Common Fixed Point Results in Complex Valued Metric Spaces with Application to Integral Equations

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1 Filomat 8:7 (04), DOI 0.98/FIL407363H Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: Common Fixed Point Results in Complex Valued Metric Spaces with Application to Integral Equations N. Hussain a, Akbar Azam b, Jamshaid Ahmad b, Muhammad Arshad c a Department of Mathematics, King Abdulaziz University P.O. Box 8003, Jeddah 589, Saudi Arabia b Department of Mathematics COMSATS Institute of Information Technology, Chack Shahzad, Islamabad , Pakistan c Department of Mathematics, International Islamic University, H-0, Islamabad , Pakistan Abstract. In this paper, common fixed point of six mappings satisfying a contractive condition involving rational inequality in the framework of complex valued metric space are obtained. Moreover, some examples and applications to integral equations are given here to illustrate the usability of the obtained results.. Introduction and Preliminaries There exist a number of generalizations of metric spaces, and one of them is the cone metric space initiated by Huang and Zhang [9]. They described the convergence in cone metric spaces, introduced the notion of completeness and proved some fixed point theorems of contractive mappings on these spaces. Then several authors [, 4 8,, 4 6] obtained fixed points in different generalized metric spaces. The problem of existence of common fixed points to a pair of nonlinear mappings is now a classical theme. The applications to differential and integral equations made it more interesting. A considerable importance has been attached to common fixed point theorems in ordered spaces [, ]. Azam et al.[3] introduced the concept of complex valued metric space and obtained the existence and uniqueness of common fixed points involving rational expressions. Then Rouzkard et al. [7] and Sintunavarat et al. [8] generalized the concept of Azam et al [3]. The aim of this paper is to extend and generalize common fixed point theorems for six self-maps of Jankovic et al. [3] from cone metric space to complex valued metric space of contractive type mappings involving rational inequality. We will illustrate this fact by proving the existence of nonnegative integrable solutions for an implicit integral equation in complex valued metric spaces. Consistent with Azam, Fisher and Khan [3], the following definitions and results will be needed in what follows. Let C be the set of complex numbers and z, z C. Define a partial order on C as follows: z z if and only if Re (z ) Re (z ), Im (z ) Im (z ). 00 Mathematics Subject Classification. Primary 47H0; Secondary 46S40 Keywords. Compatible and weakly compatible maps, common fixed point, complex valued metric space Received: 0 June 03; Accepted: 9 January 04 Communicated by Ljubomir Ćirić addresses: nhusain@kau.edu.sa (N. Hussain), akbarazam@yahoo.com (Akbar Azam), jamshaid_jasim@yahoo.com (Jamshaid Ahmad), marshadzia@iiu.edu.pk (Muhammad Arshad)

2 N. Hussain et al. / Filomat 8:7 (04), It follows that z z if one of the following conditions is satisfied: (i) Re (z ) = Re (z ), Im (z ) < Im (z ), (ii) Re (z ) < Re (z ), Im (z ) = Im (z ), (iii) Re (z ) < Re (z ), Im (z ) < Im (z ), (iv) Re (z ) = Re (z ), Im (z ) = Im (z ). In particular, we will write z z if z z and one of (i), (ii) and (iii) is satisfied and we will write z z if only (iii) is satisfied. Note that 0 z z = z < z, z z, z z 3 = z z 3. Definition.. Let X be a nonempty set. Suppose that the self-mapping d : X X C satisfies:. 0 d(x, y), for all x, y X and d(x, y) = 0 if and only if x = y;. d(x, y) = d(y, x) for all x, y X 3. d(x, y) d(x, z) + d(z, y), for all x, y, z X. Then d is called a complex valued metric on X, and (X, d) is called a complex valued metric space. A point x X is called interior point of a set A X whenever there exists 0 r C such that B(x, r) = { y X : d(x, y) r } A. A point x X is called a limit point of A whenever for every 0 r C, B(x, r) (A {x}) φ. A is called open whenever each element of A is an interior point of A. Moreover, a subset B X is called closed whenever each limit point of B belongs to B. The family F = {B(x, r) : x X, 0 r} is a sub-basis for a Hausdorff topology τ on X. Let x n be a sequence in X and x X. If for every c C with 0 c there is n 0 N such that for all n > n 0, d(x n, x) c, then {x n } is said to be convergent, {x n } converges to x and x is the limit point of {x n }. We denote this by lim n x n = x, or x n x, as n. If for every c C with 0 c there is n 0 N such that for all n > n 0, d(x n, x n+m ) c, then {x n } is called a Cauchy sequence in (X, d). If every Cauchy sequence is convergent in (X, d), then (X, d) is called a complete complex valued metric space. Let X be a complete complex valued metric space and T, f : X X. The mappings T, f are said to be compatible if, for for arbitrary {x n } X such that lim n Tx n = lim n f x n = t X, and for arbitrary c C with 0 c, there exists n 0 N such that d(t f x n, f Tx n ) c, whenever n > n 0. The mappings T, f are said to be weakly compatible if they commute at their coincidence point (i. e. T f x = f Tx whenever Tx = f x). A point y X is called point of coincidence of T and f if there exists a point x X such that y = Tx = f x.we require the following lemmas: Lemma.. Let (X, d) be a complex valued metric space and let {x n } be a sequence in X. Then {x n } converges to x if and only if d(x n, x) 0 as n. Lemma.3. Let (X, d) be a complex valued metric space and let {x n } be a sequence in X. Then {x n } is a Cauchy sequence if and only if d(x n, x n+m ) 0 as n.

3 N. Hussain et al. / Filomat 8:7 (04), Lemma.4. If the pair ( f, ) of self-mappings on the complex valued metric space (X, d) is compatible, then it is weakly compatible but the converse does not holds. Proof. Let f u = u for some u X. We have to prove that f u = f u. Put x n = u for every n N. We have f x n, x n f u = u. If c C with 0 c then since the pair ( f, ) is compatible, so we have d( f x n, f x n ) = d( f u, f u) c implies that f u = f u as required.. Main Result Hussain et al. [0] proved six mappings fixed point theorem for generalized (ψ, ϕ) contractions. Recently Jankovic et al. [3] proved a common fixed point theorem for six self-mappings satisfying generalized contraction in a cone metric space. Here we improve and generalize the result of Jankovic et al. to a complex valued metric space involving a rational type inequality. Theorem.. Let (X, d) be a complete complex valued metric space and let A, B, S, T, L, M : X X be a selfmappings satisfying the conditions: (cvms ) d(lx, My)) R(x, y) () for all x, y X and [0, ), R(x, y) {d(abx, STy), d(abx, Lx), d(sty, My), (d(sty, Lx) + d(abx, My)), d(abx, Lx)d(STy, My) }; + d(abx, STy) (cvms ) L(X) ST(X); M(X) AB(X); (cvms 3 ) AB = BA; ST = TS; LB = BL; MT = TM; (cvms 4 ) the pair (L, AB) is compatible and the pair (M, ST) is weakly compatible; (cvms 5 ) either AB or L is continuous. Then A, B, S, T, L and M have a unique common fixed point. Proof. Let x 0 X be arbitrary. From the condition (cvms ), there exist x, x X such that Lx 0 = STx = y 0 and Mx = ABx = y. We can construct successively the sequences {x n } and {y n } in X as follows: y n = STx n+ = Lx n and y n+ = ABx n+ = Mx n+ () for n = 0,,,. We prove that d(y n+, y n ) d(y n, y n ), for n =,,. Now from (cvms ), we get d(y n, y n+ ) = d(lx n, Mx n+ ) R(x n, x n+ ), (3) R(x n, x n+ ) {d(abx n, STx n+ ), d(abx n, Lx n ), d(stx n+, Mx n+ ), (d(stx n+, Lx n ) + d(abx n, Mx n+ )), d(abx n, Lx n )d(stx n+, Mx n+ ) } + d(abx n, STx n+ ) = {d(y n, y n ), d(y n, y n ), d(y n, y n+ ), ((d(y n, y n ) + d(y n, y n+ )), d(y n, y n )d(y n, y n+ ) } + d(y n, y n ) = {d(y n, y n ), d(y n, y n+ ), d(y n, y n+ ), d(y n, y n )d(y n, y n+ ) }. (4) + d(y n, y n )

4 By (3) and (4), we have possible four cases that are N. Hussain et al. / Filomat 8:7 (04), d(y n, y n+ ) d(y n, y n ), (5) and d(y n, y n+ ) d(y n, y n+ ) < d(y n, y n+ ) (6) which is a contradiction. And d(y n, y n+ ) d(y n, y n+ ) d(y n, y n ) + d(y n, y n+ ). As 0 <, so d(y n, y n+ ) d(y n, y n ) + d(y n, y n+ ), which implies that And d(y n, y n+ ) d(y n, y n ). d(y n, y n+ ) d(y n, y n )d(y n, y n+ ) + d(y n, y n ) d(y n, y n ) d(y n, y n+ ), + d(y n, y n ) since + d(y n, y n ) > d(y n, y n ), so we d(y n, y n+ ) d(y n, y n+ ) < d(y n, y n+ ) (7) which is a contradiction. Thus d(y n, y n+ ) d(y n, y n ). (8) Similarly from (cvms ), we get d(y n+, y n+ ) = d(mx n+, Lx n+ ) = d(lx n+, Mx n+ ) R(x n+, x n+ ), (9) R(x n+, x n+ ) {d(abx n+, STx n+ ), d(abx n+, Lx n+ ), d(stx n+, Mx n+ ), (d(stx n+, Lx n+ ) + d(abx n+, Mx n+ )), d(abx n+, Lx n+ )d(stx n+, Mx n+ ) } + d(abx n+, STx n+ ) {d(y n+, y n ), d(y n+, y n+ ), d(y n, y n+ ), (d(y n, y n+ ) + d(y n+, y n+ )), d(y n+, y n+ )d(y n, y n+ ) } + d(y n+, y n ) {d(y n, y n+ ), d(y n+, y n+ ), d(y n, y n+ ), d(y n+, y n+ )d(y n, y n+ ) }. + d(y n+, y n ) By (9) and (0), we have possible four cases that are d(y n+, y n+ ) d(y n, y n+ ). ()

5 N. Hussain et al. / Filomat 8:7 (04), and d(y n+, y n+ ) d(y n+, y n+ ) < d(y n+, y n+ ). () which is a contradiction. And d(y n+, y n+ ) d(y n, y n+ ) d(y n, y n+ ) + d(y n+, y n+ ). As 0 <, so And d(y n+, y n+ ) d(y n, y n+ ) + d(y n+, y n+ ) d(y n, y n+ ). d(y n+, y n+ ) d(y n+, y n+ ) d(y n, y n+ ), + d(y n, y n+ ) since + d(y n, y n+ ) > d(y n, y n+ ), so we have d(y n+, y n+ ) d(y n+, y n+ ) < d(y n+, y n+ ) (3) which is a contradiction. Thus d(y n+, y n+ ) d(y n, y n+ ). (4) It follows that d(y n, y n+ ) d(y n, y n ) n d(y 0, y ). (5) Using (5) and triangle inequality, for m > n, we have: d(y n, y m ) d(y n, y n+ ) + d(y n+, y n+ ) + + d(y m, y m ) [ n + n+ + + m ] d(y 0, y ) [ n ] d(y 0, y ) 0 as n. By lemmas. and.3, it follows that {y n } is a Cauchy sequence. Since X is complete, so there exists some z X such that y n z as n. For its subsequences we also have Mx n+ z, STx n+ z, Lx n z and ABx n z. From the condition (cvms 5 ), we have two cases. Case 0. If AB is continuous. As AB is continuous, then ABABx n ABz and ABLx n ABz, as n. Also, since the pair (L, AB) is compatible, this implies that LABx n ABz. Indeed Now d(labx n, ABz) d(labx n, ABLx n ) + d(ablx n, ABz). d(labx n, ABz) d(labx n, ABLx n ) + d(ablx n, ABz) 0 as n. (a) We first prove that ABz = z. We suppose on the contrary that ABz z. Then d(abz, z) 0. Now from the triangular inequality, we get d(abz, z) d(abz, LABx n ) + d(labx n, Mx n+ ) + d(mx n+, z). (6)

6 N. Hussain et al. / Filomat 8:7 (04), Applying the condition (cvms 3 ) to x = ABx n, y = x n+, we get d(labx n, Mx n+ ) R(ABx n, x n+ ), R(ABx n, x n+ ) {d(ababx n, STx n+ ), d(ababx n, LABx n ), d(stx n+, Mx n+ ), (d(stx n+, LABx n ) + d(ababx n, Mx n+ )), d(ababx n, LABx n )d(stx n+, Mx n+ ) }. + d(ababx n, STx n+ ) Now, we have the following five cases: (i) d(labx n, Mx n+ ) d(ababx n, STx n+ ) from (6), we get d(abz, z) Taking the limit as n, we get d(abz, z) 0. d(ababx n, ABz) + d(abz, z) + d(z, STx n+ ), d(abz, LABx n) + d(ababx n, ABz) + d(z, STx n+) + d(mx n+, z). That is d(abz, z) = 0, a contradiction. Thus by lemma., we get ABz = z. (ii) d(labx n, Mx n+ ) d(ababx n, LABx n ) using (6), we get d(ababx n, ABz) + d(abz, LABx n ), d(abz, z) ( + ) d(abz, LABx n ) + d(ababx n, ABz) + d(mx n+, z). Now taking the limit as n, we get d(abz, z) 0. That is d(abz, z) = 0, a contradiction. Thus by lemma., we get ABz = z. (iii) d(labx n, Mx n+ ) d(stx n+, Mx n+ ) from (6), we get d(stx n+, z) + d(z, Mx n+ ), d(abz, z) d(abz, LABx n ) + ( + ) d(mx n+, z) + d(stx n+, z). Now taking the limit as n and since the pair (L, AB) is compatible, so we get d(abz, z) 0. That is d(abz, z) = 0, a contradiction. Thus by lemma., we get ABz = z.

7 N. Hussain et al. / Filomat 8:7 (04), (iv) d(labx n, Mx n+ ) (d(stx n+, LABx n ) + d(ababx n, Mx n+ )), d(labx n, Mx n+ ) ((d(stx n+, z) + d(z, ABz) + d(abz, LABx n )) From (6), we get d(abz, z)} + (d(ababx n, ABz) + d(abz, z) + d(z, Mx n+ )) ((d(stx n+, z) + d(z, Mx n+ )) + (d(ababx n, ABz) +d(abz, LABx n )) + d(abz, z). d(abz, LABx n) + ( ) ( (d(stx n+, z) + d(z, Mx n+ ) ) + ( ) ( d(ababx n, ABz) + d(abz, LABx n ) ) + d(mx n+, z). That is d(abz, z) = 0, a contradiction. Thus ABz = z. (v) d(labx n, Mx n+ ) d(ababx n, LABx n )d(stx n+, Mx n+ ), + d(ababx n, STx n+ ) from (6), we get d(abz, z) d(abz, LABx n ) + d(ababx n, LABx n ) d(stx n+, Mx n+ ) + d(ababx n, STx n+ ) Now taking the limit as n, we get d(abz, z) 0. + d(mx n+, z). That is d(abz, z) = 0, a contradiction. Thus ABz = z. Hence, in all cases ABz = z. (b) Now we prove that Lz = z. We suppose on the contrary that Lz z. Then d(lz, z) 0. From the triangular inequality, we get d(lz, z) d(lz, Mx n+ ) + d(mx n+, z). (7) Applying the condition (iii) to x = z, y = x n+, we get d(lz, Mx n+ ) R(ABx n, x n+ ), R(z, x n+ ) {d(abz, STx n+ ), d(abz, Lz), d(stx n+, Mx n+ ), (d(stx n+, Lz) + d(abz, Mx n+ )), We have the following five cases: (i) d(abz, Lz)d(STx n+, Mx n+ ) }. + d(abz, STx n+ ) d(lz, Mx n+ ) d(abz, STx n+ ) = d(z, STx n+ ). from (7), we get d(lz, z) d(z, STx n+ ) + d(mx n+, z),

8 Now taking the limit as n, we get d(lz, z) 0. That is d(lz, z) = 0, a contradiction. Thus Lz = z. (ii) d(lz, Mx n+ ) d(abz, Lz) = d(z, Lz), from (7), we get N. Hussain et al. / Filomat 8:7 (04), d(lz, z) d(z, Lz) + d(mx n+, z) d(mx n+, z) 0 as n. Which implies that d(lz, z) 0, a contradiction. Thus Lz = z. (iii) d(lz, Mx n+ ) d(stx n+, Mx n+ ) from (7), we get d(stx n+, z) + d(z, Mx n+ ), d(lz, z) ( + ) d(mx n+, z) + d(stx n+, z) 0 as n. Which implies that d(lz, z) = 0, a contradiction. Thus Lz = z. (iv) d(lz, Mx n+ ) (d(stx n+, Lz) + d(abz, Mx n+ )) d(lz, Mx n+ ) ((d(stx n+, z) + d(z, Lz)) + d(z, Mx n+), from (7), we get d(lz, z) ((d(stx n+, z) + d(z, Lz)) + ( + )d(z, Mx n+). Which implies that d(lz, z) ( (d(stx n+, z) + d(z, Lz) ) + ( + ) d(z, Mx n+). Now taking the limit as n, we get d(lz, z) = 0, a contradiction. Thus Lz = z. (v) d(lz, Mx n+ ) d(abz, Lz)d(STx n+, Mx n+ ), + d(abz, STx n+ ) from (7), we get d(lz, z) d(abz, Lz) d(stx n+, Mx n+ ) + d(abz, STx n+ ) Taking the limit as n, we get d(lz, z) = 0, a contradiction. + d(mx n+, z),

9 N. Hussain et al. / Filomat 8:7 (04), Thus Lz = z. Thus in all cases we have Lz = z. (c) Now we prove that Bz = z. We suppose on the contrary that Bz z. Then d(bz, z) 0. Now using the triangular inequality, we get d(bz, z) = d(blz, z) = d(lbz, z) d(lbz, Mx n+ ) + d(mx n+, z), (8) From (), we get d(lbz, Mx n+ ) R(Bz, x n+ ), R(Bz, x n+ ) {d(abbz, STx n+ ), d(abbz, LBz), d(stx n+, Mx n+ ), (d(stx n+, Lz) + d(abz, Mx n+ )), d(abbz, LBz)d(STx n+, Mx n+ ) } + d(abbz, STx n+ ) R(Bz, x n+ ) {d(babz, STx n+ ), d(babz, BLz), d(stx n+, Mx n+ ), (d(stx n+, Lz) + d(abz, Mx n+ )), d(babz, BLz)d(STx n+, Mx n+ ) } + d(babz, STx n+ ) R(Bz, x n+ ) {d(bz, STx n+ ), d(bz, Bz), d(stx n+, Mx n+ ), (d(stx n+, z) + d(z, Mx n+ )), d(bz, Bz)d(STx n+, Mx n+ ) }. + d(bz, STx n+ ) Now, we have the following five cases: (i) d(lbz, Mx n+ ) d(bz, STx n+ ) from (8), we get d(bz, z) + d(z, STx n+ ), d(bz, z) d(bz, z) + d(z, STx n+ ) + d(mx n+, z) d(bz, z) d(z, STx n+) + d(mx n+, z). Taking limit as n in the above inequality, we get d(bz, z) = 0, a contradiction. Thus Bz = z. (ii) d(lbz, Mx n+ ) d(bz, Bz), from (8), we get d(bz, z) d(mx n+, z). Taking limit as n in the above inequality, we get d(bz, z) = 0, a contradiction. Thus Bz = z. (iii) d(lbz, Mx n+ ) d(stx n+, Mx n+ ) from (8), we get d(stx n+, z) + d(z, Mx n+ ) d(bz, z) d(stx n+, z) + d(z, Mx n+ ) + d(mx n+, z). Taking limit as n in the above inequality, we get d(bz, z) = 0, a contradiction. Thus Bz = z.

10 N. Hussain et al. / Filomat 8:7 (04), (iv) d(lbz, Mx n+ ) (d(stx n+, z) + d(z, Mx n+ )), from (8), we get d(lbz, z) d(stx n+, z) + ( + )d(mx n+, z), which implies that d(lbz, z) d(stx n+, z) + ( + ) d(mx n+, z). Taking limit as n in the above inequality, we get d(bz, z) = 0, a contradiction. Thus Bz = z. (v) d(lbz, Mx n+ ) d(bz, Bz)d(STx n+, Mx n+ ), + d(bz, STx n+ ) from (8), we get d(bz, z) d(bz, Bz) d(stx n+, Mx n+ ) + d(mx n+, z) + d(bz, STx n+ ) d(mx n+, z), Taking limit as n in the above inequality, we get d(bz, z) = 0, a contradiction. Thus Bz = z. Thus in all cases we have Bz = z. (d) As L(X) ST(X), so there exists v X such that z = Lz = STv. First, we shall show that STv = Mv. For this we have d(stv, Mv) = d(lz, Mv) R(z, v), (9) R(z, v) {d(abz, STv), d(abz, Lz), d(stv, Mv), (d(stv, Lz) + d(abz, Mv)), d(abz, Lz)d(STv, Mv) }; + d(abz, STv) R(z, v) {d(z, z), d(z, z), d(stv, Mv), (d(z, z) + d(z, Mv)), This implies that d(z, z)d(stv, Mv) }. + d(z, z) R(z, v) {0, d(stv, Mv), d(stv, Mv)}. (0) From (9) and (0), it follows that d(stv, Mv) = 0. That is STv = Mv = z. As the pair (M, ST) is weakly compatible, so we have STMv = MSTv. Thus STz = Mz.

11 N. Hussain et al. / Filomat 8:7 (04), (e) Now we prove that Mz = z. Now we have d(z, Mz) = d(lz, Mz) R(z, z), () R(z, z) {d(abz, STz), d(abz, Lz), d(stz, Mz), (d(stz, Lz) + d(abz, Mz)), d(abz, Lz)d(STz, Mz) }; + d(abz, STz) R(z, z) {d(z, Mz), d(z, z), d(mz, Mz), R(z, z) {0, d(z, Mz)}. From () and (), we get d(z, z)d(mz, Mz) (d(mz, z) + d(z, Mz)), } + d(z, Mz) () that is d(z, Mz) = 0, Mz = z. (f) Now we prove that Tz = z. Now we have d(z, Tz) = d(lz, TMz) = d(lz, MTz). From (cvms ), we get d(z, Tz) = d(lz, MTz) R(z, Tz), (3) R(z, Tz) {d(abz, STTz), d(abz, Lz), d(sttz, MTz), (d(sttz, Lz) + d(abz, MTz)), d(abz, Lz)d(STz, Mz) } + d(abz, STTz) R(z, Tz) {d(z, TSTz), d(z, z), d(tstz, TMz), (d(tstz, Lz) + d(abz, TMz)), d(z, z)d(tstz, TMz) } + d(z, TSTz) R(z, Tz) {d(z, Tz), d(z, z), d(tz, Tz), R(z, Tz) {0, d(z, Tz)}, from (3) and (4), we get d(z, z)d(tz, Tz) (d(tz, z) + d(z, Tz)), } + d(z, Tz) (4) d(z, Tz) = 0, that is Tz = z. Since STz = z,

12 N. Hussain et al. / Filomat 8:7 (04), it follows that Sz = z. Thus if AB is continuous then we proved that Az = Bz = Sz = Tz = Lz = Mz = z. Hence, the six self mappings have a common fixed point in the case when AB is continuous. Case 0. If L is continuous. As L is continuous, then L x n Lz and LABx n Lz, as n. As the pair (L, AB) is compatible, we have ABLx n Lz, as n. Indeed Now d(ablx n, Lz) d(ablx n, LABx n ) + d(labx n, Lz), d(ablx n, Lz) d(ablx n, LABx n ) + d(labx n, Lz) 0, as n. (a) First we prove that Lz = z. By triangular inequality, we get d(lz, z) d(lz, L x n ) + d(l x n, Mx n+ ) + d(mx n+, z). (5) Now putting x = Lx n and y = x n+, in (cvms ), we get d(l x n, Mx n+ ) R(Lx n, x n+ ), R(Lx n, x n+ ) {d(ablx n, STx n+ ), d(ablx n, L x n ), d(stx n+, Mx n+ ), (d(stx n+, L x n ) + d(ablx n, Mx n+ )), d(ablx n, LLx n )d(stx n+, Mx n+ ) }; + d(ablx n, STx n+ ) Now, we have the following five cases: (i) d(l x n, Mx n+ ) d(ablx n, STx n+ ) from (5), we get Now d(ablx n, Lz) + d(lz, z) + d(z, STx n+ ), d(lz, z) d(lz, L x n ) + d(ablx n, Lz) + d(lz, z) + d(z, STx n+ ) + d(mx n+, z). d(lz, z) d(lz, L x n ) + d(ablx n, Lz) + d(lz, z) + d(z, STx n+ ) + d(mx n+, z), which implies that d(lz, z) d(lz, L x n ) + d(ablx n, Lz) + d(z, STx n+) + d(mx n+, z). Taking limit as n in the above inequality, we get d(lz, z) = 0.

13 Thus Lz = z. (ii) d(l x n, Mx n+ ) d(ablx n, L x n ) from (5), we get Now N. Hussain et al. / Filomat 8:7 (04), d(ablx n, Lz) + d(lz, L x n ), d(lz, z) d(lz, L x n ) + d(ablx n, Lz) + d(lz, L x n ) + d(mx n+, z). d(lz, z) d(lz, L x n ) + d(ablx n, Lz) + d(lz, L x n ) + d(mx n+, z). Taking limit as n in the above inequality, we get d(lz, z) = 0. Thus Lz = z. (iii) d(l x n, Mx n+ ) d(stx n+, Mx n+ ) from (5), we get Now d(stx n+, z) + d(z, Mx n+ ), d(lz, z) d(lz, L x n ) + d(stx n+, z) + d(z, Mx n+ ) + d(mx n+, z) d(lz, L x n ) + d(stx n+, z) + ( + )d(z, Mx n+ ). d(lz, z) d(lz, L x n ) + d(stx n+, z) + ( + ) d(z, Mx n+ ). Taking limit as n in the above inequality, we get d(lz, z) = 0. Thus Lz = z. (iv) d(l x n, Mx n+ ) (d(stx n+, L x n ) + d(ablx n, Mx n+ )) d(l x n, Mx n+ ) (d(stx n+, z) + d(z, Lz) + d(lz, L x n )) + (d(ablx n, Lz) + d(lz, z) + d(z, Mx n+ )), from (5), we get d(lz, z) d(lz, L x n ) + (d(stx n+, z) + d(z, Lz) + d(lz, L x n )) + (d(ablx n, Lz) + d(lz, z) + d(z, Mx n+ )) + d(mx n+, z) d(lz, z) d(lz, L x n ) + + ( ) d(ablx n, Lz) + ( ) d(stx n+, z) + ( ) d(lz, L x n ) ( ) d(z, Mx n+) + d(mx n+, z).

14 This implies that d(lz, z) N. Hussain et al. / Filomat 8:7 (04), d(lz, L x n ) + + ( ) d(ablx n, Lz) + ( ) d(stx n+, z) + Taking limit as n in the above inequality, we get d(lz, z) 0, a contradiction. Thus d(lz, z) = 0 implies Lz = z. (v) d(l x n, Mx n+ ) d(ablx n, LLx n )d(stx n+, Mx n+ ), + d(ablx n, STx n+ ) from (5), we get Now d(lz, z) d(lz, L x n ) + d(ablx n, LLx n )d(stx n+, Mx n+ ) + d(ablx n, STx n+ ) ( ) d(lz, L x n ) ( ) d(z, Mx n+) + d(mx n+, z). d(lz, z) d(lz, L x n ) + d(ablx n, LLx n ) d(stx n+, Mx n+ ) + d(ablx n, STx n+ ) Taking limit as n in the above inequality, we get d(lz, z) = 0. + d(mx n+, z), + d(mx n+, z). Thus Lz = z. Now, using steps (d),(e) and (f), and continuing the step (f) give us Mz = Sz = Tz = z. (b) As M(X) AB(X), so there exists w X such that z = Mz = ABw. We shall show that Lw = ABw = z. For this we have that is d(lw, ABw) = d(lw, Mz) R(w, z), R(w, z) {d(abw, STz), d(abw, Lw), d(stz, Mz), (d(stz, Lw) d(abw, Lw)d(STz, Mz) +d(abw, Mz)), } + d(abw, STz) R(w, z) {0, (d(z, Lw)},

15 N. Hussain et al. / Filomat 8:7 (04), which implies that d(lw, ABw) = d(lw, Mz) = d(lw, z) (d(z, Lw) i.e Lw = ABw = z. Which implies that Lw = ABw = z. As the pair (L, AB) is compatible, so it must be weakly compatible, we have Lz = ABz. Further, Bz = z follows from step (c). Thus, Az = Bz = Lz = z and we obtain that z is the common fixed point of six mappings in this case too. Now we prove the uniqueness of these six mappings. Let z be another common fixed point of A, B, S, T, L and M; then Az = Bz = Sz = Tz = Lz = Mz = z. Putting x = z, y = z in (cvms ), we get d(z, z ) = d(lz, Mz )) R(z, z ), (6) R(z, z ) {d(abz, STz ), d(abz, Lz), d(stz, Mz ), (d(stz, Lz) R(z, z ) {0, d(z, z )}. From (6) and (7), we get +d(abz, Mz )), d(abz, Lz)d(STz, Mz ) + d(abz, STz } ) (7) d(z, z ) = 0. Which implies that z = z. Thus z is the unique common fixed point of A, B, S, T, L and M. In Theorem., put B = T = I X, the identity mapping on X, to obtain the following result: Corollary.. Let (X, d) be a complete complex valued metric space and let A, S, L, M : X X be a self-mappings satisfying the conditions: (cvms 6 ) d(lx, My)) R(x, y) for all x, y X and [0, ), R(x, y) {d(ax, Sy), d(ax, Lx), d(sy, My), d(ax, Lx)d(Sy, My) {d(ax, My) + d(sy, Lx)}, }; + d(ax, Sy) (cvms 7 ) L(X) S(X); M(X) A(X); (cvms 8 ) the pair (L, A) is compatible and the pair (M, S) is weakly compatible; (cvms 9 ) either A or L is continuous. Then A, S, L and M have a unique common fixed point. Putting L = M = F and A = B = S = T = I X in Theorem., we get the following corollary.

16 N. Hussain et al. / Filomat 8:7 (04), Corollary.3. Let (X, d) be a complete complex valued metric space and let F : X X be a self-mappings satisfying the conditions: (cvms 0 ) d(fx, Fy)) R(x, y) for all x, y X and [0, ), R(x, y) {d(x, y), d(x, Fx), d(y, Fy), Then F has a unique fixed point. d(x, Fx)d(y, Fy) {d(x, Fy) + d(y, Fx)}, }. + d(x, y) By setting L, M = F and A, B, S, T = in Theorem., following example illustrates of our main result. Example.4. Let X = {(, ), (, 3), (3, 4), (4, 5), (5, 6)} and define a mapping d : X X C as d(z, z ) = 4 3 x x + i y y z = x + iy, z = x + iy, then (X, d) is a complete complex valued metric space. Set L = M = F, A = B = S = T = and define the self mappings F and on X (with z = x + iy) as Fz = x y + i x y for all z X and z = { x + iy for z = (, ) x y + 3i x y for z {(, 3), (3, 4), (4, 5), (5, 6)}. By a routine calculation, one can easily verify that F and satisfy the contraction condition (). Notice that the point (, ) X a unique common fixed point of F and. The following example illustrates our corollary.3. Example.5. Consider X = {z C : Rez, Imz = 0} X = {z C : Imz, Rez = 0} and let X = X X. Then with z = x + iy. Set L = M = F and A = B = S = T = I (identity mapping). Define F : X X as follows { i Fz= x if z X y if z X If d u is usual metric on X then F is not contractive as d u (Fz, Fz ) = y y = d u (z, z ) z, z X. Therefore, the Banach contraction theorem is not valid to find the unique fixed point 0 of F. To apply the Theorem., consider a complex valued metric d : X X C as follows d(z, z ) = x x + i 3 x x, if z, z X 3 y y + i 3 y y, if z, z X ( x + 3 y ) + i( 3 x + 3 y ), if z X,z X ( 3 y + x ) + i( 3 y + 3 x ) if z X,z X z = x + iy, z = x + iy X. Then (X, d) is a complete complex valued metric space. By a routine calculation, one can easily verify that F satisfies the contraction condition (). Notice that the point 0 X remains fixed under F is indeed unique.

17 N. Hussain et al. / Filomat 8:7 (04), Application Fixed point theorems for operators in ordered Banach spaces are widely investigated and have found various applications in differential and integral equations (see [, ] and references therein). Theorem 3.. Let X = C([a, b], R n ), a > 0 and d : X X C be defined as follows: d(x, y) = max x (t) y (t) + a e i tan a. t [a,b] Consider the Urysohn integral equations x(t) = x(t) = b a b a t [a, b] R, x,, h X. K (t, s, x(s))ds + (t), (8) K (t, s, x(s))ds + h(t), (9) Suppose that K, K : [a, b] [a, b] R n R n are such that F x, G x X for each x X,, F x (t) = b a K (t, s, x(s))ds, G x (t) = b If there exists 0 < h < such that for every x, y X a K (t, s, x(s))ds for all t [a, b]. F x (t) G y (t) + (t) h(t) + a e i tan a hr(x, y)(t),, R(x, y)(t) { A ( x, y ) (t), B ( x, y ) (t), C ( x, y ) (t), D ( x, y ) (t), E ( x, y ) (t) }, A ( x, y ) (t) = x(t) y(t) + a e i tan a, B ( x, y ) (t) = Fx (t) + (t) x(t) + a e i tan a, C ( x, y ) (t) = Gy (t) + h(t) y(t) + a e i tan a ( Gy (t) + h(t) x(t) + Fx (t) + (t) x(t) ) + a e i tan a D ( x, y ) (t) = E ( x, y ) (t) = F x (t) + (t) x(t) Gy (t) + h(t) y(t) + max t [a,b] A ( x, y ) (t) + a e i tan a then the system of integral equations (8) have a unique common solution. Proof. Define L, M : X X by Lx = F x +, Mx = G x + h. Then d ( Lx, My ) = max F x (t) G y (t) + (t) h(t) + a e i tan a, t [a,b]

18 d ( x, y ) = max t [a,b] A ( x, y ) (t), d (x, Lx) = max t [a,b] B ( x, y ) (t), d ( y, My ) = max t [a,b] C ( x, y ) (t) d(x, Ly) + d ( y, Mx ) d(x, Lx)d(y, My) + d(x, y) N. Hussain et al. / Filomat 8:7 (04), = max t [a,b] D ( x, y ) (t) = max t [a,b] E ( x, y ) (t). It is easily seen that d(lx, My) hr(x, y), { R(x, y) d(x, y), d(x, Lx), d(y, My), } d(x, Lx)d(y, My) {d(x, My) + d(y, Lx)}, + d(x, y) for every x, y X. By Theorem., the Urysohn integral equations (8) and (9) have a unique common solution. References [] Ravi P. Agarwal, Nawab Hussain and Mohamed-Aziz Taoudi, Fixed point theorems in ordered Banach spaces and applications to nonlinear integral equations, Abstract and Applied Analysis, vol. 0, Article ID 4587, 5 pp. [] M. Arshad, A. Azam and P. Vetro, Some common fixed point results in cone metric spaces, Fixed Point and Theory Appl., (009), Article ID , pp. [3] A. Azam, B. Fisher and M. Khan, Common fixed point theorems in complex valued metric spaces, Num. Funct. Anal. and Optimization, 3(3)(0), [4] A. Azam, M. Arshad, Common fixed points of generalized contractive maps in cone metric spaces, Bull. Iranian. Math. Soc., 35()(009), [5] Lj. Ćirić, A. Razani, S. Radenovic and J.S. Ume, Common fixed point theorems for families of weakly compatible maps, Comput. Math. Appl. 55(008) [6] Lj. Ćirić, A generalization of Banach s contraction principle, Proc. Amer. Math. Soc. 45(974), [7] Lj. Ćirić, N. Hussain, and N. Cakic, Common fixed points for Ciric type f weak contraction with applications, Publ. Math. Debrecen, 76 (00), [8] Lj. Ćirić, M. Abbas, R. Saadati, N. Hussain, Common fixed points of almost generalized contractive mappings in ordered metric spaces, Appl. Math. Comput, 7(0), [9] L.G. Huang and X. Zhang, Cone metric spaces and fixed point theorems of contractive mappings, J. Math. Anal. Appl. 33(007), [0] N. Hussain, M.H. Shah, and S. Radenović, Fixed points of weakly contractions through occasionally weak compatibility, Journal of Computational Analysis and Applications, 3(0), [] N. Hussain, A.R. Khan and Ravi P. Agarwal, Krasnoselskii and Ky Fan type fixed point theorems in ordered Banach spaces, J. Nonlinear and Convex Analysis, Vol., Number 3, (00), [] G. Jungck, Compatible mappings and common fixed points, Int. J. Math. Sci. (986) [3] S. Janković, Z. Golubović and S. Radenović, Compatible and weakly compatible mappings in cone metric spaces, Mathematical and Computer Modelling. 5(00) [4] E. Karapinar, Fixed point theorems in cone Banach spaces, Fixed Point Theory Appl., vol 009, 9 pp. [5] N. Hussain, M. H. Shah, A. Amini-Harandi and Z. Akhtar, Common fixed point theorems for generalized contractive mappings with applications, Fixed Point Theory and Applications, 03, 03:69. [6] A. Razani and M. Shirdaryazdi, A common fixed point theorem of compatible maps in Menger space, Chaos, Solitons, Fractals, 3(007)6 34. [7] F. Rouzkard and M. Imdad, Some common fixed point theorems on complex valued metric spaces, Comp. Math. Appls., doi:0.06/j.camwa [8] W. Sintunavarat and P. Kumam, Generalized common fixed point theorems in complex valued metric spaces and applications, Journal Inequalities and Applications, 0, 0:84.