International Journal of Analysis and Applications Volume 19, Number 6 (2021), 915-928 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-915 COMMON FIXED POINT OF FOUR MAPS IN Sm-METRIC SPACE V. SRINIVAS1,∗, K. MALLAIAH2 1Department of Mathematics, UCS, Saifabad, OU,Hyderabad, Telangana, India 2Lecturer in Mathematics, JNGP, Ramanthapur, Hyderabad, Telangana, India ∗Corresponding author: srinivasmaths4141@gmail.com Abstract. In this paper, first, we deal with new metric space Sm-metric space that combines multiplica- tive metric space and S-metric space. We generate a common fixed point theorem in a Sm-metric space using the notions of reciprocally continuous mappings, faintly compatible mappings and occasionally weakly compatible mappings (OWC). We are also studying the well-posedness of Sm metric space. Further, some examples are presented to support our outcome. 1. Introduction The idea of Multiplicative metric space(MMS for short ) was first introduced by Bashirove [1] in 2008.Ozak- sar and Cevical [2] investigated and proved the properties of MMS. Following that, several theorems like [3] and [4] in this area of MMS were developed. Sedhi.S et al. [5] introduced a new structure of S-metric space and developed some fixed point theorems. Pant et al. [6] used the concept of reciprocally continuous map- pings which is weaker than continuous mappings. In this article, we use the multiplicative metric space and S metric space and generated a new Sm-metric space [7]. We used the concept of occasionally weakly compatible (OWC for shot) [9]mappings, reciprocally continuous and faintly compatible mappings [10] to generate a common fixed point theorem in Sm-metric space. We also discuss the well-posedness property [11] in Sm-metric space. Furthermore, some examples are provided to support our new findings. Received August 23rd, 2021; accepted October 13th, 2021; published November 5th, 2021. 2010 Mathematics Subject Classification. 54H25. Key words and phrases. Multiplicative metric space; S-metric space; Sm-metric space; occasionally weakly compatible; reciprocally continuous; faintly compatible mappings; well-posed 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. 915 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-915 Int. J. Anal. Appl. 19 (6) (2021) 916 2. Mathematical Preliminaries: Definition 2.1. [1] ” Let X 6= φ. An operator δ : X2 → R+ be a multiplicative metric space (MMS) holding in the conditions below: (M1) δ(α,β) ≥ 1, and δ(α,β) = 1 ⇐⇒ α = β (M2) δ(α,β) = δ(β,α) (M3) δ(α,β) ≤ δ(α,γ)δ(γ,β),∀α,β,γ ∈ X. Mapping δ together with X, (X,δ) is called a MMS”. A three-dimensional metric space was proposed by Sedghi et al . [5], and it is called S-metric space. Definition 2.2. [5] ” Let X 6= φ defined on a function S : X3 → [0,∞) satisfying: (S1) S(α,β,γ) ≥ 0; (S2) S(α,β,γ) = 0; ⇐⇒ α = β = γ, (S3) S(α,β,γ) ≤ S(α,α,ω) + S(β,β,ω) + S(γ,γ,ω),∀α,β,γ,ω ∈ X. The pair (X,S) is known as S-metric space on X ”. We now present the concept of Sm-metric space which is consolidation of multiplicative metric space defined by Bashirov [1] and S-metric space defined by Sedgi [5] by as follows Definition 2.3. [7] ” Let X 6= φ . A function Sm : X3 → R+ holding the conditions below: (MS1) Sm(α,β,γ) ≥ 1 (MS2) Sm(α,β,γ) = 1 ⇐⇒ α = β = γ (MS3) Sm(α,β,γ) ≤ Sm(α,α,ω)Sm(β,β,ω)Sm(γ,γ,ω),∀α,β,γ,ω ∈ X. Mapping Sm together with X, (X,Sm) is known as Sm-metric space.” Example 2.1. ” Let X 6= φ,Sm : X3 → [0,∞) by Sm(α,β,γ) = a |α−γ|+|β−γ|, where α,β,γ,a ∈ X, then (X,Sm) is a Sm -metric space on X.” Example 2.2. Let X=R+ ,define Sm : X3 → [0,∞) by Sm(α,β,γ) = a |β+γ−2α|+|β−γ|, where α,β,γ ∈ X, then (X,Sm) is a Sm -metric space on X. Now we present some definitions in Sm -metric space. Definition 2.4. [7] Suppose (X,Sm) is a Sm-metric space, a sequence {αk}∈ X is called ( 2.4.1) cauchy sequence ⇐⇒ Sm(αk,αk,αl) → 1, for all k,l →∞; Int. J. Anal. Appl. 19 (6) (2021) 917 ( 2.4.2) convergent ⇐⇒ ∃α ∈ X such that Sm(αk,αk,α) → 1 as k →∞; ( 2.4.3) is complete if every cauchy sequence is convergent. Definition 2.5. [8] ” The mappings G and I be compatible mappings in Sm-metric space if Sm(GIαk,GIαk,IGαk) = 1, whenever a sequence {αk} in X such that limk→∞Gαk = limk→∞Iαk = η for some η ∈ X. ” Definition 2.6. [8] ”Let G and I be weakly compatible mappings in Sm-metric space if for all η ∈ X, Gη = Iη =⇒ GIη = IGη”. Definition 2.7. [9] ” Suppose G and I are mappings in Sm-metric are said to be occasionally weakly compatible (OWC for shot) iff ∃η ∈ X such that Gη = Iη =⇒ GIη = IGη. ” Example 2.3. Let X = [0,∞) is a Sm-metric space on X , Sm(α,β,γ) = a |α−β|+|β−γ|+|γ−α|, for every α,β,γ ∈ X. Construct two self maps G and I as G(α) = 3α− 2 and I(α) = α2. Consider a sequence {αk} given by αk = 2 + 1k for k ≥ 0. G(αk) = 3(2 + 1 k ) − 2 = 4 and I(αk) = (2 + 1k ) 2 = 4 as k →∞ Therefore Gαk = Iαk = 4 6= φ. Moreover, GI(βk) = GI(2 + 1 k ) = G(2 + 1 k )2 = G(4 + 4 1 k + 1 k2 ) = 3(4 + 4 1 k + 1 k2 ) − 2 =10 and IG(βk) = IG(2 + 1 k ) = I(3(2 + 1 k ) − 2) = I(4 + 3 k ) = (4 + 3 k )2 = 16 as k →∞. This gives Sm(GIαk,GIαk,IGαk) = Sm(10, 10, 16) 6= 1. Hence, (G,I) is not compatible. Now G(1)= I(1)=1 also GI(1)= IG(1)=1 =⇒ GI(1)= IG(1). G(2)=4, I(2)=4 also GI(2)=10 , IG(2)=16 =⇒ GI(2) 6= IG(2). As a result G and I have OWC, but not weakly compatible. Definition 2.8. [10] ” Two self maps G and I in Sm-metric space as conditionally- compatible if there exists a sequence {αk}∈ X such that Gαk = Iαk 6= φ,∃ a sequence {βk}∈ X such that Gβk = Iβk → η for some η ∈ X and Sm(GIβk,GIβk,IGη) = 1 as k →∞. ” Definition 2.9. [10] ” Two self maps G and I in Sm-metric space are called as faintly compatible iff (G , I )is conditionally- compatible and G and I commute on a non -empty subset of the set of coincidence points if the collection of coincidence points is non- empty.” Int. J. Anal. Appl. 19 (6) (2021) 918 Definition 2.10. [6] ”A reciprocally continuous mappings G and I of a Sm-metric space is defined as Sm(GIαk,GIαk,Iη) = 1andSm(IGαk,IGαk,Gη) = 1 letting k → ∞ if there exists a sequence {αk} ∈ X such that limk→∞Gαk = limk→∞Iαk = η as η ∈ X. ” Example 2.4. Let X = [0,∞) is a Sm-metric space on X , Sm(α,β,γ) = a |α−β|+|β−γ|+|γ−α|, for every α,β,γ ∈ X. Construct two self maps G and I as G(α) = α2 − 3α + 2 and I(α) = 3α2 − 7α + 2. Consider a sequence {αk} given by αk = 2 + 1k for k ≥ 0. then G(αk) = (2 + 1 k )2 − 3(2 + 1 k ) + 2 = 0 and I(αk) = 3(2 + 1 k )2 − 7(2 + 1 k ) + 2 = 0 as k → ∞ therefore limk→∞Gαk = limk→∞Iαk = 0 6= φ. Moreover, GI(αk) = GI(2 + 1 k ) = G[3(2 + 1 k )2 −7(2 + 1 k ) + 2] = G( 3 k2 + 5 k ) = ( 3 k2 + 5 k )2 −3( 3 k2 + 5 k ) + 2 = 2 and IG(αk) = IG(2 + 1 k ) = I[(2 + 1 k )2−3(2 + 1 k ) + 2] = I( 1 k2 + 1 k ) = 3( 1 k2 + 1 k )2−7( 1 k2 + 1 k ) + 2 = 2 as k →∞. =⇒ Sm(GIαk,GIαk,IGαk) = Sm(2, 2, 2) = 1. Hence, (G,I) is compatible. Consider another sequence {βk} given by βk = 1k for k ≥ 0. G(βk) = ( 1 k2 − 3 k + 2) = 2 and I(βk) = ( 3 k2 − 7 k + 2) = 2 as k →∞ therefore limk→∞Gβk = limk→∞Iβk = 2. Further GI(βk) = GI( 1 k ) = G( 3 k2 − 7 k + 2) = ( 3 k2 − 7 k + 2)2 − 3( 3 k2 − 7 k + 2) + 2 = 0 and IG(βk) = IG( 1 k ) = I( 1 k2 − 3 k + 2) = 3( 1 k2 − 3 k + 2) − 7( 1 k2 − 3 k + 2) + 2 = −6 as k →∞. This gives Sm(GIαk,IGαk,η) = Sm(0, 0,−6) 6= 1. the pair (G,I) is not compatible Hence (G,I) is conditionally compatible. Compatibility is distinct from the concept of conditional compatibility, Now G(2)=0,I(2)=0 and GI(2)=2,IG(2)=2. Also G(0)=I(0)=2 and GI(0)=IG(0)=0. Hence the pair (G,I) is faintly compatible. As a result, the mappings G and I have faintly compatible, but they are not compatible. Definition 2.11. [11] ”The mappings G and I of a Sm-metric space are called well-posed if • G and I have a unique common fixed point η in X • If αk ∈ X such that Sm(Gαk,Gαk,αk) = 1 and Sm(Iαk,Iαk,αk) = 1 as k → ∞ we have Sm(αk,αk,η) = 1 as k →∞.” Int. J. Anal. Appl. 19 (6) (2021) 919 3. MAIN THEOREM Theorem: Suppose G, H, I and J are self-mapping in a complete Sm-metric space X, suppose that there exist λ ∈ (0, 12 ) such that the conditions (3.1.1) G(X) ⊆ J(X) and H(X) ⊆ I(X) (3.1.2) Sm(Gα,Gα,Hβ) ≤ { max[Sm(Gα,Gα,Iα)Sm(Hβ,Hβ,Jβ),Sm(Gα,Gα,Jβ)Sm(Iα,Iα,Hβ), Sm(Gα,Gα,Jβ)Sm(Hβ,Hβ,Jβ),Sm(Gα,Gα,Iα)Sm(Hβ,Hβ,Iα)] }λ (3.1.3) the pair (H,J) is OWC (3.1.4) and the pair (G,I) is reciprocally continuous and faintly compatible. Then the common fixed point problem of G, H, I and J is Well-posed. Proof: We begin with using (3.1.1), then there is a point α0 ∈ X, such that Gα0 = Jα1 = β0. For this point α1then there ∃α2 ∈ X such that Hα1 = Iα2 = β1. In general, by induction choose αk+1 so that β2k = Gα2k = Jα2k+1 and β2k+1 = Hα2k+1 = Iα2k+2 for k ≥ 0. We show that {βk} is a cauchy sequence in Sm - metric space . Indeed, it follows that Sm(β2k,β2k,β2k+1) = Sm(Gα2k,Gα2k,Hα2k+1) ≤ max { Sm(Gα2k,Gα2k,Iα2k)Sm(Hα2k+1,Hα2k+1,Jα2k+1), Sm(Gα2k,Gα2k,Jα2k+1)Sm(Hα2k+1,Hα2k+1,Iα2k), Sm(Gα2k,Gα2k,Jα2k+1)Sm(Hα2k+1,Hα2k+1,Jα2k+1), Sm(Gα2k,Gα2k,Iα2k)Sm(Hα2k+1,Hα2k+1,Iα2k) }λ Int. J. Anal. Appl. 19 (6) (2021) 920 Sm(β2k,β2k,β2k+1) ≤ max { Sm(β2k,β2k,β2k−1)Sm(β2k+1,β2k+1,β2k), Sm(β2k,β2k,β2k)Sm(β2k+1,β2k+1,β2k−1), Sm(β2k,β2k,β2k)Sm(β2k+1,β2k+1,β2k), Sm(β2k,β2k,β2k−1)Sm(β2k+1,β2k+1,β2k−1) }λ on simplification Sm(β2k,β2k,β2k+1) ≤ Sm(β2k−1,β2k−1,β2k+1)λ. Sm(β2k,β2k,β2k+1) ≤{Sm(β2k−1,β2k−1,β2k)Sm(β2k,β2k,β2k+1)}λ. S1−λm (β2k,β2k,β2k+1) ≤ S λ m(β2k−1,β2k−1,β2k). Sm(β2k,β2k,β2k+1) ≤ S λ 1−λ m (β2k−1,β2k−1,β2k). Sm(β2k,β2k,β2k+1) ≤ Spm(β2k−1,β2k−1,β2k). where p = λ 1 −λ Now this gives Sm(βk,βk,βk+1) ≤ Spm(βk−1,βk−1,βk) ≤ S p2 m (βk−2,βk−2,βk−1) ≤ ···S pn m (β0,β0,βn) By using triangular inequality, Sm(βk,βk,βn) ≤ Sp k m (β0,β0,βl) ≤ S pk+1 m (β0,β0,βn) ≤ ···S pn−1 m (β0,β0,βn) Hence {βk} is a cauchy sequence in Sm-metric space. Now X being complete in Sm-metric space ∃η ∈ X such that limk→∞βk → η . Consequently, the sub sequences {Gα2k}, {Iα2k}, {Jα2k+1} and {Hα2k+1} of {βk} also converges to the point η ∈ X. Since the pair (G,I) is faintly compatible mappings, so that ∃ another sequence νk ∈ X such that limk→∞Gνk = limk→∞Iνk = ω for ω ∈ X satisfying limk→∞S(GIνk,GIνk,IGνk) = 1 and the pair (G,I) is reciprocally continuous Sm(GIνk,GIνk,Iω) = 1, and Sm(IGνk,IGνk,Gω) = 1. as k →∞. (3.1) Gω = Iω Int. J. Anal. Appl. 19 (6) (2021) 921 On putting α = ω and β = α2k+1 in (3.1.2) we get Sm(Gω,Gω,Hα2k+1) ≤ { max[Sm(Gω,Gω,Iω)Sm(Hα2k+1,Hα2k+1,Jα2k+1), Sm(Gω,Gω,Jα2k+1)Sm(Iω,Iη,Hα2k+1), Sm(Gω,Gω,Jα2k+1)Sm(Hα2k+1,Hα2k+1,Jα2k+1), Sm(Gω,Gω,Iω)Sm(Hα2k+1,Hα2k+1,Iω)] }λ and Sm(Gω,Gω,η) ≤ { max[Sm(Gω,Gω,Iω)Sm(η,η,η),Sm(Gω,Gω,η)Sm(Iω,Iω,η), Sm(Gω,Gω,η)S ∗(η,η,η),Sm(Gω,Gω,Iω)Sm(η,η,Iω)] }λ which gives Sm(Gω,Gω,η) ≤ { max[Sm(Gω,Gω,Gω)Sm(η,η,η),Sm(Gω,Gω,η)Sm(Gω,Gω,η), Sm(Gω,Gω,η)Sm(η,η,η),Sm(Gω,Gω,Gω)Sm(η,η,Gω)] }λ implies Sm(Gω,Gω,η) ≤ { max[1,S2m(Gω,Gω,η),Sm(Gω,Gω,η),Sm(Gω,Gω,η)] }λ this gives Sm(Gω,Gω,η) ≤ { S2λm (Gω,Gω,η) } this implies Gω = η. therefore Gω = Iω = η.(3.2) Since the pair (G,I) is faintly compatible, so that Gω = Iω this gives GIω = IGω this implies Gη = Iη. By using the inequality (3.1.2) on putting α = η and β = α2k+1 we get Sm(Gη,Gη,Hα2k+1) ≤ { max[Sm(Gη,Gη,Iη)Sm(Hα2k+1,Hα2k+1,Jα2k+1), Sm(Gη,Gη,Jα2k+1)Sm(Iη,Iη,Hα2k+1), Sm(Gη,Gη,Jα2k+1)Sm(Hα2k+1,Hα2k+1,Jα2k+1), Sm(Gη,Gη,Iη)Sm(Hα2k+1,Hα2k+1,Iη)] }λ Int. J. Anal. Appl. 19 (6) (2021) 922 and Sm(Gη,Gη,η) ≤ { max[Sm(Gη,Gη,Iη)Sm(η,η,η),Sm(Gη,Gη,η)Sm(Iη,Iη,η), Sm(Gη,Gη,η)Sm(η,η,η),Sm(Gη,Gη,Iη)Smη,η,Iη)] }λ which gives Sm(Gη,Gη,η) ≤ { max[Sm(Gη,Gη,Gη)Sm(η,η,η),Sm(Gη,Gη,η)Sm(Gη,Gη,η), Sm(Gη,Gη,η)Sm(η,η,η),Sm(Gη,Gη,Gη)Sm(η,η,Gη)] }λ implies Sm(Gη,Gη,η) ≤ { max[1,Sm 2(Gη,Gη,η),Sm(Gη,Gη,η),Sm(Gη,Gη,η)] }λ which implies Sm(Gη,Gη,η) ≤ { Sm 2λ(Gη,Gη,η) } =⇒ Gη = η.(3.3) Gη = Iη = η.(3.4) =⇒ η = Gη ∈ G(X) ⊆ J(X) =⇒ Gη = Jv for some v ∈ X. Gη = Iη = Jv = η.(3.5) Using the inequality (3.1.2) on putting α = η and β = v we have Sm(Gη,Gη,Hv) ≤ { max[Sm(Gη,Gη,Iη)Sm(Hv,Hv,Jv),Sm(Gη,Gη,Jv)Sm(Iη,Iη,Hv), Sm(Gη,Gη,Jv)Sm(Hv,Hv,Jv),Sm(Gη,Gη,Iη)Sm(Hv,Hv,Iη)] }λ this implies Sm(η,η,Hv) ≤ { max[Sm(η,η,η)Sm(Hv,Hv,η),Sm(η,η,η)Sm(η,η,Hv), Sm(η,η,η)Sm(Hv,Hv,η),Sm(η,η,η)Sm(Hv,Hv,η)] }λ which implies Sm(η,η,Hv) ≤ { max[Sm(Hv,Hv,η),Sm(η,η,Hv), Sm(Hv,Hv,η),Sm(Hv,Hv,η)] }λ Int. J. Anal. Appl. 19 (6) (2021) 923 this gives Sm(η,η,Hv) ≤ { Sm(Hv,Hv,η) }λ which gives Hv = η. Gη = Iη = Jv = Hv = η.(3.6) Again (H,J) is OWC with v ∈ X so that Hv = Jv =⇒ HJv = JHv which implies that Hη = Jη. Using the inequality (3.1.2) and take α = η and β = η we get Sm(Gη,Gη,Hη) ≤ { max[Sm(Gη,Gη,Iη)Sm(Hη,Hη,Jη),Sm(Gη,Gη,Jη)Sm(Iη,Iη,Hη), Sm(Gη,Gη,Jη)Sm(Hη,Hη,Jη),Sm(Gη,Gη,Iη)Sm(Hη,Hη,Iη)] }λ this implies Sm(η,η,Hη) ≤ { max[Sm(η,η,η)Sm(Hη,Hη,η),Sm(η,η,η)Sm(η,η,Hη), Sm(η,η,η)Sm(Hη,Hη,η),Sm(η,η,η)Sm(Hη,Hη,η)] }λ where Sm(η,η,Hη) ≤ { max[Sm(Hη,Hη,η),Sm(η,η,Hη),Sm(Hη,Hη,η),Sm(Hη,Hη,η)] }λ this gives Sm(η,η,Hη) ≤ { Sm(Hη,Hη,η) }λ this gives Hη = η. Hη = Jη = η.(3.7) From( 3.4) and ( 3.7) Gη = Iη = Jη = Hη = η.(3.8) =⇒ η is a common fixed point for the mappings G,H,I and J. For the proof of well-posed property Suppose ρ(ρ 6= η) is one more fixed point of G,I,H and J Int. J. Anal. Appl. 19 (6) (2021) 924 i.e Gρ = Iρ = Hρ = Jρ = ρ. Using the inequality (3.1.2) take α = ρ and β = η we have Sm(Gρ,Gρ,Hη) ≤ { max[Sm(Gρ,Gρ,Iη)Sm(Hη,Hη,Jη),Sm(Gρ,Gρ,Jη)Sm(Iρ,Iρ,Hη), Sm(Gρ,Gρ,Jη)Sm(Hη,Hη,Jη),Sm(Gρ,Gρ,Iρ)Sm(Hη,Hη,Iρ)] }λ this gives Sm(ρ,ρ,η) ≤ { max[Sm(ρ,ρ,η)Sm(η,η,η),Sm(ρ,ρ,η)Sm(ρ,ρ,η), Sm(ρ,ρ,η)Sm(η,η,η),Sm(ρ,ρ,ρ)Sm(η,η,ρ)] }λ this gives Sm(ρ,ρ,η) ≤ { max[1,Sm(ρ,ρ,η), 1, 1] }λ which gives ∴ Sm(ρ,ρ,η) ≤ Sm(ρ,ρ,η)λ This gives ρ = η. Hence η is the unique common fixed point of G,H,I and J Suppose {αk} be a sequence in X such that Sm(Gαk,Gαk,αk) = Sm(Iαk,Iαk,αk) = 1 and Sm(Hαk,Hαk,αk) = Sm(Jαk,Jαk,αk) = 1 as k →∞. We have to show that Sm(αk,αk,η) = 1, Sm(αk,αk,η) ≤ Sm(Gαk,Gαk,η)Sm(Gαk,Gαk,αk) Sm(αk,αk,η) ≤{ max[Sm(Gαk,Gαk,Iαk)Sm(Hη,Hη,Jη),Sm(Gαk,Gαk,Jη)Sm(Iαk,Iαk,Hη), Sm(Gαk,Gαk,Jη)Sm(Hη,Hη,Jη),Sm(Gαk,Gαk,Iαk)Sm(Hη,Hη,Iαk)] }λ Sm(Gαk,Gαk,αk) this gives Sm(αk,αk,η) ≤{ max[Sm(Gαk,Gαk,Gη)Sm(Iαk,Iαk,Iη)Sm(Hη,Hη,Jη),Sm(Gαk,Gαk,Jη)Sm(Iαk,Iαk,Hη), Sm(Gαk,Gαk,Jη)Sm(Hη,Hη,Jη),Sm(Gαk,Gαk,Iαk)Sm(Hη,Hη,Iαk)] }λ Sm(Gαk,Gαk,αk) Int. J. Anal. Appl. 19 (6) (2021) 925 which gives Sm(αk,αk,η) ≤{ max[Sm(αk,αk,η)Sm(αk,αk,η))Sm(η,η,η),Sm(αk,αk,η)Sm(αk,αk,η), Sm(αk,αk,η)Sm(η,η,η),Sm(αk,αk,αk)Sm(η,η,αk)] }λ Sm(Gαk,Gαk,αk) therefore Sm(αk,αk,η) ≤ { Sm(αk,αk,η) }λ Sm(Gαk,Gαk,αk) Sm(αk,αk,η) 1−λ ≤ Sm(Gαk,Gαk,αk) Sm(αk,αk,η) ≤ S 1 1−λ m (Gαk,Gαk,αk) Sm(αk,αk,η) = 1 as k →∞. Thus G,H,I and J is well-posed. 4. Example Suppose X = [0, 1] ,Sm- metric space by Sm(α,β,γ) = e |α−β|+|β−γ|+|γ−α|, when α,β,γ ∈ X. Define G ,I ,H J:XxX → X as follows G(α) =   1−α 2 if 0 ≤ α ≤ 1 3 ; 3−2α 4 if 1 3 < α ≤ 1. J(α) =   1+4α 7 if 0 ≤ α ≤ 1 3 ; 2α+1 4 if 1 3 < α ≤ 1. and H(α) =   2−α 5 if 0 ≤ α ≤ 1 3 ; 3−2α 4 if 1 3 < α ≤ 1. I(α) =   2α+1 5 if 0 ≤ α ≤ 1 3 ; α+5 11 if 1 3 < α ≤ 1. Then G(X)=( 7 12 , 1 4 ] and J(X) =[ 1 7 , 1 3 ] ∪ ( 5 12 , 3 4 ] . And also H(X)=( 7 21 , 1 4 ] and I(X)=[ 1 5 , 1 3 ] ∪ ( 16 33 , 6 11 ] thisimpliesimpliesG(X) ⊆ J(X) and H(X) ⊆ I(X) hence the inequality (3.1.1) holds. Take a sequence {αk} as αk = 12 + 1 k as k ≥ 0. Now G(αk) = G( 1 2 + 1 k )= 3−2( 1 2 + 1 k ) 4 = 1 2 and J(αk) = J( 1 2 + 1 k )= 2( 1 2 + 1 k )+1 4 = 1 2 . And also H(αk) = H( 1 2 + 1 k )= 3−2( 1 2 + 1 k ) 4 = 1 2 and I(αk) = I( 1 2 + 1 k )= 5+( 1 2 + 1 k ) 11 = 1 2 . ∴ Gαk=Jαk= 1 2 and Hαk=Jαk= 1 2 as k →∞. then GI(αk) = GI( 1 2 + 1 k ) = G( 1 2 + 1 11k ) = 3−2( 1 2 + 1 11k ) 4 = 1 2 and IG(αk) = IG( 1 2 + 1 k ) = I( 1 2 − 1 2k ) = 5+( 1 2 − 1 2k ) 11 = 1 2 as k →∞. HJ(αk) = HJ( 1 2 + 1 k ) = H( 5+( 1 2 + 1 k ) 11 ) = H( 1 2 + 1 11k ) = 3−2( 1 2 + 1 11K ) 4 = 1 2 Int. J. Anal. Appl. 19 (6) (2021) 926 and JH(αk) = JH( 1 2 + 1 k ) = J( 3−2( 1 2 + 1 k ) 4 ) = I( 1 2 − 1 2k ) = 1+2( 1 2 − 1 2k ) 4 = 1 2 as k →∞. limk→∞Sm(GIαk,GIαk,IGαk) = Sm( 1 2 , 1 2 , 1 2 ) = 1 and limk→∞Sm(HJαk,HJαk,JHαk) = Sm( 1 2 , 1 2 , 1 2 ) = 1 Hence the pairs (G,I) and (H,J) are satisfies compatible property . Take another sequence {βk} as βk = 13 − 1 k as k ≥ 0. Now G(βk) = G( 1 3 − 1 k )= 1−( 1 3 −1 k ) 2 = 1 3 = and I(βk) = I( 1 3 − 1 k )= 2+( 1 3 −1 k ) 5 = 1 3 . And also H(βk) = H( 1 3 − 1 k )= 2−( 1 3 −1 k ) 5 = 1 3 and J(βk) = J( 1 3 − 1 k )= 4( 1 3 −1 k )+1 7 = 1 3 as k →∞. ∴ Gβk=Iβk= 1 3 = η similarly Hβk=Jβk= 1 3 = η as k →∞. Further GI(βk) = GI( 1 3 − 1 k )=G( 2( 1 3 −1 k )+1 5 )=G( 1 3 − 2 5k )= 1−( 1 3 − 2 5k ) 2 = 1 3 and IG((βk) = IG( 1 3 − 1 k )=I( 1−( 1 3 −1 k 2 )=I( 1 3 + 1 2k )= 5+( 1 3 + 1 2k ) 11 = 16 33 HJ(βk) = HJ( 1 3 − 1 k ) = H( 4( 1 3 −1 k )+1 7 = H( 1 3 − 4 7k ) = 2−( 1 3 + 4 7k ) 11 = 1 3 and JH(βk) = JH( 1 3 − 1 k ) = J( 2−( 1 3 −1 k 5 ) = J( 1 3 + 1 5k ) = 2( 1 3 + 1 5k )+1 4 = 5 12 as k →∞. this implies limk→∞Sm(GIβk,GIβk,IGβk) = Sm( 1 3 , 1 3 , 66 33 ) 6= 1 which shows that the pairs (G,I) is faintly compatible mappings. Moreover I( 1 2 ) = 1 2 and G( 1 2 ) = 1 2 and also limk→∞Sm(GIαk,GIαk,Iw) = Sm( 1 2 , 1 2 , 1 2 ) = 1 limk→∞Sm(IGαk,IGαk,Gw) = Sm( 1 2 , 1 2 , 1 2 ) = 1 this shows that the pairs (G,I) is reciprocally continuous. The inequity (3.1.4) holds. Further H( 1 2 ) = 5+ 1 2 11 = 1 2 and J( 1 2 ) = 2 1 2 +1 4 = 1 2 . ∴ H( 1 2 ) = J( 1 2 ) = ( 1 2 ) where 1 2 ∈ X. And also HJ( 1 2 ) = 1 2 and JH( 1 2 ) = 1 2 , =⇒ HJ( 1 2 ) = JH( 1 2 ) = 1 2 . Moreover, H( 1 3 ) = 2−1 3 5 = 1 3 and J( 1 3 ) = 4 1 3 +1 7 = 1 3 . ∴ H( 1 3 ) = J( 1 3 ) = ( 1 3 ) where 1 3 ∈ X. And also HJ( 1 3 ) = 1 3 and JH( 1 3 ) = 1 3 , =⇒ HJ( 1 3 ) = JH( 1 3 ) = 1 3 . Which shows that the pair(H,J) satisfies OWC. So that the inequity (3.1.3) holds. Further more Sm(Gβk,Gβk,βk) = Sm( 1 3 , 1 3 , 1 3 ) = 1,Sm(Iβk,Iβk,βk) = Sm( 1 3 , 1 3 , 1 3 ) = 1, when k → ∞. Which implies limk→∞Sm(βk,βk,η) = limk→∞Sm( 1 3 , 1 3 , 1 3 ) = 1, CASE-I Let α,β ∈ [0, 1 3 ],while we have Sm(α,β,γ) = e |α−γ|+|β−γ| In the inequality(3.1.2) putting α = 1 3 and β = 1 4 implies Sm(0.375, 0.375, 0.36) ≤{ max[Sm(0.375, 0.375, 0.28)Sm(0.36, 0.36, 0.286)Sm(0.375, 0.375, 0.286)Sm(0.36, 0.36, 0.28), Sm(0.375, 0.375, 0.286)Sm(0.36, 0.36, 0.286),Sm(0.375, 0.375, 0.28)Sm(0.36, 0.36, 0.28)] }λ =⇒ e0.03 ≤ { max[e0.19e0.15,e0.18e0.16,e0.18e0.15,e0.19e0.16] }λ Int. J. Anal. Appl. 19 (6) (2021) 927 e0.03 ≤{ max[e0.34,e0.34,e0.33,e0.35]}λ =⇒ e0.03 ≤ e0.35λ which gives λ = 0.08, where λ ∈ (0, 1 2 ). CASE-II Let α,β ∈ [ 1 2 , 1], then Sm(α,β,γ) = e |α−γ|+|β−γ| In the inequality(3.1.2) putting α = 2 3 and β = 3 4 implies Sm(0.42, 0.42, 0.375) ≤{ max[Sm(0.42, 0.42, 0.52)Sm(0.375, 0.375, 0.47)Sm(0.42, 0.42, 0.47)Sm(0.375, 0.375, 0.52), Sm(0.42, 0.42, 0.47)Sm(0.375, 0.375, 0.47),Sm(0.42, 0.42, 0.52)Sm(0.375, 0.375, 0.52)] }λ =⇒ e0.09 ≤ { max[e0.2e0.19,e0.1e0.29,e0.1e0.19,e0.2e0.29] }λ e0.09 ≤{ max[e0.39,e0.39,e0.29,e0.49]}λ =⇒ e0.09 ≤ e0.49λ which gives λ = 0.18, where λ ∈ (0, 1 2 ). Hence the inequality(3.1.2) holds. The verification in the remaining intervals is also simple. It can be observed that 1 2 is a unique common fixed point of G,H,I and J. 5. CONCLUSION: This article, aimed to prove a common fixed point theorem in Sm-metric space using conditions OWC, reciprocally continuous and faintly compatible mappings. Also proved the well- posed property. Further our result is supported with a suitable example. 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