On the C-class functions of fixed point and best proximity point results for generalised cyclic-coupled mappings

Abstract

Existence of fixed point for C-class functions was first proved by Ansari in 2014. Then, many authors gave interesting results using C-class functions. In this paper, we prove the existence of strong coupled proximity point for generalised cyclic-coupled proximal maps. Our result generalises the results of Kadwin and Marudai.

Keywords:

• cyclic-coupled contraction
• best proximity point
• multivalued mapping
• fixed point
• C-class function

AMS Mathematics subject classifications:

• Primary 47H10
• Secondary 54H25

Public Interest Statement

Many mathematical problems can be formulated as a fixed point equation of the form $T(x)=x$, where T is a self-mapping in some suitable framework. However, if T is non-self mapping, the above-mentioned equation does not necessarily have a solution. In such case, it is worthy to determine an approximate solution x such that the error $\text{d}(x,T(x))$ is minimum.

1. Introduction and mathematical preliminaries

Initially, in 1922, Banach proved the existence and uniqueness of fixed point for contraction mapping. Later, among several interesting results given by various authors, Kannan (1969) introduced a kind of mapping which has its own significance, as it also admits fixed point on discontinuous maps. In spite of many authors proving the existence of fixed point on self-mappings, it has been proved by Kirk, Srinivasan and Veeramani (2003) that fixed points do exist on a special kind of map called cyclic maps.

Let A and B be two non-empty subsets of metric space (X, d). A mapping $T:A\cup B\to A\cup B$ is said to be cyclic if $T(A)\subset B$ and $T(B)\subset A$.

Meanwhile, another class of mappings called coupled maps were introduced by Lakshmikantham and Ciric (2009) to find coupled fixed point which has wide range of applications to partial differential equations and boundary value problems.

Definition 1.1

An element $(x,y)\in X\times X$ in a non-empty set X is said to be a coupled fixed point for a mapping $F:X\times X\to X$ if $F(x,y)=x$ and $F(y,x)=y$.

These kind of maps were later generalised by Kumam, Pragadeeswarar, Marudai and Sitthithakerngkiet (2014) finding out coupled best proximity points for coupled proximal maps with respect to A and B as non-empty closed subsets of metric space (X, d) with $A\cap B=\mathrm{\varnothing }$. Very recently, (Choudhury & Maity, 2014) extended concept of cyclic maps by introducing cyclic-coupled Kannan-type contraction as follows.

Definition 1.2

A mapping $T:X\times X\to X$ is said to be cyclic with respect to A and B if $T(A,B)\subset B$ and $T(B,A)\subset A$.

Definition 1.3

Let A and B be two non-empty subsets of a metric space (X, d). A mapping $F:X\times X\to X$ is called cyclic-coupled Kannan-type mapping if F is cyclic with respect to A and B satisfying, for some $k\in \left(0,\frac{1}{2}\right)$, the inequality$$\begin{array}{c}\hfill \text{d}(F(x,y),F(u,v))\le k[\text{d}(x,F(x,y))+\text{d}(u,F(u,v))].\end{array}$$

where $x,v\in A$ and $y,u\in B$.

Definition 1.4

Let X be a non-empty set. An element $(x,x)\in X\times X$ is said to be strong coupled fixed point if $F(x,x)=x$.

The following theorem was proved by Choudhury and Maity (2014).

Theorem 1.5

Let A and B be two non-empty closed subsets of a complete metric space (X, d) with $A\cap B\ne \mathrm{\varnothing }$ and $F:X\times X\to X$ be a cyclic-coupled Kannan-type mapping with respect to A and B with $A\cap B\ne \mathrm{\varnothing }$. Then F has a strong coupled fixed point on $A\cap B$.

Immediately, Udo-utun (2014) extended the result of (Choudhury & Maity, 2014) using Ciric-type contractions.

The existence and convergence of best proximity points is an interesting topic on optimisation theory on which several interesting results were published (Abkar & Gabeleh, 2013; Aydi & Felhi, 2016; Aydi, Felhi, & Karapinar, 2016; Eldred & Veeramani, 2006; Gupta, Rajput, & Kaurav, 2014; Latif, Abbas, & Hussain, 2016; Mursaleen, Srivastava, & Sharma, 2016). Such results may sometimes assume a sequential property on metric spaces called UC-property.

Definition 1.6

Let A and B be non-empty subsets of a metric space (X, d). Then (A, B) is said to satisfy the UC property if $\{x_n\}$ and $\{z_n\}$ are sequences in A and $\{y_n\}$ is a sequence in B such that $li{m}_{n\to \infty }\text{d}(x_n,y_n)=\text{d}(A,B)$ and $li{m}_{n\to \infty }\text{d}(z_n,y_n)=\text{d}(A,B)$, then $li{m}_{n\to \infty }\text{d}(x_n,z_n)=0$.

In 2014, the concept of C-class functions was introduced by Ansari (2014). Using this concept, we can generalise many fixed point theorems in the literature.

Definition 1.7

   Ansari (2014) A mapping $f:{[0,\infty )}^2\to \mathbb{R}$ is called C-class function if it is continuous and satisfies the following axioms:

(1)	$f(s,t)\le s$;
(2)	$f(s,t)=s$ implies that either $s=0$ or $t=0$; for all $s,t\in [0,\infty )$.

Note for some f we have that $f(0,0)=0$.

We denote C-class functions as $\mathcal{C}$.

Example 1.8

   Ansari (2014), Liu, Ansari, Chandok, and Park (2016) The following functions $f:{[0,\infty )}^2\to \mathbb{R}$ are elements of $\mathcal{C}$, for all $s,t\in [0,\infty )$:

(1)	$f(s,t)=s-t$, $f(s,t)=s\Rightarrow t=0$;
(2)	$f(s,t)=ms$, $0<m<1$, $f(s,t)=s\Rightarrow s=0$;
(3)	$f(s,t)=\frac{s}{{(1+t)}^r}$; $r\in (0,\infty )$, $f(s,t)=s$$\Rightarrow$$s=0$ or $t=0$;
(4)	$f(s,t)=log(t+a^s)/(1+t)$, $a>1$, $f(s,t)=s$$\Rightarrow$$s=0$ or $t=0$;
(5)	$f(s,t)=ln(1+a^s)/2$, $a>e$, $f(s,1)=s$$\Rightarrow$$s=0$;
(6)	$f(s,t)={(s+l)}^{(1/{(1+t)}^r)}-l$, $l>1,r\in (0,\infty )$, $f(s,t)=s$$\Rightarrow$$t=0$;
(7)	$f(s,t)=s{log}_{t+a}a$, $a>1$, $F(s,t)=s\Rightarrow$$s=0$ or $t=0$;
(8)	$f(s,t)=s-\left(\frac{1+s}{2+s}\right)\left(\frac{t}{1+t}\right)$, $f(s,t)=s\Rightarrow t=0$;
(9)	$f(s,t)=s\mathit{\beta }(s)$, $\mathit{\beta }:[0,\infty )\to [0,1)$, and is continuous, $f(s,t)=s\Rightarrow s=0$;
(10)	$f(s,t)=s-\frac{t}{k+t},f(s,t)=s\Rightarrow t=0$;
(11)	$f(s,t)=s-\mathit{\phi }(s),f(s,t)=s\Rightarrow s=0,$ here $\mathit{\phi }:[0,\infty )\to [0,\infty )$ is a continuous function such that $\mathit{\phi }(t)=0\iff t=0$;
(12)	$f(s,t)=sh(s,t),f(s,t)=s\Rightarrow s=0,$ here $h:[0,\infty )\times [0,\infty )\to [0,\infty )$ is a continuous function such that $h(t,s)<1$ for all $t,s>0$;
(13)	$f(s,t)=s-\left(\frac{2+t}{1+t}\right)t$, $f(s,t)=s\Rightarrow t=0$. (8) $f(s,t)=s-\left(\frac{1+s}{2+s}\right)\left(\frac{t}{1+t}\right)$, $f(s,t)=s\Rightarrow t=0$;
(14)	$f(s,t)=\sqrt[n]{ln(1+s^n)}$, $f(s,t)=s\Rightarrow s=0$;
(15)	$f(s,t)=\mathit{\varphi }(s),f(s,t)=s\Rightarrow s=0,$ here $\mathit{\varphi }:[0,\infty )\to [0,\infty )$ is a upper semi-continuous function such that $\mathit{\varphi }(0)=0,$ and $\mathit{\varphi }(t)<t$ for $t>0,$
(16)	$f(s,t)=\frac{s}{{(1+s)}^r}$; $r\in (0,\infty )$, $f(s,t)=s$$\Rightarrow$$s=0$ ;
(17)	$f(s,t)=\mathit{\vartheta }(s)$; $\mathit{\vartheta }:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to \mathbb{R}$ is a generalised Mizoguchi?Takahashi-type function, $f(s,t)=s$$\Rightarrow$$s=0$;
(18)	$f(s,t)=\frac{s}{\mathrm{\Gamma }(1/2)}{\int }_0^{\infty }\frac{e^{-x}}{\sqrt{x}+t}\text{d}x$, where $\mathrm{\Gamma }$ is the Euler gamma function.

Let $\mathrm{\Phi }$ denote the set of all functions $\mathit{\phi }:[0,+\infty )\to [0,+\infty )$ that satisfy the following conditions:

(1)	$\mathit{\phi }$ is lower semi-continuous on $[0,+\infty )$,
(2)	$\mathit{\phi }(0)=0$,
(3)	$\mathit{\phi }(s)>0$ for each $s>0$.

Let ${\mathrm{\Phi }}_1$ denote the set of all functions $\mathit{\phi }:[0,+\infty )\to [0,+\infty )$ that satisfy the following conditions:

(1)	$\mathit{\phi }$ is lower semi-continuous on $[0,+\infty )$,
(2)	$\mathit{\phi }(0)\ge 0$,
(3)	$\mathit{\phi }(s)>0$ for each $s>0$

Let $\mathrm{\Psi }$ denote all the functions $\mathit{\psi }:[0,\infty )\to [0,\infty )$ which satisfy

(i)	$\mathit{\psi }(t)=0$ if and only if $t=0$,
(ii)	$\mathit{\psi }$ is continuous,
(iii)	$\mathit{\psi }(s)\le s$, $\forall s>0$.

Definition 1.9

Let A and B be two non-empty disjoint subsets of a metric space (X, d). A mapping $F:X\times X\to X$ is called cyclic-coupled proximal mappings of type I if F is cyclic with respect to A and B satisfying the inequality$$\begin{array}{c}\hfill \text{d}(F(x,y),F(u,v))\le kmax[\text{d}(x,F(x,y)),\text{d}(u,F(u,v))]+(1-k)\text{d}(A,B)\end{array}$$

where $x,v\in A$ and $y,u\in B$ for some $k\in (0,1)$.

Definition 1.10

Let A and B be two non-empty disjoint subsets of a metric space (X, d). A mapping $F:X\times X\to X$ is called cyclic-coupled proximal mappings of type II if F is cyclic with respect to A and B satisfying the inequality$$\begin{array}{c}\hfill \text{d}(F(x,y),F(u,v))\le k[\text{d}(x,F(x,y))+\text{d}(u,F(u,v))]+(1-2k)\text{d}(A,B)\end{array}$$

where $x,v\in A$ and $y,u\in B$ for some $k\in \left(0,\frac{1}{2}\right)$.

In this paper, we define new generalised cyclic-coupled mappings using C-class functions and prove the existence of strong coupled proximity points.

2. Best proximity points for cyclic-coupled mappings

In this part, we introduce cyclic-coupled proximal maps and prove the existence of proximity points for those maps under suitable conditions.

Definition 2.1

Let A and B be two non-empty disjoint subsets of a metric space (X, d). A mapping $F:X\times X\to X$ is called cyclic-coupled proximal mappings of type $I_{f\mathit{\phi }.}$ if F is cyclic with respect to A and B satisfying the inequality$$\begin{array}{cc}\hfill \text{d}(F(x,y),F(u,v))& \le f\left(max\{\text{d}(x,F(x,y)),\text{d}(u,F(u,v))\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(x,F(x,y)),\text{d}(u,F(u,v))\}-\text{d}(A,B)\right)\right)+\text{d}(A,B),\hfill \end{array}$$

where $x,v\in A$ and $y,u\in B$ for some $\mathit{\phi }\in \mathrm{\Phi }$, (or $\mathit{\phi }\in {\mathrm{\Phi }}_1)$, $f\in \mathbb{C}$.

Remark 2.2

With choice $F(s,t)=ks$, $0<k<1,$ in Definition 2.1 we obtain Definition 1.9.

Definition 2.3

Let A and B be two non-empty disjoint subsets of a metric space (X, d). A mapping $F:X\times X\to X$ is called cyclic-coupled proximal mappings of type $I{I}_{f\mathit{\phi }}$ if F is cyclic with respect to A and B satisfying the inequality$$\begin{array}{cc}\hfill \text{d}(F(x,y),F(u,v))& \le f\left(\frac{\text{d}(x,F(x,y))+\text{d}(u,F(u,v))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(x,F(x,y))+\text{d}(u,F(u,v))}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B),\hfill \end{array}$$

where $x,v\in A$ and $y,u\in B$ for some $\mathit{\phi }\in \mathrm{\Phi }$,(or $\mathit{\phi }\in {\mathrm{\Phi }}_1)$, $f\in \mathbb{C}$.

Remark 2.4

With choice $F(s,t)=ks$, $0<k<1,$ in Definition 2.3 we obtain Definition 1.10.

Definition 2.5

Let (X, d) be a metric space. An element $(x,y)\in X\times X$ is said to be strong coupled proximal points if $\text{d}(x,F(x,y))=\text{d}(y,F(y,x))=\text{d}(x,y)=d(A,B)$.

Theorem 2.6

Let (X, d) be a complete metric space and A, B be two non-empty closed disjoint subsets of X. Let $F:X\times X\to X$ be cyclic-coupled proximal mapping of type $I_{f\mathit{\phi }.}$. Then F has strong coupled proximal point if (A, B) satisfies UC property.

Proof

Let $x_0\in A,y_0\in B$ be any two arbitrary elements of X. Let $\{x_n\}$ and $\{y_n\}$ be two sequences defined as $F(x_n,y_n)=y_{n+1}$ and $F(y_n,x_n)=x_{n+1}$. Then, for n = 1,$$\begin{array}{cc}\hfill \text{d}(x_1,y_2)& =\text{d}(F(y_0,x_0),F(x_1,y_1))\hfill \\ \hfill & \le f\left(max\{\text{d}(y_0,F(y_0,x_0)),\text{d}(x_1,F(x_1,y_1))\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_0,F(y_0,x_0)),\text{d}(x_1,F(x_1,y_1))\}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le f\left(max\{\text{d}(y_0,x_1)),\text{d}(x_1,y_2)\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_0,x_1)),\text{d}(x_1,y_2)\}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \end{array}$$

and$$\begin{array}{cc}\hfill \text{d}(y_1,x_2)& =\text{d}(F(x_0,y_0),F(y_1,x_1))\hfill \\ \hfill & \le f\left(max\{\text{d}(x_0,F(x_0,y_0)),\text{d}(y_1,F(y_1,x_1))\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(x_0,F(x_0,y_0)),\text{d}(y_1,F(y_1,x_1))\}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le f\left(max\{\text{d}(x_0,y_1)),\text{d}(y_1,x_2)\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(x_0,y_1)),\text{d}(y_1,x_2)\}-\text{d}(A,B)\right)\right)+\text{d}(A,B).\hfill \end{array}$$

Similarly, for $n=2$, we get$$\begin{array}{cc}\hfill \text{d}(x_2,y_3)& \le f\left(max\{\text{d}(y_1,x_2)),\text{d}(x_2,y_3)\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_1,x_2)),\text{d}(x_2,y_3)\}-\text{d}(A,B)\right)\right)+\text{d}(A,B).\hfill \\ \hfill \text{d}(y_2,x_3)& \le f\left(max\{\text{d}(x_1,y_2)),\text{d}(y_2,x_3)\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(x_1,y_2)),\text{d}(y_2,x_3)\}-\text{d}(A,B)\right)\right)+\text{d}(A,B).\hfill \end{array}$$

In general, we have(2.1) $$\begin{array}{cc}\hfill \text{d}(x_n,y_{n+1})& \le f\left(max\{\text{d}(y_{n-1},x_n)),\text{d}(x_n,y_{n+1})\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_{n-1},x_n)),\text{d}(x_n,y_{n+1})\}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \end{array}$$(2.1)

and(2.2) $$\begin{array}{cc}\hfill \text{d}(y_n,x_{n+1})& \le f\left(max\{\text{d}(x_{n-1},y_n)),\text{d}(y_n,x_{n+1})\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(x_{n-1},y_n)),\text{d}(y_n,x_{n+1})\}-\text{d}(A,B)\right)\right)+\text{d}(A,B).\hfill \end{array}$$(2.2)

For, $\text{d}(y_{n-1},x_n))\le \text{d}(x_n,y_{n+1})$. Then, Equation (2.1) reduces to(2.3) $$\begin{array}{c}\hfill \text{d}(x_n,y_{n+1})-\text{d}(A,B)\le f\left(\text{d}(x_n,y_{n+1})-\text{d}(A,B),\mathit{\phi }\left(\text{d}(x_n,y_{n+1})-\text{d}(A,B)\right)\right).\end{array}$$(2.3)

By definition of f, $\text{d}(x_n,y_{n+1})-\text{d}(A,B)=0$ or $\mathit{\phi }(\text{d}(x_n,y_{n+1})-\text{d}(A,B))=0.$ Therefore, $\text{d}(x_n,y_{n+1})=\text{d}(A,B).$

For, $\text{d}(x_n,y_{n+1})\le \text{d}(y_{n-1},x_n))$. Then, Equation (2.1) reduces to(2.4) $$\begin{array}{cc}\hfill \text{d}(x_n,y_{n+1})-\text{d}(A,B)& \le f\left(\text{d}(y_{n-1},x_n)-\text{d}(A,B),\mathit{\phi }\left(\text{d}(y_{n-1},x_n)-\text{d}(A,B)\right)\right)\hfill \\ \hfill & \le \text{d}(y_{n-1},x_n)-\text{d}(A,B).\hfill \end{array}$$(2.4)

Thus, from (2.3) and (2.4), we conclude that(2.5) $$\begin{array}{c}\hfill \text{d}(x_n,y_{n+1})\le \text{d}(y_{n-1},x_n).\end{array}$$(2.5)

Similarly, Equation (2.2) reduces as(2.6) $$\begin{array}{cc}\hfill \text{d}(y_n,x_{n+1})-\text{d}(A,B)& \le f\left(\text{d}(x_{n-1},y_n)-\text{d}(A,B),\mathit{\phi }\left(\text{d}(x_{n-1},y_n)-\text{d}(A,B)\right)\right)\hfill \\ \hfill & \le \text{d}(x_{n-1},y_n)-\text{d}(A,B)\hfill \end{array}$$(2.6)

and hence(2.7) $$\begin{array}{c}\hfill \text{d}(y_n,x_{n+1})\le \text{d}(x_{n-1},y_n).\end{array}$$(2.7)

Hence, using (2.5), $\{\text{d}(x_{2n},y_{2n+1})\}$ and $\{\text{d}(y_{2n-1},x_{2n})\}$ are decreasing sequences and converge to some $r\ge 0$ as n$\to \infty$. Therefore, the Equation (2.3) reduces to$$\begin{array}{c}\hfill r-\text{d}(A,B)\le f\left(r-\text{d}(A,B),\mathit{\phi }\left(r-\text{d}(A,B)\right)\right)\le r-\text{d}(A,B).\end{array}$$

By the definition of f, $r-\text{d}(A,B)=0$ or $\mathit{\phi }(r-\text{d}(A,B))=0.$ Therefore, $r=d(A,B)$ and hence $\left\{\left\{\text{d}(x_{2n},y_{2n+1})\right\}\right.$ and $\left.\left\{\text{d}(y_{2n-1},x_{2n})\right\}\right\}$ converge to $\text{d}(A,B)$. Similarly, using (2.7) $\{\text{d}(y_{2n},x_{2n+1})\}$ and $\{\text{d}(x_{2n-1},y_{2n})\}$ are decreasing sequences and converge to some $s\ge 0$ as n$\to \infty .$ Therefore, the Equation (2.6) reduces to$$\begin{array}{c}\hfill s-\text{d}(A,B)\le f\left(s-\text{d}(A,B),\mathit{\phi }\left(s-\text{d}(A,B)\right)\right)\le s-\text{d}(A,B).\end{array}$$

By the definition of f, $s-\text{d}(A,B)=0$ or $\mathit{\phi }(s-\text{d}(A,B))=0$. Therefore, $s=\text{d}(A,B)$ and hence $\left\{\{\text{d}(y_{2n},x_{2n+1})\}\right.$ and $\left.\{\text{d}(x_{2n-1},y_{2n})\}\right\}$ converge to $\text{d}(A,B)$.

Using the above arguments, we conclude that

${lim}_{n\to \infty }\text{d}(x_n,y_{n+1})={lim}_{n\to \infty }\text{d}(y_n,x_{n+1})=\text{d}(A,B)$.

Now, using UC-property we get(2.8) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\text{d}(x_n,x_{n+1})=\underset{n\to \infty }{lim}\text{d}(y_n,y_{n+1})=0.\end{array}$$(2.8)

Claim: $\{x_n\}$ is a Cauchy sequence.

Let $m<n$ and$$\begin{array}{cc}\hfill \text{d}(x_n,y_m)& =\text{d}(F(y_{n-1},x_{n-1}),F(x_{m-1},y_{m-1})).\hfill \\ \hfill & \le f\left(max\{\text{d}(y_{n-1},x_n),d(x_{m-1},y_m)\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_{n-1},x_n),\text{d}(x_{m-1},y_m)\}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \end{array}$$

and$$\begin{array}{cc}\hfill \text{d}(x_m,y_m)& =\text{d}(F(y_{m-1},x_{m-1}),F(x_{m-1},y_{m-1})).\hfill \\ \hfill & \le f\left(max\{\text{d}(y_{m-1},x_m),\text{d}(x_{m-1},y_m)\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_{m-1},x_m),\text{d}(x_{m-1},y_m)\}-\text{d}(A,B)\right)\right)+\text{d}(A,B).\hfill \end{array}$$

Therefore, ${lim}_{m\to \infty }$ Claim: $\{x_n\}$ is a Cauchy sequence $(x_n,y_m)=\text{d}(A,B)$ and ${lim}_{m\to \infty }\text{d}(x_m,y_m)=\text{d}(A,B)$

Therefore, using UC-property, we get ${lim}_{n\to \infty }\text{d}(x_n,x_m)=0$ (i.e. for given $\mathit{ϵ}>0,\exists n_0\in \mathbb{N}$ such that $\forall m>n>n_0$, $\text{d}(x_n,x_m)<\mathit{ϵ}$). Hence, $\{x_n\}$ is a Cauchy sequence and converges to some point $x\in A$. Similarly, it can be proved that $\{y_n\}$ is a Cauchy sequence and converges to some point $y\in B$.

Since ${lim}_{m\to \infty }(x_m,y_m)=\text{d}(A,B)$ and d is uniformly continuous, $\text{d}(x,y)=\text{d}(A,B)$.

Now, we consider$$\begin{array}{cc}\hfill \text{d}(x,F(x,y))& \le \text{d}(x,x_{n+1})+\text{d}(x_{n+1},F(x,y)),\forall n\in \mathbb{N}.\hfill \\ \hfill & =\text{d}(x,x_{n+1})+\text{d}(F(y_n,x_n),F(x,y)),\forall n\in \mathbb{N}.\hfill \\ \hfill & \le \text{d}(x,x_{n+1})+f\left(max\{\text{d}(y_n,x_{n+1}),\text{d}(x,F(x,y))\}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max\{\text{d}(y_n,x_{n+1}),\text{d}(x,F(x,y))\}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le \text{d}(x,x_{n+1})+max\{\text{d}(y_n,x_{n+1}),\text{d}(x,F(x,y))\},\forall n\in \mathbb{N}.\hfill \end{array}$$

Thus, as $n\to \infty$, $\text{d}(x,F(x,y))\le \text{d}(A,B)$. That is $\text{d}(x,F(x,y))=\text{d}(A,B)$.

Similarly, we can prove that $\text{d}(y,F(y,x))=\text{d}(A,B)$ which concludes that (x, y) is the strong coupled proximal point of F.

Corollary 2.7

Let (X, d) be a complete metric space and A, B be two non-empty closed subsets of X such that $A\cap B=\mathrm{\varnothing }$. Let $F:X\times X\to X$ be cyclic-coupled proximal mapping of type I. Then F has strong coupled proximal point if (A, B) satisfies UC property.

Proof

Letting, $f(s,t)=ks,0<k<1$, we have$$\begin{array}{cc}\hfill \text{d}(F(x,y),F(u,v))& \le kmax[\text{d}(x,F(x,y)),\text{d}(u,F(u,v))]+(1-k)\text{d}(A,B)\hfill \\ \hfill & =k\left[max[\text{d}(x,F(x,y)),\text{d}(u,F(u,v))]-\text{d}(A,B)\right]+\text{d}(A,B)\hfill \\ \hfill & =f\left(max[\text{d}(x,F(x,y)),\text{d}(u,F(u,v))]-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(max[\text{d}(x,F(x,y)),\text{d}(u,F(u,v))]-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \end{array}$$

Now, using previous theorem, we have strong coupled proximal point of F on $A\times B$. $\square$

Theorem 2.8

Let (X, d) be a complete metric space and A, B be two non-empty, closed, disjoint subsets of X. Let $F:X\times X\to X$ be cyclic-coupled proximal mapping of type $I{I}_{f\mathit{\phi }.}$. Then F has strong coupled proximal point if (A, B) satisfies UC property and(2.9) $$\begin{array}{c}\hfill \text{d}(u,v)+\text{d}(A,B)\le \text{d}(u,F(a,b))\end{array}$$(2.9)

whenever $v=F(b,a)$ and $\{u,v\}$ belongs to set A.

Proof

Let $x_0\in A,y_0\in B$ be any two arbitrary elements of X. Define $F(x_n,y_n)=y_{n+1}$ and $F(y_n,x_n)=x_{n+1}$.

Then, for $n\in \mathbb{N}$,(2.10) $$\begin{array}{cc}\hfill \text{d}(x_n,y_{n+1})& =\text{d}(F(y_{n-1},x_{n-1}),F(x_n,y_n))\hfill \\ \hfill & \le f\left(\frac{\text{d}(y_{n-1},F(y_{n-1},x_{n-1}))+\text{d}(x_n,F(x_n,y_n))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(y_{n-1},F(y_{n-1},x_{n-1}))+\text{d}(x_n,F(x_n,y_n))}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le f\left(\frac{\text{d}(y_{n-1},x_n)+\text{d}(x_n,y_{n+1}))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(y_{n-1},x_n)+\text{d}(x_n,y_{n+1}))}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le \frac{\text{d}(y_{n-1},x_n)+\text{d}(x_n,y_{n+1}))}{2}.\hfill \end{array}$$(2.10)

Similarly,(2.11) $$\begin{array}{cc}\hfill \text{d}(y_n,x_{n+1})& =\text{d}(F(x_{n-1},y_{n-1}),F(y_n,x_n))\hfill \\ \hfill & \le f\left(\frac{\text{d}(x_{n-1},F(x_{n-1},y_{n-1}))+\text{d}(y_n,F(y_n,x_n))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(x_{n-1},F(x_{n-1},y_{n-1}))+\text{d}(y_n,F(y_n,x_n))}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le f\left(\frac{\text{d}(x_{n-1},y_n)+\text{d}(y_n,x_{n+1})}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(x_{n-1},y_n)+\text{d}(y_n,x_{n+1})}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le \frac{\text{d}(x_{n-1},y_n)+\text{d}(y_n,x_{n+1})}{2}.\hfill \end{array}$$(2.11)

From above, we have(2.12) $$\begin{array}{c}\hfill \text{d}(x_n,y_{n+1})\le \text{d}(y_{n-1},x_n)\text{and}\end{array}$$(2.12) (2.13) $$\begin{array}{c}\hfill \text{d}(y_n,x_{n+1})\le \text{d}(x_{n-1},y_n)\end{array}$$(2.13)

Hence, using (2.12) $\{\text{d}(x_{2n},y_{2n+1})\}$ and $\{\text{d}(y_{2n-1},x_{2n})\}$ are decreasing sequences and converge to some $r\ge 0$ as n$\to \infty$. Therefore, the Equation (2.10) reduces to$$\begin{array}{c}\hfill r-\text{d}(A,B)\le f\left(r-\text{d}(A,B),\mathit{\phi }\left(r-\text{d}(A,B)\right)\right).\end{array}$$

By the definition of f, $r-\text{d}(A,B)=0$ or $\mathit{\phi }(r-\text{d}(A,B))=0.$ Therefore, $r=\text{d}(A,B)$ and hence $\left\{\{d(x_{2n},y_{2n+1})\}\right.$ and $\left.\{d(y_{2n-1},x_{2n})\}\right\}$ converge to $\text{d}(A,B)$. Similarly, using (2.13)$\{\text{d}(y_{2n},x_{2n+1})\}$ and $\{\text{d}(x_{2n-1},y_{2n})\}$ are decreasing sequences and converge to some $s\ge 0$ as n$\to \infty$. Therefore, Equation (2.11) reduces to$$\begin{array}{c}\hfill s-\text{d}(A,B)\le f\left(s-\text{d}(A,B),\mathit{\phi }\left(s-\text{d}(A,B)\right)\right).\end{array}$$

By the definition of f, $s-\text{d}(A,B)=0$ or $\mathit{\phi }(s-\text{d}(A,B))=0$. Therefore, $s=\text{d}(A,B)$ and hence $\left\{\{d(y_{2n},x_{2n+1})\}\right.$ and $\left.\{d(x_{2n-1},y_{2n})\}\right\}$ converge to $\text{d}(A,B)$.

Using the above arguments, we conclude that

${lim}_{n\to \infty }\text{d}(x_n,y_{n+1})={lim}_{n\to \infty }\text{d}(y_n,x_{n+1})=\text{d}(A,B).$ Now, using UC-property, we get(2.14) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\text{d}(x_n,x_{n+1})=\underset{n\to \infty }{lim}\text{d}(y_n,y_{n+1})=0.\end{array}$$(2.14)

Claim: $\{x_n\}$ is a Cauchy sequence.

Letting $m<n$ and using (2.9), we get$$\begin{array}{cc}\hfill \text{d}(x_n,x_m)& \le \text{d}(F(y_{n-1},x_{n-1}),F(x_{m-1},y_{m-1}))-\text{d}(A,B).\hfill \\ \hfill & \le f\left(\frac{\text{d}(y_{n-1},x_n)+\text{d}(x_{m-1},y_m)}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{d(y_{n-1},x_n)+\text{d}(x_{m-1},y_m)}{2}-\text{d}(A,B)\right)\right)\hfill \\ \hfill & \le \frac{\text{d}(y_{n-1},x_n)+\text{d}(x_{m-1},y_m)}{2}-\text{d}(A,B).\hfill \end{array}$$

Hence, as ${lim}_{n\to \infty }d(x_n,x_m)=0$.

Therefore, $\{x_n\}$ is a Cauchy sequence and hence converges to some point $x\in A$.

Similairly, $\{y_n\}$ is a Cauchy sequence and hence converges to some point $y\in B$. Therefore, $\text{d}(x,y)=\text{d}(A,B)$.

Now, we observe that$$\begin{array}{cc}\hfill \text{d}(x,F(x,y))& \le \text{d}(x,F(y_n,x_n))+\text{d}(F(y_n,x_n),F(x,y))\hfill \\ \hfill & \le \text{d}(x,x_{n+1})+f\left(\frac{\text{d}(y_n,F(y_n,x_n)+\text{d}(x,F(x,y))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(y_n,F(y_n,x_n)+\text{d}(x,F(x,y))}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le \mathit{d}(x,x_{n+1})+f\left(\frac{\text{d}(y_n,x_{n+1})+\text{d}(x,F(x,y))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\phi }\left(\frac{\text{d}(y_n,x_{n+1})+\text{d}(x,F(x,y))}{2}-\text{d}(A,B)\right)\right)+\text{d}(A,B)\hfill \\ \hfill & \le \text{d}(x,x_{n+1})+\frac{\text{d}(y_n,x_{n+1})+\text{d}(x,F(x,y))}{2}+\text{d}(A,B).\hfill \end{array}$$

Now letting $n\to \infty$, the above inequality reduces to$$\begin{array}{cc}\hfill \text{d}(x,F(x,y))-\text{d}(A,B)& \le f\left(\frac{\text{d}(x,F(x,y))-\text{d}(A,B)}{2},\mathit{\phi }\left(\frac{\text{d}(x,F(x,y))-\text{d}(A,B)}{2}\right)\right)\hfill \\ \hfill & \le \text{d}(x,F(x,y))-\text{d}(A,B).\hfill \end{array}$$

So, $\frac{\text{d}(x,F(x,y))-\text{d}(A,B)}{2}=0,$ or $\mathit{\phi }(\frac{\text{d}(x,F(x,y))-\text{d}(A,B)}{2})=0.$ Therefore, $\text{d}(x,F(x,y))=\text{d}(A,B)$ and similarly $\text{d}(y,F(y,x))=\text{d}(A,B)$ with $\text{d}(x,y)=\text{d}(A,B)$ conclude that (x, y) is the strong coupled proximal point of F. $\square$

Corollary 2.9

Let (X, d) be a complete metric space and A, B be two non-empty closed subsets of X such that $A\cap B=\mathrm{\varnothing }$. Let $F:X\times X\to X$ be cyclic-coupled proximal mapping of type I. Then F has strong coupled proximal point if (A, B) satisfies UC property.

Proof

Letting, $f(s,t)=k^{\prime }s,0<k<1$, we have$$\begin{array}{cc}\hfill \text{d}(F(x,y),F(u,v))& \le k[\text{d}(x,F(x,y))+\text{d}(u,F(u,v))]+(1-2k)\text{d}(A,B)\hfill \\ \hfill & =k^{\prime }\left[\frac{\text{d}(x,F(x,y))+\text{d}(u,F(u,v))}{2}-\text{d}(A,B)\right]+\text{d}(A,B)\hfill \\ \hfill & =\left[f\left(\frac{\text{d}(x,F(x,y))+\text{d}(u,F(u,v))]}{2}-\text{d}(A,B),\right.\right.\hfill \\ \hfill & \left.\left.\mathit{\phi }\left(\frac{\text{d}(x,F(x,y))+\text{d}(u,F(u,v))]}{2}+\text{d}(A,B)\right)\right)\right]+\text{d}(A,B).\hfill \end{array}$$

Now, using previous theorem, we have strong coupled proximal point of F on $A\times B$.

Example 2.10

Consider $A=\{[-2.5,a]$ where $a\in [-1,0]\}$ and $B=\{[2.5,b]$ where $b\in [-1,0]\}$ on ${\mathbf{R}}^2$ under $1-$norm with $d(A,B)=5$. Also the sets satisfies UC-property.

Define $f(s,t)=\frac{2}{3}s$ and$$\begin{array}{c}\hfill F(a^{\prime },b^{\prime })=\left\{\begin{array}{c}\left(2.5,\frac{ab}{3}\right)\text{if}(a^{\prime },b^{\prime })\in A\times B,\text{where}{a}^{\prime }=(-2.5,a)\text{and}{b}^{\prime }=(2.5,b).\hfill \\ \left(-2.5,\frac{ab}{3}\right)\text{if}(a^{\prime },b^{\prime })\in B\times A,\text{where}{a}^{\prime }=(2.5,a)\text{and}{b}^{\prime }=(-2.5,b).\hfill \end{array}\right.\end{array}$$

Let $x^{\prime }=(-2.5,x)$, $v^{\prime }=(-2.5,v)$ be elements of $A_0$ and $y^{\prime }=(2.5,y)$, $u^{\prime }=(2.5,u)$ be elements of $B_0$.

Now, we compute$$\begin{array}{cc}\hfill \text{d}(F(x^{\prime },y^{\prime }),F(u^{\prime },v^{\prime }))& =\text{d}\left(\left(2.5,\frac{xy}{3}\right),\left(-2.5,\frac{uv}{3}\right)\right)\hfill \\ \hfill & \le \text{d}\left(\left(2.5,\frac{xy}{3}\right),\left(2.5,\frac{uv}{3}\right)\right)+\text{d}(A,B)\hfill \\ \hfill & =\left|\frac{xy}{3}-\frac{uv}{3}\right.∥+\text{d}(A,B)\hfill \\ \hfill & \le \frac{1}{3}[|x|+|u|]+\text{d}(A,B)\hfill \\ \hfill & \le \frac{1}{3}\left[\left|x-\frac{xy}{3}\right|+\left|u-\frac{uv}{3}\right|\right]+\text{d}(A,B)\hfill \\ \hfill & \le \frac{1}{3}\left[\left|x-\frac{xy}{3}\right|+|5|-|5|+\left|u-\frac{uv}{3}\right|+|5|-|5|\right]+\text{d}(A,B)\hfill \\ \hfill & \le \frac{1}{3}\left[\left[\left|x-\frac{xy}{3}\right|\right]+|5|+\left[\left|u-\frac{uv}{3}\right|\right]+|5|\right]-2\left(\frac{5}{3}\right)+\text{d}(A,B)\hfill \\ \hfill & \le \frac{1}{3}\left[\left[\left|x-\frac{xy}{3}\right|\right]+|5|+\left[\left|u-\frac{uv}{3}\right|+|5|\right]\right]-\frac{2}{3}\text{d}(A,B)+\text{d}(A,B)\hfill \\ \hfill & \le \frac{2}{3}\left[\frac{\text{d}(x^{\prime },F(x^{\prime },y,))+\text{d}(u^{\prime },F(u^{\prime },v^{\prime }))}{2}-\text{d}(A,B)\right]+\text{d}(A,B)\hfill \\ \hfill & =f\left[\frac{\text{d}(x^{\prime },F(x^{\prime },y,))+\text{d}(u^{\prime },F(u^{\prime },v^{\prime }))}{2}-\text{d}(A,B),\right.\hfill \\ \hfill & \left.\mathit{\varphi }\left(\frac{\text{d}(x^{\prime },F(x^{\prime },y,))+\text{d}(u^{\prime },F(u^{\prime },v^{\prime }))}{2}-\text{d}(A,B)\right)\right]+\text{d}(A,B).\hfill \end{array}$$

Therefore, the problem satisfies all conditions of Corollary (3.2) and $((-2.5,0),(2.5,0))\in {\mathbf{R}}^2\times {\mathbf{R}}^2$ is the coupled proximity pair of F.

Additional information

Funding

This project was supported by the Theoretical and Computational Science (TaCS) Center under Computational and Applied Science for Smart Innovation Research Cluster (CLASSIC), Faculty of Science, KMUTT [grant number 2559]. Also, this article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah.

Notes on contributors

Arslan Hojat Ansari

Arslan Hojat Ansari is a research scholar and PhD student at the Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran. His area of research includes fixed point theory and special functions.

Geno Kadwin Jacob

Geno Kadwin Jacob is a research scholar at the Department of Mathematics, Bharathidasan University, India. His area of research includes metric fixed point theory and linear complementarity problem.

Muthiah Marudai

Dr Muthiah Marudai is a professor and the chair at the Department of Mathematics, Bharathidasan University, India. His area of research includes metric fixed point theory and fuzzy analysis. He had published many research articles in international journals.

Poom Kumam

Dr Poom Kumam is the head of Theoretical and Computational Science (TaCS) Center and KMUTT-Fixed Point Theory and Applications Research Group. His area of research is fixed point theory with applications. He had published more than 350 research articles in international journals around the world.