FIXED POINT POINTS OF RATIONAL TYPE CONTRACTIONS IN MULTIPLICATIVE METRIC SPACES. Dong-A University Busan 49315, KOREA 2 Department of Mathematics

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1 International Journal of Pure Applied Mathematics Volume 106 No , ISSN: (printed version); ISSN: (on-line version) url: doi: /ijpam.v106i2.22 PAijpam.eu FIXED POINT POINTS OF RATIONAL TYPE CONTRACTIONS IN MULTIPLICATIVE METRIC SPACES Young Chel Kwun 1, Poonam Nagpal 2, Sudhir Kumar Garg 2, Sanjay Kumar 2, Shin Min Kang 3 1 Department of Mathematics Dong-A University Busan 49315, KOREA 2 Department of Mathematics Deenbhu Chhotu Ram University of Science Technology Murthal, Sonipat , Haryana, INDIA 3 Department of Mathematics RINS Gyeongsang National University Jinju 52828, KOREA Abstract: In this paper, we prove common fixed point results for mappings satisfying some contractive condition in multiplicative metric spaces. We also give examples in support of our results. AMS Subject Classification: 47H10, 54H25 Key Words: multiplicative metric spaces, weakly compatible mappings 1. Introduction Preliminaries ItiswellknowthatthesetofpositiverealnumbersR + isnotcompleteaccording to the usual metric. To overcome this problem, in 2008, Bashirov et al. [2] introduced the concept of multiplicative metric spaces as follows: Received: November 23, 2015 Published: February 18, 2016 Correspondence author c 2016 Academic Publications, Ltd. url:

2 594 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang Definition 1.1. Let X be a nonempty set. A multiplicative metric is a mapping d : X X R + satisfying the following conditions: (i) d(x,y) 1 for all x,y X d(x,y) = 1 if only if x = y; (ii) d(x,y) = d(y,x) for all x,y X; (iii) d(x, y) d(x, z) d(z, y) for all x, y, z X (multiplicative triangle inequality). Then the mapping d together with X, that is, (X, d) is a multiplicative metric space. Example 1.2. ([6]) Let R n + be the collection of all n-tuples of positive real numbers. Let d : R n + R n + R be defined as follows: d (x,y) = x 1 x 2 x n, y 1 where x = (x 1,...,x n ), y = (y 1,...,y n ) R n + : R + R + is defined by a a if a 1; = 1 a if a < 1. Then it is obvious that all conditions of a multiplicative metric are satisfied. Therefore (R n +,d ) is a multiplicative metric space. Example 1.3. ([7]) Let d : R R [1, ) be defined as d(x,y) = a x y for all x,y R, where a > 1. Then d is a multiplicative metric (R,d) is a multiplicative metric space. We may call it usual multiplicative metric spaces. Remark 1.4. We note that the Example 1.2 is valid for positive real numbers Example 1.3 is valid for all real numbers. Example 1.5. ([7]) Let (X,d) be a metric space. Define a mapping d a on X by d a (x,y) = a d(x,y) 1 if x = y, = a if x y for all x,y X, where a > 1. Then d a is a multiplicative metric (X,d a ) is known as the discrete multiplicative metric space. Example 1.6. ([1]) Let X = C [a,b] be the collection of all real-valued multiplicative continuous functions on [a,b] R +. Then (X,d) is a multiplicative metric space with d defined by d(f,g) = sup x [a,b] f(x) g(x) for arbitrary f,g X. y 2 y n

3 FIXED POINT POINTS OF RATIONAL TYPE Remark 1.7. ([7]) We note that multiplicative metric metric spaces are independent. Indeed, the mapping d defined in Example 1.2 is multiplicative metric but not metric as it does not satisfy triangular inequality. Consider ( 1 d 3, 1 ( ) 12 )+d 2,3 = 32 ( ) = 7.5 < 9 = d,3. On the other h the usual metric on R is not multiplicative metric as it doesnt satisfy multiplicative triangular inequality, since d(2,3) d(3,6) = 3 < 4 = d(2,6). One can refer to [3, 6] for detailed multiplicative metric topology. Definition 1.8. Let (X,d) be a multiplicative metric space. Then a sequence x n in X said to be (1) a multiplicative convergent to x if for every multiplicative open ball B ǫ (x) = y d(x,y) < ǫ, ǫ > 1, there exists N N such that x n B ǫ (x) for all n N, that is, d(x n,x) 1 as n. (2) a multiplicative Cauchy sequence if for all ǫ > 1, there exists N N such that d(x n,x m ) < ǫ for all m,n N, that is, d(x n,x m ) 1 as n,m. (3) We call a multiplicative metric space complete if every multiplicative Cauchy sequence in it is multiplicative convergent to x X. Remark 1.9. The set of positive real numbers R + is not complete according to the usual metric. Let X = R + the sequence x n = 1 n. It is obvious x n is a Cauchy sequence in X with respect to usual metric X is not a complete metric space, since 0 / R +. In case of a multiplicative metric space, we take a sequence x n = an, 1 where a > 1. Then x n is a multiplicative Cauchy sequence since for n m, d(x n,x m ) = x n = a 1 n = 1 a n 1 m a if a 1, x m a 1 m a 1 m 1 n < a 1 m < ǫ if m > loga logǫ, where a = 1 a if a < 1. Also, x n 1 as n 1 R +. Hence (X,d) is a complete multiplicative metric space. In 2012, Özavsar Çevikel [6] gave the concept of multiplicative contraction mappings proved some fixed point theorems of such mappings in a multiplicative metric space.

4 596 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang Definition Let f be a mapping of a multiplicative metric space (X, d) into itself. Then f is said to be a multiplicative contraction if there exists a real constant λ [0,1) such that d(fx,fy) d λ (x,y) for all x,y X. Also they proved the Banach Contraction Principle in the setting of multiplicative metric spaces as follows: Theorem Let f be a multiplicative contraction mapping of a complete multiplicative metric space (X, d) into itself. Then f has a unique fixed point. In 1996, Jungck [4] introduce the notion of weakly compatible mappings in a metric space. Now, we introduce the notions in multiplicative metric spaces Definition Let f g be two mappings of a multiplicative metric space (X,d) into itself. Then f g are said to be weakly compatible if they commute at coincidence points, that is, if ft = gt for some t X implies that fgt = gft. Khan et al. [5] initiated the use of the control function as follows: Definition A function φ : [1, ) [1, ) is called an alternating distance function if the following properties are satisfied: (1) φ is increasing continuous, (2) φ(t) = 1 if only if t = 1. In our results we will use the following class of functions. Φ = φ : [1, ) [1, ) : φ is an alternating distance function, Ψ = ψ : [1, ) [1, ) : for any sequence x n in [1, ) with x n t > 1, lim ψ(x n) > 1. We note that Ψ is non-empty since ψ(t) = e t for t [1, ). Thus ψ Ψ. Remark Clearly for ψ Ψ, ψ(t) > 1 for t > 1 ψ(1) need not be equal to Main Results Now we prove some fixed point results as follows:

5 FIXED POINT POINTS OF RATIONAL TYPE Theorem 2.1. Let f be a continuous mapping of a complete multiplicative metric space (X,d) into itself such that for all x,y X where φ Φ ψ Ψ, φ(d(fx,fy)) φ(m(x,y)) ψ(n(x,y)), (2.1) d(x,fx)d(y,fy) M(x,y), d(y,fx)d(x,fy),d(x,y) d(x, y) d(x, y) d(x,fx)d(y,fy) N(x,y),d(x,y). d(x, y) Then f has a unique fixed point. Proof. Let x 0 X be arbitrary. Then there exists x 1 X such that x 1 = fx 0. So we can define a sequence x n in X such that x n+1 = fx n for n 0. If there exists some n N such that x ( n+1) = x n. Then we have x n+1 = fx n = x n, which implies that x n is a fixed point of f. Suppose that x n+1 x n, that is, d(x n+1,x n ) 1 for all n. Let R n = d(x n+1,x n ) for all n 0. From (2.1), we have where M(x n 1,x n ) d(xn 1,fx n 1 )d(x n,fx n ) d(x n 1,x n ) d(xn 1,x n )d(x n,x n+1 ) d(x n 1,x n ) φ(d(x n,x n+1 ) = φ(d(fx n 1,fx n )) φ(m(x n 1,x n )) ψ(n(x n 1,x n )), maxd(x n,x n+1 ),d(x n,x n+1 ),d(x n 1,x n ) d(x n,x n+1 ),d(x n 1,x n ) R n,r n 1, d(x n,fx n 1 )d(x n 1,fx n ) d(x n 1,x n ),d(x n 1,x n ), d(x n,x n )d(x n 1,x n+1 ),d(x n 1,x n ) d(x n 1,x n )

6 598 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang d(xn 1,fx n 1 )d(x n,fx n ) N(x n 1,x n ),d(x n 1,x n ) d(x n 1,x n ) d(xn 1,x n )d(x n,x n+1 ),d(x n 1,x n ) d(x n 1,x n ) Therefore, we have maxd(x n,x n+1 ),d(x n 1,x n ) d(x n,x n+1 ),d(x n 1,x n ) R n,r n 1. If R n > R n 1, then from (2.2), we have φ(r n ) φ(maxr n,r n 1 ) ψ(maxr n,r n 1 ). (2.2) φ(r n ) φ(r n) ψ(r n ), that is, ψ(r n ) 1, which is a contradiction. So R n R n 1, that is, R n is a decreasing sequence. Then the inequality (2.2) yields that φ(r n ) φ(r n 1) ψ(r n 1 ). (2.3) Since R n is a decreasing sequence of positive real numbers it is bounded below, there exists r 1 such that R n = d(x n+1,x n ) r (2.4) as n. Now we shall show that r = 1. Assume that r > 1. Taking limit on both sides (2.3) using (2.4), the property of ψ the continuity of φ, we get φ(r) φ(r) lim ψ(r n 1), which implies that lim ψ(r n 1) 1, which, by the property of ψ, is a contradiction. Therefore, R n = d(x n+1,x n ) 1 (2.5) as n. Next to show that x n is a multiplicative Cauchy sequence. Suppose that x n is not a multiplicative Cauchy sequence. Then there exists an ǫ > 1 for

7 FIXED POINT POINTS OF RATIONAL TYPE which we can find two sequences of positive integers m(k) n(k) such that for all positive integers k, n(k) > m(k) k d(x m(k),x n(k) ) ǫ. Assume that n(k) is the smallest such positive integer, we get n(k) > m(k) k, Now, d(x m(k),x n(k) ) ǫ d(x m(k),x n(k) 1 ) < ǫ. ǫ d(x m(k),x n(k) ) d(x m(k),x n(k) 1 ) d(x n(k) 1,x n(k) ). Letting k using (2.5), we have Again, lim d(x m(k),x n(k) ) = ǫ. (2.6) k d(x m(k) 1,x n(k) 1 ) d(x m(k) 1,x m(k) ) d(x m(k),x n(k) ) d(x n(k),x n(k) 1 ) d(x m(k),x n(k) ) d(x m(k),x m(k) 1 ) d(x m(k) 1,x n(k) 1 ) d(x n(k) 1,x n(k) ). Letting k in above inequalities using (2.5) (2.6), we have Again lim d(x m(k) 1,x n(k) 1 ) = ǫ. (2.7) k d(x n(k) 1,x m(k) ) d(x m(k) 1,x n(k) 1 ) d(x m(k) 1,x m(k) ) d(x m(k) 1,x n(k) 1 ) d(x m(k) 1,x m(k) ) d(x n(k) 1,x m(k) ). Letting k in the above inequalities using (2.5) (2.7), we have Similarly, we have lim d(x n(k) 1,x m(k) ) = ǫ. (2.8) k lim d(x m(k) 1,x n (k)) = ǫ. (2.9) k

8 600 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang Let M(x n(k) 1,x m(k) 1 ) d(xn(k) 1,fx n(k) 1 )d(x m(k) 1,fx m(k) 1 ), d(x m(k) 1,x n(k) 1 ) d(x n(k) 1,fx m(k) 1 )d(x m(k) 1,fx n(k) 1 ),d(x d(x m(k) 1,x n(k) 1 ) m(k) 1,x n(k) 1 ) d(xn(k) 1,x n(k) )d(x m(k) 1,x m(k) ), d(x m(k) 1,x n(k) 1 ) d(x n(k) 1,x m(k) )d(x m(k) 1,x n(k) ),d(x d(x m(k) 1,x n(k) 1 ) m(k) 1,x n(k) 1 ) (2.10) N(x n(k) 1,x m(k) 1 ) d(xn(k) 1,fx n(k) 1 )d(x m(k) 1,fx m(k) 1 ),d(x d(x m(k) 1,x n(k) 1 ) m(k) 1,x n(k) 1 ) d(xn(k) 1,x n(k) )d(x m(k) 1,x m(k) ),d(x d(x m(k) 1,x n(k) 1 ) m(k) 1,x n(k) 1 ). (2.11) Letting k in (2.10) (2.11), using (2.5), (2.7), (2.8) (2.9), we have lim M(x n(k) 1,x m(k) 1 ) 1,ǫ,ǫ = ǫ (2.12) k lim N(x n(k) 1,x m(k) 1 ) 1,ǫ = ǫ. (2.13) k From (2.1), using (2.10) (2.11), we have φ(d(x m(k),x n(k) )) = φ(d(fx m(k) 1,fx n(k) 1 )) φ(m(x n(k) 1,x m(k) 1 )) ψ(n(x n(k) 1,x m(k) 1 )). Taking limit on both the sides using (2.6), (2.12) (2.13), the property of ψ the continuity of φ, we have φ(ǫ) φ(ǫ) lim ψ(n(x n(k) 1,x m(k) 1 )), k

9 FIXED POINT POINTS OF RATIONAL TYPE that is, lim k ψ(n(x n(k) 1,x m(k) 1 )) 1, which is a contradiction by property of ψ. Thus x n is a multiplicative Cauchy sequence in X. Since X is complete, there exists u X such that lim x n = u. Then using continuity of f, we get ( ) fu = f lim x n = lim x n+1 = u. Hence u is a fixed point of f. Finally, we shall prove the uniqueness of the fixed point of f. Suppose that u v (u v) be two fixed points of f. Consider where φ Φ, ψ Ψ φ(d(u,v)) = φ(d(fu,fv)) φ(m(u,v)) ψ(n(u,v)), d(u,fu)d(v,fv) M(u,v) d(u, v) d(u,u)d(v,v) Therefore, we have = d(u,v) d(u, v), d(v,fu)d(u,fv) d(u, v), d(v,u)d(u,v),d(u,v) d(u, v),d(u,v) d(u,fu)d(v,fv) N(u,v),d(u,v) = d(u,v). d(u, v) φ(d(u,v)) = φ(d(fu,fv)) φ(d(u,v)) ψ(d(u,v)), which implies that ψ(d(u,v)) 1, which is a contraction by definition of ψ. Hence u = v. Therefore f has a unique fixed point. This completes the proof. Next we prove the following result without the condition of continuity of f. Theorem 2.2. Let f be a mapping of a complete multiplicative metric space (X,d) into itself such that for all x,y X φ(d(fx,fy)) φ(m(x,y)) ψ(n(x,y)),

10 602 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang where φ Φ ψ Ψ, d(x,fx)d(y,fy) M(x,y), d(y,fx)d(x,fy),d(x,y) d(x, y) d(x, y) Then f has a unique fixed point. d(x,fx)d(y,fy) N(x,y),d(x,y). d(x, y) Proof. From the proof of Theorem 2.1 x n is a multiplicative Cauchy sequence hence there exists u X such that lim x n = u. Suppose that fu u, that is, d(u,fu) > 1. Consider φ(d(fx n,fu)) φ(m(x n,u)) ψ(n(x n,u)), (2.14) where M(x n,u) d(xn,fx n )d(u,fu), d(u,fx n)d(x n,fu) d(x n,u) d(x n,u) d(xn,x n+1 )d(u,fu) d(x n,u),d(x n,u), d(u,x n+1)d(x n,fu),d(x n,u) d(x n,u) d(xn,fx n )d(u,fu) N(x n,u),d(x n,u) d(x n,u) d(xn,x n+1 )d(u,fu),d(x n,u). d(x n,u) Letting n in (2.15) (2.16),we have (2.15) (2.16) lim M(x n,u) = d(u,fu) > 1 lim N(x n,u) = d(u,fu) > 1. Again letting n in (2.14), using (2.15), (2.16) property of φ ψ, we have φ(d(u,fu)) φ(d(u,fu)) lim ψ(n(x n,u)), which implies that lim ψ(n(x n,u)) 1, which is a contradiction by property of ψ. Therefore, fu = u hence u is a fixed point of f. Uniqueness easily follows from Theorem 2.1. This completes the proof.

11 FIXED POINT POINTS OF RATIONAL TYPE Corollary 2.3. Let f be a mapping of a complete multiplicative metric space (X,d) into itself such that for all x,y X φ(d(fx,fy)) φ(n(x,y)) ψ(n(x,y)), where φ Φ, ψ Ψ, d(x,fx)d(y,fy) N(x,y),d(x,y). d(x, y) Then f has a unique fixed point. Corollary 2.4. Let f be a mapping of a complete multiplicative metric space (X,d) into itself such that for all x,y X for some k (0,1) d(x,fx)d(y,fy) φ(d(fx,fy)) kmax,d(x,y). d(x, y) Then f has a unique fixed point. Example 2.5. Let X = [0,1] d : X X R + be a multiplicative metric defined as d(x,y) = a x y, where a > 1. Then (X,d) is a complete multiplicative metric space. Define a mapping f : X X as follows fx = 1 for all x X. Define φ,ψ : [1, ) [1, ) as φ(t) = ψ(t) = t. Clearly φ beincreasing φ(1) = 1.Alsoforanysequencex n in[1, ) withx n t > 1, lim ψ(x n) > 1. Also φ(d(fx,fy)) φ(m(x,y)) ψ(n(x,y)) holds. Hence all the conditions of Theorem 2.1 Theorem 2.2 are satisfied. Also f has a fixed point 1. Theorem 2.6. Let f g be mappings of a multiplicative metric space (X,d) into itself such that f(x) g(x) for all x,y X φ(d(fx,fy)) φ(m(x,y)) ψ(n(x,y)), where φ Φ ψ Ψ, d(fx,gx)d(fy,gy) M(x,y), d(gy,fx)d(gx,fy),d(gx,gy) d(gx, gy) d(gx, gy) d(fx,gx)d(fy,gy) N(x,y),d(gx,gy). d(gx, gy) Assume that g(x) is complete. Then f g have a coincidence fixed point. Moreover if f g are weakly compatible, then f g have a unique common fixed point.

12 604 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang Proof. Let x 0 X be arbitrary. Since f(x) g(x), there exists x 1 X such that fx 0 = gx 1. Again for x 1 X, there exists x 2 X such that fx 1 = gx 2. Continuing like this, we can define a sequence y n in X such that y n = fx n = gx n+1. If y n = y n+1, then clearly f g have a coincidence point. Since y n = fx n = gx n+1 = y n+1 = fx n+1 = gx n+2, which implies that x n+1 is a coincidence point of f g. Now we assume that y n y n+1, that is, d(y n,y n+1 ) > 1. Consider φ(d(y n,y n+1 )) = φ(d(fx n,fx n+1 )) φ(m(x n,x n+1 )) ψ(n(x n,x n+1 )), where φ Φ, ψ Ψ, M(x n,x n+1 ) d(fxn,gx n )d(fx n+1,gx n+1 ), d(gx n,gx n+1 ) d(gx n+1,fx n )d(gx n,fx n+1 ),d(gx n,gx n+1 ) d(gx n,gx n+1 ) d(yn,y n 1 )d(y n+1,y n ), d(y n,y n )d(y n 1,y n+1 ),d(y n 1,y n ) d(y n 1,y n ) d(y n 1,y n ) d(y n+1,y n ),d(y n+1,y n ),d(y n 1,y n ) d(fxn,gx n )d(fx n+1,gx n+1 ) N(x n,x n+1 ),d(gx n,gx n+1 ) d(gx n,gx n+1 ) d(yn,y n 1 )d(y n+1,y n ),d(y n 1,y n ) d(y n 1,y n ) d(y n+1,y n ),d(y n 1,y n ). Ifd(y n+1,y n ) > d(y n 1,y n ),thenm(x n,x n+1 ) = N(x n,x n+1 ) = d(y n+1,y n ). Then we have φ(d(y n,y n+1 )) φ(d(y n+1,y n )) ψ(d(y n+1,y n )),

13 FIXED POINT POINTS OF RATIONAL TYPE which implies that ψ(d(y n+1,y n )) 1, which is a contradiction. Therefore, M(x n,x n+1 ) = N(x n,x n+1 ) = d(y n 1,y n ) we have which implies that φ(d(y n,y n+1 )) φ(d(y n 1,y n )) ψ(d(y n 1,y n )), (2.17) φ(d(y n,y n+1 )) φ(d(y n 1,y n )) ψ(d(y n 1,y n )) φ(d(y n 1,y n )). Since φ is increasing, therefore d(y n,y n+1 ) < d(y n 1,y n ), which implies that d(y n,y n+1 ) is decreasing sequence bounded below. Hence it converges to some positive number r 1. Now we prove that r = 1. Letting n in (2.17), usingthe continuity of φ, we have φ(r) φ(r) lim ψ(d(y n 1,y n )), which implies that lim ψ(d(y n 1,y n )) 1, which is is a contradiction by property of ψ. Hence r = 1. Therefore, we have lim d(y n,y n+1 ) = 1. (2.18) Next to show that y n is a multiplicative Cauchy sequence. Suppose that y n is not a multiplicative Cauchy sequence. Then there exists an ǫ > 1 for which we can find two sequences of positive integers m(k) n(k) such that forall positive integers k, n(k) > m(k) k d(y m(k),y n(k) ) ǫ. Assume that n(k) is the smallest such positive integer, we get n(k) > m(k) k, Now, d(y m(k),y n(k) ) ǫ d(y m(k),y n(k) 1 ) < ǫ. ǫ d(y m(k),y n(k) ) d(y m(k),y n(k) 1 ) d(y n(k) 1,y n(k) ). Letting k using (2.18), we have Again, lim d(y m(k),y n(k) ) = ǫ. (2.19) d(y m(k) 1,y n(k) 1 ) d(y m(k) 1,y m(k) ) d(y m(k),y n(k) ) d(y n(k),y n(k) 1 )

14 606 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang d(y m(k),y n(k) ) d(y m(k),y m(k) 1 ) d(y m(k) 1,y n(k) 1 ) d(y n(k) 1,y n(k) ). Letting k in above inequality using (2.18) (2.19), we have Again lim d(y m(k) 1,y n(k) 1 ) = ǫ. (2.20) d(y n(k) 1,y m(k) ) d(y m(k) 1,y n(k) 1 ) d(y m(k) 1,y m(k) ) d(y m(k) 1,y n(k) 1 ) d(y m(k) 1,y m(k) ) d(y n(k) 1,y m(k) ). Letting k in the above inequalities using (2.20), we have lim d(y n(k) 1,y m(k) ) = ǫ. k Similarly, we have Consider lim d(y m(k) 1,y n(k) ) = ǫ. k where φ(d(y n(k),y m(k) )) = φ(d(fx n(k),fx m(k) )) φ(m(x n(k),x m(k) )) ψ(n(x n(k),x m(k) )), (2.21) M(x n(k),x m(k) ) d(fxn(k),gx n(k) )d(fx m(k),gx m(k) ), d(gx n(k),gx m(k) ) d(gx m(k),fx n(k) )d(gx n(k),fx m(k) ),d(gx d(gx n(k),gx m(k) ) n(k),gx m(k) ) d(yn(k),y n(k) 1 )d(y m(k),y m(k) 1 ), d(y n(k) 1,y m(k) 1 ) d(y m(k) 1,y n(k) )d(y n(k) 1,y m(k) ),d(y d(y n(k) 1,y m(k) 1 ) n(k) 1,y m(k) 1 ) (2.22)

15 FIXED POINT POINTS OF RATIONAL TYPE N(x n(k),x m(k) ) d(fxn(k),gx n(k) )d(fx m(k),gx m(k) ) d(gx n(k),gx m(k) ) d(yn(k),y n(k) 1 )d(y m(k,y m(k) 1 ) d(y n(k) 1,y m(k) 1 ),d(gx n(k),gx m(k) ),d(y n(k) 1,y m(k) 1 ). (2.23) Letting n in (2.22) (2.23), we have lim M(x n(k),x m(k) ) = ǫ lim N(x n(k),x m(k) ) = ǫ. k k Letting k in (2.21), we have φ(ǫ) φ(ǫ) lim ψ(n(x n(k),x m(k) )), which implies that lim k ψ(n(x n(k),x m(k) )) 1, which is a contradiction. Hence y n is a multiplicative Cauchy sequence. Since g(x) is complete, there exists z g(x) such that y n z as n. Let u X such that gu = z. Now consider φ(d(fx n,fu)) φ(m(x n,u)) ψ(n(x n,u)), (2.24) where M(x n,u) d(fxn,gx n )d(fu,gu) d(gx n,gu) d(yn,y n 1 )d(fu,gu) d(y n 1,gu), d(gu,fx n)d(gx n,fu) d(gx n,gu),d(gx n,gu), d(gu,y n)d(y n 1,fu),d(y n 1,gu) d(y n 1,gu) (2.25) d(fxn,gu)d(fu,gu) N(x n,u),d(gx n,gu) d(gx n,gu) d(yn,y n 1 )d(fu,gu),d(y n 1,gu). d(y n 1,gu) Letting n in (2.25) (2.26), we have (2.26) lim M(x n,u) = d(gu,fu) lim N(x n,u) = d(gu,fu).

16 608 Y.C. Kwun, P. Nagpal, S.K. Garg, S. Kumar, S.M. Kang Letting n in (2.24), we have φ(d(gu,fu)) φ(d(gu,fu)) lim ψ(n(x n,u)), which implies that lim ψ(n(x n,u)) 1, which is a contradiction. Thus d(gu,fu) = 1, that is, gu = fu. Hence u is a coincidence point of f g. Now we prove that the coincidence point of f g is unique. Suppose u v (u v) is two coincidence points of f g. Consider φ(d(fv,fu)) φ(m(v,u)) ψ(n(v,u)), (2.27) where M(x n,u) d(fv,gv)d(fu,gu), d(gu,fv)d(gx n,fu),d(gv,gu) d(gv, gu) d(gv, gu) = d(v,u) d(fv,gu)d(fu,gu) N(v,u),d(gv,gu) d(gv, gu) = d(v,u). Using (2.28) (2.29) in (2.27), we have (2.28) (2.29) φ(d(v,u)) φ(d(v,u)) ψ(d(v,u)), which implies that ψ(d(v, u)) 1, which is a contradiction. Hence u = v. Therefore the coincidence point of f g is unique. Finally suppose that f g are weakly compatible. Since fu = gu = p, hence fgu = gfu, that is, fp = gp. Using uniqueness of the coincidence point of f g, we have p = u. Thus u = fu = gu. Hence u is a unique common fixed point of f g. This completes the proof. Remark 2.7. In Theorem 2.6, if g = I (: the identity mapping), then we obtain Theorem 2.2.

17 FIXED POINT POINTS OF RATIONAL TYPE Acknowledgments This study was supported by research funds from Dong-A University. References [1] M. Abbas, B. Ali, Y.I. Suleiman, Common fixed points of locally contractive mappings in multiplicative metric spaces with application, Int. J. Math. Math. Sci., 2015 (2015), Article ID , 7 pages. doi: /2015/ [2] A.E. Bashirov, E.M. Kurplnara, A. Ozyapici, Multiplicative calculus its applications, J. Math. Anal. Appl., 337 (2008), doi: /j.jmaa [3] X. He, M. Song, D. Chen, Common fixed points for weak commutative mappings on a multiplicative metric space, Fixed Point Theory Appl., 48 (2014), 9 pages. doi: / [4] G. Jungck, Common fixed points for non-continuous non-self mappings on non-metric spaces, Far East J. Math. Sci., 2 (1996), [5] M.S. Khan, M. Swaleh, S. Sessa, Fixed points theorems by altering distances between the points, Bull. Austral. Math. Soc., 30 (1984) 1-9. [6] M. Özavsar, A.C. Çevikel, Fixed points of multiplicative contraction mappings on multiplicative metric spaces, arxiv: v1 [math.gm], [7] M. Sarwar, R. Badshah-e, Some unique fixed point theorems in multiplicative metric space, arxiv: v2 [math.gm], 2014.

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