i=1}^p{B}_i,$ where

$$B_i=\{e_{pn+i},n\in {\mathbb{N}}_0\},i=1,2,3,\dots ,p,$$

and $\left\{e_n\right\}$ is a sequence whose nth term is 1 and all the other terms are zero. Then, $B_1,B_2,\dots ,B_p$ are closed subsets of $l^{\infty }$ and hence $(M,\rho )$ is complete. Further, $dist(B_i,B_{i+1})=1$ for all $i=1,2,\dots ,p$. Since

$$\parallel e_{pn+i}-e_{pm+i+1}\parallel =1=dist(B_i,B_{i+1}),n,m\in {\mathbb{N}}_0,i=1,2,\dots ,p.$$

Then, the sequence ${\left\{e_n\right\}}_{n=1}^{\infty }$ is a p-cyclic Cauchy sequence in $X.$ However, none of the subsequence ${\left\{e_{pn+i}\right\}}_{n=0}^{\infty }$ of ${\left\{e_n\right\}}_{n=1}^{\infty }$ converges in $B_i$ for all $i=1,2,\dots ,p$. Hence, $(M,\rho )$ is not p-cyclic complete.

The following Proposition is an example of two-cyclic complete metric space.

Proposition 3.

Let $A_1$ and $A_2$ be subsets of a uniformly convex Banach space X, which are non-empty and closed. If either $A_1$ or $A_2$ is convex, then $A_1\cup A_2$ is two-cyclic complete.

Proof. 

Let $\left\{x_n\right\}$ be a two-cyclic Cauchy sequence in $A_1\cup A_2$. Then, ${\left\{x_{2n+1}\right\}}_{n=0}^{\infty }\subseteq A_1$ and ${\left\{x_{2n}\right\}}_{n=1}^{\infty }\subseteq A_2.$ Assume that $A_1$ is convex. Since $\left\{x_n\right\}$ is a two-cyclic Cauchy sequence in $A_1\cup A_2$, for $ϵ>0$, there exists an $n_0\in \mathbb{N}$ such that

$$\parallel x_{2m+1}-x_{2n+2}\parallel <dist(A_1,A_2)+ϵ,m>n\ge n_0.$$

(3)

In addition, since $\parallel x_{2n+1}-x_{2n+2}\parallel <dist(A_1,A_2)+ϵ,n\ge n_0$, we have $\parallel x_{2n+1}-x_{2n+2}\parallel ⟶dist(A_1,A_2)asn\to \infty .$ Let $n=n_1=n_0+1$ and $m=m_1=n_0+s+1$ where $s\in \mathbb{N}$ be the first element satisfying Equation (3). Then, ${\left\{x_{2n_i+1}\right\}}_{i=1}^{\infty }$ and ${\left\{x_{2m_i+1}\right\}}_{i=1}^{\infty }$ are two sequences in $A_1$ and ${\left\{x_{2n_i+2}\right\}}_{i=1}^{\infty }$ is a sequence in $A_2$ satisfying the following:

(i)

For given $ϵ>0$ there exists an $N_1\in \mathbb{N}$ such that

$$\parallel x_{2m_i+1}-x_{2n_i+2}\parallel <dist(A_1,A_2)+ϵ,m_i>n_i\ge N_1.$$

(ii)

$\parallel x_{2n_i+1}-x_{2n_i+2}\parallel ⟶dist(A_1,A_2)asi\to \infty .$

Thus, Conditions (i) and (ii) of Lemma 2 are satisfied. Since $A_1$ is convex, by Lemma 2, there exists $N_2\in \mathbb{N}$ such that

$$\parallel x_{2n_i+1}-x_{2m_i+1}\parallel \le ϵ,m_i>n_i\ge N_2.$$

Choosing $N_2>n_0+s$, we have

$$\parallel x_{2n+1}-x_{2m+1}\parallel \le ϵ,m>n\ge N_2.$$

Hence, ${\left\{x_{2n+1}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence in $A_1$. Therefore, ${\left\{x_{2n+1}\right\}}_{n=1}^{\infty }$ converges in $A_1$. It yields that ${\left\{x_n\right\}}_{n=1}^{\infty }$ is two-cyclic convergent and hence $A_1\cup A_2$ is two-cyclic complete.

Similarly, we can prove that $A_1\cup A_2$ is two-cyclic complete if $A_2$ is convex. □

5. p-Cyclic Strict Contraction Maps

We introduce a notion of p-cyclic strict contraction, which is a generalization of strict contraction in the usual sense.

Definition 7.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. A p-cyclic map T is said to be p-cyclic strict contraction if, for all $x\in B_i$,$y\in B_{i+1}$, $1\le i\le p$:

(i) 

$\rho (x,y)>dist(B_i,B_{i+1})\Rightarrow \rho (Tx,Ty)<\rho (x,y)$; and

(ii) 

$\rho (x,y)=dist(B_i,B_{i+1})\Rightarrow \rho (Tx,Ty)=\rho (x,y).$

Remark 3.

Note that, if $B_i=A$, for all $i=1,2,\dots ,p$, then p-cyclic strict contraction is a strict contraction in the usual sense. It is clear that the p-cyclic strict contraction also forms a p-cyclic non-expansive map.

The following Proposition proves an important property of p-cyclic strict contraction map.

Proposition 4.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Let $x\in B_i$ ($1\le i\le p$). Suppose that $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is a p-cyclic strict contraction map and if for all $ϵ>0$, there exists an $n_0\in \mathbb{N}$ such that

$$d(T^{pn}x,T^{pm+1}x)<dist(B_i,B_{i+1})+ϵ,n,m\ge n_0,$$

(4)

then for a given $ϵ>0$, there exists an $n_1\in \mathbb{N}$ such that

$$d(T^{pn+k}x,T^{pm+k+1}x)<dist(B_{i+k},B_{i+k+1})+ϵ,n,m\ge n_1,k\in \{1,2,\cdots ,p\}.$$

Proof. 

Let $x\in B_i$ be such that $\left\{T^{pn}x\right\}$ satisfies Equation (4)$(1\le i\le p)$ and let $ϵ>0$ be given. Since T is p-cyclic non-expansive, for any $m,n\in \mathbb{N}$ with $m>n$, $k\in \{1,2,\dots ,p\}$, we have

$$\begin{array}{ccc}\hfill d(T^{pn+k}x,T^{pm+k+1}x)& \le & d(T^{pn}x,T^{pm+1}x)\hfill \\ & <& dist(B_i,B_{i+1})+ϵ,n,m\ge n_0\left(byEquation(4)\right)\hfill \\ & =& dist(A_{i+k},A_{i+k+1})+ϵ,n,m\ge n_0\left(byEquation(1)\right).\hfill \end{array}$$

 □

6. $\mathsf{\Omega }$ Class of Mappings

Many p-cyclic maps with various contractive conditions posses some common properties listed in the following definition. Thus, we introduce a notion of class $\mathsf{\Omega }$.

Definition 8.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. A p-cyclic map $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is said to belong to the class Ω if

1. 

T is p-cyclic strict contraction.

2. 

If $x,y\in B_i,$ then $\underset{n\to \infty }{lim}\rho (T^{pn}x,T^{pn+1}y)=dist(B_i,B_{i+1}),1\le i\le p.$

We list some p-cyclic maps that belong to the class $\mathsf{\Omega }$. First, we prove that a p-cyclic contraction map, which is defined in [10] (Definition 2), belongs to the class $\mathsf{\Omega }$.

Example 3.

Let $B_1,B_2,\dots ,B_p$ be non-empty subsets of a metric space $(M,\rho )$. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is a p-cyclic contraction map. Then, $T\in \mathsf{\Omega }$.

Proof. 

Because the the map T is a p-cyclic contraction, we have

$$\rho (Tx,Ty)\le k\rho (x,y)+(1-k)dist(B_i,B_{i+1}),x\in B_i,y\in B_{i+1},i=1,2,\dots ,p,$$

for some $k\in (0,1).$ If $\rho (x,y)=dist(B_i,B_{i+1}),then\rho (Tx,Ty)=\rho (x,y).$ In addition, if $\rho (x,y)>dist(B_i,B_{i+1})$, then

$$\begin{array}{ccc}\hfill \rho (Tx,Ty)& <& k\rho (x,y)+(1-k)\rho (x,y)\hfill \\ & =& \rho (x,y).\hfill \end{array}$$

Therefore, T is p-cyclic strict contraction. The second condition of Definition 8 follows from Lemma 3.3 in [10]. Hence, $T\in \mathsf{\Omega }$. □

Next, we prove that the p-cyclic Meir–Keeler map (p-cyclic MK-map) introduced in [9] (Definition 3) belongs to the class $\mathsf{\Omega }$.

Example 4.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic MK-contraction map. Then, $T\in \mathsf{\Omega }$.

Proof. 

From Remark 3.6 in [9], we know that T is a p-cyclic MK-contraction if and only if there is an L-function $\psi$ such that for all $x\in B_i$, $y\in B_{i+1}$ and $1\le i\le p,$ the following conditions hold:

(i)

If $\rho (x,y)-dist(B_i,B_{i+1})>0,$ then

$$\rho (Tx,Ty)-dist(B_i,B_{i+1})<\psi (\rho (x,y)-dist(B_i,B_{i+1}))\le \rho (x,y)-dist(B_i,B_{i+1}).$$

(ii)

If $\rho (x,y)-dist(B_i,B_{i+1})=0,$ then $\rho (Tx,Ty)-dist(B_i,B_{i+1})=0.$

Therefore, T is p-cyclic strict contraction and Condition (2) of Definition 8 follows from Lemma 3.8 in [9]. Thus, $T\in \mathsf{\Omega }$. □

Now, we prove that cyclic $\phi$ - contraction introduced in [3] (Definition 5) belongs to the class $\mathsf{\Omega }$.

Example 5.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $C,D$ are non-empty subsets of M. Let $T:C\cup D⟶C\cup D$ be a cyclic φ-contraction map. Then, $T\in \mathsf{\Omega }$.

Proof. 

Let $x\in C$ and $y\in D$. If $\rho (x,y)>dist(C,D),$ since $\phi$ is strictly increasing,

$$\begin{array}{ccc}\hfill \rho (Tx,Ty)& \le & \rho (x,y)-\phi \left(\rho \right(x,y\left)\right)+\phi \left(dist\right(C,D\left)\right)\hfill \\ & <& \rho (x,y)-\phi \left(\rho \right(x,y\left)\right)+\phi \left(\rho \right(x,y\left)\right)\hfill \\ & =& \rho (x,y).\hfill \end{array}$$

If $\rho (x,y)=dist(C,D)$, then $\rho (Tx,Ty)\le \rho (x,y)-\phi \left(\rho \right(x,y\left)\right)+\phi \left(\rho \right(x,y\left)\right)=\rho (x,y)$. Therefore, we get $\rho (x,y)=dist(C,D)\le \rho (Tx,Ty)\le \rho (x,y).$ It yields that $\rho (Tx,Ty)=\rho (x,y)$. Thus, T is a two-cyclic strict contraction and Condition (2) of Definition 8 follows from Theorem 3 in [3] (by putting $d_n=\rho (T^{2n}x,T^{2n+1}y),x,y\in Corx,y\in D$). Hence, $T\in \mathsf{\Omega }$. □

Next, we establish an example of p-cyclic map satisfying a contraction condition of Geraghty’s type [18] and show that it belongs to the class $\mathsf{\Omega }$. Here, we use a class of functions $\mathbb{S}$ introduced by Geraghty [18], where, if $\mathbb{S}$ is the class of all functions $\vartheta :[0,\infty )\to [0,1)$ that satisfies $\vartheta \left(t_n\right)\to 1$, then $t_n\to 0,t_n\in [0,\infty )forn\in \mathbb{N}$.

Example 6.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic map such that for some $\vartheta \in \mathbb{S},$

$$\rho (Tx,Ty)\le \vartheta \left(\rho (x,y)\right)\rho (x,y)+(1-\vartheta \left(\rho (x,y)\right))dist(B_i,B_{i+1}),x\in B_i,y\in B_{i+1}.$$

Then,

(a) 

T is a p-cyclic strict contraction.

(b) 

$\underset{n\to \infty }{lim}\rho (T^{pn}x,T^{pn+1}y)=dist(B_i,B_{i+1}),x,y\in B_i.$

Proof. 

(a) Let $x\in B_i,y\in B_{i+1}.$

Case (1): If $\rho (x,y)>\mathrm{dist}(B_i,B_{i+1}),$ we have

$$\begin{array}{ccc}\hfill \rho (Tx,Ty)& \le & \vartheta \left(\rho (x,y)\right)\rho (x,y)+(1-\vartheta \left(\rho (x,y)\right))\mathrm{dist}(B_i,B_{i+1})\hfill \\ & =& \vartheta \left(\rho (x,y)\right)\rho (x,y)+\mathrm{dist}(B_i,B_{i+1})-\vartheta \left(\rho (x,y)\right)\mathrm{dist}(B_i,B_{i+1})\hfill \\ & =& \vartheta \left(\rho (x,y)\right)[\rho (x,y)-\mathrm{dist}(B_i,B_{i+1})]+\mathrm{dist}(B_i,B_{i+1}),(*)\hfill \\ & <& \rho (x,y)-\mathrm{dist}(B_i,B_{i+1})+\mathrm{dist}(B_i,B_{i+1})(\mathrm{because}\vartheta \left(\rho (x,y)\right)<1)\hfill \\ & =& \rho (x,y).\hfill \end{array}$$

Case (2): If $\rho (x,y)=\mathrm{dist}(B_i,B_{i+1}),$ then from (*),

$$\begin{array}{ccc}\hfill \rho (Tx,Ty)& \le & \rho (x,y).\hfill \end{array}$$

By Equation (1), $\rho (x,y)=dist(B_i,B_{i+1})=dist(B_{i+1},B_{i+2})\le \rho (Tx,Ty)\le \rho (x,y),$ therefore

$$\rho (Tx,Ty)=\rho (x,y).$$

Hence, T is p-cyclic strict contraction.

(b) Let $x,y\in B_i.$ Since T is p-cyclic non-expansive, $\left\{\rho (T^{pn}x,T^{pn+1}y)\right\}$ is a decreasing sequence and is bounded below by $\mathrm{dist}(B_i,B_{i+1}).$ Therefore,

$$\rho (T^{pn}x,T^{pn+1}y)⟶rasn\to \infty \mathrm{and}r\ge \mathrm{dist}(B_i,B_{i+1}),$$

where $r=\underset{n\ge 1}{inf}\rho (T^{pn}x,T^{pn+1}y).$

Claim: $r=\mathrm{dist}(B_i,B_{i+1}).$

If $\rho (T^{pn}x,T^{pn+1}y)=\mathrm{dist}(B_i,B_{i+1})$ for some $n,$ then by the p-cyclic non-expansiveness of T,

$$\rho (T^{pn+k}x,T^{pn+k+1}y)=\rho (T^{pn}x,T^{pn+1}y),k=1,2,\dots .$$

Hence, we have

$$\rho (T^{pn}x,T^{pn+1}y)\to \mathrm{dist}(B_i,B_{i+1})asn\to \infty .$$

Let us assume that $\rho (T^{pn}x,T^{pn+1}y)>\mathrm{dist}(B_i,B_{i+1}),n\in \mathbb{N}.$

Suppose that $r>\mathrm{dist}(B_i,B_{i+1}).$ Since T is p-cyclic non expansive,

$$\begin{array}{ccc}\hfill \rho (T^{p(n+1)}x,T^{p(n+1)+1}y)& \le & \rho (T^{pn+1}x,T^{pn+2}y)\hfill \\ & \le & \vartheta \left(\rho (T^{pn}x,T^{pn+1}y)\right)\rho (T^{pn}x,T^{pn+1}y)\hfill \\ & & +(1-\vartheta \left(\rho (T^{pn}x,T^{pn+1}y)\right))dist(B_i,B_{i+1}),\hfill \\ \hfill \rho (T^{p(n+1)}x,T^{p(n+1)+1}y)-dist(B_i,B_{i+1})& \le & \vartheta \left(\rho (T^{pn}x,T^{pn+1}y)\right)\hfill \\ & & [\rho (T^{pn}x,T^{pn+1}y)-dist(B_i,B_{i+1})],\hfill \end{array}$$

$$\mathrm{Since}\vartheta \in \mathbb{S},\frac{\rho (T^{p(n+1)}x,T^{p(n+1)+1}y)-\mathrm{dist}(B_i,B_{i+1})}{\rho (T^{pn}x,T^{pn+1}y)-\mathrm{dist}(B_i,B_{i+1})}\le \vartheta \left(\rho (T^{pn}x,T^{pn+1}y)\right)<1.$$

(5)

Since $r=\underset{n\to \infty }{lim}\rho (T^{p(n+1)}x,T^{p(n+1)+1}y)>dist(B_i,B_{i+1})$ by our assumption, letting $n\to \infty$ in Equation (5), we get

$$1\le \underset{n\to \infty }{lim}\vartheta \left(\rho (T^{pn}x,T^{pn+1}y)\right)\le 1,$$

that is, $\underset{n\to \infty }{lim}\vartheta \left(\rho (T^{pn}x,T^{pn+1}y)\right)=1$. However, $\underset{n\to \infty }{lim}\rho (T^{pn}x,T^{pn+1}y)=r>0,$ which contradicts $\vartheta \in \mathbb{S}$. Hence, $r=dist(B_i,B_{i+1}).$ This proves Part (b). □

Finally, an example of a p-cyclic map that is of Boyd–Wong type [19] and belongs to the class $\mathsf{\Omega }$ is given:

Example 7.

Let M be a non-empty set equipped with a metric ρ. Suppose that $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Suppose that $\psi :[0,\infty )\to [0,\infty )$ is upper semi-continuous from the right and satisfies $\psi \left(t\right)<tfort>0$ and $\psi \left(0\right)=0.$ Suppose also that $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ is a p-cyclic map. Suppose

$$\rho (Tx,Ty)\le \psi (\rho (x,y)-dist(B_i,B_{i+1}))+dist(B_i,B_{i+1}),$$

(6)

$x\in B_i,y\in B_{i+1},1\le i\le p$.

Then, the following conditions hold:

(a) 

T is p-cyclic strict contraction.

(b) 

For $x,y\in B_i,\rho (T^{pn}x,T^{pn+1}y)\to dist(B_i,B_{i+1}),$ as $n\to \infty .$

Proof. 

(a) Let $x\in B_i$ and $y\in B_{i+1}.$

Case (i): If $\rho (x,y)>\mathrm{dist}(B_i,B_{i+1})$. Since $\psi \left(t\right)<t$ for $t>0$, by Equation (6), we have

$$\begin{array}{ccc}\hfill \rho (Tx,Ty)& <& \rho (x,y)-\mathrm{dist}(B_i,B_{i+1})+\mathrm{dist}(B_i,B_{i+1})\hfill \\ & =& \rho (x,y).\hfill \end{array}$$

Case (ii): If $\rho (x,y)=\mathrm{dist}(B_i,B_{i+1}),$ then $\psi (\rho (x,y)-dist(B_i,B_{i+1}))=0$. Therefore, by Equation (6), $\rho (Tx,Ty)=\mathrm{dist}(B_i,B_{i+1})$. That is, $\rho (Tx,Ty)=\rho (x,y).$

(b) Let $x,y\in B_i.$ Note that

$$\rho (T^{p(n+1)}x,T^{p(n+1)+1}y)\le \rho (T^{pn}x,T^{pn+1}y),n\in \mathbb{N}.$$

Then, the sequence ${\left\{\rho (T^{pn}x,T^{pn+1}y)\right\}}_{n=1}^{\infty }$ is bounded below by $\mathrm{dist}(B_i,B_{i+1}$) and non-increasing sequence. Hence, $\rho (T^{pn}x,T^{pn+1}y)\to rasn\to \infty \mathrm{and}r\ge \mathrm{dist}(B_i,B_{i+1}),$ where ${\displaystyle r=\underset{n\ge 1}{inf}\left\{\rho (T^{pn}x,T^{pn+1}y)\right\}.}$

Claim: $r=\mathrm{dist}(B_i,B_{i+1}).$

Case (1): If $\rho (T^{pn}x,T^{pn+1}y)=\mathrm{dist}(B_i,B_{i+1})$ for some $n\in \mathbb{N}$.

Then, by the p-cyclic non-expansiveness of T,

$$\rho (T^{pn+k}x,T^{pn+k+1}y)=\mathrm{dist}(B_i,B_{i+1}),k=1,2,\dots .$$

Thus, $\rho (T^{pn}x,T^{pn+1}y)\to \mathrm{dist}(B_i,B_{i+1}),asn\to \infty .$

Case (2): If $\rho (T^{pn}x,T^{pn+1}y)>\mathrm{dist}(B_i,B_{i+1})$ for all $n\in \mathbb{N}$.

Since T is p-cyclic non-expansive,

$$\begin{array}{ccc}\hfill \rho (T^{p(n+1)}x,T^{p(n+1)+1}y)& \le & \rho (T^{pn+1}x,T^{pn+2}y)\hfill \\ & \le & \psi (\rho (T^{pn}x,T^{pn+1}y)-\mathrm{dist}(B_i,B_{i+1}))\hfill \\ & & +\mathrm{dist}(B_i,B_{i+1}),\hfill \\ \hfill \rho (T^{p(n+1)}x,T^{p(n+1)+1}y)-dist(B_i,B_{i+1})& \le & \psi (\rho (T^{pn}x,T^{pn+1}y)-\mathrm{dist}(B_i,B_{i+1})).\hfill \end{array}$$

Taking the lim sup on both sides,

$$\underset{n\to \infty }{lim}\rho (T^{p(n+1)}x,T^{p(n+1)+1}y)-dist(B_i,B_{i+1})\le \underset{n\to \infty }{lim\; sup}\psi (\rho (T^{pn}x,T^{pn+1}y)-dist(B_i,B_{i+1})).$$

Since $\rho (T^{p(n+1)}x,T^{p(n+1)+1}y)-dist(B_i,B_{i+1})\downarrow r-\mathrm{dist}(B_i,B_{i+1})$ and $\psi$ is upper semi-continuous from the right,

$$r-\mathrm{dist}(B_i,B_{i+1})\le \psi (r-\mathrm{dist}(B_i,B_{i+1})).$$

Let $t=r-\mathrm{dist}(B_i,B_{i+1})$. If $r>\mathrm{dist}(B_i,B_{i+1})$, then $t>0$ and also $t\le \psi \left(t\right)$, which is a contradiction to the definition of $\psi .$ Hence, $r=\mathrm{dist}(B_i,B_{i+1}).$ □

In a similar way, one may check whether some other known/unknown p-cyclic maps belong to the class $\mathsf{\Omega }$.

7. Best Proximity Point Results of $\mathsf{\Omega }$ Class of Mappings

Below, we give a convergence result for $\mathsf{\Omega }$ class of mappings.

Theorem 1.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M.

Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic map that belongs to the class$\mathsf{\Omega }$. Assume for some $k\in \mathbb{N}$ and $x\in B_i,(1\le i,k\le p)$, $\left\{T^{pn+k}x\right\}$ converges to $\xi \in B_{i+k}$. Then, ξ is a best proximity point of T in $B_{i+k}$.

Proof. 

Let $x\in B_i$ be as given in the theorem. By Equation (1), for each $n\in \mathbb{N}$, we have,

$$\begin{array}{ccc}\hfill dist(B_{i+k},B_{i+k+1})& =& dist(B_{i+k-1},B_{i+k})\hfill \\ & \le & \rho (T^{pn+k-1}x,\xi )\hfill \\ & \le & \rho (T^{pn+k-1}x,T^{pn+k}x)+\rho (T^{pn+k}x,\xi ).\hfill \end{array}$$

Since $T\in \mathsf{\Omega },$

$$\underset{n\to \infty }{lim}(\rho (T^{pn+k-1}x,T^{pn+k}x)+\rho (T^{pn+k}x,\xi ))=dist(B_{i+k-1},B_{i+k}).$$

Therefore,

$$\underset{n\to \infty }{lim}\rho (T^{pn+k-1}x,\xi )=dist(B_{i+k-1},B_{i+k})=dist(B_{i+k},B_{i+k+1}).$$

(7)

Now,

$$\begin{array}{ccc}\hfill dist(B_{i+k},B_{i+k+1})& \le & \rho (\xi ,T\xi )\hfill \\ & =& \underset{n\to \infty }{lim}\rho (T^{pn+k}x,T\xi )\hfill \\ & \le & \underset{n\to \infty }{lim}\rho (T^{pn+k-1}x,\xi )\hfill \\ & =& dist(B_{i+k},B_{i+k+1}),\left(by\left(7\right)\right).\hfill \end{array}$$

Hence, $\rho (\xi ,T\xi )=dist(B_{i+k},B_{i+k+1}).$ □

Now, we prove the existence of best proximity point for mappings which belong to class $\mathsf{\Omega }$ defined on a p-cyclic complete metric space.

Theorem 2.

For a non-empty set M, suppose $\rho :M\times M\to [0,\infty )$ forms a metric and $B_1,B_2,\dots ,B_p(p\ge 2)$ are non-empty subsets of M. Suppose that $M={\cup }_{i=1}^p{B}_i$ and ${\cup }_{i=1}^p{B}_i$ is p-cyclic complete. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic map which belongs to the class Ω. Then, there exists a best proximity point of T in $B_j$ for some $j\in \{1,2,\dots ,p\}$.

Proof. 

Let $x\in B_i$, $1\le i\le p$.

Define a sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ in $(M,\rho )$ by

$$x_n:=T^{n}xforn\in \mathbb{N}.$$

Claim: ${\left\{T^{n}x\right\}}_{n=1}^{\infty }$ is a p-cyclic Cauchy sequence.

Let $m,n\in \mathbb{N}$ be such that $m>n,$

$$\begin{array}{ccc}\hfill \rho (T^{pm}x,T^{pn+1}x)& =& \rho (T^{p(n+r)}x,T^{pn+1}x),\mathrm{where}m=n+r,r\in \mathbb{N}\hfill \\ & =& \rho (T^{pn}y,T^{pn+1}x),\mathrm{where}y=T^{pr}x\in B_i\hfill \\ & \to & dist(B_i,B_{i+1}),asn\to \infty (\mathrm{because}T\in \mathsf{\Omega }).\hfill \end{array}$$

This implies that, for all $ϵ>0,$ there exists an $n_0\in \mathbb{N}$ such that

$$\rho (T^{pm}x,T^{pn+1}x)<ϵ+dist\{(B_i,B_{i+1}\},m,n\ge n_0.$$

By Proposition 4, for a given $ϵ>0,$ there exists an $n_1\in \mathbb{N}$ such that

$$d(T^{pm+k}x,T^{pn+k+1}x)<ϵ+d(A_{i+k},A_{i+k+1}),m,n\ge n_1,k\in \{1,2,\dots ,p\}.$$

Therefore, the sequence $\left\{T^{n}x\right\}$ is a p-cyclic Cauchy sequence in $(M,\rho )$. Since $(M,\rho )$ is p-cyclic complete, there exists a $k\in \{1,2,\dots ,p\}$ such that $\left\{T^{pn+k}x\right\}$ converges to $z\in A_{i+k}$. By Theorem 1, z is best proximity point of T in $B_j$, where $j=i+k$. □

Remark 4.

In Theorem 2, suppose that $dist(B_i,B_{i+1})=0$, for some $i\in \{1,2,\dots ,p\}$. Then, by Equation (1), $dist(B_k,B_{k+1})=0,1\le k\le p$. This implies that $Tz=z$, that is, z is a fixed point of T. Since T is p-cyclic, $z\in {\cap }_{i=1}^p{B}_i$ and hence ${\cap }_{i=1}^p{B}_i$ is non-empty.

To prove the uniqueness of fixed point of T, let $\alpha ,\beta \in B_i$ be such that $\alpha =T\alpha$,$\beta =T\beta$ and $x\alpha \ne \beta$. Then, $\rho (\alpha ,\beta )>0=dist(B_i,B_{i+1})$. Since T is p-cyclic strict contraction,

$$\rho (\alpha ,\beta )=\rho (T\alpha ,T\beta )<\rho (\alpha ,\beta ),$$

which is a contradiction. Hence, $\alpha =\beta$. This shows that there exists a unique fixed point for Ω class of mappings in a p-cyclic complete metric space.

Theorem 3.

Let X be a strictly convex normed linear space. Let $B_1,B_2,\dots ,B_p(p\ge 2)$ be non-empty, closed, convex subsets of X such that $X={\cup }_{i=1}^p{B}_i$ and ${\cup }_{i=1}^p{B}_i$ is p-cyclic complete. Let $T:{\cup }_{i=1}^p{B}_i⟶{\cup }_{i=1}^p{B}_i$ be a p-cyclic map which belongs to the class Ω . Then, for each $j\in \{1,2,\dots ,p\}$, there exists a unique best proximity point $z_j\in A_j$. In addition, $z_j$ is a unique periodic point of T in $B_j$ and $T^k{z}_j$ is the unique best proximity point of T in $B_{j+k}$, $k=0,1,2,\dots ,(p-1)$.

Proof. 

From the proof of Theorem 2, for any $x\in B_i$,$i\in \{1,2,\dots ,p\}$, $\left\{T^{pn+q}x\right\}$ converges to $z\in B_j,j=i+q$, for some $q\in \{1,2,\dots ,p\}$ and by Theorem 1, z is a best proximity point of T in $B_j$. Since X is a strictly convex, by Proposition 1, z is unique.

By Lemma 1, since T is p-cyclic non expansive map, $T^{k}z$ is the best proximity point of T in $B_{j+k}$, for $k=1,2,\dots ,(p-1)$.

To prove z is a periodic point of T in $B_j$, it is enough to prove $\parallel T^{p}z-Tz\parallel =dist(B_j,B_{j+1})$. Suppose $\parallel T^{p}z-Tz\parallel >dist(B_j,B_{j+1})$. Since T is p-cyclic strict contraction,

$$\begin{array}{ccc}\hfill \parallel T^{p+1}z-T^{2}z\parallel & <& \parallel T^{p}z-Tz\parallel \hfill \\ & \le & \parallel T^{p-1}z-z\parallel \hfill \\ & =& \underset{n}{lim}\parallel T^{p-1}z-T^{pn+q}x\parallel \hfill \\ & \le & \underset{n}{lim}\parallel z-T^{(pn+q)-(p-1)}x\parallel \hfill \\ & =& \underset{n}{lim}\parallel z-T^{p(n-1)+q+1}x\parallel \hfill \\ & =& \underset{n}{lim}\parallel T^{p(n-1)+q}x-T^{p(n-1)+q+1}x\parallel \hfill \\ & =& dist(B_{i+q},B_{i+q+1}),(\mathrm{because}T\in \mathsf{\Omega })\hfill \\ & =& dist(B_i,B_{i+1}),\left(byEquation\left(1\right)\right)\hfill \\ & =& dist(B_{j+1},B_{j+2})\hfill \end{array}$$

which is a contradiction. Hence, $\parallel T^{p}z-Tz\parallel =dist(B_j,B_{j+1})$. Since X is strictly convex, $B_j$ is convex and $\parallel z-Tz\parallel =dist(B_j,B_{j+1})$, then $T^{p}z=z$, that is z is a periodic point of T in $B_j$. To prove the uniqueness of z, let $\xi \in B_j$ be such that $T^p\xi =\xi$, then $\left\{T^{pn}\xi \right\}$ converges to $\xi \in B_j$. By Theorem 1, $\xi$ is a best proximity point of T in $B_j$. Since z is the unique best proximity point of T in $A_j$, we have $\xi =z$. Hence, the theorem holds. □

8. Conclusions

In this paper, we have introduced a notion of p-cyclic Cauchy sequence, which is weaker than the notion of Cauchy sequence in the usual sense. If one subsequence of a p-cyclic Cauchy sequence converges, then we say that the p-cyclic sequence is p-cyclic convergent. If all p-cyclic Cauchy sequences converge, then we call the underlying metric space as p-cyclic complete metric space. We have shown that a complete metric space need not be p-cyclic complete. A class of mappings called $\mathsf{\Omega }$ is introduced. The existence of fixed point and best proximity point of mappings of class $\mathsf{\Omega }$ is guaranteed in a p-cyclic complete metric space. Many p-cyclic maps with various contractive conditions introduced in the literature fall under class $\mathsf{\Omega }$, where the best proximity point for such maps were obtained in a uniformly convex Banach space, whereas we have obtained a unique best proximity point in a strictly convex norm linear space. Thus, our main result is a natural generalization of main results of Al-Thagafi [3], Boyd [19], Eldred [5], Geraghty [18], Karpagam [10], Karpagam [9], Kirk [2], Meir [20].