Research Article Employing Common Limit Range Property to Prove Unified Metrical Common Fixed Point Theorems

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1 International Analysis Volume 2013, Article ID , 10 pages Research Article Employing Common Limit Range Property to Prove Unified Metrical Common Fixed Point Theems Mohammad Imdad 1 and Sunny Chauhan 2 1 Department of Mathematics, Aligarh Muslim University, Aligarh , India 2 Near Nehru Training Centre, H. No. 274, Nai Basti B-14, Bijn, Uttar Pradesh , India Crespondence should be addressed to Sunny Chauhan; sun.gkv@gmail.com Received 29 January 2013; Accepted 4 April 2013 Academic Edit: Chuanxi Qian Copyright 2013 M. Imdad and S. Chauhan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the iginal wk is properly cited. The purpose of this paper is to emphasize the role of common limit range property to ascertain the existence of common fixed point in metric spaces satisfying an implicit function essentially due to the paper of Ali and Imdad (2008). As an application to our main result, we derive a fixed point theem f four finite families of self-mappings which can be utilized to derive common fixed point theems involving any finite number of mappings. Our results improve and extend a host of previously known results including the ones contained in the paper of Ali and Imdad (2008). We also furnish some illustrative examples to suppt our main results. 1. Introduction and Preliminaries The well-known Banach fixed point theem is often referred to as Banach contraction principle which appeared in its explicit fm in the thesis of Banach in 1922 [1]. Owing to its simplicity and usefulness, it became a very popular and powerful tool in solving existence problems in pure and applied sciences which include biology, medicine, physics, and computer science (see [2, 3]). This natural theem asserts that every contraction mapping defined on a complete metricspacehasauniquefixedpointandthatfixedpointcan be explicitly obtained as limit of repeated iteration of the mapping at any point of the underlying space. Evidently, every contraction mapping is a continuous but not conversely. A result on the existence and uniqueness of common fixed point in metric spaces generally involves conditions on commutativity, continuity, and contraction along with a suitable condition on the containment of range of one mapping into the range of other. Hence, one is always required to improve one me of these conditions to prove a new fixed point theem. In the recent past, several auths have contributed to the vigous development of metric fixed point they (e.g., [4 18]). In metric fixed point they, recently implicit relations are utilized to cover several contraction conditions in one go rather than proving a separate theem f each contraction condition. The first ever attempt to coin an implicit relation can be traced back to Popa [19]. In 2008, Ali and Imdad [20] introduced a new class of implicit functions which covers several classes of contraction conditions such as Ćirić quasi-contractions, generalized contractions, ψ-type contractions, rational inequalities, and among others. They (i.e., Ali and Imdad [20]) proved some fixed point theems f weakly compatible mappings satisfying common property (E.A). Thereafter, many researchers utilized various implicit relations to prove a number of fixed point theems in different settings (e.g., [21 27]). In this paper, utilizing the implicit function of Ali and Imdad [20], we prove some common fixed point theems f two pairs of weakly compatible mappings employing common limit range property. In process, many known results (especially the ones contained in Ali and Imdad [20]) are enriched and improved. Some related results are also derived besides furnishing illustrative examples.

2 2 International Analysis Definition 1. Let A, S : X X be two self-mappings of a metric space (X, d).thenthepair(a, S) is said to be (1) commuting if ASx = SAx,fallx X, (2) weakly commuting (cf. [28]) if d(asx, SAx) d(ax, Sx),fallx X, (3) compatible (cf. [29]) if lim n d(asx n,sax n ) = 0 f each sequence x n } in X such that lim n Ax n = lim n Sx n, (4) noncompatible (cf. [30]) if there exists a sequence x n } in X such that lim n Ax n = lim n Sx n but lim n d(asx n,sax n ) is either nonzero nonexistent, (5) weakly compatible (cf. [31]) if they commute at their coincidence points, that is, ASu = SAu whenever Au = Su,fsomeu X. F me details on systematic comparisons and illustrations of earlier described notions, we refer to Murthy [32]. Definition 2 (cf. [33]). A pair (A, S) of self-mappings of a metric space (X, d) is said to satisfy the property (E.A) if there exists a sequence x n } in X such that f some t X. lim n Ax n = lim n Sx n =t, (1) Hence any pair of compatible as well as noncompatible self-mappings of a metric space (X, d) satisfy the property (E.A) but a pair of mappings satisfying the property (E.A) need not be noncompatible (see [34, Example 1]). In 2005, Liu et al. [35] defined the notion of common property (E.A) f hybrid pairs of mappings which contains the property (E.A). Definition 3. Two pairs (A, S) and (B, T) of self-mappings of ametricspace(x, d) are said to satisfy the common property (E.A), if there exist two sequences x n } and y n } in X such that lim n Ax n = lim n Sx n = lim n By n = lim n Ty n =t, (2) f some t X. Motivated by the observation that the majity of the fixed point results always require the closedness of the underlying subspaces f the existence of common fixed point under the property (E.A) and common property (E.A), Sintunavarat and Kumam [36] coined the idea of common limit range property (also see [37]). Recently, Imdad et al. [38] extended the notion of common limit range property to two pairs of self-mappings which relaxes the requirements on closedness of the underlying subspaces. Definition 4 (cf. [36]). A pair (A, S) of self-mappings of a metric space (X, d) is said to satisfy the common limit range property with respect to S, denoted by (CLR S ), if there exists asequencex n } in X such that where t S(X). lim n Ax n = lim n Sx n =t, (3) Thus, one can infer that a pair (A, S) satisfying the property (E.A) along with closedness of the subspace S(X) always enjoys the (CLR S ) propertywithrespecttothemappings (see [38, Examples ]). Definition 5. Two pairs (A, S) and (B, T) of self-mappings of ametricspace(x, d) are said to satisfy the common limit rangepropertywithrespecttomappingss and T, denoted by (CLR ST ), if there exist two sequences x n } and y n } in X such that lim Ax n n = lim Sx n n = lim By n n = lim Ty n n =t, (4) where t S(X) T(X). Definition 6 (cf. [39]). Two families of self-mappings A i } m i=1 and S k } n k=1 are said to be pairwise commuting if (1) A i A j =A j A i f all i,j 1,2,...,m}, (2) S k S l =S l S k f all k,l 1,2,...,n}, (3) A i S k = S k A i f all i 1,2,...,m} and k 1,2,...,n}. 2. Implicit Functions In an attempt to cover Ćirić s quasi-contractions, Ali and Imdad [20] introduced a new class of implicit functions which includes a variety of contraction classes such as Ćirić quasi-contractions, generalized contractions, ψ-type contractions, rational inequalities, and among others. Let Φ be the family of lower semicontinuous functions φ:r 6 R satisfying the following conditions: (φ 1 ): φ(t, 0, t, 0, 0, t) > 0,fallt>0, (φ 2 ): φ(t, 0, 0, t, t, 0) > 0,fallt>0, (φ 3 ): φ(t, t, 0, 0, t, t) > 0,fallt>0. The following examples furnished in Ali and Imdad s [20] establishes the utility of the preceeding definition. Example 7. Define :R 6 R as =t 1 kmax t 2,t 3,t 4,t 5,t 6 }, where k [0, 1). Example 8. Define :R 6 R as =t 1 kmax t 2,t 3,t 3 t 5,t 4 t 6 }, where k [0, 1). (5) (6)

3 International Analysis 3 Example 9. Define :R 6 R as =t 1 k[max t 2 2,t 3t 4,t 5 t 6,t 3 t 5,t 4 t 6 }] 1/2, where k [0, 1). (7) Example 10. Define :R 6 R as =t 1 α[βmax t 2,t 3,t 4,t 5,t 6 } where α [0,1)and β 0. +(1 β)(max t 2 2,t 3t 4,t 5 t 6,t 3 t 6,t 4 t 5 }) 1/2 ], (8) Example 11. Define :R 6 R as =t 2 1 αmax t2 2,t2 3,t2 4 } βmax t 3t 5,t 4 t 6 } γt 5 t 6, (9) where α, β, γ 0 and α+γ<1. Example 12. Define :R 6 R as =(1+αt 2 )t 1 αmax t 3 t 4,t 5 t 6 } where α 0and β [0,1). βmax t 2,t 3,t 4,t 5,t 6 }, Example 13. Define :R 6 R as (10) =t 1 αt 2 βmax t 3,t 4 } γmax t 3 +t 4,t 5 +t 6 }, (11) where α, β, γ 0 and α+β+2γ<1. Example 14. Define :R 6 R as φ (t 1,t 2,t 3,t 4,t 5,t 6 ) =t 1 ψ(max t 2,t 3,t 4,t 5,t 6 }), (12) where ψ:r R is an upper semicontinuous function such that ψ(0) = 0 and ψ(t) < t f all t>0. Example 15. Define :R 6 R as =t 1 ψ(t 2,t 3,t 4,t 5,t 6 ), (13) where ψ:r 5 R is an upper semicontinuous function such that maxψ(0,t,0,0,t), ψ(0, 0, t, t, 0), ψ(t, 0, 0, t, t)} < t f each t>0. Example 16. Define :R 6 R as =t 2 1 ψ(t2 2,t 3t 4,t 5 t 6,t 3 t 6,t 4 t 5 ), (14) where ψ:r 5 R is an upper semicontinuous function such that maxψ(0,0,0,t,0), ψ(0, 0, 0, 0, t), ψ(t, 0, t, 0, 0)} < t f each t>0. Example 17. Define :R 6 R as t = 1 αt 2 β t2 3 +t2 4 γ(t t 3 +t 5 +t 6 ), if t 3 +t 4 =0; 4 t 1, if t 3 +t 4 =0, (15) where α, β, γ 0 and β+γ<1. Example 18. Define :R 6 R as t p = 1 ktp 2 t 3t p 4 +t 5t p 6, if t t 3 +t 3 +t 4 =0; 4 t 1, if t 3 +t 4 =0, where p 1and 0 k<. Example 19. Define :R 6 R as (16) t = 1 αt 2 β t2 5 +t2 6 γ(t t 5 +t 3 +t 4 ), if t 5 +t 6 =0; 6 t 1, if t 5 +t 6 =0, (17) where α, β, γ 0 and β+γ<1. Example 20. Define :R 6 R as where 0 k<. = t 1 kt 2 t 3t 4 +t 5 t 6, if t t 5 +t 5 +t 6 =0; 6 t 1, if t 5 +t 6 =0, Example 21. Define :R 6 R as t 1 kt 2 t 3t 4 +t 5 t 6 t 3t 5 +t 4 t 6, t 3 +t 4 t 5 +t 6 = if t 3 +t 4 =0, t 5 +t 6 =0; t 1, if t 3 +t 4 =0, t 5 +t 6 =0, where 0 k<. (18) (19)

4 4 International Analysis Example 22. Define :R 6 R as =t 1 t 3t 4 +t 5 t 6 1+t 2. (20) Example 23. Define :R 6 R as =t 1 αt 2 β t 3 +t 4 1+t 5 t 6, (21) where α, β [0, 1). Example 24. Define :R 6 R as =t 2 1 αt2 2 β t 5 t 6 1+t3 2, (22) +t2 4 where α, β 0 and α+β<1. Example 25. Define :R 6 R as =t 3 1 t2 3 t2 4 +t2 5 t2 6 1+t 2. (23) Example 26. Define :R 6 R as =t 3 1 αt2 1 t 2 βt 1 t 3 t 4 γt 2 5 t 6 ηt 5 t 2 6, (24) where α,β,γ,η 0and α+γ+η<1. Apart from earlier stated definitions, still there are many contractive definitions which meet the requirements (φ 1 ), (φ 2 ) and (φ 3 ) butduetopaucityofthespacewehavenotopted to include me examples. 3. Main Results Firstly, we observe the following. Lemma 27. Let A, B, S, andt beself-mappingsofametric space (X, d).supposethefollowing: (1) the pair (A, S) satisfies the (CLR S ) property ( (B, T) satisfies the (CLR T ) property), (2) A(X) T(X) ( B(X) S(X)), (3) T(X) ( S(X)) is a closed subset of X, (4) there exists φ Φsuch that (f all x, y X) where t S(X).SinceA(X) T(X), feachsequence x n } there exists a sequence y n } in X such that Ax n =Ty n. Therefe, due to closedness of T(X), lim Ty n n = lim Ax n n =t, (27) where t S(X) T(X).Thusinall,wehaveAx n t, Sx n t and Ty n t as n.now,weshowthatby n t as n. On using inequality (25) withx=x n, y=y n,we have φ (d (Ax n,by n ),d(sx n,ty n ), d (Ax n,sx n ), d(by n,ty n ),d(sx n,by n ),d(ty n,ax n )) 0. (28) Let on contrary that By n l( =t)as n.then,on taking limit as n in the preceeding inequality, we get φ (d (t, l),d(t, t),d(t, t),d(l, t),d(t, l),d(t, t)) 0, (29) φ (d (t, l),0,0,d(l, t),d(t, l),0) 0, (30) which is a contradiction to (φ 2 ).Hence,By n t which shows that the pairs (A, S) and (B, T) share the (CLR ST ) property. This concludes the proof. Remark 28. In general, the converse of Lemma 27 is not true (see [38, Example 3.5]). Now, we state and prove our main result f two pairs of weakly compatible mappings satisfying the (CLR ST ) property. Theem 29. Let A, B, S, andt be self-mappings of a metric space (X, d) satisfying the inequality (25) of Lemma 27. Ifthe pairs (A, S) and (B, T) share the (CLR ST ) property, then (A, S) and (B, T) have a coincidence point each. Meover, A, B, S and T haveauniquecommonfixedpointprovidedthatbothpairs (A, S) and (B, T) are weakly compatible. Proof. Since the pairs (A, S) and (B, T) satisfy the (CLR ST ) property, there exist two sequences x n } and y n } in X such that lim Ax n n = lim Sx n n = lim By n n = lim Ty n n =t, (31) where t S(X) T(X).Sincet S(X), there exists a point u Xsuch that Su = t. WeshowthatAu = Su. Ifnot,then using inequality (25)withx=u, y=y n,weget φ(d(ax,by),d(sx,ty),d(ax, Sx),d(By,Ty), d(sx,by),d(ty,ax)) 0. (25) φ(d(au,by n ),d(su,ty n ),d(au, Su),d(By n,ty n ), d(su,by n ),d(ty n,au)) 0, (32) Then the pairs (A, S) and (B, T) share the (CLR ST ) property. Proof. If the pair (A, S) enjoys the (CLR S ) property with respect to mapping S, then there exists a sequence x n } in X such that lim Ax n n = lim Sx n n =t, (26) which onmakingn reduces to φ (d (Au, t),d(t, t),d(au, t),d(t, t),d(t, t),d(t, Au)) 0, (33) φ (d (Au, t),0,d(au, t),0,0,d(t, Au)) 0, (34)

5 International Analysis 5 a contradiction to (φ 1 ).Hence,Au = Su = t. Therefe, u is a coincidence point of the pair (A, S). Also t T(X), there exists a point V Xsuch that TV =t. We assert that BV =TV.Ifnot,thenusinginequality(25)with x=u, y=v,weget φ (d (Au, BV),d(Su, TV),d(Au, Su),d(BV,TV), d (Su, BV),d(TV,Au)) 0, which reduces to (35) φ (d (t, BV),d(t, t),d(t, t),d(bv,t),d(t, BV),d(t, t)) 0, (36) φ (d (t, BV),0,0,d(BV,t),d(t, BV),0) 0, (37) a contradiction to (φ 2 ).Hence,BV =TV =t,whichshows that V is a coincidence point of the pair (B, T). Since the pair (A, S) is weakly compatible and Au = Su, hence At = ASu = SAu = St.Now,weshowthatt is a common fixed point of the pair (A, S). SupposethatAt =t, using inequality (25) withx=t, y=v, wehave φ (d (At, BV),d(St, TV),d(At, St),d(BV,TV), d (St, BV),d(TV,At)) 0, (38) φ (d (At, t),d(at, t),0,0,d(at, t),d(t, At)) 0, (39) a contradiction to (φ 3 );wegetat=t=stwhich shows that t is a common fixed point of the pair (A, S). Also the pair (B, T) is weakly compatible and BV =TV, then Bt = BTV =TBV =Tt.SupposethatBt =t,thenusing inequality (25)withx=u, y=t,wehave φ (d (Au, Bt),d(Su, Tt),d(Au, Su),d(Bt, Tt), d (Su, Bt),d(Tt, Au)) 0, (40) φ (d (t, Bt),d(t, Bt),0,0,d(t, Bt),d(Bt, t)) 0, (41) a contradiction to (φ 3 ). Therefe, Bt = t = Tt which shows that t is a common fixed point of the pair (B, T) and in all t is a common fixed point of both pairs (A, S) and (B, T). Uniqueness of common fixed point is an easy consequence of the inequality (25) in view of condition (φ 3 ). This concludes the proof. Remark 30. Theem 29 improves the cresponding results contained in Ali and Imdad s [20, Theem 3.1] as closedness of the underlying subspaces is not required. Now, we present an example which demonstrates the validity of the hypotheses and degree of generality of our main result over comparable ones from the existing literature. Example 31. X = [5, 21) equipped with the usual metric. Define four self-mappings A, B, S, andt by 5, if x 5} (9, 21) ; A (x) = 20, if x (5, 9], 5, if x 5} (9, 21) ; B (x) = 13, if x (5, 9], 5, if x=5; S (x) = 10, if x (5, 9] ; x+1, if x (9, 21), 2 (42) 5, if x=5; T (x) = 18, if x (5, 9] ; x 4, if x (9, 21). Also, define implicit function : (R + ) 6 R as =t 1 kmax t 2,t 3,t 4,t 5,t 6 } (43) kmax d (Sx, Ty), d (Ax, Sx),d(By,Ty), d (Sx, By), d (Ty, Ax)}, (44) f all x, y X and k [0,1).Byaroutinecalculation,one can verify the inequality (44)withk=1/2.Ifwechoosetwo sequences as x n }=9+(1/n)} n N, y n }=5}( x n }=5}, y n }=9+(1/n)} n N ), then the pairs (A, S) and (B, T) satisfy the (CLR ST ) property: lim Ax n n = lim Sx n n = lim By n n (45) = lim Ty n n =5 S(X) T(X). Also, A(X) = 5, 20} [5, 17) 18} = T(X) and B(X) = 5, 13} [5, 11) = S(X). Thusalltheconditions of Theem 29 are satisfied and 5 is a unique common fixed pointofthepairs(a, S) and (B, T) which also remains a point of coincidence as well. Here, one may notice that all the involved mappings are discontinuous at their unique common fixed point 5. Notice that the subspaces S(X) and T(X) are not closed subspaces of X; therefe, Theem 29 of Ali and Imdad [20] cannot be used in the context of this example which establishes the genuineness of our extension. Theem 32. Let A, B, S, andt be self-mappings of a metric space (X, d) satisfying all the hypotheses of Lemma 27. Then A, B, S,andT have a unique common fixed point provided that both pairs (A, S) and (B, T) are weakly compatible. Proof. In view of Lemma 27,thepairs(A, S) and (B, T) share the (CLR ST ) property so that there exist two sequences x n } and y n } in X such that lim n Ax n = lim n Sx n = lim n Ty n = lim n By n =t, (46)

6 6 International Analysis where t S(X) T(X).Therestoftheproofcanbecompleted on the lines of the proof of Theem 29. Thiscompletesthe proof. The following example demonstrates the utility of Theem 32 over Theem 29. Example 33. In the setting of Example 31, replacetheselfmappings A and B by the following besides retaining the rest (i.e., S and T): 5, if x 5} (9, 21) ; A (x) = 15, if x (5, 9], 5, if x 5} (9, 21) ; B (x) = 10, if x (5, 9], 5, if x=5; S (x) = 11, if x (5, 9] ; x+1, if x (9, 21), 2 5, if x=5; T (x) = 17, if x (5, 9] ; x 4, if x (9, 21). (47) Then, like earlier example, the pairs (A, S) and (B, T) satisfy the (CLR ST ) property. the implicit function :(R + ) 6 R as =t 1 kmax t 2,t 3,t 4,t 5,t 6 }, (48) which implies k max d (Sx, Ty), d (Ax, Sx),d(By,Ty), d (Sx, By), d (Ty, Ax)}, (49) f all x, y X and k [0,1).Byaroutinecalculation, one can verify inequality (49) with k=1/2.also,a(x) = 5, 15} [5, 17] = T(X) and B(X) = 5, 10} [5, 11] = S(X). The pairs (A, S) and (B, T) commute at 5 which is also their common coincidence point as well. Thus all the conditions of Theems 32 are satisfied and 5 is a unique common fixed point of the involved mappings A, B, S,andT. Here, it can be pointed out that Theem 29 is not applicable to this example as both S(X) and T(X) are closed subsets of X which demonstrates the situational utility of Theem 32 over Theem 29. Collary 34. The conclusions of Lemma 27, Theems 29, and 32 remain true if inequality (25) is replaced by one of the following contraction conditions. F all x, y X, k max d (Sx, Ty), d (Ax, Sx),d(By,Ty), d(ax,by) kmax d (Sx, Ty), d (Ax, Sx),d(Ax, Sx) d(sx,by), where k [0,1). d(by,ty)d(ty,ax)}, k[max d 2 (Sx, Ty), d (Ax, Sx) d(by,ty), where k [0,1). (51) d(sx,by)d(ty,ax),d(ax, Sx) d(sx,by), d(by,ty)d(ty,ax)}] 1/2, α[βmax d (Sx, Ty), d (Ax, Sx),d(By,Ty), d (Sx, By), d (Ty, Ax)} (52) +(1 β)(max d 2 (Sx, Ty), d (Ax, Sx) d(by,ty), where α [0,1)and β 0. d 2 (Ax, By) d(sx,by)d(ty,ax), d (Ax, Sx) d(ty,ax), d(by, Ty)d(Sx, By)}) 1/2 ], αmax d 2 (Sx, Ty), d 2 (Ax, Sx),d 2 (By, Ty)} (53) +βmax d (Ax, Sx) d (Sx, By), d (By, Ty) d (Ty, Ax)} +γd(sx,by)d(ty,ax), where α, β, γ 0 and α+γ<1. (1 + αd (Sx, Ty)) αmax d (Ax, Sx) d(by,ty),d(sx,by)d(ty,ax)} +βmax d (Sx, Ty), d (Ax, Sx),d(By,Ty), (54) d (Sx, By), d (Ty, Ax)}, (50) d (Sx, By), d (Ty, Ax)}, (55) where k [0,1). where α 0and β [0,1).

7 International Analysis 7 αd(sx,ty)+βmax d (Ax, Sx), d (By, Ty)} +γmax d (Ax, Sx) +d(by,ty), d (Sx, By) + d (Ty, Ax)}, where α, β, γ 0 and α+β+2γ<1. d(ax,by) ψ(max d (Sx, Ty), d (Ax, Sx),d(By,Ty), (56) (57) where p 1and 0 k<, d(ax,by) αd (Sx, Ty) + β d2 (Sx, By) + d 2 (Ty, Ax) d (Sx, By) + d (Ty, Ax) +γ (d (Ax, Sx) +d(by,ty)), if D 1 =0; 0, if D 1 =0, (62) where α, β, γ 0 and β+γ<1, d(sx,by),d(ty,ax)}), where ψ:r + R is an upper semicontinuous function such that ψ(0) = 0 and ψ(t) < t f all t>0. ψ (d (Sx, Ty), d (Ax, Sx),d(By,Ty), (58) d (Sx, By), d (Ty, Ax)), where ψ:(r + ) 5 R is an upper semicontinuous function such that maxψ(0,t,0,0,t), ψ(0, 0, t, t, 0), ψ(t, 0, 0, t, t)} < t f each t>0. d 2 (Ax, By) ψ(d 2 (Sx, Ty), d (Ax, Sx) d(by,ty), d(sx,by)d(ty,ax), d (Ax, Sx) d(ty,ax),d(by,ty)d(sx,by)), (59) where ψ:(r + ) 5 R is an upper semicontinuous function such that maxψ(0,0,0,t,0), ψ(0, 0, 0, 0, t), ψ(t, 0, t, 0, 0)} < t f each t>0. In the following contraction conditions, one denotes D= d(ax, Sx) + d(by, Ty) and D 1 = d(sx, By) + d(ty, Ax): αd (Sx, Ty) + β d2 (Ax, Sx) +d 2 (By, Ty) d (Ax, Sx) +d(by,ty) +γ (d (Sx, By) + d (Ty, Ax)), if D =0; 0, if D=0, (60) where α, β, γ 0 and β+γ<1, d p (Ax, By) kd p (Sx, Ty) + d (Ax, Sx) dp (By, Ty) + d (Sx, By) d p (Ty, Ax), d (Ax, Sx) +d(by,ty) if D =0; 0, if D=0, (61) kd (Sx, Ty) d (Ax, Sx) d (By, Ty) + d (Sx, By) d (Ty, Ax) +, d(sx,by)+d(ty,ax) if D 1 =0; 0, if D 1 =0, (63) where 0 k<, kd (Sx, Ty) d (Ax, Sx) d (By, Ty) + d (Sx, By) d (Ty, Ax) + d (Ax, Sx) +d(by,ty) d (Ax, Sx) d (Sx, By) + d (By, Ty) d (Ty, Ax) +, d (Sx, By) + d (Ty, Ax) if D = 0, D 1 =0; 0, if D=0, D 1 =0, (64) where 0 k< d (Ax, Sx) d (By, Ty) + d (Sx, By) d (Ty, Ax), 1+d(Sx,Ty) d (Ax, Sx) +d(by,ty) d(ax,by) αd(sx,ty)+β 1+d(Sx,By)d(Ty,Ax), where α, β [0, 1), d 2 (Ax, By) (65) αd 2 d(sx,by)d(ty,ax) (Sx, Ty) + β 1+d 2 (Ax, Sx) +d 2 (By, Ty), (66)

8 8 International Analysis where α, β 0 and α+β<1, d 3 (Ax, By) d2 (Ax, Sx) d 2 (By, Ty) + d 2 (Sx, By) d 2 (Ty, Ax), 1+d(Sx,Ty) d 3 (Ax, By) αd 2 (Ax,By)d(Sx,Ty) +βd(ax,by)d(ax, Sx) d(by,ty) +γd 2 (Sx, By) d (Ty, Ax) + ηd (Sx, By) d 2 (Ty, Ax), (67) where α, γ, η, β 0 and α+γ+η<1. Proof. The proof of each inequality (50) (67) easilyfollows from Theem 29 in view of Examples Remark 35. Collary 34 improves and generalizes a multitude of well-known results especially those contained in [12, 14, 29, 30, 35, 40 42]. By choosing A, B, S, and T suitably, we can deduce collaries involving two as well as three self-mappings. F the sake of naturality, we only derive the following collary involving a pair of self-mappings. Collary 36. Let A and S beself-mappingsofametricspace (X, d).supposethefollowing: (1) the pair (A, S) satisfies the (CLR S ) property, (2) there exists φ Φsuch that φ(d(ax,ay),d(sx,sy),d(ax, Sx),d(Ay,Sy), d(sx,ay),d(sy,ax)) 0, f all x, y X. (68) Then (A, S) has a coincidence point. Meover, if the pair (A, S) is weakly compatible, then the pair has a unique common fixed point in X. As an application of Theem 29, wehavethefollowing result involving four finite families of self-mappings. Theem 37. Let A i } m i=1, B j} n r=1, S k} p k=1,andt l} q l=1 be four finitefamiliesofself-mappingsofametricspace(x, d) with A = A 1 A 2 A m, B = B 1 B 2 B n, S = S 1 S 2 S p and T=T 1 T 2 T q satisfying the condition (25). Suppose that the pairs (A, S) and (B, T) enjoy the (CLR ST ) property, then (A, S) and (B, T) have a point of coincidence each. Meover, A i } m i=1,b j} n j=1,s k} p k=1, and T l} q l=1 have a unique common fixed point if the families (A i }, S k }) and (B r }, T h }) commute pairwise wherein i 1,2,...,m}, k 1,2,...,p}, j 1,2,...,n},andl 1,2,...,q}. Proof. The proof of this theem can be completed on the lines of Theem 2.2 of Imdad et al. [10,Theem2.2]. Remark 38. (1) A result similar to Theem 37 can be outlined in respect of Theem 29. (2) Theem 37 improves and extends the result of Ćirić et al. [5] and Pathak et al. [25]. Now, we indicate that Theem 37 can be utilized to derive common fixed point theems f any finite number of mappings. As a sample, f five mappings, we can derive the following by setting one family of two members while the remaining three of single members. Collary 39. Let A, B, R, S, andt be self-mappings of a metric space (X, d).suppose the following: (1) the pairs (A, SR) and (B, T) share the (CLR (SR)(T) ) property, (2) there exists φ Φsuch that φ(d(ax,by),d(srx,ty),d(ax, SRx),d(By,Ty), d(srx,by),d(ty,ax)) 0, f all x, y X. (69) Then (A, SR) and (B, T) have a coincidence point each. Meover, A, B, R, S,andT have a unique common fixed point provided that both pairs (A, SR) and (B, T) commute pairwise, that is, AS = SA, AR = RA, SR = RS, BT = TB. Similarly, we can derive a common fixed point theem f six mappings by setting two families of two members whiletheresttwoofsinglemembers. Collary 40. Let A, B, H, R, S,andT be self-mappings of a metric space (X, d).suppose that (1) the pairs (A, SR) and (B, TH) share the (CLR (SR)(TH) ) property, (2) there exists φ Φsuch that φ(d(ax,by),d(srx,thy),d(ax, SRx),d(By,THy), d (SRx, By), d (THy, Ax)) 0, f all x, y X. (70) Then (A, SR) and (B, TH) have a coincidence point each. Meover, A, B, H, R, S, andt have a unique common fixed point provided that both pairs (A, SR) and (B,TH) commute pairwise, that is, AS = SA, AR = RA, SR = RS, BT = TB, BH = HB,andTH=HT. By setting A 1 =A 2 = =A m =A, B 1 =B 2 = = B n =B, S 1 =S 2 = =S p =Sand T 1 =T 2 = =T q =Tin Theem 37,we deduce the following. Collary 41. Let A, B, S, andt be self-mappings of a metric space (X, d).supposethefollowing: (1) the pairs (A m,s p ) and (B n,t q ) share the (CLR S p,t q) property,

9 International Analysis 9 (2) there exists φ Φsuch that φ(d(a m x, B n y), d (S p x, T q y),d(a m x, S p x), (71) d(b n y, T q y), d (S p x, B n y),d(t q y, A m x)) 0, f all x, y X where m, n, p, q are fixed positive integers. Then A, B, S, andt have a unique common fixed point provided that AS = SA and BT = TB. Remark 42. Collary41 is a slight but partial generalization of Theem 29 as the commutativity requirements (i.e., AS = SA and BT = TB) inthiscollaryarerelativelystrongeras compared to weak compatibility in Theem 29. Remark 43. Results similar to Collary 41 can be derived from Theem 29 and Collary 34. Remark 44. It may be noticed that the earlier proved results, namely, Theems (also Collaries 36 41) remain validinsymmetricspace(x, d) whenever d is continuous. Acknowledgment The auths wish to dedicate this wk to Profess Mohammad Saeed Khan f his crucial role in their initial development in general and his fruitful observations on the first draft of this paper in particular. 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