International Journal of Analysis and Applications Volume 19, Number 5 (2021), 794-811 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-794 COMMON FIXED POINT THEOREMS FOR SIX SELF-MAPPINGS ON S− METRIC SPACES THANGJAM BIMOL SINGH1,∗, G. A. HIRANKUMAR SHARMA2, Y. MAHENDRA SINGH2 AND M. RANJIT SINGH3 1Department of Mathematics, Jadonang Memorial College, Longmai(Noney)-795159, Manipur, India 2Department of Basic Sciences and Humanities, Manipur Institute of Technology (A Constituent College of Manipur University), Takyelpat -795004, Manipur, India 3Department of Mathematics, Manipur University, Canchipur-795003, Manipur, India ∗Corresponding author: btsalun29@gmail.com Abstract. In this paper, we introduce the concepts of common property −(E.A) and common limit range property for six self-mappings and prove common fixed point theorems of such mappings satisfying (ψ,ϕ)− weak contraction on an S−metric space. Examples are given to illustrate our results. 1. Introduction and Preliminaries In 2006, Mustafa and Sims [21] introduced G− metric space to overcome fundamental flaws in B. C. Dhage’s theory of generalized metric spaces ( [10–12]) and discussed the topological properties of G− metric spaces. In 2012, Sedghi et al. [26] introduced the concept of S− metric space as a modification of D∗− metric space [27] and G− metric space [21]. But, in 2014, Dung et al. [14] showed by giving examples that the class of S− metric spaces and the class of G− metric spaces are distinct. Received July 5th, 2021; accepted August 23rd, 2021; published September 20th, 2021. 2010 Mathematics Subject Classification. 47H10, 54H25. Key words and phrases. fixed point; coincidence point; (A,B,C)(ψ,ϕ)− weak contraction; property −(E.A); common limit range property. ©2021 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 794 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-794 Int. J. Anal. Appl. 19 (5) (2021) 795 Before going to our main work, let us recall some basic definitions, lemmas, and preliminaries that will be used in this paper. Definition 1.1. [26] Let X be a non-empty set. A function S : X ×X ×X → [0,∞) is said to be an S− metric on X if it satisfies the following properties: (S1) S(x,y,z) = 0 if and only if x = y = z; (S2) S(x,y,z) ≤ S(x,x,a) + S(y,y,a) + S(z,z,a), for all x,y,z,a ∈ X. The pair (X,S) is called an S− metric space. Example 1.1. [26] Let X = Rn and ‖ · ‖ be a norm on X. Define S(x,y,z) = ‖2x−y −z‖ + ‖y −z‖, for all x,y,z ∈ X. Then (X,S) is an S− metric space. Example 1.2. [26] Let X = R. Define S(x,y,z) = |x−z| + |y −z|, for all x,y,z ∈ X. Then (X,S) is an S− metric space. Definition 1.2. [26] Let (X,S) be an S− metric space. (i) A sequence {xn} in X is called a Cauchy sequence if and only if S(xn,xn,xm) → 0 as n,m →∞. (ii) A sequence {xn} in X converges to x ∈ X if and only if S(xn,xn,x) → 0 as n → ∞. In this case, we write lim n→∞ xn = x. (iii) The S− metric space (X,S) is said to be complete if every Cauchy sequence in it is convergent. Lemma 1.1. [26] In an S− metric space, we have S(x,x,y) = S(y,y,x). Lemma 1.2. [26] Let (X,S) be an S− metric space. If sequence {xn} in X converges to x, then x is unique. Lemma 1.3. [26] Let (X,S) be an S− metric space. If sequence {xn} in X converges to x, then {xn} is a Cauchy sequence. Lemma 1.4. [26] Let (X,S) be an S− metric space. If there exist sequences {xn} and {yn} such that lim n→∞ xn = x and lim n→∞ yn = y, then lim n→∞ S(xn,xn,yn) = S(x,x,y). Definition 1.3. [3] Let X 6= ∅ and P,Q : X → X be two self-mappings. If u = Px = Qx, for some x ∈ X, then x is called a coincidence point of P and Q, and u is called a point of coincidence (briefly, poc) of P and Q. Lemma 1.5. [3] Suppose that P and Q be weakly compatible self-mappings on a non-empty set X. If P and Q have a unique point of coincidence u = Px = Qx, then u is the unique common fixed point P and Q. Int. J. Anal. Appl. 19 (5) (2021) 796 In 1997, Alber and Guere-Delabriere [5] introduced the concept of weak contraction, wherein the authors introduced the following notion for mappings defined on a Hilbert space X. Consider the following set of real functions Φ = { ϕ : [0,∞) → [0,∞) : ϕ is a lower semi-continuous and ϕ(t) = 0 if and only if t = 0 } . A mapping T : X → X is called a ϕ− weak contraction if there exists a function ϕ ∈ Φ such that d ( T x,T y ) ≤ d(x,y) −ϕ ( d(x,y) ) , for all x,y ∈ X. Dutta and Choudhury [15] proved a fixed point theorem for a self-mapping satisfying (ψ,ϕ)−weak contractive condition as follows. Theorem 1.1. Let (X,d) be a complete metric space and T : X → X be a self-mapping satisfying ψ ( d(T x,T y) ) ≤ ψ ( d(x,y) ) −ϕ ( d(x,y) ) , for some ϕ ∈ Φ and ψ ∈ Ψ = { ψ : [0,∞) → [0,∞) : ψ is continuous non-decreasing and ψ(0) = 0 } . Then, T has a common fixed point in X. Many researchers utilized (ψ,ϕ)− weak contractive conditions to prove a number of metrical fixed point theorems (e.g., [2, 4–9, 13], [20], [30]). Recently, Singh and Bimol Singh [29] proved some coincidence and common fixed point theorems involving ψ ∈ Ψ and ϕ ∈ Φ in S− metric spaces. Definition 1.4. [28] A pair (A,B) of self-mappings of an S− metric space (X,S) is said to be compatible if lim n→∞ S(ABxn,ABxn,BAxn) = 0, whenever {xn} is a sequence in X such that lim n→∞ Axn = lim n→∞ Bxn = t, for some t ∈ X. In 1998, Jungck and Rhoades [18] introduced the following concept of weakly compatibility. Definition 1.5. A pair (A,B) of self-mappings of an S− metric space (X,S) is said to be weakly compatible if they commute at each coincidence point (i.e., ABx = BAx, x ∈ X whenever Ax = Bx). In 2002, Aamri and Moutawakil [1] introduced the concept of property −(E.A) in metric spaces. In the same line, we use this concept in S− metric space as follows. Definition 1.6. A pair (A,P) of self-mappings of an S− metric space (X,S) is said to satisfy the property −(E.A) if there exists a sequence {xn} in X such that lim n→∞ Axn = lim n→∞ Pxn = t, for some t ∈ X. Int. J. Anal. Appl. 19 (5) (2021) 797 Any pair of compatible as well as non-compatible self-mappings of an S− metric space (X,S) satisfy the property −(E.A), but a pair of mappings satisfying the property −(E.A) need not be non-compatible (see Example 1 of [16]). In 2005, Liu et al. [19] introduced the notion of common property −(E.A) for hybrid pairs of mappings, which contain the property −(E.A). For more details on various type of compatible mappings and their relation, one may refer to ( [8], [22–25], [31], [32]) and references therein. Definition 1.7. Two pairs (A,P) and (B,Q) of self-mappings of an S− metric space (X,S) are said to satisfy the common property −(E.A) if there exist two sequences {xn} and {yn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = t, for some t ∈ X. In a similar way, we define the notion of common property −(E.A) for six self-mappings on S−metric space. Definition 1.8. Three pairs (A,P), (B,Q) and (C,R) of self-mappings of an S− metric space (X,S) are said to satisfy the common property −(E.A) if there exist three sequences {xn}, {yn} and {zn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = t, for some t ∈ X. It can be observed that the fixed point results usually require closeness of the underlying subspaces for the existence of common fixed points under the property −(E.A) and common property −(E.A). In 2011, Sintunavarat and Kumam [33] coined the idea of ‘common limit range property’. In 2012, Imdad et al. [17] extended the notion of common limit range property to two pairs of self-mappings of a metric space which relax the closeness requirements of the underlying subspaces. Definition 1.9. A pair (A,P) of self-mappings of an S− metric space (X,S) is said to satisfy the common limit range property with respect to P, (briefly, (CLRP)− property), if there exists a sequence {xn} in X such that lim n→∞ Axn = lim n→∞ Pxn = t, where t ∈PX. Thus, one can infer that a pair (A,P) satisfying the property −(E.A) along with the closeness of the subspace PX always enjoys the (CLRP)− property with respect to the mapping P (see Examples 2.16–2.17 of [17]). Definition 1.10. Two pairs (A,P) and (B,Q) of self-mappings of an S− metric space (X,S) are said to satisfy the common limit range property (briefly, (CLRPQ)− property) with respect to mappings P and Q, if there exist two sequences {xn} and {yn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = t, where t ∈PX ∩QX. Int. J. Anal. Appl. 19 (5) (2021) 798 Example 1.3. [20] Let X = [0, 12) endow with S− metric S(x,y,z) = |x−z|+|y−z|. Define self-mappings A,B,P,Q : X → X by Ax =   6, 0 ≤ x ≤ 6 9, 6 < x < 12 ; Bx =   0, 0 ≤ x < 6 6, 6 ≤ x < 12 ; Px =   6, 0 ≤ x ≤ 6 3, 6 < x < 12 ; Qx =   4, 0 ≤ x < 6 12 −x, 6 ≤ x < 12. Consider two sequences {xn} and {yn} of X such that xn = 1n and yn = 6+ 1 n , n ∈ N. Note that PX = {3, 6} and QX = (0, 6]. Also, we have lim n→∞ Axn = lim n→∞ Pxn = 6 ∈ X and lim n→∞ Byn = lim n→∞ Qyn = 6 ∈QX. It follows that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = t, where t = 6 ∈PX ∩QX. Therefore the pairs (A,P) and (B,Q) satisfy (CLRPQ)− property. In a similar mode, we give the concept of the common limit range property for six self-mappings as follows. Definition 1.11. Three pairs (A,P), (B,Q) and (C,R) of self-mappings of an S−metric space (X,S) are said to satisfy the common limit range property with respect to mappings P, Q and R (briefly, (CLRPQR)− property), if there exist three sequences {xn}, {yn} and {zn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = t, where t ∈PX ∩QX ∩RX, for some t ∈ X. Example 1.4. Let X = [0, 5]. Define a mapping S : X3 → [0,∞) by S(x,y,z) = |x−y|+|y −z| , ∀x,y,z ∈ X. Clearly, (X,S) is an S−metric space. Let A,B,C,P,Q,R : X → X be six self-mappings defined by Ax =   1, if x = [0, 1] 2, if x ∈ (1, 5] ; Bx =   0, if x = [0, 1) 1, if x ∈ [1, 5] ; Cx =   1, if x = [0, 1] 5, if x ∈ (1, 5] ; Px =   1, if x = [0, 1] 3, if x ∈ (1, 5] ; Qx =   1 2 , if x = [0, 1) 1, if x ∈ [1, 5] ; Rx =   1, if x = [0, 1] 4, if x ∈ (1, 5]. Consider the three sequences {xn} = { 1 n } , {yn} = { 1 + 1 2n } , {zn} = { 1 − 1 n } ,∀n ∈ N. Now, we have lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = 1 ∈ PX ∩QX ∩RX. The pairs (A,P), (B,Q) and (C,R) satisfy the (CLRPQR)−property. Int. J. Anal. Appl. 19 (5) (2021) 799 Definition 1.12. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be six self-mappings. Then the mappings A,B,C,P,Q and R are called an (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R) if there exist two functions ψ ∈ Ψ and ϕ ∈ Φ such that ψ ( M(x,y,z) ) ≤ ψ ( ∆(x,y,z) ) −ϕ ( ∆(x,y,z) ) , (1.1) for all x,y,z ∈ X, where M(x,y,z) = max { S(Ax,Ax,By),S(By,By,Cz) } and ∆(x,y,z) = max { S(Px,Px,Qy),S(Ax,Ax,Rz),S(Px,Px,By),S(Qy,Qy,Cz) } . In the present paper, we discuss some common fixed point theorems for three pairs of self-mappings employing the common property −(E.A) and common limit range property in S−metric spaces. 2. Main results Before we start to prove our main theorems, we discuss the following lemmas. Lemma 2.1. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be an (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R) satisfying the following conditions: (i) BX ⊂RX (resp. AX ⊂RX); (ii) the pairs (A,P) and (B,Q) satisfy the common property −(E.A). Then the pairs (A,P), (B,Q) and (C,R) share the common property −(E.A). Proof. Suppose the pair (A,P) and (B,Q) satisfy the common property −(E.A), then there exist two sequences {xn} and {yn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = t, for some t ∈ X. Since BX ⊂RX and lim n→∞ Byn = t, then there exist n0 ∈ N∪{0} and a sequence {zn} in RX such that Byn = Rzn, for all n ≥ n0. Therefore lim n→∞ Byn = lim n→∞ Rzn = t. Now we claim that lim n→∞ Czn = t. On contrary, we suppose that lim n→∞ Czn 6= t, then there exists ε > 0 and k ≥ n0 for all k ∈ N∪{0} such that lim k→∞ S(t,t,Cznk ) = ε. For this, from (1.1), we obtain ψ ( M(xnk,ynk,znk ) ) ≤ ψ ( ∆(xnk,ynk,znk ) ) −ϕ ( ∆(xnk,ynk,znk ) ) , where M(xnk,ynk,znk ) = max { S(Axnk,Axnk,Bynk ),S(Bynk,Bynk,Cznk ) } Int. J. Anal. Appl. 19 (5) (2021) 800 and ∆(xnk,ynk,znk ) = max { S(Pxnk,Pxnk,Qynk ),S(Axnk,Axnk,Rznk ),S(Pxnk,Pxnk,Bynk ), S(Qynk,Qynk,Cznk ) } Taking limit as n →∞, we obtain lim k→∞ ψ ( M(xnk,ynk,znk ) ) ≤ lim k→∞ ψ ( ∆(xnk,ynk,znk ) ) − lim k→∞ ϕ ( ∆(xnk,ynk,znk ) ) , where lim k→∞ M(xnk,ynk,znk ) = lim k→∞ max{S(t,t,t),S(t,t,Cznk )} = lim k→∞ S(t,t,Cznk ) = ε and lim k→∞ ∆(xnk,ynk,znk ) = max{0, 0, 0,ε} = ε. Since ϕ is lower semi-continuous function, so we obtain ϕ(ε) ≤ lim k→∞ inf ϕ ( ∆(xnk,ynk,znk ) ) . Consequently, we obtain ψ(ε) ≤ ψ(ε) −ϕ(ε)), gives ϕ(ε)) = 0 implies ε = 0. This is a contradiction. � Lemma 2.2. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be an (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R) satisfying the following conditions: (i) BX ⊂RX and RX is closed; (ii) the pairs (A,P) and (B,Q) satisfy the (CLRPQ)− property. Then the pairs (A,P), (B,Q) and (C,R) share the common property −(E.A). Proof. By Lemma 2.1, the pairs (A,P), (B,Q) and (C,R) satisfy the common property −(E.A). Then there exist three sequences {xn}, {yn} and {zn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = t, for some t ∈PX ∩QX. Also by (ii), we obtain t ∈RX. This completes the proof. � Theorem 2.1. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be six self-mappings. Suppose the mappings A,B,C,P,Q, and R be (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R) satisfying the following conditions: (i) the pairs (A,P), (B,Q) and (C,R) share the common property −(E.A); (ii) PX, QX and RX are closed subsets of X. Int. J. Anal. Appl. 19 (5) (2021) 801 Then the pairs (A,P), (B,Q) and (C,R) have their coincidence points in X. Further, A,B,C,P,Q and R have a unique common fixed point, provided the pairs (A,P) (B,Q) and (C,R) are weakly compatible. Proof. From (i), the pairs (A,P), (B,Q) and (C,R) share the common property −(E.A), then there exist three sequences {xn}, {yn} and {zn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = t, for some t ∈ X. Since PX is a closed subset of X and lim n→∞ Pxn = t, then there exists a point u ∈ X such that Pu = t. Now, we assert that Au = Pu. Using inequality (1.1) with x = u, y = yn and z = zn, we get ψ ( M(u,yn,zn) ) ≤ ψ ( ∆(u,yn,zn) ) −ϕ ( ∆(u,yn,zn) ) , (2.1) where M(u,yn,zn) = max{S(Au,Au,Byn),S(Byn,Byn,Czn)} and ∆(u,yn,zn) = max { S(Pu,Pu,Qyn),S(Au,Au,Rzn),S(Pu,Pu,Byn), S(Qyn,Qyn,Czn) } . Taking the limit as n →∞ in (2.1), we obtain ψ ( S(Au,Au,t) ) ≤ lim n→∞ ψ ( ∆(u,yn,zn) ) − lim n→∞ ϕ ( ∆(u,yn,zn) ) , (2.2) where lim n→∞ M(u,yn,zn) = max { S(Au,Au,t),S(t,t,t) } = S(Au,Au,t) and lim n→∞ ∆(u,yn,zn) = max { S(Pu,Pu,t),S(Au,Au,t),S(Pu,Pu,t),S(t,t,t) } (2.3) = max { 0,S(Au,Au,t), 0, 0 } =S(Au,Au,t). Since ϕ is lower semi-continuous, we obtain ϕ ( S(Au,Au,t) ) ≤ lim n→∞ inf ϕ ( ∆(u,yn,zn) ) . (2.4) From (2.2), (2.3) and (2.4), we obtain ψ ( S(Au,Au,t) ) ≤ ψ ( S(Au,Au,t) ) − lim n→∞ inf ϕ ( ∆(u,yn,zn) ) (2.5) ≤ ψ ( S(Au,Au,t) ) −ϕ ( S(Au,Au,t) ) . Int. J. Anal. Appl. 19 (5) (2021) 802 Consequently, ϕ ( S(Au,Au,t) ) = 0 implies S(Au,Au,t) = 0. Hence Au = t = Pu. This shows that the pair (A,P) has a coincidence point in X. Since QX is a closed subset of X, then lim n→∞ Qyn = t ∈ QX. Then there exists a point v ∈ X such that Qv = t. Now, we assert that Bv = Qv. Otherwise from (1.1) with x = u, y = v and z = zn, we obtain ψ ( M(u,v,zn) ) ≤ ψ ( ∆(u,v,zn) ) −ϕ ( ∆(u,v,zn) ) (2.6) where M(u,v,zn) = max { S(Au,Au,Bv),S(Bv,Bv,Czn) } and ∆(u,v,zn) = max { S(Pu,Pu,Qv),S(Au,Au,Rzn),S(Pu,Pu,Bv), S(Qv,Qv,Czn) } Taking the limit as n →∞ in (2.6), we get lim n→∞ ψ ( M(u,v,zn) ) ≤ lim n→∞ ψ ( ∆(u,v,zn) ) − lim n→∞ ϕ ( ∆(u,v,zn) ) (2.7) where lim n→∞ M(u,v,zn) = max { S(t,t,Bv),S(Bv,Bv,t) } = S(t,t,Bv) and lim n→∞ ∆(u,v,zn) = max { S(t,t,t),S(t,t,t),S(t,t,Bv),S(t,t,t) } (2.8) = S(t,t,Bv) Moreover, lower semi-continuity of ϕ, we have ϕ ( S(t,t,Bv) ) ≤ lim n→∞ ϕ ( ∆(u,v,zn) ) (2.9) From (2.7), (2.8) and (2.9), we obtain ψ ( S(t,t,Bv) ) ≤ ψ ( S(t,t,Bv) ) −ϕ ( S(t,t,Bv) ) , so ϕ(S(t,t,Bv)) = 0 and it implies S(t,t,Bv) = 0. Hence Bv = Qv = t. This shows that v is a coincidence point of the pair (B,Q) in X. Also since RX is a closed subset of X and lim n→∞ Rzn = t. Then there exists a point w ∈ X such that Rw = t. We show that Rw = Cw. Using inequality (1.1) with x = u, y = v and z = w, we get ψ ( M(u,v,w) ) ≤ ψ ( ∆(u,v,w) ) −ϕ ( ∆(u,v,w) ) , Int. J. Anal. Appl. 19 (5) (2021) 803 where M(u,v,w) = max { S(Au,Au,Bv),S(Bv,Bv,Cw) } = max { S(t,t,t),S(t,t,Cw) } = S(t,t,Cw) and ∆(u,v,w) = max { S(Pu,Pu,Qv),S(Au,Au,Rw),S(Pu,Pu,Bv),S(Qv,Qv,Cw) } = max { S(t,t,t),S(t,t,t),S(t,t,t),S(t,t,Cw) } = S(t,t,Cw). From the above inequality, we obtain ψ ( S(t,t,Cw) ) ≤ ψ ( S(t,t,Cw) ) −ϕ ( S(t,t,Cw) ) . So ϕ ( S(t,t,Cw) ) = 0, then S(t,t,Cw) = 0. Hence Cw = t = Rw. This shows that w is a coincidence point of the pair (C,R). Thus the pairs (A,P), (B,Q) and (C,R) have their coincidence points in X. It remains to prove that the pairs (A,P), (B,Q) and (C,R) have a unique common fixed point in X. Since the pairs (A,P), (B,Q) and (C,R) are weakly compatible. Then Au = Pu = t implies At = APu = PAu = Pt. Similarly, Bt = BQv = QBv = Qt and Ct = CRw = RCw = Rt. Therefore, t is a coincidence point of the pairs (A,P), (B,Q) and (C,R). One can show that At = Pt = t by taking x = t,y = v and z = w in (1.1). Also At = Bt, this can be proved by putting x = y = t and z = w in (1.1). Similarly, by putting x = u,y = v and z = t in (1.1), we obtain Bt = Ct. Thus, At = Bt = Ct = Pt = Qt = Rt. Now, we show that the point of coincidence of the pairs (A,P), (B,Q) and (C,R) is unique. If the point of coincidence of the pairs (A,P), (B,Q) and (C,R) is not unique, then there exist ξ,ξ∗ ∈ X,ξ 6= ξ∗ such that At = Pt = Bt = Qt = ξ and Ct = Rt = ξ∗. Using inequality (1.1), we obtain ψ ( M(t,t,t) ) ≤ ψ ( ∆(t,t,t) ) −ϕ ( ∆(t,t,t) ) . where M(t,t,t) = max { S(At,At,Bt),S(Bt,Bt,Ct) } = max { S(ξ,ξ,ξ),S(ξ,ξ,ξ∗) } = S(ξ,ξ,ξ∗) Int. J. Anal. Appl. 19 (5) (2021) 804 and ∆(t,t,t) = max { S(Pt,Pt,Qt),S(At,At,Rt),S(Pt,Pt,Bt),S(Qt,Qt,Ct) } = max { S(ξ,ξ,ξ),S(ξ,ξ,ξ∗),S(ξ,ξ,ξ),S(ξ,ξ,ξ∗) } =S(ξ,ξ,ξ∗) Therefore, the above inequality becomes ψ ( S(ξ,ξ,ξ∗) ) ≤ ψ ( S(ξ,ξ,ξ∗) ) −ϕ ( S(ξ,ξ,ξ∗) ) , so ϕ ( S(ξ,ξ,ξ∗) ) = 0 i.e., S(ξ,ξ,ξ∗) = 0 which implies ξ = ξ∗. Therefore, the point of coincidence of the pairs (A,P), (B,Q) and (C,R) is unique and hence by Lemma 1.5, the pairs (A,P), (B,Q) and (C,R) have a unique common fixed point in X. � Example 2.1. Let X = [0, 1]. Define a mapping S : X3 → [0,∞) by S(x,y,z) =   0, if x = y = z max{x,y,z}, otherwise for all x,y,z ∈ X. Clearly, (X,S) is an S− metric space. Consider the self-mappings Ax = x 4 , Bx = x 4 , Cx = x 4 , Px = x, Qx = Rx = x 2 , for all x ∈ X. Setting ψ(t) = t and ϕ(t) = t 4 for t ∈ [0,∞). (a) In order to check the inequality (1.1), consider the following four cases: (i) x = y = z, (ii) x ≤ y < z, (iii) x ≤ z < y, (iv) y ≤ z < x. Case (i): If x = y = z, we get M(x,y,z) = 0, so the condition is trivially satisfied. Case (ii): If x ≤ y < z. Then, we have M(x,y,z) = max { S (x 4 , x 4 , y 4 ) ,S (y 4 , y 4 , z 4 )} = z 4 and ∆(x,y,z) = max { S ( x,x, y 2 ) ,S (x 4 , x 4 , z 2 ) ,S ( x,x, y 4 ) ,S (y 2 , y 2 , z 4 )} = x or z 2 If x < z 2 , then ψ (z 4 ) = z 4 ≤ 3z 8 = ψ (z 2 ) −ϕ (z 2 ) If z 2 < x =⇒ z 4 < x 2 , so ψ (z 4 ) < ψ (x 2 ) ≤ 3x 4 = ψ(x) −ϕ(x). Similarly, the inequality (1.1) is also satisfied for case (iii). Case (iv): If y ≤ z < x, we have M(x,y,z) = x 4 and ∆(x,y,z) = x, so the inequality (1.1) reduces to ψ (x 4 ) = x 4 ≤ 3x 4 = ψ(x) −ϕ(x). Int. J. Anal. Appl. 19 (5) (2021) 805 Thus, for all x,y,z ∈ X, we obtain ψ ( M(x,y,z) ) ≤ ψ ( ∆(x,y,z) ) −ϕ ( ∆(x,y,z) ) . (b) Now, let us show that the pairs (A,P), (B,Q) and (C,R) are weakly compatible. For this, let Ax = Px =⇒ x 4 = x =⇒ x = 0. Now, AP0 = A0 = 0 = P0 = PA0. Therefore, (A,P) is weakly compatible. Similarly, (B,Q) and (C,R) are also weakly compatible mappings. (c) Now, we show that the pairs (A,P), (B,Q) and (C,R) share the common property −(E.A). For this, let xn = 1 n , yn = 1 n + 2 and zn = 1 2n + 3 for n ∈ N. Clearly, {xn}, {yn} and {zn} are in X. Then, we have S(Axn,Axn, 0) = S ( 1 4n , 1 4n , 0 ) = max { 1 4n , 1 4n , 0 } = 1 4n → 0 as n →∞. Also, S(Pxn,Pxn, 0) = S ( 1 n , 1 n , 0 ) = max { 1 n , 1 n , 0 } = 1 n → 0 as n →∞. Similarly, we get that Byn, Qyn, Czn and Rzn → 0 as n →∞. Therefore, there exist three sequences {xn}, {yn} and {zn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = t, Therefore, (A,P), (B,Q) and (C,R) share the common property −(E.A). (d) As PX = [0, 1], QX = RX = [0, 1 2 ], then PX, QX and RX are closed subsets of X. Therefore, all the conditions of Theorem 2.1 are satisfied and 0 is the unique common fixed point of the self-mappings. Theorem 2.2. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be an (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R). If the pairs (A,P), (B,Q) and (C,R) satisfy the (CLRPQR)− property, then (A,P), (B,Q) and (C,R) have their coincidence points. Moreover, A,B,C,P,Q and R have a unique common fixed point provided the pairs (A,P), (B,Q) and (C,R) are weakly compatible. Proof. Suppose the pairs (A,P), (B,Q) and (C,R) satisfy the (CLRPQR)− property, then there exist three sequences {xn}, {yn} and {zn} in X such that lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = t, for some t ∈ PX ∩QX ∩RX. It follows that t ∈ PX and there exists u ∈ X such that Pu = t. Now we assert that Au = Pu. Using inequality (1.1) with x = u, y = yn, z = zn, we get ψ ( M(u,yn,zn) ) ≤ ψ ( ∆(u,yn,zn) ) −ϕ ( ∆(u,yn,zn) ) , (2.10) Int. J. Anal. Appl. 19 (5) (2021) 806 where M(u,yn,zn) = max { S(Au,Au,Byn),S(Byn,Byn,Czn) } ∆(u,yn,zn) = max { S(Pu,Pu,Qyn),S(Au,Au,Rzn),S(Pu,Pu,Byn), S(Qyn,Qyn,Czn) } . Taking the limit as n →∞ in (2.10), we get lim n→∞ ψ ( M(u,yn,zn) ) ≤ lim n→∞ ψ ( ∆(u,yn,zn) ) − lim n→∞ ϕ ( ∆(u,yn,zn) ) where lim n→∞ M(u,yn,zn) = max { S(Au,Au,t),S(t,t,t) } = S(Au,Au,t) lim n→∞ ∆(u,yn,zn) = max { S(t,t,t),S(Au,Au,t),S(t,t,t),S(t,t,t), } = S(Au,Au,t). From the above inequality, we obtain ψ ( S(Au,Au,t) ) ≤ ψ ( S(Au,Au,t) ) −ϕ ( S(Au,Au,t) ) , so ϕ ( S(Au,Au,t) ) = 0, i.e., S(Au,Au,t) = 0. Hence Au = t = Pu, which shows that u is a coincidence point of the pair (A,P). As t ∈QX, there exists a point v ∈ X such that Qv = t. We show that Bv = Qv. Using inequality (1.1) with x = u, y = v and z = zn, we have ψ ( M(u,v,zn) ) ≤ ψ ( ∆(u,v,zn) ) −ϕ ( ∆(u,v,zn) ) (2.11) where M(u,v,zn) = max { S(Au,Au,Bv),S(Bv,Bv,Czn) } = max { S(t,t,Bv),S(Bv,Bv,Czn) } and ∆(u,v,zn) = max { S(Pu,Pu,Qv),S(Au,Au,Rzn),S(Pu,Pu,Bv), S(Qv,Qv,Czn) } = max { S(t,t,t),S(t,t,Rzn),S(t,t,Bv),S(t,t,Czn) } Taking the limit as n →∞ in (2.11), we get lim n→∞ ψ ( M(u,v,zn) ) ≤ lim n→∞ ψ ( ∆(u,v,zn) ) − lim n→∞ ϕ ( ∆(u,v,zn) ) Int. J. Anal. Appl. 19 (5) (2021) 807 where lim n→∞ M(u,v,zn) = max { S(t,t,Bv),S(Bv,Bv,t) } = S(Bv,Bv,t) and lim n→∞ ∆(u,v,zn) = max { S(t,t,t),S(t,t,t),S(t,Bv,Bv),S(t,t,t) } = S(Bv,Bv,t), The above equation gives ψ ( S(Bv,Bv,t) ) ≤ ψ ( S(Bv,Bv,t) ) −ϕ ( S(Bv,Bv,t) ) , so ϕ(S(Bv,Bv,t)) = 0, i.e., S(Bv,Bv,t) = 0. Hence, Bv = Qv = t, which shows that v is a coincidence point of the pair (B,Q). As t ∈ RX, there exists a point w ∈ X such that Rw = t. We show that Rw = Cw. Using inequality (1.1) with x = u, y = v and z = w, we get ψ ( M(u,v,w) ) ≤ ψ ( ∆(u,v,w) ) −ϕ ( ∆(u,v,w) ) where M(u,v,w) = max { S(Au,Au,Bv),S(Bv,Bv,Cw) } = S(t,t,Cw) and ∆(u,v,w) = max { S(Pu,Pu,Qv),S(Au,Au,Rw),S(Pu,Pu,Bv),S(Qv,Qv,Cw) } = max { S(t,t,t),S(t,t,t),S(t,t,t),S(t,t,Cw) } = S(t,t,Cw). Follows from the above inequality, we obtain ψ ( S(t,t,Cw) ) ≤ ψ ( S(t,t,Cw) ) −ϕ ( S(t,t,Cw) ) , so ϕ ( S(t,t,Cw) ) = 0, i.e., S(t,t,Cw) = 0. Hence, Cw = t = Rw, which shows that w is a point of coincidence of the pair (C,R). Thus the pairs (A,P), (B,Q) and (C,R) have their coincidence points in X. It remains to prove that the pairs (A,P), (B,Q) and (C,R) have a unique common fixed point in X. Since the pairs (A,P), (B,Q) and (C,R) are weakly compatible. Then Au = Pu = t implies At = APu = PAu = Pt. Similarly, Bt = BQv = QBv = Qt and Ct = CRw = RCw = Rt. Therefore, t is a coincidence point of the pairs (A,P), (B,Q) and (C,R). Following the same steps as in Theorem 2.1, one can show that At = Bt = Ct = Pt = Qt = Rt. Now, we show that the point of coincidence of the pairs (A,P), (B,Q) and (C,R) is unique. Int. J. Anal. Appl. 19 (5) (2021) 808 If the point of coincidence of the pairs (A,P), (B,Q) and (C,R) is not unique, then there exist ξ,ξ∗ ∈ X,ξ 6= ξ∗ such that At = Pt = Bt = Qt = ξ and Ct = Rt = ξ∗. Using inequality (1.1), we obtain ψ ( M(t,t,t) ) ≤ ψ ( ∆(t,t,t) ) −ϕ ( ∆(t,t,t) ) , where M(t,t,t) = max { S(At,At,Bt),S(Bt,Bt,Ct) } = max { S(ξ,ξ,ξ),S(ξ,ξ,ξ∗) } = S(ξ,ξ,ξ∗) and ∆(t,t,t) = max { S(Pt,Pt,Qt),S(At,At,Rt),S(Pt,Pt,Bt),S(Qt,Qt,Ct) } = max { S(ξ,ξ,ξ),S(ξ,ξ,ξ∗),S(ξ,ξ,ξ),S(ξ,ξ,ξ∗) } =S(ξ,ξ,ξ∗) Therefore, the above inequality becomes ψ ( S(ξ,ξ,ξ∗) ) ≤ ψ ( S(ξ,ξ,ξ∗) ) −ϕ ( S(ξ,ξ,ξ∗) ) , so ϕ ( S(ξ,ξ,ξ∗) ) = 0 i.e., S(ξ,ξ,ξ∗) = 0 which implies ξ = ξ∗. Therefore, the point of coincidence of the pairs (A,P), (B,Q) and (C,R) is unique and hence by Lemma 1.5, the pairs (A,P), (B,Q) and (C,R) have a unique common fixed point in X. � Example 2.2. Let X = [0, 20]. Define a mapping S : X3 → [0,∞) by S(x,y,z) = |x−y|+|y −z| , ∀x,y,z ∈ X. Clearly, (X,S) is an S−metric space. Let A,B,C,P,Q,R : X → X be six self-mappings defined by Ax =   2, if x ∈ [0, 2]3, if x ∈ (2, 20] ; Bx =   1, if x ∈ [0, 2)2, if x ∈ [2, 20] ; Cx =   2, if x ∈ [0, 2]1, if x ∈ (2, 20] Px =   2, if x ∈ [0, 2]6, if x ∈ (2, 20] , Qx =   4, if x ∈ [0, 2)2, if x ∈ [2, 20] ; Rx =   2, if x ∈ [0, 2]8, if x ∈ (2, 20]. Consider three sequences {xn} = {2 − 1 n }, {yn} = {2 + 1 n + 1 }, {zn} = { 1 n },∀n ∈ N. lim n→∞ Axn = lim n→∞ Pxn = lim n→∞ Byn = lim n→∞ Qyn = lim n→∞ Czn = lim n→∞ Rzn = 2, where 2 ∈PX ∩QX ∩RX. Therefore, the pairs (A,P), (B,Q) and (C,R) satisfy (CLRPQR)− property. Consider ψ(t) = t and ϕ(t) = t 4 . In order to check the inequality (1.1), we have the following eight cases: Int. J. Anal. Appl. 19 (5) (2021) 809 (i) x,z ∈ [0, 2],y ∈ [0, 2), (ii) x ∈ [0, 2],y ∈ [0, 2), z ∈ (2, 20], (iii) x ∈ [0, 2], y ∈ [2, 20], z ∈ [0, 2], (iv) x ∈ [0, 2], y ∈ [2, 20], z ∈ (2, 20], (v) x ∈ (2, 20],y ∈ [0, 2),z ∈ [0, 2], (vi) x ∈ (2, 20]y ∈ [0, 2),z ∈ (2, 20], (vii) x ∈ (2, 20],y ∈ [2, 20],z ∈ [0, 2], (viii) x ∈ (2, 20],y ∈ [2, 20],z ∈ (2, 20], In case (i), we have M(x,y,z) = 1 and ∆(x,y,z) = 2, so the inequality (1.1) reduces to ψ(1) = 1 ≤ 3 2 = ψ(2) −ϕ(2) In case (ii) and (vi), we have M(x,y,z) = 1 and ∆(x,y,z) = 6, so (1.1) reduces to ψ(1) = 1 ≤ 9 2 = ψ(6) −ϕ(6). In case (iii), we have M(x,y,z) = 0, so the inequality (1.1) is trivially satisfied. In case (v) and (vi), we have M(x,y,z) = 2 and ∆(x,y,z) = 5, so the inequality (1.1) reduces to ψ(2) = 2 ≤ 15 4 = ψ(5) −ϕ(5) In case (vii), we have M(x,y,z) = 1 and ∆(x,y,z) = 4, so the inequality (1.1) reduces to ψ(1) = 1 ≤ 3 = ψ(4) −ϕ(4) In case (viii), we have M(x,y,z) = 1 and ∆(x,y,z) = 5, so the inequality (1.1) reduces to ψ(1) = 1 ≤ 15 4 = ψ(5) −ϕ(5) Thus, the inequality (1.1) holds true for all x,y,z ∈ X. Hence, all the conditions of Theorem 2.2 are satisfied, and 2 is a unique common fixed point of the pairs (A,P), (B,Q) and (C,R) which also remains a point of coincidence. Here, one may notice that all the involved mappings are discontinuous at their unique common fixed point 2. Theorem 2.3. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be an (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R) satisfying the following conditions: (i) BX ⊂RX (resp. AX ⊂RX); (ii) the pairs (A,P) and (B,Q) satisfy the common property −(E.A); (iii) PX, QX and RX are closed subsets of X. Then the pairs (A,P), (B,Q) and (C,R) have their coincidence points in X. Further, A,B,C,P,Q and R have a unique common fixed point, provided the pairs (A,P), (B,Q) and (C,R) are weakly compatible. Proof. It follows from Lemma 2.1 and Theorem 2.1. � Theorem 2.4. Let (X,S) be an S− metric space and A,B,C,P,Q,R : X → X be an (A,B,C)(ψ,ϕ)− weak contraction with respect to (P,Q,R) satisfying the following conditions: (i) BX ⊂RX and RX is closed; Int. J. Anal. Appl. 19 (5) (2021) 810 (ii) the pairs (A,P) and (B,Q) satisfy the (CLRPQ)− property. Then the pairs (A,P), (B,Q) and (C,R) have their coincidence points in X. Further, A,B,C,P,Q and R have a unique common fixed point, provided the pairs (A,P), (B,Q) and (C,R) are weakly compatible. Proof. It follows from Lemma 2.2 and Theorem 2.2. � 2.1. Conclusion. 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