NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE PROPERTY IN SYMMETRIC SPACES AND FIXED POINTS

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1 TWMS J. App. Eng. Math. V.8, N.1, 018, pp NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE PROPERTY IN SYMMETRIC SPACES AND FIXED POINTS SUMIT CHANDOK 1, DEEPAK KUMAR, Abstract. In this paper, some sufficient conditions are provided for the existence of fixed points for two pairs of non-self mappings satisfying CLR property without the condition of continuity on mappings in the framework of symmetric spaces. Several interesting corollaries are also deduced. Some examples are also provided which illustrate the usability of the results obtained. Keywords: weak compatible mappings, (φ, ψ)-weak rational type contraction mappings, metric space. AMS Subject Classification: 47H10; 54H5; 46J10; 46J Introduction The classical Banach s contraction principle given by S. Banach proved to be an effective tool which assures the existence and uniqueness of fixed points in complete metric spaces. Besides offering a constructive procedure to compute the fixed points of the underlying mappings this principle has played a major role in the development of nonlinear analysis. A vast amount of mathematical activites has been carried out in various branches of mathematics and in economics, life sciences, physical sciences, engineering, computer science, and others by using Banach s contraction principle by the several authors. Fixed point theorems for various types of nonlinear contractive mappings in different abstract spaces have been scrutinized extensively by various researchers (see [1]-[17] and the references cited therein). Sometimes one may come across situations the full force of metric requirements are not used in the proofs of certain metrical fixed point theorems. Motivated by this fact, several authors obtained fixed point and common fixed point results in symmetric and semi-metric space. In this paper, we establish some theorems of common fixed point for two pairs of non-self weakly compatible mappings using CLR (common limit range) property satisfying a generalized rational type contractive condition in the setting of symmetric spaces. To discuss the importance of the results obtained some illustrative examples have been discussed at the last.. Preliminaries In this section, definitions and notations are defined which are required to establish the results. Definition.1. A symmetric function d 1 defined on a nonempty set X as d 1 : X X [0, ) satisfies the following conditions: (1) d 1 (s 1, t 1 ) = 0 if and only if s 1 = t 1, for s 1, t 1 X; 1 School of Mathematics, Thapar University, Patiala , Punjab, India. sumit.chandok@thapar.edu; ORCID: IKG Punjab Technical University, Jalandhar-Kapurthala Highway, Kapurthala , Punjab, India. Department of Mathematics, Lovely Professional University, Phagwara, Punjab , India. deepakanand@live.in; ORCID: Manuscript received: October 10, 0; accepted: January 17, 017. TWMS Journal of Applied and Engineering Mathematics, Vol.8, No.1 c Işık University, Department of Mathematics, 018; all rights reserved. 0

2 S. CHANDOK, D. KUMAR : NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE... 1 () d 1 (s 1, t 1 ) = d 1 (t 1, s 1 for s 1, t 1 X. A symmetric space (X, d 1 ) is a toplogical space whose topology is induced by symmetric d 1. The main difference between a metric and a symmetric space is of triangle inequality. Since a symmetric space may not be Hausdorff, therefore to prove fixed point theorems we required some additional axioms. which are discussed in [1, 3, 4, 5, 6, 17]. From now on (X, d 1 ) stands for a symmetric space, as X denotes an arbitrary non-empty set. Then (W 3 ) [17] for given t n, s and t in X, lim n d 1 (t n, s) = 0 and lim n d 1 (t n, t) = 0 imply s = t; (W 4 ) [17] for given t n, s n and s in X, lim n d 1 (t n, s) = 0 and lim n d 1 (t n, s n ) = 0 imply lim n d 1 (s n, s) = 0; (HE) [1] for given t n, s n and s in X, lim n d 1 (t n, s) = 0 and lim n d 1 (s n, s) = 0 imply lim n d 1 (t n, s n ) = 0; (1C) [5] a symmetric d 1 is said to be 1-continuous if lim n d 1 (t n, s) = 0 implies lim n d 1 (t n, s) = d 1 (s, t) t n is a sequence in X and s, t X; (CC) [5] a symmetric d 1 is said to be continuous if lim n d 1 (t n, s) = 0 and lim n d 1 (s n, t) = 0 implies lim n d 1 (t n, s n ) = d 1 (s, t) t n, s n are sequences in X and s, t X. Here, it can be seen that (1C) (W 3 ) and (W 4 ) (W 3 ) but the converse implications i.e (W ) (1C) and (W 3 ) (W 4 ) are not true. In general, all other possible implications amongst (W 3 (1C) and (HE) are not true. However, (CC) implies (W 3 (W 4 (HE) and (1C). For detail, we refer to an interesting note written by Cho et al.[4] which contains few illustrative examples. Employing these axioms, several authors have proved some common fixed point theorems in the framework of semi-metric and symmetric spaces (see, e.g.,[1, 3, 4, 5, 6, 17]). It is noted that if (X, d 1 ) is a cone metric space over a normal cone and d = d 1 then (X, d ) is a symmetric space which satisfies axioms (CC) but not in general a metric space (see [10] ). Some results on fixed point were obtained in the framework of cone symmetric spaces.[15]. Definition.. Let (X, d 1 ) be symmetric space. The mappings P and Q: X X are said to be (1) commuting if P Qt = QP t for all t X; () compatible [11] if lim n d 1 (P Qt n, QP t n ) = 0 for each sequence t n in X such that lim n P t n = lim n Qt n ; (3) non-compatible [14] if a sequence t n exists in X such that lim n P t n = lim n Qt n, but lim n d 1 (P Qt n, QP t n ) is either nonzero or nonexistent; (4) weakly compatible [1] if P and Q commute at their coincidence points, that is, P Qt = QP t whenever P t = Qt, for some t X. Definition.3. [9] Two families P i m i=1 and Q k n k=1 of self mappings are said to pairwise commuting if (1) P i P j = P j P i for all i, j 1,,..., m; () Q k Q l = Q l Q k for all k, l 1,,..., n; (3) P i Q k = Q k P i for all i 1,,..., m and k 1,,..., n. Some definitions for non-self mappings are given below; Definition.4. Let Y be an arbitrary set, (X, d) be a symmetric space and let P, Q, R, S be mappings from Y into X. Then (1) the pair (P, R) is said to satisfy (E.A) property [] if there exists a sequence t n in Y such that lim n P t n = lim n Rt n = z 1, for some z 1 X; () the pairs (P, R) and (Q, S) are said to share the common (E.A) property [13], if there exists two sequences t n and s n in Y such that lim n P t n = lim n Rt n = lim n Qs n = lim n Ss n = z 1, for some z 1 X; (3) the pairs (P, R) is said to have (CLR R )) property [] if there exists a sequence t n in Y such that lim n P t n = lim n Rt n = z, for some z 1 R(Y );

3 TWMS J. APP. ENG. MATH. V.8, N.1, 018 (4) the pairs (P, R) and (Q, S) are said to share (CLR RS ) property [7] if there exit two sequences t n and s n in Y such that lim n P t n = lim n Rt n = lim n Qs n = lim n Ss n = z 1, for some z 1 R(Y ) S(Y ). Remark.1. Note that (1) If we set P = Q and R = S, then condition (4) reduces to condition (3). () Evidently, (CLR RS ) property implies the common (E.A) property but converse is not true. 3. Main Results To begin our main section, we have some notations which will be required in the sequel to establish the results. Let Ψ be the collection of all functions ψ : [0, ) [0, ) satisfying the following conditions: (i) ψ is continuous as well as monotonically nondecreasing function on [0, ); (ii) ψ(t 1 ) = 0 if and only if t 1 = 0. Let Θ be the collection of all functions α : [0, ) [0, ) satisfying the following conditions: (i) α is continuous as well as monotonically non decreasing on [0, ) ; (ii) α(t 1 ) = 0 if and only if t 1 = 0 and α(t 1 ) t 1 for all t 1 > 0. Let Φ be the collection of all functions ϕ : [0, ) [0, ) satisfying the following conditions: (i) ϕ is continuous on [0, ); (ii) ϕ(t 1 ) = 0 if and only if t 1 = 0. we begin with the following lemma. Lemma 3.1. Let (X, d 1 ) be symmetric space d 1 satisfies the condition (CC) while Y be an arbitrary non- empty set with P, Q, R, S : Y X. Suppose that the following hypotheses hold: (1) the pair (P, R) satisfies the (CLR R ) property [respectively the pair (Q, S) satisfies the (CLR S ) property]; () P (Y ) S(Y ) [respectively Q(Y ) R(Y )]; (3) S(Y ) [respectively, R(Y )] is a closed subset of X; (4) Qy n converges for every sequence y n in Y such that Sy n converges [repectively P x n converges for every sequence x n in Y such that Rx n converges]; (5) for all t 0, ψ(t) ϕ(t) 0 and ψ(t) ϕ(t) = 0 t = 0, for some ψ Ψ, ϕ Φ, α Θ, (6) the mappings P, Q, R and S satisfy, for some ψ Ψ, ϕ Φ, α Θ, ψ(d 1 (P x, Qy)) α(m(x, y)) β(ϕ(d 1 (P x, Rx) ϕ(d 1 (P x, Sy))) (3.1) M(x, y) = max ϕ(d 1 (P x, Qy) ϕ(d 1 (P x, Rx) ϕ(d 1 (Qy, Sy) ϕ( d 1(P x, Sy) + d 1 (Rx, Qy) (3.) ϕ( 1 + d 1(P x, Rx) 1 + d 1 (P x, Qy) d 1(Qy, Sy) ϕ( 1 + d 1(Qy, Sy) 1 + d 1 (P x, Qy) d 1(P x, Rx) ϕ( 1 + d 1(P x, Sy) + d 1 (Rx, Qy) 1 + d 1 (P x, Rx) + d 1 (Qy, Sy) d 1(P x, Rx)) and β : [0, ) [0, ) is a continuous mapping such that β(x, y) = 0 if and only if x = 0, y = 0 for all x, y X. Then the pair (P, R) and (Q, S) share the (CLR RS ) property. Proof. Suppose that (CLR R ) property for the pair (P, R) holds. So, there exists a sequence of x n in Y such that lim n P x n = lim n Rx n = z, z R(Y ). As S(Y ) is a closed subset of X, P (Y ) S(Y ) and for each x n Y there corresponds a sequence y n Y such that P x n = Sy n. Therefore, lim n Sy n = lim n P x n = z, z R(Y ) S(Y ). Thus, we have lim n d 1 (P x n, z) = lim n d 1 (Rx n, z) = lim n d 1 (Sy n, z) = 0 Therefore, we have lim n d 1 (P x n, Rx n ) = 0 and lim n d 1 (Rx n, Sy n ) = 0.

4 S. CHANDOK, D. KUMAR : NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE... 3 By (4 sequence Qy n converges. Now, we need to show that Qy n z as n. By (CC we get lim n d 1 (P x n, Qy n ) = lim n d 1 (z, lim n Qy n ), lim n d 1 (Rx n, Qy n ) = lim n d 1 (z, lim n Qy n ) and lim n d 1 (Qy n, Sy n ) = d 1 (lim n Qy n z). By inserting x = x n and y = y n in equation (3.1) we get M(x n, y n ) = max ψ(d 1 (P x n, Qy n )) α(m(x n, y n )) β(ϕ(d 1 (P x n, Rx n ) ϕ(d 1 (P x n, Sy n )) (3.3) ϕ(d 1 (P x n, Qy n ) ϕ(d 1 (P x n, Rx n d 1 (Qy n, Sy n ) ϕ( d 1(P x n, Sy n ) + d 1 (Rx n, Qy n ) Taking the limit n in equation (3.3 we have ϕ( 1 + d 1(P x n, Rx n ) 1 + d 1 (P x n, Qy n ) d 1(Qy n, Sy n ) ϕ( 1 + d 1(Qy n, Sy n ) 1 + d 1 (P x n, Qy n ) d 1(P x n, Rx n ) ϕ( 1 + d 1(P x n, Sy n ) + d 1 (Rx n, Qy n ) 1 + d 1 (P x n, Rx n ) + d 1 (Qy n, Sy n ) d 1(P x n, Rx n )) ψ(d 1 (z, lim n Qy n)) α( lim n M(x n, y n )) β(ϕ(d 1 (z, z) ϕ(d 1 (z, z)) (3.4) = max 1+d 1(z,z) lim n M(x n, y n ) d1(z,z)+d1(z,limn Qyn) ϕ(d 1 (z, lim n Qy n ) ϕ(d 1 (z, z) ϕ(d 1 (lim n Qy n, z) ϕ( ϕ( 1+d 1(z,lim n Qy d n) 1(lim n Qy n, z) ϕ( 1+d 1(z,lim n Qy d n) 1(z, z) ϕ( 1+d 1(z,z)+d 1(lim n Qy d n,z) 1(z, z)) 1+d1(limn Qyn,z) 1+d1(z,z)+d1(z,limn Qyn) = max ϕ(d 1 (z, lim n) ϕ(0 ϕ(d 1 ( lim n, z) ϕ( d 1(z, lim n Qy n ) n n 1 ϕ( 1 + d 1 (z, lim n Qy n ) d 1( lim n, z) ϕ(0 ϕ(0) n Hence equation (3.4) implies = ϕ(d 1 (z, lim n Qy n)) ψ(d 1 (z, lim n Qy n)) α(ϕ(d 1 (z, lim n Qy n)) β(ϕ(0 ϕ(0) ψ(d 1 (z, lim Qy n)) α(ϕ(d 1 (z, lim Qy n) n n Using property of Θ, we get Using condition (5 we have hence ψ(d 1 (z, lim n Qy n)) ϕ(d 1 (z, lim n Qy n)) ψ(d 1 (z, lim n Qy n)) ϕ(d 1 (z, lim n Qy n)) 0 ψ(d 1 (z, lim n Qy n)) ϕ(d 1 (z, lim n Qy n)) 0 ψ(d 1 (z, lim n Qy n)) ϕ(d 1 (z, lim n Qy n)) = 0 Again by using condition (5 we have d 1 (z, lim n Qy n ) = 0, Hence lim n Qy n = z which shows that the pair (P, R) and (Q, S) share the (CLR RS ) property.

5 4 TWMS J. APP. ENG. MATH. V.8, N.1, 018 Theorem 3.. Let (X, d 1 ) be a symmetric space the symmetric d 1 satisfies the condition (HE) and (1C) and Y be a non-empty set with P, Q, R, S : Y X. Suppose that the conditions (5) and (6) of Lemma 3.1 holds. If the pair (P, R) and (Q, S) share the (CLR RS ) property, then (P, R) and (Q, S) have a coincidence point each. Moreover if Y = X both the pair (P, R) and (Q, S) are weakly compatible then P, Q, R and S have a unique common fixed point. Proof. Since the pair (P, R) and (Q, S) share (CLR RS ) property. Therefore there exist two sequences x n and y n in Y such that lim n P x n = lim n Rx n = lim n Sy n = lim n Qy n = z, z R(Y ) S(Y ). Since z R(Y there exists a pair u Y such that Ru = z. Putting x = u and y = y n in equation (3.1) we get M(u, y n ) = max ψ(d 1 (P u, Qy n )) α(m(u, y n )) β(ϕ(d 1 (P u, Ru) ϕ(d 1 (P u, Sy n ))) (3.5) ϕ(d 1 (P u, Qy n ) ϕ(d 1 (P u, Ru) ϕ(d 1 (Qy n, Sy n ) ϕ( d 1(P u, Sy n ) + d 1 (Ru, Qy n ) ϕ( 1 + d 1(P u, Ru) 1 + d 1 (P u, Qy n ) d 1(Qy n, Sy n ) ϕ( 1 + d 1(Qy n, Sy n ) 1 + d 1 (P u, Qy n ) d 1(P u, Ru) ϕ( 1 + d 1(P u, Sy n ) + d 1 (Ru, Qy n ) 1 + d 1 (P u, Ru) + d 1 (Qy n, Sy n ) d 1(P u, Ru)) Letting n (taking the limit) in equation (3.5) and using the properties (1C) and (HE we have ψ(d 1 (P u, z)) α( lim n M(u, y n)) β( lim n (ϕ(d 1(P u, Ru) ϕ(d 1 (P u, Sy n ))) (3.6) lim M(u, y n) = max n ϕ(d 1 (P u, z) ϕ(d 1 (P u, z) ϕ(d 1 (z, z) ϕ( d 1(P u, z) + d 1 (z, z) ϕ( 1 + d 1(P u, z) 1 + d 1 (P u, z) d 1(z, z) ϕ( 1 + d 1(z, z) 1 + d 1 (P u, z) d 1(P u, z) ϕ( 1 + d 1(P u, z) + d 1 (z, z) 1 + d 1 (P u, z) + d 1 (z, z) d 1(P u, z)) = max ϕ(d 1 (P u, z) ϕ(d 1 (P u, z) ϕ(0 ϕ( d 1(P u, z) From equation (3.6 we obtain = ϕ(d 1 (P u, z)) ψ(d 1 (P u, z)) α(ϕ(d 1 (P u, z))) β(ϕ(d 1 (P u, z) ϕ(d 1 (P u, z))) 1 ϕ(0 ϕ( 1 + d 1 (P u, z) d 1(P u, z) ϕ(d 1 (P u, z)) Using property of Θ, we get ψ(d 1 (P u, z)) α(ϕ(d 1 (P u, z))) ψ(d 1 (P u, z)) ϕ(d 1 (P u, z)) ψ(d 1 (P u, z)) ϕ(d 1 (P u, z)) 0 Using condition 5 of lemma, we have ψ(d 1 (P u, z)) ϕ(d 1 (P u, z)) 0 Therefore ψ(d 1 (P u, z)) ϕ(d 1 (P u, z)) = 0 and hence it follows easily that P u = z. Therefore P u = Ru = z, which shows that u is a coincidence point of the pair (P, R).

6 S. CHANDOK, D. KUMAR : NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE... 5 As z S(Y there exists a point v Y such that Sv = z. Putting x = u and y = v in equation (3.1 we have ψ(d 1 (z, Qv)) = ψ(d 1 (P u, Qv)) α(m(u, v)) β(ϕ(d 1 (z, z) ϕ(d 1 (z, z)) (3.7) M(u, v) = max ϕ(d 1 (P u, Qv) ϕ(d 1 (P u, Ru) ϕ(d 1 (Qv, Sv) ϕ( d 1(P u, Sv) + d 1 (Ru, Qv) ϕ( 1 + d 1(P u, Ru) 1 + d 1 (P u, Qv) d 1(Qv, Sv) ϕ( 1 + d 1(Qv, Sv) 1 + d 1 (P u, Qv) d 1(P u, Ru) ϕ( 1 + d 1(P u, Sv) + d 1 (Ru, Qv) 1 + d 1 (P u, Ru) + d 1 (Qv, Sv) d 1(P u, Ru)) = max ϕ(d 1 (z, Qv) ϕ(d 1 (z, z) ϕ(d 1 (Qv, z) ϕ( d 1(z, z) + d 1 (z, Qv) ϕ( 1 + d 1(z, z) 1 + d 1 (z, Qv) d 1(Qv, z) ϕ( 1 + d 1(Qv, z) 1 + d 1 (z, Qv) d 1(z, z))ϕ( 1 + d 1(z, z) + d 1 (z, Qv) 1 + d 1 (z, z) + d 1 (Qv, z) d 1(z, z)) = max ϕ(d 1 (z, Qv) ϕ(0 ϕ(d 1 (Qv, z) ϕ( d 1(z, Qv) Hence the equation (3.7) implies Using property of Θ, we get Using condition 5 of lemma, we have Hence = ϕ(d 1 (z, Qv)) ψ(d 1 (z, Qv)) α(ϕ(d 1 (z, Qv))) β(ϕ(0 ϕ(0) ψ(d 1 (z, Qv)) α(ϕ(d 1 (z, Qv))) ψ(d 1 (z, Qv)) ϕ(d 1 (z, Qv)) ψ(d 1 (z, Qv)) ϕ(d 1 (z, Qv)) 0, ψ(d 1 (z, Qv)) ϕ(d 1 (z, Qv)) 0, ψ(d 1 (z, Qv)) ϕ(d 1 (z, Qv)) = 0, 1 ϕ( 1 + d 1 (z, Qv) d 1(Qv, z) ϕ(0 ϕ(0) and it follows easily that z = Qv. Thus Qv = Sv = z, which shows that v is a coincidence point of the pair (Q, S). Suppose now that Y = X. Since the pair (P, R) and (Q, S) are weakly compatible, P u = Ru and Qv = Sv, therefore P z = P Ru = RP u = Rz and Qz = QSv = SQv = Sz. Putting x = z and y = v in equation (3.1 we have ψ(d 1 (P z, z)) = ψ(d 1 (P z, Qv)) α(m(z, v)) β(ϕ(d 1 (P z, Rz) ϕ(d 1 (P z, Sv)) (3.8) M(z, v) = max ϕ(d 1 (P z, Qv) ϕ(d 1 (P z, Rz) ϕ(d 1 (Qv, Sv) ϕ( d 1(P z, Sv) + d 1 (Rz, Qv) ϕ( 1 + d 1(P z, Rz) 1 + d 1 (P z, Qv) d 1(Qv, Sv) ϕ( 1 + d 1(Qv, Sv) 1 + d 1 (P z, Qv) d 1(P z, Rz) ϕ( 1 + d 1(Az, T v) + d 1 (Sz, Bv) 1 + d 1 (Az, Sz) + d 1 (Bv, T v) d 1(Az, Sz))

7 6 TWMS J. APP. ENG. MATH. V.8, N.1, 018 = max ϕ(d 1 (P z, z) ϕ(d 1 (P z, P z) ϕ(d 1 (z, z) ϕ( d 1(P z, z) + d 1 (P z, z) ϕ( 1 + d 1(P z, P z) 1 + d 1 (P z, z) d 1(z, z) ϕ( 1 + d 1(z, z) 1 + d 1 (P z, z) d 1(P z, P z) ϕ( 1 + d 1(P z, z) + d 1 (P z, z) 1 + d 1 (P z, z) + d 1 (z, z) d 1(P z, P z)) = max ϕ(d 1 (P z, z) ϕ(0 ϕ(0 ϕ(d 1 (P z, z) ϕ(0 ϕ(0 ϕ(0) = ϕ(d 1 (P z, z)) From equation (3.8 we get ψ(d 1 (P z, z)) α(ϕ(d 1 (P z, z))) β(ϕ(d 1 (P z, P z) ϕ(d 1 (P z, z))) ψ(d 1 (P z, z)) α(ϕ(d 1 (P z, z))) β(ϕ(0 ϕ(d 1 (P z, z))) Using property of Θ, we get ψ(d 1 (P z, z)) α(ϕ(d 1 (P z, z))) ψ(d 1 (P z, z)) ϕ(d 1 (P z, z)) Using condition 5 of lemma, we have Hence, ψ(d 1 (P z, z)) ϕ(d 1 (P z, z)) 0 ψ(d 1 (P z, z)) ϕ(d 1 (P z, z)) 0 ψ(d 1 (P z, z)) ϕ(d 1 (P z, z)) = 0 Using condition 5 of Lemma 3.1 it follows easily that z = P z = Rz and therefore z is a common fixed point of the pair (P, R). Putting x = u and y = z in equation (3.1 we have ψ(d 1 (z, Qz)) = ψ(d 1 (P u, Qz)) α(m(u, z)) β(ϕ(d 1 (P u, Ru) ϕ(d 1 (P u, Sz))) (3.9) M(u, z) = max ϕ(d 1 (P u, Qz) ϕ(d 1 (P u, Ru) ϕ(d 1 (Qz, Sz) ϕ( d 1(P u, Sz) + d 1 (Ru, Qz) ϕ( 1 + d 1(P u, Ru) 1 + d 1 (P u, Qz) d 1(Qz, Sz) ϕ( 1 + d 1(Qz, Sz) 1 + d 1 (P u, Qz) d 1(P u, Ru) ϕ( 1 + d 1(P u, Sz) + d 1 (Ru, Qz) 1 + d 1 (P u, Ru) + d 1 (Qz, Sz) d 1(P u, Ru)) = max ϕ(d 1 (z, Qz) ϕ(d 1 (z, z) ϕ(d 1 (Qz, Qz) ϕ( d 1(z, Qz) + d 1 (z, Qz) ϕ( 1 + d 1(z, z) 1 + d 1 (z, Qz) d 1(Qz, Qz) ϕ( 1 + d 1(Qz, Qz) 1 + d 1 (z, Qz) d 1(z, z) ϕ( 1 + d 1(z, Qz) + d 1 (z, Qz) d 1 (z, z)) 1 + d 1 (z, z) + d 1 (z, z) = max ϕ(d 1 (z, Qz) ϕ(0 ϕ(0 ϕ(d 1 (z, Qz) ϕ(0 ϕ(0 ϕ(0) = ϕ(d 1 (z, Qz)) From equation (3.9) we get ψ(d 1 (z, Qz)) α(ϕ(d 1 (z, Qz))) β(ϕ(d 1 (z, z) ϕ(d 1 (z, Qz))) ψ(d 1 (z, Qz)) α(ϕ(d 1 (z, Qz)))

8 S. CHANDOK, D. KUMAR : NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE... 7 Using property of Θ, we get ψ(d 1 (z, Qz)) ϕ(d 1 (z, Qz)) Using condition 5 of lemma, we have Hence, ψ(d 1 (z, Qz)) ϕ(d 1 (z, Qz)) 0 ψ(d 1 (z, Qz)) ϕ(d 1 (z, Qz)) 0 ψ(d 1 (z, Qz)) ϕ(d 1 (z, Qz)) = 0 Using condition 5 of Lemma 3.1 it follows easily that z = Qz. Therefore Qz = Sz = z and we can conclude that z is a common fixed point of P, Q, R and S. The uniqueness of a common fixed point can be easily checked using the equation (3.1). Theorem 3.3. Let (X, d 1 ) be symmetric space, the symmetric d 1 satisfies the condition (CC and Y be an arbitrary non-empty set with P, Q, R, S : Y X. Suppose that the conditions (1 6) of Lemma 3.1 hold. Then (P, R) and (Q, S) have coincidence point each. Moreover, if Y = X and both the pairs (P, R) and (Q, S) are weakly compatible then P, Q, R and S have a unique common fixed point. Proof. In view of Lemma 3.1, the pairs (P, R) and (Q, S) share the (CLR RS ) property, therefore there exist two sequences x n and y n in Y such that lim n P x n = lim n Rx n = lim n Sy n = lim n Qy n = z, z R(Y ) S(Y ). The rest proof can be done on the same lines of Theorem 3.. Theorem 3.4. Let (X, d 1 ) be symmetric space, the symmetirc d 1 satisfies the conditions (HE) and (1C and let Y be an arbitrary non-empty set with P, Q, R, S : Y X. Suppose that the conditions (5) and (6) of Lemma 3.1 and the following hypotheses hold: (1) the pairs (P, R) and (Q, S) satisfy the common property (E.A); () R(Y ) and S(Y ) are closed subsets of X. Then (P, R) and (Q, S) have coincidence point each. Moreover, if Y = X and both the pairs (P, R) and (Q, S) are weakly compatible then P, Q, R and S have a unique common fixed point. Proof. If the pairs (P, R) and (Q, S) share the common property (E.A then there exist two sequences x n and y n in Y such that lim n P x n = lim n Rx n = lim n Sy n = lim n Qy n = z, for some z X. Since R(Y ) is closed, lim n Rx n = z = Ru for some u Y. Also, since S(Y ) is closed, then lim n Sy n = z = Sν for some ν Y. The rest proof can be done on the same lines of Theorem 3.. Corollary 3.5. The conclusion of Theorem 3.4 remain same on replacing condition () by the following: ( ) P (Y ) S(Y ) and Q(Y ) R(Y P (Y ) and Q(Y ) denote the closure of ranges of the mappings P and Q. Corollary 3.6. The conclusion of Theorem 3.4 remain same on replacing condition () by the following: ( ) P (Y ) and Q(Y ) are closed subsets of X, and P (Y ) S(Y Q(Y ) R(Y ). Corollary 3.7. Let (X, d 1 ) be symmetric space the symmetric d 1 satisfies the conditions (HE) and (1C and Y be an arbitrary non- empty set with P, R : Y X. Suppose that the given hypotheses hold:

9 8 TWMS J. APP. ENG. MATH. V.8, N.1, 018 (1) the pair (P, R) satisfies the (CLR R ) property; () for some ψ Ψ, ϕ Φ, α Θ and all x, y Y, satisfies ψ(d 1 (P x, P y)) α(m(x, y)) β(ϕ(d 1 (P x, Rx) ϕ(d 1 (P x, Ry))) (3.10) M(x, y) = max ϕ(d 1 (P x, P y) ϕ(d 1 (P x, Rx) ϕ(d 1 (P y, Ry) ϕ( d 1(P x, Ry) + d 1 (Rx, P y) ϕ( 1 + d 1(P x, Rx) 1 + d 1 (P x, P y) d 1(P y, Ry) ϕ( 1 + d 1(P y, Ry) 1 + d 1 (P x, P y) d 1(P x, Rx) ϕ( 1 + d 1(P x, Ry) + d 1 (Rx, P y) 1 + d 1 (P x, Rx) + d 1 (P y, Ry) d 1(P x, Rx)) and β : [0, ) [0, ) is a continuous mapping such that β(x, y) = 0 if and only if x = 0, y = 0 for all x, y Y Then the pair (P, R) has a coincidence point. Moreover, if Y = X and the pair (P, R) is weakly compatible then P and R have a unique common fixed point. Corollary 3.8. Let (X, d 1 ) be symmetric space the symmetric d 1 satisfies the conditions (HE) and (1C and Y be an arbitrary non- empty set with P, Q, T, R, S : Y X. Suppose that the following hypotheses hold: (1) the pairs (P, RT ) and (Q, S) satisfies the (CLR (RT )(S) ) property. () for some ψ Ψ, ϕ Φ, α Θ and all x, y Y, satisfies ψ(d 1 (P x, Qy)) α(m(x, y)) β(ϕ(d 1 (P x, RT x) ϕ(d 1 (P x, Sy))) (3.11) M(x, y) = max ϕ(d 1 (P x, Qy) ϕ(d 1 (P x, RT x) ϕ(d 1 (Qy, Sy) ϕ( d 1(P x, Sy) + d 1 (RT x, Qy) ϕ( 1 + d 1(P x, RT x) 1 + d 1 (P x, Qy) d 1(Qy, Sy) ϕ( 1 + d 1(Qy, Sy) 1 + d 1 (P x, Qy) d 1(P x, RT x) ϕ( 1 + d 1(Ax, T y) + d 1 (SRx, By) 1 + d 1 (Ax, SRx) + d 1 (By, T y) d 1(Ax, SRx)) and β : [0, ) [0, ) is a continuous mapping such that β(x, y) = 0 if and only if x = 0, y = 0 for all x, y Y Then the pairs (P, RT ) and (Q, S) have a coincidence point each. Moreover, if Y = X and both the pairs (P, RT ) and (Q, S) commute pairwise, that is P R = RP, P T = T P, RT = T R, QS = SQ then P, Q, T, R and S have a unique common point. Corollary 3.9. Let P, Q, R and S be self mappings of a symmetric space (X, d 1 ) satisfying the conditions (HE) and (1C). Suppose that, for fixed positive integer m, n, p, q. (1) the pair (P m, R p ) and (Q n, S q ) share the (CLR Rp Sq) property; () for some ψ Ψ, ϕ Φ, α Θ and all x, y Y, satisfies ψ(d 1 (P m x, Q n y)) α(m(x, y)) β(ϕ(d 1 (P m x, R p x) ϕ(d 1 (P m x, S q y))) (3.1)

10 S. CHANDOK, D. KUMAR : NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE... 9 M(x, y) = max ϕ(d 1 (P m x, Q n y) ϕ(d 1 (P m x, R p x) ϕ(d 1 (Q n y, S q y) ϕ( d 1(P m x, S q y) + d 1 (R p x, Q n y) ϕ( 1 + d 1(P m x, R p x) 1 + d 1 (P m x, Q n y) d 1(Q n y, S q y) ϕ( 1 + d 1(Q n y, S q y) 1 + d 1 (P m x, Q n y) d 1(P m x, R p x)), ϕ( 1 + d 1(P m x, S q y) + d 1 (R p x, Q n y) 1 + d 1 (P m x, R p x) + d 1 (Q n y, S q y) d 1(P m x, R p x)) and β : [0, ) [0, ) is a continuous mapping such that β(x, y) = 0 if and only if x = 0, y = 0 for all x, y Y If P R = RP and QS = SQ then P, Q, R and S have a unique common fixed point. Example Let Y = [, 11) [1, ) = X equipped with the symmetric d 1 (s, t) = (s t) for all s, t Y which also satisfies (1C) and (HE). Define the mappings P, Q, R, S : Y X by if s (4, 11) if s (4, 11) P (s) =, Q(s) = 1 if s (, 4] 14 if s (, 4] if s = if s = R(s) = 11 if s (, 4], S(s) = 11 if s (, 4] 3s+ 7 ifs (4, 11) s if s (4, 11) Then we have P (Y ) =, 9 [, 9) 11 = S(Y ) and Q(Y ) =, 5 [, 5) 11 = R(Y ). Consider two sequences s n = n n N and t n = n N. Then the pair (P, R) and (Q, S) satisfies the (CLR RS ) property. Indeed we have lim P s n = lim Rs n = lim Qs n = lim St n = n n n n R(Y ) S(Y ); however, R(Y ) and S(Y ) are not closed subsets of X. Now define φ Φ by φ(t 1 ) = t 1, ψ Φ by ψ(t 1 ) = t1, α Θ by α(t 1) = t 1 and β : [0, ) [0, ) by β(t 1, s 1 ) = t1+s1. In order to verify the inequality (3.1 we distinguish the following possible cases: (1) If s = t =, then we get d 1 (P s, Qt) = 0 and hence ψ(0) = 0 and (3.1) is trivially satisfied; () If s =, t (, 4], then we get d 1 (P s, Qt) = 144 and hence ψ(d 1 (P s, Qt)) = 7, M(s, t) = 144, α(m(s, t)) = 144, β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = 81, therefore R.H.S of the inequality (3.1) is 3 so L.H.S R.H.S; (3) If s =, t (4, 11 d 1 (P s, Qt) = 0 and hence ψ(d 1 (P s, Qt)) = 0, M(s, t) = (4 t), α(m(s, t)) = (4 t), β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = (4 t), therefore R.H.S of the inequality (3.1) is 15(4 t) so L.H.S R.H.S; (4) If s (, 4], t =,then d 1 (P s, Qt) = 100 and hence ψ(d 1 (P s, Qt)) = 50, M(s, t) = 100, α(m(s, t)) = 100, β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = 101, therefore R.H.S of the inequality (3.1) is 1499 so L.H.S R.H.S; (5) If s, t (, 4], then d 1 (P s, Qt) = 4 and hence ψ(d 1 (P s, Qt)) =, M(s, t) = 9, α(m(s, t)) = 9, β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = , therefore R.H.S of the inequality (3.1) is 8 so L.H.S R.H.S;

11 30 TWMS J. APP. ENG. MATH. V.8, N.1, 018 (6) If s (, 4], t (4, 11then d 1 (P s, Qt) = 100 and hence ψ(d 1 (P s, Qt)) = 50, M(s, t) = 100 and β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = 1+(14 t), therefore the R.H.S of the inequality (3.1) is 1599 (4 s) so L.H.S R.H.S; (7) If s (4, 11 t = then, d 1 (P s, Qt) = 0 and hence ψ(d 1 (P s, Qt)) = 0, M(s, t) = (1 3s) 49, α(m(s, t)) = (1 3s) 49, β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = (1 3s) 98, therefore R.H.S of the inequality (3.1) is (1 3s) 98 so L.H.S R.H.S; (8) If s (4, 11 t (, 4] then, d 1 (P s, Qt) = 144 and hence ψ(d 1 (P s, Qt)) = 7, M(s, t) = 144 and β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = (1 3s) , thereforer.h.s of the inequality (3.1) is (1 3s) 784 so L.H.S R.H.S; (9) If s, t (4, 11) then, d 1 (P s, Qt) = 0 and hence ψ(d 1 (P s, Qt)) = 0, M(s, t) = (49+(1 3s) )(4 s) and β(φ(d 1 (P s, Rs φ(d 1 (P s, St)))) = (1 3s) +49(4 s) 98, therefore, in all the cases inequality (3.1) holds; Thus, the above example satisfies all the conditions of Theorem 3., except for Y = X, but a unique common fixed point of the pair (P, R) and (Q, S) is. Here, one may check that all the involved mappings are discontinuous at their unique fixed point. Example In the setting of Example 3.10, replace the mappings R and T by the following, beside retaining the rest: if t = R(t) = S(t) = if t (, 4] 9t+6 7 ift (4, 11) Then we have P (Y ) =, 1 [, ] = S(Y ) and Q(Y ) =, 14 [, ] = R(Y ); Now R(Y ) and S(Y ) are closed subsets of X. Thus, all the conditions of Theorem 3.3 are satisfied, except Y = X; however the unique common fixed point of the pairs (P, R) and (Q, S) is. Example 3.1. In the setting of Example 3.10, replace the mappings R and S by the following, beside retaining the rest: if t = R(t) = S(t) = if t (, 4] 8t+17 7 ift (4, 11) Then we have P (Y ) =, 1 [, 15) = S(Y ) and Q(Y ) =, 14 [, 15) = R(Y ). Now, R(Y ) and S(Y ) are not closed subsets of X, but the conditions ( ) and ( ) of Corollaries 3.5 and 3.6 are satisfied, except Y = X; however the unique common fixed point of the pairs (P, R) and (Q, S) is. Acknowledgements. Authors are thankful to the learned referees for the valuable suggestions. References [1] A. Aliouche, A common fixed point theorem for weakly compatible mappings in symmetric spaces satisfying a contrative condition of integral type, J. Math. Anal. Appl. 3()( [] M. Aamri, D. EI Moutawakil, Common fixed points under contractive conditions in symmetric spaces, Appl. Math. E-notes 3( [3] I.D. Arandjelović, D. Kečkić, Symmetric spaces approach to some fixed point result, Nonlinear Anal. 75( [4] S.H. Cho, G.Y.Lee, J.S. Bae, On coincidence and fixed-point theorems in symmetric spaces, Fixed Point Theory Appl. (008) Article ID 56130, 9 pages. [5] F. Galvin, S.D. Shore, Completeness in semi-metric, Pacific. J. Math. 113(1)(

12 S. CHANDOK, D. KUMAR : NON-SELF MAPPINGS UNDER COMMON LIMIT RANGE [6] M. Imdad, A.H. Soliman, Some common fixed point theorems for a pair of tangential mappings in symmetic spaces, Appl. Math. Lett. 3(4)( [7] M. Imdad, B.D. Pant, S. Chauhan, Fixed points theorems in Menger spaces using (CLRST ) property and applications. J. Nonlinear Anal. Optim. 3()( [8] M. Imdad, S. Chauhan, Z. Kadelburg, C. Vetro, Fixed point theorems for non-self mappings in symmetric spaces under φ-weak contractive conditions and an application to functional equations in dynamic programming, Appl. Math. Comp. 7( [9] M. Imdad, J. Ali, M. Tanveer, Coincidence and common fixed point theorems for nonlinear contractions in Menger PM spaces, Choas Solitons Fractals 4 (5)( [10] M. Jovanovic, Z. Kadelburg, S. Radenovic, Common fixed point results in metric-type spaces, Fixed Point Theory Appl. (010 Article ID pages. [11] G. Jungck, Compatible mappings and common fixed points. Int. J. Math. Math. Sci. 9(4)( [1] G. Jungck, B.E. Rhodes, Fixed points for set valued functions without continuity, Indian J. Pure Appl. Math. 9(3)( [13] Y. Liu, J. Wu, Z. Li, Common fixed points of single-valued and multivalued maps, Int. J. Math. Math. Sci. 19( [14] R.P. Pant, Noncompatible mappings and common fixed points, Soochow J. Math. 6(1)( [15] S. Radenovic, Z. Kadelburg, Quasi-contractions on symmetric and cone symmetric spaces, Banach J. Math. Anal. 5( [] W. Sintunavarat, P. Kumam, Common fixed point theorems for a pair of weakly compatible mappings in fuzzy metric sapces, J. Appl. Math.(011) Article ID pages. [17] W.A. Wilson, On semi-metric space, Am. J. Math. 53( Sumit Chandok obtained his Ph.D. Degree from Guru Nanak Dev University, Amritsar, Punjab, India. He published a book and more than 70 research papers in the International/National journals of high repute on Fixed Point Theory, Approximation Theory and Nonlinear Analysis. Deepak Kumar obtained his M.Sc. Degree from Guru Nanak Dev University, Amritsar, Punjab, India. He published research papers in the International/National journals of high repute on Fixed Point Theory and Nonlinear Analysis.