A General Fixed Point Theorem for Mappings Satisfying a Cyclical Contractive Condition in S-Metric Spaces

Abstract

In this paper we introduce a new type of implicit relation in S-metric spaces. Our aim is to prove a general fixed point theorem for mappings satisfying the cyclical contractive condition, which extends several results from the literature.

1. Introduction

Banach’s contraction principle [1] was extended to a case of cyclical mappings by Kirk et al. [2].

Many fundamental metrical fixed point theorems are extended to cyclic contractive mappings in [3], and other results are presented in [4,5,6,7,8,9] and in other papers.

A new class of generalized metric space, named D-metric spaces, is introduced in [10,11], by Dhage.

It has been shown, by Mustafa and Sims [12,13], that most of the statements about fundamental topological structures on D-metric spaces are incorrect. Thus, the authors introduced the appropriate notion of generalized metrics space, called G-metric space. In fact, Mustafa, Sims and other authors obtained numerous fixed point results for self mappings in G-metric spaces. Fixed points for cyclical contractive mappings in G-metric spaces are stated in [14,15,16] and in other papers.

Recently, in [17], the authors introduced a new type of generalized metric spaces, named S-metric spaces, as a “generalization” of G-metric spaces.

In [18], the authors proved that the notion of S-metric space is not a generalization of G-metric space or vice versa. Hence, the notions of G-metric space and S-metric space are independent.

The study of cyclical contractive mapping in S-metric spaces is introduced in [19].

By considering a general condition given by an implicit relation, various classical (common) fixed point theorems in metric spaces have been unified into [20,21]. In [22,23] begins the study of fixed points for mappings that satisfy an implicit relation in G-metric spaces.

The study of cyclical implicit contractive conditions is initiated in [24] and in G-metric spaces in [25].

The study of fixed points for mappings satisfying implicit relation in S-metric spaces is initiated in [26,27].

The purpose of this paper is to prove, using a new type of implicit relation, one general fixed point theorem for mappings which satisfy a cyclical contractive condition, thus extending some results from [26,28,29,30,31].

2. Preliminaries

Definition 1

([17,26]). Let X be a nonempty set. An S-metric on X is a function $S:X^3\mapsto {\mathbb{R}}_{+}$ such that:

$\left(S_1\right)$:

$S\left(x,y,z\right)=0$ if and only if $x=y=z$;

$\left(S_2\right)$:

$S\left(x,y,z\right)\le S\left(x,x,a\right)+S\left(y,y,a\right)+S\left(z,z,a\right)$ for all $x,y,z,a\in X$.

The pair $\left(X,S\right)$ is called an S-metric space.

Example 1.

Let $X=\mathbb{R}$ and $S\left(x,y,z\right)=\left|x-z\right|+\left|y-z\right|$. Then, $S\left(x,y,z\right)$ is an S-metric on $\mathbb{R}$, which is named the usual S - metric on X.

Lemma 1

( [17]). If S is an S-metric on a nonempty set X, then

$$S\left(x,x,y\right) = S\left(y,y,x\right)\mathrm{for}\mathrm{all}x,y\in X.$$

Definition 2.

Let $\left(X,S\right)$ be a S-metric space. For $r>0$ and $x\in X$ we define the open ball with center in x and radius r, the set

$$B_S\left(x,r\right) = \left\{y\in X:S\left(x,x,y\right)<r\right\}.$$

The topology induced by the S-metric is the topology determined by the base of all open balls in X.

Definition 3.

(a) 

A sequence $\left\{x_n\right\}$ in $\left(X,S\right)$ is convergent to x, denoted ${lim}_{n\to \infty }{x}_n=x$ or $x_n\to x$, if $S\left(x_n,x_n,x\right)\to 0$ as $n\to \infty$.

(b) 

A sequence $\left\{x_n\right\}$ in $\left(X,S\right)$ is a Cauchy sequence if $S\left(x_n,x_n,x_m\right)\to 0$ as $n,m\to \infty$.

(c) 

$\left(X,S\right)$ is complete if every Cauchy sequence is convergent.

Example 2.

$\left(X,S\right)$ from Example 1 is complete.

Lemma 2

([17,26]). Let $\left(X,S\right)$ be a S - metric space. If $x_n\to x$ and $y_n\to y$, then $S\left(x_n,x_n,y_n\right)\to S\left(x,x,y\right)$.

Lemma 3

([17,26]). Let $\left(X,S\right)$ be a S - metric space and $x_n\to x$. Then ${lim}_{n\to \infty }{x}_n$ is unique.

Lemma 4

([19]). Let $\left(X,S\right)$ be a S-metric space and A is a nonempty subset of X. If A is S-closed, then for all convergent sequence $x_n$ to x, $x\in A$.

Definition 4

([2]). Let $\left(X,d\right)$ be a metric space. Let $p>1$, $p\in \mathbb{N}$ be, T a self mapping on X and ${\left\{A_i\right\}}_{i=1}^p$ nonempty closed subsets of X. The mapping T is said to be cyclical if

$$T\left(A_i\right)\subset A_{i+1},i=1,2,\dots ,p,where{A}_{p+1}=A_1.$$

(1)

The following theorems are proved in [3,29].

Theorem 1

([3]). Let ${\left\{A_i\right\}}_{i=1}^p$ be nonempty closed subsets of a complete metric space $\left(X,d\right)$ and $f:{\displaystyle \bigcup _{i=1}^p}{A}_i\to {\displaystyle \bigcup _{i=1}^p}{A}_i$ satisfying (1) such that

$$d\left(fx,fy\right) \le \alpha max\left\{d\left(x,y\right),d\left(x,fx\right),d\left(y,fy\right),d\left(y,fx\right),d\left(x,fy\right)\right\}$$

(2)

for all $x\in A_i,y\in A_{i+1},i=1,2,\dots ,p$ and some $\alpha \in \left(0,{\displaystyle \frac{1}{2}}\right)$. Then f has a unique fixed point in ${\displaystyle \bigcap _{i=1}^p}{A}_i$.

Theorem 2

([29]). Let $\left(X,d\right)$ be a complete metric space, $p\in \mathbb{N}$, ${\left\{A_i\right\}}_{i=1}^p$ a finite family of nonempty closed subsets of X and $T:{\displaystyle \bigcup _{i=1}^p}{A}_i\to {\displaystyle \bigcup _{i=1}^p}{A}_i$ satisfying (1) and

$$d\left(Tx,Ty\right) \le kmax\left\{d\left(x,y\right),d\left(x,Tx\right),d\left(y,Ty\right),{\displaystyle \frac{d\left(x,Ty\right)+d\left(y,Tx\right)}{2}}\right\},$$

(3)

where $k\in \left(0,1\right)$, for all $x\in A_i,y\in A_{i+1},i=1,2,\dots ,p$. Then T has a unique fixed point in ${\displaystyle \bigcap _{i=1}^p}{A}_i$.

3. $\mathbf{S}$-Implicit Relations

Let ${\mathcal{F}}_S$ be the set of all lower semi-continuous functions $F:{\mathbb{R}}_{+}^6\mapsto \mathbb{R}$ satisfying the following conditions:

$\left(F_1\right)$:

F is nondecreasing in $t_6$;

$\left(F_2\right)$:

there exists $h\in \left[0,1\right)$ such that for all $u,v\ge 0$, $F\left(u,v,v,u,0,2u+v\right) \le 0$ implies $u\le hv$;

$\left(F_3\right)$:

$F\left(t,t,0,0,t,t\right)>0,\forall t>0$.

Example 3.

$F\left(t_1,\dots ,t_6\right)=t_1-a{t}_2-bmax\left\{t_3,t_4,t_5,t_6\right\}$, where $a,b\ge 0$ and $a+3b<1$.

$\left(F_1\right)$:

obvious.

$\left(F_2\right)$:

Let $u,v\ge 0$ and $F\left(u,v,v,u,0,2u+v\right)=u-av-bmax\left\{u,v,2u+v\right\}\le 0$. If $u>v$, then $u\left[1-\left(a+3b\right)\right]\le 0$, a contradiction. Hence, $u\le v$, which implies $u\le hv$, where $0\le h=a+3b<1$.

$\left(F_3\right)$:

$F\left(t,t,0,0,t,t\right)=t\left[1-\left(a+3b\right)\right]>0,\forall t>0$.

In the following examples, since the proofs are similar, we will omit them.

Example 4.

$F\left(t_1,\dots ,t_6\right) = t_1-a{t}_2-b{t}_3-c{t}_4-dmax\left\{t_5,t_6\right\}$, where $a,b,c,d\ge 0$ and $a+b+c+3d<1$.

Example 5.

$F\left(t_1,\dots ,t_6\right) = t_1-a{t}_2-dmax\left\{t_3,t_4\right\}-b{t}_5-c{t}_6$, where $a,b,c,d\ge 0$, $a+3c+d<1$ and $a+b+c<1$.

Example 6.

$F\left(t_1,\dots ,t_6\right) = t_1-a{t}_2-b{t}_3-e{t}_4-c{t}_5-d{t}_6-fmax\left\{t_2,t_3,t_4,t_5,t_6\right\}$, where $a,b,c,d,e,f\ge 0$, $a+b+e+3d+3f<1$ and $a+d+c+f<1$.

Example 7.

$F\left(t_1,\dots ,t_6\right) = t_1-a\left(t_5+t_6\right)-b{t}_2-cmax\left\{t_3,t_4\right\}$, where $a,b,c\ge 0$ and $3a+b+c<1$.

Example 8.

$F\left(t_1,\dots ,t_6\right) = t_1-a\left(t_3+t_4\right)-b{t}_2-cmax\left\{t_5,t_6\right\}$, where $a,b,c\ge 0$ and $2a+b+3c<1$.

Example 9.

$F\left(t_1,\dots ,t_6\right) = t_1-amax\left\{t_4+t_5,t_3+t_6\right\}-b{t}_2$, where $a,b\ge 0$ and $4a+b<1$.

Example 10.

$F\left(t_1,\dots ,t_6\right) = t_1^2-t_1\left(a{t}_2+b{t}_3+c{t}_4\right)-d{t}_5{t}_6$, where $a,b,c\ge 0$, $a+b+c<1$ and $a+d<1$.

Example 11.

$F\left(t_1,\dots ,t_6\right) = t_1^2-a{t}_1{t}_2-b{t}_3{t}_4-c{t}_5{t}_6$, where $a,b,c\ge 0,a+b<1$ and $a+c<1$.

Example 12.

$F\left(t_1,\dots ,t_6\right) = t_1^3-a{t}_1{t}_2{t}_3-b{t}_2{t}_3{t}_4-c{t}_3{t}_4{t}_5-d{t}_4{t}_5{t}_6$, where $a,b,c\ge 0$and$a+b<1$.

Example 13.

$F\left(t_1,\dots ,t_6\right) = t_1-a{t}_2-b{t}_3-c{t}_4-d{t}_5-e{t}_6$, where $a,b,c,d,e\ge 0$, $a+b+c+3e<1$ and $a+d+e<1$.

Example 14.

$F\left(t_1,\dots ,t_6\right) = t_1-kmax\left\{t_2,t_3,t_4,t_5,t_6\right\}$, where $k\in \left[0,{\displaystyle \frac{1}{3}}\right)$.

Example 15.

$F\left(t_1,\dots ,t_6\right) = t_1-kmax\left\{t_2,t_3,t_4,{\displaystyle \frac{t_5+t_6}{3}}\right\}$, where $k \in \left[0,1\right)$.

Remark 1.

Examples 10–12 are not of the type of the examples from [18].

4. Main Result

Theorem 3.

Let $\left(X,S\right)$ be a complete S-metric space and ${\left\{A_i\right\}}_{i=1}^p$ be a family of nonempty closed subsets of X. Let $Y={\displaystyle \bigcup _{i=1}^p}{A}_i$ and let $T:Y\mapsto Y$ satisfying

$$T\left(A_i\right)\subset A_{i+1},i=1,2,\dots ,p,$$

(4)

where $A_{p+1}=A_1$.

If the inequality

$$F\left(\begin{array}{c}S\left(Tx,Ty,Ty\right),S\left(x,y,y\right),S\left(x,Tx,Tx\right),\\ S\left(y,Ty,Ty\right),S\left(y,Tx,Tx\right),S\left(x,Ty,Ty\right)\end{array}\right)\le 0$$

(5)

holds for all $x\in A_i$,$y\in A_{i+1},i=1,2,\dots ,p$ and $F\in {\mathcal{F}}_S$, then T has a unique fixed point in ${\displaystyle \bigcap _{i=1}^p}{A}_i$.

Proof. 

Let $x_0$ be an arbitrary point of $A_1$. We define $x_n=T{x}_{n-1},n=1,2,\dots$. By (5) for $x_0\in A_1$ and $x_1\in A_2$, $x_{p-1}=T{x}_{p-2}\in A_p$, $x_p=T{x}_{p-1}\in A_{p+1}=A_1,x_{p+1}=T{x}_p\in A_2$, we have

$$F\left(\begin{array}{c}S\left(T{x}_0,T{x}_1,T{x}_1\right),S\left(x_0,x_1,x_1\right),S\left(x_0,T{x}_0,T{x}_0\right),\\ S\left(x_1,T{x}_1,T{x}_1\right),S\left(x_1,T{x}_0,T{x}_0\right),S\left(x_0,T{x}_1,T{x}_1\right)\end{array}\right)\le 0,$$

$$F\left(\begin{array}{c}S\left(x_1,x_2,x_2\right),S\left(x_0,x_1,x_1\right),S\left(x_0,x_1,x_1\right),\\ S\left(x_1,x_2,x_2\right),0,S\left(x_0,x_2,x_2\right)\end{array}\right)\le 0.$$

(6)

By $\left(S_2\right)$ and Lemma 1 we have

$$\begin{array}{ccc}\hfill S\left(x_0,x_2,x_2\right)& =& S\left(x_2,x_2,x_0\right)\le 2S\left(x_2,x_2,x_1\right)+S\left(x_0,x_0,x_1\right)\hfill \\ & =& 2S\left(x_1,x_2,x_2\right)+S\left(x_0,x_1,x_1\right).\hfill \end{array}$$

By (6) and $\left(F_1\right)$ we obtain

$$F\left(\begin{array}{c}S\left(x_1,x_2,x_2\right),S\left(x_0,x_1,x_1\right),S\left(x_0,x_1,x_1\right),\\ S\left(x_1,x_2,x_2\right),0,2S\left(x_1,x_2,x_2\right)+S\left(x_0,x_1,x_1\right)\end{array}\right)\le 0.$$

By $\left(F_2\right)$ we obtain

$$S\left(x_1,x_2,x_2\right)\le hS\left(x_0,x_1,x_1\right).$$

Similarly, we obtain

$$S\left(x_{p-2},x_{p-1},x_{p-1}\right)\le hS\left(x_{p-3},x_{p-2},x_{p-2}\right).$$

Again, we get

$$S\left(x_{p-1},x_p,x_p\right)\le hS\left(x_{p-2},x_{p-1},x_{p-1}\right).$$

Hence, we have

$$S\left(x_n,x_{n+1},x_{n+1}\right) \le hS\left(x_{n-1},x_n,x_n\right) \le \dots \le h^{n}S\left(x_0,x_1,x_1\right).$$

Using $\left(S_2\right)$ it follows that $\left\{x_n\right\}$ is a Cauchy sequence in X. Since $\left(X,S\right)$ is complete, it follows that $x_n$ is convergent to a point z. Then, the sequence

$$\left\{x_{np+1}\in A_1,x_{np+2}\in A_2,...,x_{np+p}\in A_p,n=0,1,2,...\right\}$$

also converges to z and $z\in {\displaystyle \bigcap _{i=1}^p}{A}_i$, because by Lemma 4, $z\in A_i,i=1,2,\dots ,p$.

We have to prove that z is a fixed point of T.

For $x=x_n$ and $y=z$, by (5) we obtain

$$F\left(\begin{array}{c}S\left(T{x}_n,Tz,Tz\right),S\left(x_n,z,z\right),S\left(x_n,T{x}_n,T{x}_n\right),\\ S\left(z,Tz,Tz\right),S\left(z,T{x}_n,T{x}_n\right),S\left(x_n,Tz,Tz\right)\end{array}\right)\le 0,$$

$$F\left(\begin{array}{c}S\left(x_{n+1},Tz,Tz\right),S\left(x_n,z,z\right),S\left(x_n,x_{n+1},x_{n+1}\right),\\ S\left(z,Tz,Tz\right),S\left(z,x_{n+1},x_{n+1}\right),S\left(x_n,Tz,Tz\right)\end{array}\right)\le 0.$$

Letting n tend to infinity, by Lemma 2 we obtain

$$F\left(S\left(z,Tz,Tz\right),0,0,S\left(z,Tz,Tz\right),0,S\left(z,Tz,Tz\right)\right)\le 0.$$

By $\left(F_1\right)$ we get

$$F\left(S\left(z,z,Tz\right),0,0,S\left(z,z,Tz\right),0,2S\left(z,Tz,Tz\right)\right)\le 0.$$

By $\left(F_2\right)$ we obtain $S\left(z,Tz,Tz\right) = 0$ and by $\left(S_1\right)$ it follows that $z=Tz$. Hence, z is a fixed point and $z\in {\displaystyle \bigcap _{i=1}^p}{A}_i$.

Suppose that there exists another fixed point $z^{\prime }\in {\displaystyle \bigcap _{i=1}^p}{A}_i$. By (5) for $x=z$ and $y=z^{\prime }$ we obtain

$$F\left(\begin{array}{c}S\left(Tz,T{z}^{\prime },T{z}^{\prime }\right),S\left(z,z^{\prime },z^{\prime }\right),S\left(z,Tz,Tz\right),\\ S\left(z^{\prime },T{z}^{\prime },T{z}^{\prime }\right),S\left(z^{\prime },Tz,Tz\right),S\left(z,T{z}^{\prime },T{z}^{\prime }\right)\end{array}\right)\le 0,$$

$$F\left(S\left(z,z^{\prime },z^{\prime }\right),S\left(z,z^{\prime },z^{\prime }\right),0,0,S\left(z^{\prime },z,z\right),S\left(z,z^{\prime },z^{\prime }\right)\right) \le 0.$$

By Lemma 1, $S\left(z^{\prime },z,z\right) = S\left(z,z^{\prime },z^{\prime }\right)$. Hence,

$$F\left(S\left(z,z^{\prime },z^{\prime }\right),S\left(z,z^{\prime },z^{\prime }\right),0,0,S\left(z,z^{\prime },z^{\prime }\right),S\left(z,z^{\prime },z^{\prime }\right)\right) \le 0,$$

a contradiction of $\left(F_3\right)$ if $S\left(z,z^{\prime },z^{\prime }\right)>0$. Hence, $S\left(z,z^{\prime },z^{\prime }\right) = 0$ and by $\left(S_1\right)$, $z=z^{\prime }$. Hence, z is the unique fixed point of T and $z\in {\displaystyle \bigcap _{i=1}^p}{A}_i$. □

5. Final Remarks

In this paper, using a new type of implicit relation, we have proved a general fixed point theorem for mappings that satisfy a cyclical contractive condition. Our result extends certain results from [26,28,29,30,31]:

(i)

Theorem 3 and Example 3 extend Corollary 2.21 [26] to cyclical form;

(ii)

Theorem 3 and Examples 3–9 extend Theorem 3.1 [28] to cyclical form;

(iii)

Theorem 3 and Example 15 extend Theorem 2 to cyclical form in S-metric spaces;

(iv)

Theorem 3 and Example 13 extend Corollary 2.19 [26], Theorems 2.3 and 2.4 [30], Theorems 3.2–3.4 [31] to cyclical form.