Ratio Mathematica Volume 47, 2023 Relation-theoretic contraction principle in metric spaces using multiplicative contraction Radha Yadav* Balbir Singh† Abstract Alam and Imdad have presented a novel application of the Banach contraction principle on a complete metric spaces with a binary rela- tion. We have extended the concept of binary relation with the mul- tiplicative contraction in a complete metric spaces. We have also in- cluded corollary to demonstrate our results. Keywords: fixed point, metric spaces, binary relation, multiplicative contraction 2020 AMS subject classifications: 47H10, 54H25.1 1 Introduction In many scientific domains, particularly in fixed point theory, the concept of a metric space is extremely useful. This notion has been generalised in numerous di- rections in recent years, and many notions of a metric-type space have been intro- duced (b-metric, dislocated space, generalised metric space, quasi-metric space, symmetric space, etc.).The Banach contraction principle’s [3]contraction condi- tion has been generalised to numerous forms in the last fifty years. Furthermore, the metric space in the Banach contraction principle has been generalised to a va- riety of generalised metric spaces. Many authors researched other sorts of fixed point theorems in metric spaces later on, as seen by the and references therein. *Research scholar (Starex University, Gurugram, India); radha97yadav@gmail.com. †Professor (Starex University, Gurugram, India); balbir.vashist007@gmail.com. 1Received on March 28, 2022. Accepted on December 28, 2022. Published online on February 6, 2023. doi:10.23755/rm.v39i0.730. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This paper is published under the CC-BY licence agreement. 30 Radha Yadav and Balbir Singh 2 Preliminaries We give the required background material needed to prove our results in this part to make our exposition self-contained. In what follows, N, N0, Q and R de- note the sets of positive integers, non-negative integers, rational numbers and real numbers, respectively. Aftab and Alam [1] proved new relation-theoretic fixed point theorems on metric spaces in 2015, and then inferred comparable findings in metric spaces. Metric spaces are sets in which there is a defined a notion of ’distance between pair of points’. The concept of metric spaces was formulated in 1906 by M.Frechet [7] ,though the definition presently in use given by the German mathematician, Fe- lix Hausdorff. Definition 2.1. Let M be a non empty arbitrary set and d be a real function from M × M into R+ such that for all u, v, w ∈ M we have 1. d(u, v) ≥ 0, 2. d(u, v) = 0 ⇐⇒ u = v, 3. d(u, v) = d(v, u) and 4. d(u, w) ≤ d(u, v) + d(v, w), Here (M, d) is called a metric in R and (R, d) is a metric space. Example 2.1. 1. d(u, v) = |u − v| is a metric space in R. 2. If d(u, v) defined by d(u, v) = {1 if u ̸= v 0 if u = v} Definition 2.2. [10] Let M be a nonempty set. A subset R of M2 is called a binary relation on M. Notice that for each pair u, v ∈ M, one of the following conditions holds: 1. (u, v) ∈ R; which amounts to saying that “u is R-related to v” or “u re- lates to v under R”. Sometimes, we write uRv instead of (u, v) ∈ R ; 2. (u, v) /∈ R; which means that “u is not R-related to v” or “u does not relate to v under R”. 31 Relation-theoretic contraction principle in metric spaces using multiplicative contraction Definition 2.3. [10] Let R be a binary relation defined on a nonempty set M and u, v ∈ M. We say that u and v are R -comparative if either (u, v) ∈ R or (v, u) ∈ R. We denote it by [u, v] ∈ R. Definition 2.4. [6, 10, 11, 12, 15] A binary relation R defined on a nonempty set M is called 1. reflexive if (u, u) ∈ R for all u ∈ M, 2. irreflexive if (u, u) /∈ R for all u ∈ M, 3. symmetric if (u, v) ∈ R implies (v, u) ∈ R, 4. antisymmetric if (u, v) ∈ R implies (v, u) /∈ R, 5. transitive if (u, w) ∈ R and (w, v) ∈ R implies (u, v) ∈ R, 6. complete, connected or dichotomous if [l, n] ∈ R for all l, m ∈ M, 7. weakly complete, weakly connected or trichotomous if [u, v] ∈ R or u = v for all u, v ∈ M. 8. strict order or sharp order if R is irreflexive and transitive, 9. near-order if R is antisymmetric and transitive, 10. pseudo-order if R is reflexive and antisymmetric, 11. quasi-order or preorder if R is reflexive and transitive, 12. partial order if R is reflexive, antisymmetric and transitive, 13. simple order if R if weakly complete strict order, 14. weak order if R is complete preorder, 15. total order, linear order or chain if R is complete partial order, 16. tolerance if R is reflexive and symmetric, 17. equivalence if R is reflexive, symmetric and transitive. Definition 2.5. [4] Let M be a nonempty set and R a binary relation on M. A sequence {un} ⊂ M is called R-preserving if (un, un+1) ∈ R ∀n ∈ N0 32 Radha Yadav and Balbir Singh The notion of d- self closeness of a partial order ⪯ defined by Turinici [16] is extended to an arbitrary binary relation in the following lines. Now, we state and prove our main result, which is as follows: Theorem 2.1. Let (M, d) be a complete metric space, R a binary relation on M and T a self-mapping on M. Suppose that the following conditions hold: a) M(f; R) is nonempty, b) R is f-closed, c) either f is continuous or R is p-self-closed, d) there exists λ ∈ [0, 1)d(f(u), f(v)) ≤ d(u, v)λ forallu, v ∈ M with(u, v) ∈ R Then f has a fixed point. Moreover, if e) Υ(u, v, Rs) is nonempty, for each u, v ∈ M, then f has a unique fixed point. Proof. Consider a point u0 ∈ M. Now we define a sequence {un} of Picard iterates, i.e., un = fun−1 for n = 1, 2, ... From the multiplicative contraction property [13] of f for all n ∈ N0. As (u0, fu0) ∈ R, using condition (b), we get (fu0, f 2u0), (f 2u0, f 3u0), ..., (f nu0, f n+1u0), ... ∈ R so that (un, un+1) ∈ R n ∈ N0. (1) Thus the sequence {un} is R-preserving. Applying the contractivity condition (d) to equation (1), we deduce, for all n ∈ N0, that d(un+1, un) ≤ d(un, un−1)λ ≤ d(un−1, un−2)λ 2 ≤ ... ≤ d(u1, u0)λ n . which by induction yields that d(un+1, un+2) ≤ d(u0, fu0)λ n+1 n ∈ N0. (2) Using equation (2) and triangular inequality, for all n ∈ N0, p ∈ N, p ≥ 2, we have d(un+1, un+p) ≤ d(un+1, un+2) + d(un+2, un+3) + ... + d(un+p−1, un+p) ≤ d(u1, u0)λ n+1+...+λp ≤ d(u1, u0) λp 1−λ This implies that d(un, up) → 1 as (n, p → ∞). Hence the sequence (xn) = (fnu0) is multiplicative Cauchy. By the completeness of M, there is z ∈ M such that un → z as n → ∞. Moreover, 33 Relation-theoretic contraction principle in metric spaces using multiplicative contraction d(fz, z) ≤ d(fun, fz).d(fun, z) ≤ d(un, z)λ.d(fun, z) → 1 as n → ∞, which implies d(fz, z) = 0. Therefore this says that z is a fixed point of f; that is fz = z. Now, if there is another point y such that fy = y, then d(z, y) = d(fz, fy) ≤ d(z, y)λ. Therefore d(z, y) = 0 and y = z. This implies that z is the unique fixed point of f. Alternatively, let us assume that R is d-self-closed. As un is an R-preserving sequence and un →d u, there exists a subsequence {unk} of {un} with [unk, u] ∈ R k ∈ N0 Using (d), [unk, u] ∈ R and un → d u, we obtain d(unk+1, fu) = d(funk, fu) ≤ d(unk, u) λ → 1 as k → ∞ so that unk+1 → d f(u). Again, owing to the uniqueness of limit, we get f(u) = u so that u is a fixed point of f. To prove uniqueness, take u, v ∈ F(f), i.e., f(u) = u and f(v) = v. (3) By assumption (d), there exists a path (say {z0, z1, z2, ..., zk}) of some finite length k in Rs from u to v so that z0 = u, zk = v, [zi, zi+1] ∈ R foreach i(0 ≤ i ≤ k − 1). (4) As R is f-closed, we have [fnzi, f nzi+1] ∈ R foreach i(0 ≤ i ≤ k − 1) and for each n ∈ N0. (5) Making use of equations (3),(4),(5),, triangular inequality and assumption (d), we obtain d(u, v) = d(fnz0, f nzk) ≤ ∑i=0 k−1 d(f nzi, f nzi+1) ≤ ∑i=0 k−1 d(f n−1zi, f n−1zi+1) λ ≤ ∑i=0 k−1 d(f n−2zi, f n−2zi+1) λ2 34 Radha Yadav and Balbir Singh ≤ ... ≤ ∑i=0 k−1 d(zi, zi+1) λn → 0 as n → ∞ so that u = v. Hence f has a unique fixed point.2 Corolary 2.1. Let (M, d) be a complete metric space. For ϵ with ϵ > 1 and u0 ∈ M, consider the multiplicative closed ball, Bϵ(u0). Suppose the mapping f : M → M satisfies the contraction condition d(f(u), f(v)) ≤ d(u, v)λ forallu, v ∈ Bϵ(u0) where λ ∈ [0, 1) is a constant R is a relation such that d(fu0, u0) ≤ ϵ1−λ. Then f has a unique fixed point in Bϵ(u0). Corolary 2.2. Let (M, d) be a complete metric space. If a mapping f : M → M satisfies for some positive integer n, d(fnu, fnv) ≤ d(u, v)λ for all u, v ∈ M, where λ ∈ [0, 1) is a constant, then f has a unique fixed point in M. Theorem 2.2. Let (M, d) be a complete metric space, R a binary relation on M and T a self-mapping on M. Suppose that the following conditions hold: a) M(f; R) is nonempty, b) R is f-closed, c) either f is continuous or R is p-self-closed, d) there exists λ ∈ [0, 1 2 ) d(fu, fv) ≤ (d(f(u, v).d(fv, u))λforallu, v ∈ M with(u, v) ∈ R Then f has a fixed point. Moreover, if e) Υ(u, v, Rs) is nonempty, for each u, v ∈ M, then f has a unique fixed point. Proof. Consider a point u0 ∈ M. Now we define a sequence {un} of Picard iterates, i.e., un = fun−1 for n = 1, 2, ... From the multiplicative contraction property of f for all n ∈ N0. As (u0, fu0) ∈ R, using condition 2, we get (fu0, f 2u0), (f 2u0, f 3u0), ..., (f nu0, f n+1u0), ... ∈ R we have d(un+1, un) = d(fun, fun−1) ≤ (d(fun, un).d(fun−1, un−1))λ = (d(un+1, un).d(un, un−1)) λ 35 Relation-theoretic contraction principle in metric spaces using multiplicative contraction Thus we have d(un+1, un) ≤ (d(un, un−1)) λ 1−λ = d(un, un−1) h, where h = λ 1−λ . For n > m, Using triangular inequality, for all n ∈ N0, m ∈ N, m ≥ m, we have d(un, um) ≤ d(un, un−1).d(un−1, un−2)...d(um+1, um) ≤ d(u1, u0)h n−1+hn−2+...+hm ≤ d(u1, u0) hm 1−h This implies d(un, um) → 1 as (n, m → ∞). Hence (un) is a Cauchy sequence. By the completeness of M, there is z ∈ M such that un → z as n → ∞. Since d(fz, z) ≤ d(fun, fz).d(fun, z) ≤ (d(fun, un).d(fz, z))λ.d(un+1, z), we have d(un+1, un) ≤ (d(un, un−1)) λ 1−λ = d(un, un−1) h, where h = λ 1−λ . For n > m, Using triangular inequality, for all n ∈ N0, m ∈ N, m ≥ m, we have d(un, um) ≤ d(un, un−1).d(un−1, un−2)...d(um+1, um) ≤ d(u1, u0)h n−1+hn−2+...+hm ≤ d(u1, u0) hm 1−h This implies d(un, um) → 1 as (n, m → ∞). Hence (un) is a Cauchy sequence. By the completeness of M, there is z ∈ M such that un → z as n → ∞. Since d(fz, z) ≤ d(fun, fz).d(fun, z) ≤ (d(fun, un).d(fz, z))λ.d(un+1, z), we have d(fz, z) ≤ (d(fun, un)λ.d(un+1, z)) 1 1−λ → 1 as n → ∞. Hence d(fz, z) = 0. This implies fz = z. Finally, it is easy to prove that the fixed point of f is unique. Alternatively, let us assume that R is d-self-closed. As {un} is an R-preserving sequence and un →d u, 36 Radha Yadav and Balbir Singh there exists a subsequence {unk} of {un} with [unk, u] ∈ R k ∈ N0 Using (d), [unk, u] ∈ R and unk → d u, we obtain d(unk+1, fu) = d(funk, fu) ≤ (d(unk+1, u).d(unk, u)) λ → 1 as k → ∞ so that unk+1 → d f(u). Again, owing to the uniqueness of limit, we obtain f(u) = u, so that u is a fixed point of f. To prove uniqueness, take u, v ∈ F(f), i.e., f(u) = u and f(v) = v. (6) By assumption (e), there exists a path (say{z0, z1, z2, ..., zk}) of some finite length k in Rs from u to v so that z0 = u, zk = v, [zi, zi+1] ∈ R for each i(0 ≤ i ≤ k − 1). (7) As R is f-closed, we have [fnzi, f nzi+1] ∈ R for each i(0 ≤ i ≤ k−1) and for each n ∈ N0. (8) Making use of equations (6),(7),(8),, triangular inequality and assumption (d) we obtain d(u, v) = d(fnz0, f nzk) ≤ ∑i=0 k−1 d(f nzi, f nzi+1) ≤ ∑i=0 k−1(d(f n−1zi, f n−1zi+1).d(f n−1zi+1, f n−1zi)) λ ≤ ∑i=0 k−1(d(f n−2zi, f n−2zi+1).d(f n−2zi+1, f n−2zi)) λ2 ≤ ... ≤ ∑i=0 k−1(d(zi, zi+1).d(zi+1, zi)) λn → 0 as n → ∞ so that u = v. Hence f has a unique fixed point.2 Theorem 2.3. Let (M, d) be a complete metric space, R a binary relation on M and T a self-mapping on M. Suppose that the following conditions hold: a) M(f; R) is nonempty, b) R is f-closed, c) either f is continuous or R is p-self-closed, d) there exists λ ∈ [0, 1 2 ) d(fu, fv) ≤ (d(f(u, u).d(fv, v))λforallu, v ∈ M with(u, v) ∈ R Then f has a fixed point. Moreover, if e) Υ(u, v, Rs) is nonempty, for each u, v ∈ M, then f has a unique fixed point. 37 Relation-theoretic contraction principle in metric spaces using multiplicative contraction Proof. Consider a point u0 ∈ M. Now we define a sequence {un} of Picard iterates, i.e., un = fun−1 for n = 1, 2, ... From the multiplicative contraction property of f for all n ∈ N0. As (u0, fu0) ∈ R, using condition 2, we get (fu0, f 2u0), (f 2u0, f 3u0), ..., (f nu0, f n+1u0), ... ∈ R we have d(un+1, un) = d(fun, fun−1) ≤ (d(fun, un).d(fun−1, un−1))λ = (d(un+1, un).d(un, un−1)) λ Thus we have d(un+1, un) ≤ d(un, un−1)) λ 1−λ = d(un, un−1) h, where h = λ 1−λ . For n > m, Using triangular inequality, for all n ∈ N0, m ∈ N, m ≥ m, we have d(un, um) ≤ d(un, un−1).d(un−1, un−2)...d(um+1, um) ≤ d(u1, u0)h n−1+hn−2+...+hm ≤ d(u1, u0) hm 1−h This implies d(un, um) → 1 as (n, m → ∞). Hence (un) is a Cauchy sequence. By the completeness of M, there is z ∈ M such that un → z as n → ∞. Since d(fz, z) ≤ d(fun, fz).d(fun, z) ≤ (d(fun, un).d(fz, z))λ.d(un+1, z), we have d(fz, z) ≤ (d(fun, un)λ.d(un+1, z)) 1 1−λ → 1 as n → ∞. Hence d(fz, z) = 0. This implies fz = z. Finally, it is easy to prove that the fixed point of f is unique. Alternatively, let us assume that R is d-self-closed. As {un} is an R-preserving sequence and un →d u, there exists a subsequence {unk} of {un} with [unk, u] ∈ R k ∈ N0 Using (d), [unk, u] ∈ R and unk → d u, we obtain 38 Radha Yadav and Balbir Singh d(unk+1, fu) = d(funk, fu) ≤ (d(unk+1, u).d(unk, u)) λ → 1 as k → ∞ so that unk+1 → d f(u). Again, owing to the uniqueness of limit, we obtain f(u) = u, so that u is a fixed point of f. To prove uniqueness, take u, v ∈ F(f), i.e., f(u) = u and f(v) = v. (9) By assumption (e), there exists a path (say{z0, z1, z2, ..., zk}) of some finite length k in Rs from u to v so that z0 = u, zk = v, [zi, zi+1] ∈ R foreach i(0 ≤ i ≤ k − 1). (10) As R is f-closed, we have [fnzi, f nzi+1] ∈ R foreach i(0 ≤ i ≤ k − 1) and for each n ∈ N0. (11) Making use of equations (9),(10),(11), triangular inequality and assumption (d), we obtain d(u, v) = d(fnz0, f nzk) ≤ ∑i=0 k−1 d(f nzi, f nzi+1) ≤ ∑i=0 k−1(d(f n−1zi, f n−1zi+1).d(f n−1zi+1, f n−1zi)) λ ≤ ∑i=0 k−1(d(f n−2zi, f n−2zi+1).d(f n−2zi+1, f n−2zi)) λ2 ≤ ... ≤ ∑i=0 k−1(d(zi, zi+1).d(zi+1, zi)) λn → 0 as n → ∞ so that u = v. 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