In this paper we obtain a unique common fixed point theorem, 
for four self maps using a generalized altering distance function in four variables , which generalizes and improves the main theorem of Choudhury [2].
We also obtain a Gregus type common fixed point theorem for four maps as a corollary

D. Vasu Babu

G. Ravi Babu

K. P. R. Rao

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Mathematical Communications 13(2008), 67-73 67Common fixed point theorems through generalizedaltering distance functionsK.P.R. Rao∗, G.Ravi Babu† and D.VasuBabu‡Abstract. In this paper we obtain a unique common fixed pointtheorem, for four self maps using a generalized altering distance func-tion in four variables, which generalizes and improves the main theoremof Choudhury [2]. We also obtain a Gregus type common fixed pointtheorem for four maps as a corollary.Key words: fixed points, semi compatible mappings, altering dis-tance functionsAMS subject classifications: 47H10, 54H25Received April 12, 2007 Accepted March 4, 20081. IntroductionM.S. Khan et al. [12] introduced the altering distances and used it for solving ﬁxedpoints problems in metric spaces. Recently, for example authors of [1, 4, 5, 8, 9, 10]used the altering distance function and obtained some ﬁxed point theorems. In2005 Choudhury [2] introduced a generalized distance function in three variablesand obtained a common ﬁxed point theorem for a pair of self maps in a completemetric space. The main aim of this paper is to prove the existence and uniqueness ofcommon ﬁxed points of two pairs of semi compatible or compatible pair of type(β)mappings by using a generalized altering distance function of four variables. Ourresults extend and generalize the main theorem of [2]. We also give an example toshow that our theorem is not valid when the generalized altering distance functionis in ﬁve variables or the second generalized altering distance function is dropped.In [14], Cho et al. introduced the non symmetrical concept of semi compatibilityof maps in d-complete topological spaces. They deﬁned a pair of self maps(S, T )to be semi compatible if (i) Sy = Ty implies STy = TSy and (ii) limn→∞Sxn =limn→∞ Txn = x for some x ∈ X implies limn→∞ STxn = Tx hold. Bijendra∗Department of Applied Mathematics, Acharya Nagarjuna University-Nuzvid Campus, Nuzvid-521201, A.P., India, e-mail:kprrao2004@yahoo.com†Department of Applied Mathematics, Acharya Nagarjuna University-Nuzvid Campus, Nuzvid-521201, A.P., India‡Department of Applied Mathematics, Acharya Nagarjuna University-Nuzvid Campus, Nuzvid-521201,A.P., India68 K.P.R. Rao, G.Ravi Babu and D.VasuBabuSingh and Jain [3] observed that (ii) implies (i). Hence they deﬁned the semicompatibility by the condition (ii) only.In [6], Pathak et. al deﬁned a pair of self maps (S, T) to be compatible of type(P ) or (β) if limn→∞ d(S2xn, T 2xn) = 0 whenever {xn} is a sequence in X suchthat limn→∞ Sxn= limn→∞ Txn = x for some x ∈ X . It is clear that in this caseSTy = TSy whenever Sy = Ty for any y ∈ X .Remark 1.1. The concepts semi compatible pair and compatible pair of type(β) maps are independent in view of the following examples.Example 1.2. Let X = [0, 2] with usual metric. Letf(x) ={2− x, if 0 ≤ x < 11, if x ≥ 1 and S(x) ={x if 0 ≤ x < 1,2 if x ≥ 1.Let xn = 1 − 1/n ,n > 1. Then fxn = 1 + 1/n → 1, Sxn = xn = 1 − 1/n → 1.fSxn = 1 + 1/n → 1 = S(1) and d(f2xn, S2xn) = d(1, 1 − 1/n) → 0. Thus (f, S)is not semi compatible and (f, S) is compatible pair of type (β).Example 1.3. Let X = [0, 2] with usual metric. Letf(x) ={2− x, if 0 ≤ x < 12, if x ≥ 1 and S(x) ={x if 0 ≤ x < 1,1 if x ≥ 1.Let xn = 1 − 1/n ,n > 1 . Then fxn = 1 + 1/n → 1, Sxn = 1 − 1/n → 1.fSxn = 1 + 1/n → 1 = S(1), d(f2xn, S2xn) = d(2, 1 − 1/n) = 1 + 1/n → 1 = 0.Thus (f, S) is semi compatible and (f, S) is not compatible pair of type (β).Let Ψn denote the set of all functions ψ : [0,∞)n → [0,∞) such that(i) ψ is continuous,(ii) ψ is monotone increasing in all the variables,(iii) ψ(t1, t2, ....., tn) = 0 if and only if t1 = t2 = .... = tn = 0.The functions in Ψn are called generalized altering distance functions.Theorem 1.4. (Theorem 1, [2]) Let (X, d) be a complete metric space and Sand T be self maps on X such thatφ1(d(Sx, T y)) ≤ ψ1(d(x, y), d(x, Sx), d(y, T y)) − ψ2(d(x, y), d(x, Sx), d(y, T y))for all x, y ∈ X, where ψ1, ψ2 ∈ Ψ3 and φ1(x) = ψ1(x, x, x)∀x ∈ [0,∞). Then Sand T have a unique common fixed point.2. Main resultsTheorem 2.1. Let (X, d) be a complete metric space and f, g, S, T : X → X besuch that(i) φ1(d(fx, gy)) ≤ ψ1(d(Sx, T y), d(Sx, fx), d(Ty, gy),12 [d(Sx, gy) + d(Ty, fx)])−ψ2(d(Sx, T y), d(Sx, fx), d(Ty, gy),12 [d(Sx, gy) + d(Ty, fx)])for all x, y ∈ X, where ψ1, ψ2 ∈ Ψ4 and φ1(x) = ψ1(x, x, x, x)∀x ∈ [0,∞),Common fixed point theorems 69(ii) one of four mappings f, g, S and T is continuous,(iii) (f, S) and (g, T ) are semi compatible pairs ,(iv) f(X) ⊆ T (X), g(X) ⊆ S(X).Then f, g, S and T have a unique common fixed point in X.Proof. Let x0 ∈ X be an arbitrary point. From (iv), construct the sequences{xn} and {yn} in X such that fx2n = Tx2n+1 = y2n, gx2n+1 = Sx2n+2 = y2n+1,n = 0, 1, 2, . . . . Let an = d(yn, yn+1). Putting x = x2n, y = x2n+1 in (i) we getφ1(a2n) ≤ ψ1(a2n−1, a2n−1, a2n,12[a2n−1 + a2n])−ψ2(a2n−1, a2n−1, a2n,12d(y2n−1, y2n+1)). (1)If a2n−1 < a2n thenφ1(a2n) ≤ ψ1(a2n, a2n, a2n, a2n)− ψ2(a2n−1, a2n−1, a2n, 12d(y2n−1, y2n+1))< φ1(a2n)which is a contradiction. Hence a2n ≤ a2n−1 , n = 1, 2, . . . . Similarly by puttingx = x2n+2, y = x2n+1 in (i) we can show that a2n+1 ≤ a2n, n = 0, 1, 2, . . . . Thusan+1 ≤ an, n = 0, 1, 2, ..... so that {an} is a decreasing sequence of non negative realnumbers and hence convergent to some a ∈ R.Let b = limn→∞12d(yn, yn+2). Letting n →∞ in (1) we getφ1(a) ≤ ψ1(a, a, a, a)− ψ2(a, a, a, b) = φ1(a)− ψ2(a, a, a, b).Thus ψ2(a, a, a, b) = 0 so that a = b = 0. Hencelimn→∞ d(yn, yn+1) = 0 (2)To show that{yn} is a Cauchy sequence, it is suﬃcient to show that the sub sequence{y2n} of {yn} is a Cauchy sequence in view of (2). If {y2n} is not Cauchy, thereexists an  > 0 and monotone increasing sequences of natural numbers {2m(k)}and {2n(k)} such that n(k) > m(k), d(y2m(k), y2n(k)) ≥  andd(y2m(k), y2n(k)−2) <  (3)From (3), ≤ d(y2m(k), y2n(k))≤ d(y2m(k), y2n(k)−2) + d(y2n(k)−2, y2n(k)−1) + d(y2n(k)−1, y2n(k))<  + d(y2n(k)−2, y2n(k)−1) + d(y2n(k)−1, y2n(k))Letting k →∞, using (2), we havelimn→∞ d(y2m(k), y2n(k)) = . (4)70 K.P.R. Rao, G.Ravi Babu and D.VasuBabuLetting k →∞, using (4), and (2) in|d(y2n(k)+1, y2m(k))− d(y2m(k), y2n(k))| ≤ d(y2n(k), y2n(k)+1)we getlimn→∞ d(y2n(k)+1, y2m(k)) = . (5)Letting k →∞ and using (4), (2) in|d(y2n(k), y2m(k)−1)− d(y2n(k), y2m(k))| ≤ d(y2m(k), y2m(k)−1)we getlimn→∞ d(y2n(k), y2m(k)−1) = . (6)Letting k →∞ and using (6) and (2) in|d(y2m(k)−1, y2n(k)+1)− d(y2m(k)−1, y2n(k))| ≤ d(y2n(k), y2n(k)+1)we getlimn→∞ d(y2m(k)−1, y2n(k)+1) = . (7)Putting x = x2m(k), y = x2n(k)−1 in (i) we haveφ1(d(y2m(k), y2n(k)+1)) ≤ ψ1⎛⎝ d(y2m(k)−1, y2n(k)), d(y2m(k)−1, y2m(k)),d(y2n(k), y2n(k)+1),12 [d(y2m(k)−1, y2n(k)+1) + d(y2n(k), y2m(k))]⎞⎠−ψ2⎛⎝ d(y2m(k)−1, y2n(k)), d(y2m(k)−1, y2m(k)),d(y2n(k), y2n(k)+1),12 [d(y2m(k)−1, y2n(k)+1) + d(y2n(k), y2m(k))]⎞⎠ .Letting k →∞ and using (2), (4), (5), (6) and (7) we getφ1() ≤ ψ1(, 0, 0, )− ψ2(, 0, 0, ) < ψ1(, , , ) = φ1().It is a contradiction. Therefore {y2n} is a Cauchy sequence and hence {yn} is aCauchy sequence from(2). Since X is complete, there exists z ∈ X such that yn → zas n →∞ .Case: Suppose S is continuous. Then Sfx2n → Sz, S2x2n → Sz. Since (f, S) issemi compatible , we have fSx2n → Sz.Step I: Putting x = Sx2n, y = x2n+1 in (i) we haveφ1(d(fSx2n, gx2n+1)) ≤ ψ1⎛⎝ d(S2x2n, Tx2n+1), d(S2x2n, fSx2n),d(Tx2n+1, gx2n+1),12 [d(S2x2n, gx2n+1) + d(Tx2n+1, fSx2n)]⎞⎠−ψ2⎛⎝ d(S2x2n, Tx2n+1), d(S2x2n, fSx2n),d(Tx2n+1, gx2n+1),12 [d(S2x2n, gx2n+1) + d(Tx2n+1, fSx2n)]⎞⎠ .Common fixed point theorems 71Letting n →∞, we getφ1(d(Sz, z)) ≤ ψ1(d(Sz, z), 0, 0, d(Sz, z))− ψ2(d(Sz, z), 0, 0, d(Sz, z))< ψ1(d(Sz, z), d(Sz, z), d(Sz, z), d(Sz, z))= φ1(d(Sz, z).It is a contradiction if Sz = z. Hence Sz = z.Step II : Putting x = z, y = x2n+1 in (i) and letting n →∞ we get fz = z.Step III: Since z = fz ∈ f(X) ⊆ T (X), there exists u ∈ X such that z = Tu .Putting x = x2n, y = u in (i) and letting n →∞, we get z = gu and hence gu = Tu.Since (g, T ) is semi compatible, we have gTu = Tgu so that gz = Tz.Step IV: Putting x = z, y = z in (i), we get gz = z so that gz = z = Tz. Thus zis a common ﬁxed point of f, g, S and T .Case: Suppose f is continuous. Then f2x2n → fz, fSx2n → fz. Since (f, S) issemi compatible, fSx2n → Sz. Hence fz = Sz. The rest of the proof follows fromSteps II,III, IV. Similarly we can show that z is a common ﬁxed point of f, g, S andT when g or T is continuous as in previous two cases. Uniqueness of common ﬁxedpoint follows easily from (i). Remark 2.2. Thoerem 2.1 is not true if the condition (i) is replaced by(A) φ1(d(fx, gy)) ≤ ψ1(d(Sx, T y), d(Sx, fx), d(Ty, gy), d(Sx, gy), d(Ty, fx))−ψ2(d(Sx, T y), d(Sx, fx), d(Ty, gy), d(Sx, gy), d(Ty, fx))for all x, y ∈ X, where ψ1, ψ2 ∈ Ψ5 and φ1(x) = ψ1(x, x, x, x, x)∀x ∈ [0,∞) in viewof the following Example even when S = T = I(Identity map).Example 2.3.(Ex.6, [7]) Let X = {1, 2, 3, 4}, d(n, n) = 0∀n ∈ X, d(1, 2) =d(3, 4) = 2, d(1, 3) = d(2, 4) = 1, d(1, 4) = d(2, 3) = 32 . f, g : X → X as f1 = f4 =2, f2 = f3 = 1 and g1 = g3 = 4, g2 = g4 = 3.Clearly,d(fx, gy) ≤ 34max{d(x, y), d(x, fx), d(y, gy), d(x, gy), d(y, fx)}∀x, y ∈ X. Let ψ1(t1, t2, t3, t4, t5) = max{t1, t2, t3, t4, t5}, ψ2 = 14ψ1. Then φ1(t) =t∀t ∈ [0,∞). All conditions of Theorem 2.1 are satisfied except (i). Here thecondition (A) is satisfied with S = T = I(identity map). But f and g have nocommon fixed point in X.Remark 2.4. Example 2.3 also shows the importance of second generalizedaltering distance function ψ2. Theorem 2.1 is not true if the condition (i) is replacedby(B) φ1(d(fx, gy)) ≤ ψ1(d(Sx, T y), d(Sx, fx), d(Ty, gy), 12 [d(Sx, gy) + d(Ty, fx)])for all x, y ∈ X, where ψ1 ∈ Ψ4 and φ1(x) = ψ1(x, x, x, x)∀x ∈ [0,∞).In Example 2.3, we have d(fx, gy) ≤ max{d(x, y), d(x, fx), d(y, gy), 12 [d(x, gy)+d(y, fx)]} for all x, y ∈ X. Let ψ1(t1, t2, t3, t4) = max{t1, t2, t3, t4} and S = T =I(Identity map). Then φ1(t) = t∀t ∈ [0,∞). All conditions of Theorem 2.1 aresatisfied except (i). Here the condition (B) is satisfied with S = T = I(identitymap). But f and g have no common fixed point in X.A number of ﬁxed point results may be obtained by assuming diﬀerent formsfor the functions ψ1 and ψ2. Here, we give some corollaries to our theorem.72 K.P.R. Rao, G.Ravi Babu and D.VasuBabuCorollary 2.5. Theorem 2.1 with the inequality (i) is replaced byd(fx, gy) ≤ kmax{d(Sx, T y), d(Sx, fx), d(Ty, gy), 12[d(Sx, gy) + d(Ty, fx)]}∀x, y ∈ X, where 0 < k < 1.Proof. Let ψ1(t1, t2, t3, t4) = max{t1, t2, t3, t4} and ψ2(t1, t2, t3, t4) = (1− k)ψ1(t1, t2, t3, t4).Then φ1(t) = t∀t ∈ [0,∞). The corollary then follows from Theo-rem 2.1. Now we give a Gregus type [11] common ﬁxed point theorem for four maps.Corollary 2.6. Theorem 2.1 with the inequality (i) is replaced byds(fx, gy) ≤ φ(ads(Sx, T y) + (1 − a)max{ds(Sx, fx), ds(Ty, gy)})∀x, y ∈ X, where s is any positive integer and φ : [0,∞) → [0,∞) is continuous,decreasing and φ(t) > 0∀t > 0.Proof. Let ψ1(t1, t2, t3, t4) = ats1 + (1 − a)max{ts2, ts3}, ψ2(t1, t2, t3, t4) =ψ1(t1, t2, t3, t4) − φ(ψ1(t1, t2, t3, t4)). Then φ1(t) = ψ1(t, t, t, t) = ts. Now thecorollary follows from Theorem 2.1. Note: Corollary 2.6 with S = T = I is a diﬀerent version of theorem of Sessaand Fisher [13].Finally we give the following theorem for compatible of type(β) pairs of map-pings.Theorem 2.7. Theorem 2.1 with condition (iii) is replaced by(iii)1 (f, S) and (g, T ) are compatible of type (β) pairs.Proof. As in Theorem 2.1, the sequence {yn} converges to z ∈ X .Case: Suppose S is continuous. Then Sfx2n → Sz, S2x2n → Sz. Since (f, S) iscompatible of type (β), we have f2x2n → Sz. Putting x = fx2n, y = x2n+1 in (i)and letting n → ∞ , we get Sz = z. Putting x = z, y = x2n+1 in (i) and lettingn →∞ , we get fz = z. As in Steps III, IV in Theorem 2.1, we have gz = z = Tz.Case: Suppose f is continuous. Then fSx2n → fz, f2x2n → fz . Since (f, S) iscompatible pair of type(β), we have S2x2n → fz.Putting x = Sx2n, y = x2n+1 in (i) and letting n → ∞, we getfz = z. Sincez = fz ∈ f(X) ⊆ T (X), there exists w ∈ Xsuch that z = Tw.Putting x = x2n, y = w in (i) and letting n → ∞ , we get z = gw so thatgw = Tw. Since (g, T ) is compatible pair of type (β), we have gz = Tz. Puttingx = x2n, y = z in (i) and letting n → ∞ , we get gz = z so that Tz = z.Since z = gz ∈ g(X) ⊆ S(X), there exists v ∈ X such that z = Sv. Puttingx = v, y = x2n+1 in (i) and letting n → ∞ , we get fv = z so that z = fv = Sv.Since (f, S) is compatible pair of type (β), we have fz = Sz so that Sz = fz = z.Similarly we can show that z is a common ﬁxed point of f, g, S and T when T or gis continuous. Uniqueness of z follows easily from (i). References[1] B. S. Choudhury, P.N. Dutta, A unified fixed point result in metric spacesinvolving a two variable function, Filomat 14(2000), 43-48.Common fixed point theorems 73[2] B. S. Choudhury, A common unique fixed point result in metric spaces involv-ing generalized altering distances, Mathematical Communications 10(2005),105-110.[3] B. Singh, S. Jain, Semi compatibility, compatibility and fixed point theoremsin fuzzy metric spaces, J. Chungcheong Math. Soc. 18(2005), 1-22.[4] G.V .R. Babu, Generalization of fixed point theorems relating to the diameterof orbits by using a control function, Tamkang J. Math. 35(2004), 159-168.[5] G.V .R. Babu, S. Ismail, A fixed point theorem by altering distances, Bull.Cal. Math. Soc. 93(2001), 393-398.[6] H.K. Pathak, Y. J. Cho, S.M. Kang, B. S. Lee, Fixed point theorems forcompatible mappings of type (p) and application to dynamic programming, LeMatematiche (Fasc.I) 50(1995), 15-33.[7] K.P. R.Sastry, S.V.R. Naidu, Fixed point theorems for generalized con-traction mappings, Yokahama Math. J. XXVIII, No.1& 2,(1980), 15-29.[8] K.P. R. Sastry, G.V.R. Babu, Fixed point theorems in metric spaces byaltering distances, Bull. Cal. Math. Soc. 90(1998), 175-182.[9] K.P. R. Sastry, G.V.R. Babu, Some fixed point theorems by altering dis-tances between the points, Indian J. Pure. Appl. Math. 30(1999), 641-647.[10] K.P. R. Sastry, S. V.R. Naidu, G.V.R. Babu, G.A. Naidu, General-ization of common fixed point theorems for weakly commuting maps by alteringdistances, Tamkang J. Math. 31(2000), 243-250.[11] M. Jr. Gregus, A fixed point theorem in Banach space, Boll. Un. Math. ItalA(5) 17(1980), 193-198.[12] M.S. Khan, M. Swaleh, S. Sessa, Fixed point theorem by altering distancesbetween the points, Bull. Austral. Math. Soc. 30(1984), 1-9.[13] S. Sessa, B. Fisher, Common fixed points of two mappings in Banach spaces,J. Math. Phy. Sci. 18(1984), 353-360.[14] Y. J. Cho, B.K. Sharma, D.R. Sahu, Semi compatibility and fixed points,Math Japonica 42(1995), 91-98.

Common fixed point theorems through generalized altering distance functions

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Mathematical Communications 13(2008), 67-73 67Common fixed point theorems through generalizedaltering distance functionsK.P.R. Rao∗, G.Ravi Babu† and D.VasuBabu‡Abstract. In this paper we obtain a unique common fixed pointtheorem, for four self maps using a generalized altering distance func-tion in four variables, which generalizes and improves the main theoremof Choudhury [2]. We also obtain a Gregus type common fixed pointtheorem for four maps as a corollary.Key words: fixed points, semi compatible mappings, altering dis-tance functionsAMS subject classifications: 47H10, 54H25Received April 12, 2007 Accepted March 4, 20081. IntroductionM. S. Khan et al. [12] introduced the altering distances and used it for solving fixedpoints problems in metric spaces. Recently, for example authors of [1, 4, 5, 8, 9, 10]used the altering distance function and obtained some fixed point theorems. In2005 Choudhury [2] introduced a generalized distance function in three variablesand obtained a common fixed point theorem for a pair of self maps in a completemetric space. The main aim of this paper is to prove the existence and uniqueness ofcommon fixed points of two pairs of semi compatible or compatible pair of type(β)mappings by using a generalized altering distance function of four variables. Ourresults extend and generalize the main theorem of [2]. We also give an example toshow that our theorem is not valid when the generalized altering distance functionis in five variables or the second generalized altering distance function is dropped.In [14], Cho et al. introduced the non symmetrical concept of semi compatibilityof maps in d-complete topological spaces. They defined a pair of self maps(S, T )to be semi compatible if (i) Sy = Ty implies STy = TSy and (ii) limn→∞Sxn =limn→∞ Txn = x for some x ∈ X implies limn→∞ STxn = Tx hold. Bijendra∗Department of Applied Mathematics, Acharya Nagarjuna University-Nuzvid Campus, Nuzvid-521201, A.P., India, e-mail:kprrao2004@yahoo.com†Department of Applied Mathematics, Acharya Nagarjuna University-Nuzvid Campus, Nuzvid-521201, A.P., India‡Department of Applied Mathematics, Acharya Nagarjuna University-Nuzvid Campus, Nuzvid-521201,A.P., India68 K.P.R. Rao, G.Ravi Babu and D.VasuBabuSingh and Jain [3] observed that (ii) implies (i). Hence they defined the semicompatibility by the condition (ii) only.In [6], Pathak et. al defined a pair of self maps (S, T) to be compatible of type(P ) or (β) if limn→∞ d(S2xn, T 2xn) = 0 whenever {xn} is a sequence in X suchthat limn→∞ Sxn= limn→∞ Txn = x for some x ∈ X . It is clear that in this caseSTy = TSy whenever Sy = Ty for any y ∈ X .Remark 1.1. The concepts semi compatible pair and compatible pair of type(β) maps are independent in view of the following examples.Example 1.2. Let X = [0, 2] with usual metric. Letf(x) ={2 − x, if 0 ≤ x < 11, if x ≥ 1 and S(x) ={x if 0 ≤ x < 1,2 if x ≥ 1.Let xn = 1 − 1/n ,n > 1. Then fxn = 1 + 1/n → 1, Sxn = xn = 1 − 1/n → 1.fSxn = 1 + 1/n → 1 = S(1) and d(f2xn, S2xn) = d(1, 1 − 1/n) → 0. Thus (f, S)is not semi compatible and (f, S) is compatible pair of type (β).Example 1.3. Let X = [0, 2] with usual metric. Letf(x) ={2 − x, if 0 ≤ x < 12, if x ≥ 1 and S(x) ={x if 0 ≤ x < 1,1 if x ≥ 1.Let xn = 1 − 1/n ,n > 1 . Then fxn = 1 + 1/n → 1, Sxn = 1 − 1/n → 1.fSxn = 1 + 1/n → 1 = S(1), d(f2xn, S2xn) = d(2, 1 − 1/n) = 1 + 1/n → 1 = 0.Thus (f, S) is semi compatible and (f, S) is not compatible pair of type (β).Let Ψn denote the set of all functions ψ : [0,∞)n → [0,∞) such that(i) ψ is continuous,(ii) ψ is monotone increasing in all the variables,(iii) ψ(t1, t2, ....., tn) = 0 if and only if t1 = t2 = .... = tn = 0.The functions in Ψn are called generalized altering distance functions.Theorem 1.4. (Theorem 1, [2]) Let (X, d) be a complete metric space and Sand T be self maps on X such thatφ1(d(Sx, T y)) ≤ ψ1(d(x, y), d(x, Sx), d(y, T y)) − ψ2(d(x, y), d(x, Sx), d(y, T y))for all x, y ∈ X, where ψ1, ψ2 ∈ Ψ3 and φ1(x) = ψ1(x, x, x)∀x ∈ [0,∞). Then Sand T have a unique common fixed point.2. Main resultsTheorem 2.1. Let (X, d) be a complete metric space and f, g, S, T : X → X besuch that(i) φ1(d(fx, gy)) ≤ ψ1(d(Sx, T y), d(Sx, fx), d(Ty, gy),12 [d(Sx, gy) + d(Ty, fx)])−ψ2(d(Sx, T y), d(Sx, fx), d(Ty, gy),12 [d(Sx, gy) + d(Ty, fx)])for all x, y ∈ X, where ψ1, ψ2 ∈ Ψ4 and φ1(x) = ψ1(x, x, x, x)∀x ∈ [0,∞),Common fixed point theorems 69(ii) one of four mappings f, g, S and T is continuous,(iii) (f, S) and (g, T ) are semi compatible pairs ,(iv) f(X) ⊆ T (X), g(X) ⊆ S(X).Then f, g, S and T have a unique common fixed point in X.Proof. Let x0 ∈ X be an arbitrary point. From (iv), construct the sequences{xn} and {yn} in X such that fx2n = Tx2n+1 = y2n, gx2n+1 = Sx2n+2 = y2n+1,n = 0, 1, 2, . . . . Let an = d(yn, yn+1). Putting x = x2n, y = x2n+1 in (i) we getφ1(a2n) ≤ ψ1(a2n−1, a2n−1, a2n,12[a2n−1 + a2n])−ψ2(a2n−1, a2n−1, a2n,12d(y2n−1, y2n+1)). (1)If a2n−1 < a2n thenφ1(a2n) ≤ ψ1(a2n, a2n, a2n, a2n) − ψ2(a2n−1, a2n−1, a2n, 12d(y2n−1, y2n+1))< φ1(a2n)which is a contradiction. Hence a2n ≤ a2n−1 , n = 1, 2, . . . . Similarly by puttingx = x2n+2, y = x2n+1 in (i) we can show that a2n+1 ≤ a2n, n = 0, 1, 2, . . . . Thusan+1 ≤ an, n = 0, 1, 2, ..... so that {an} is a decreasing sequence of non negative realnumbers and hence convergent to some a ∈ R.Let b = limn→∞12d(yn, yn+2). Letting n→ ∞ in (1) we getφ1(a) ≤ ψ1(a, a, a, a) − ψ2(a, a, a, b) = φ1(a) − ψ2(a, a, a, b).Thus ψ2(a, a, a, b) = 0 so that a = b = 0. Hencelimn→∞ d(yn, yn+1) = 0 (2)To show that{yn} is a Cauchy sequence, it is sufficient to show that the sub sequence{y2n} of {yn} is a Cauchy sequence in view of (2). If {y2n} is not Cauchy, thereexists an ε > 0 and monotone increasing sequences of natural numbers {2m(k)}and {2n(k)} such that n(k) > m(k), d(y2m(k), y2n(k)) ≥ ε andd(y2m(k), y2n(k)−2) < ε (3)From (3),ε ≤ d(y2m(k), y2n(k))≤ d(y2m(k), y2n(k)−2) + d(y2n(k)−2, y2n(k)−1) + d(y2n(k)−1, y2n(k))< ε+ d(y2n(k)−2, y2n(k)−1) + d(y2n(k)−1, y2n(k))Letting k → ∞, using (2), we havelimn→∞ d(y2m(k), y2n(k)) = ε. (4)70 K.P.R. Rao, G.Ravi Babu and D.VasuBabuLetting k → ∞, using (4), and (2) in|d(y2n(k)+1, y2m(k)) − d(y2m(k), y2n(k))| ≤ d(y2n(k), y2n(k)+1)we getlimn→∞ d(y2n(k)+1, y2m(k)) = ε. (5)Letting k → ∞ and using (4), (2) in|d(y2n(k), y2m(k)−1) − d(y2n(k), y2m(k))| ≤ d(y2m(k), y2m(k)−1)we getlimn→∞ d(y2n(k), y2m(k)−1) = ε. (6)Letting k → ∞ and using (6) and (2) in|d(y2m(k)−1, y2n(k)+1) − d(y2m(k)−1, y2n(k))| ≤ d(y2n(k), y2n(k)+1)we getlimn→∞ d(y2m(k)−1, y2n(k)+1) = ε. (7)Putting x = x2m(k), y = x2n(k)−1 in (i) we haveφ1(d(y2m(k), y2n(k)+1)) ≤ ψ1⎛⎝ d(y2m(k)−1, y2n(k)), d(y2m(k)−1, y2m(k)),d(y2n(k), y2n(k)+1),12 [d(y2m(k)−1, y2n(k)+1) + d(y2n(k), y2m(k))]⎞⎠−ψ2⎛⎝ d(y2m(k)−1, y2n(k)), d(y2m(k)−1, y2m(k)),d(y2n(k), y2n(k)+1),12 [d(y2m(k)−1, y2n(k)+1) + d(y2n(k), y2m(k))]⎞⎠ .Letting k → ∞ and using (2), (4), (5), (6) and (7) we getφ1(ε) ≤ ψ1(ε, 0, 0, ε)− ψ2(ε, 0, 0, ε) < ψ1(ε, ε, ε, ε) = φ1(ε).It is a contradiction. Therefore {y2n} is a Cauchy sequence and hence {yn} is aCauchy sequence from(2). Since X is complete, there exists z ∈ X such that yn → zas n→ ∞ .Case: Suppose S is continuous. Then Sfx2n → Sz, S2x2n → Sz. Since (f, S) issemi compatible , we have fSx2n → Sz.Step I: Putting x = Sx2n, y = x2n+1 in (i) we haveφ1(d(fSx2n, gx2n+1)) ≤ ψ1⎛⎝ d(S2x2n, Tx2n+1), d(S2x2n, fSx2n),d(Tx2n+1, gx2n+1),12 [d(S2x2n, gx2n+1) + d(Tx2n+1, fSx2n)]⎞⎠−ψ2⎛⎝ d(S2x2n, Tx2n+1), d(S2x2n, fSx2n),d(Tx2n+1, gx2n+1),12 [d(S2x2n, gx2n+1) + d(Tx2n+1, fSx2n)]⎞⎠ .Common fixed point theorems 71Letting n→ ∞, we getφ1(d(Sz, z)) ≤ ψ1(d(Sz, z), 0, 0, d(Sz, z))− ψ2(d(Sz, z), 0, 0, d(Sz, z))< ψ1(d(Sz, z), d(Sz, z), d(Sz, z), d(Sz, z))= φ1(d(Sz, z).It is a contradiction if Sz = z. Hence Sz = z.Step II : Putting x = z, y = x2n+1 in (i) and letting n→ ∞ we get fz = z.Step III: Since z = fz ∈ f(X) ⊆ T (X), there exists u ∈ X such that z = Tu .Putting x = x2n, y = u in (i) and letting n→ ∞, we get z = gu and hence gu = Tu.Since (g, T ) is semi compatible, we have gTu = Tgu so that gz = Tz.Step IV: Putting x = z, y = z in (i), we get gz = z so that gz = z = Tz. Thus zis a common fixed point of f, g, S and T .Case: Suppose f is continuous. Then f2x2n → fz, fSx2n → fz. Since (f, S) issemi compatible, fSx2n → Sz. Hence fz = Sz. The rest of the proof follows fromSteps II,III, IV. Similarly we can show that z is a common fixed point of f, g, S andT when g or T is continuous as in previous two cases. Uniqueness of common fixedpoint follows easily from (i). Remark 2.2. Thoerem 2.1 is not true if the condition (i) is replaced by(A) φ1(d(fx, gy)) ≤ ψ1(d(Sx, T y), d(Sx, fx), d(Ty, gy), d(Sx, gy), d(Ty, fx))−ψ2(d(Sx, T y), d(Sx, fx), d(Ty, gy), d(Sx, gy), d(Ty, fx))for all x, y ∈ X, where ψ1, ψ2 ∈ Ψ5 and φ1(x) = ψ1(x, x, x, x, x)∀x ∈ [0,∞) in viewof the following Example even when S = T = I(Identity map).Example 2.3.(Ex.6, [7]) Let X = {1, 2, 3, 4}, d(n, n) = 0∀n ∈ X, d(1, 2) =d(3, 4) = 2, d(1, 3) = d(2, 4) = 1, d(1, 4) = d(2, 3) = 32 . f, g : X → X as f1 = f4 =2, f2 = f3 = 1 and g1 = g3 = 4, g2 = g4 = 3.Clearly,d(fx, gy) ≤ 34max{d(x, y), d(x, fx), d(y, gy), d(x, gy), d(y, fx)}∀x, y ∈ X. Let ψ1(t1, t2, t3, t4, t5) = max{t1, t2, t3, t4, t5}, ψ2 = 14ψ1. Then φ1(t) =t∀t ∈ [0,∞). All conditions of Theorem 2.1 are satisfied except (i). Here thecondition (A) is satisfied with S = T = I(identity map). But f and g have nocommon fixed point in X.Remark 2.4. Example 2.3 also shows the importance of second generalizedaltering distance function ψ2. Theorem 2.1 is not true if the condition (i) is replacedby(B) φ1(d(fx, gy)) ≤ ψ1(d(Sx, T y), d(Sx, fx), d(Ty, gy), 12 [d(Sx, gy) + d(Ty, fx)])for all x, y ∈ X, where ψ1 ∈ Ψ4 and φ1(x) = ψ1(x, x, x, x)∀x ∈ [0,∞).In Example 2.3, we have d(fx, gy) ≤ max{d(x, y), d(x, fx), d(y, gy), 12 [d(x, gy)+d(y, fx)]} for all x, y ∈ X. Let ψ1(t1, t2, t3, t4) = max{t1, t2, t3, t4} and S = T =I(Identity map). Then φ1(t) = t∀t ∈ [0,∞). All conditions of Theorem 2.1 aresatisfied except (i). Here the condition (B) is satisfied with S = T = I(identitymap). But f and g have no common fixed point in X.A number of fixed point results may be obtained by assuming different formsfor the functions ψ1 and ψ2. Here, we give some corollaries to our theorem.72 K.P.R. Rao, G.Ravi Babu and D.VasuBabuCorollary 2.5. Theorem 2.1 with the inequality (i) is replaced byd(fx, gy) ≤ kmax{d(Sx, T y), d(Sx, fx), d(Ty, gy), 12[d(Sx, gy) + d(Ty, fx)]}∀x, y ∈ X, where 0 < k < 1.Proof. Let ψ1(t1, t2, t3, t4) = max{t1, t2, t3, t4} and ψ2(t1, t2, t3, t4) = (1 − k)ψ1(t1, t2, t3, t4).Then φ1(t) = t∀t ∈ [0,∞). The corollary then follows from Theo-rem 2.1. Now we give a Gregus type [11] common fixed point theorem for four maps.Corollary 2.6. Theorem 2.1 with the inequality (i) is replaced byds(fx, gy) ≤ φ(ads(Sx, T y) + (1 − a)max{ds(Sx, fx), ds(Ty, gy)})∀x, y ∈ X, where s is any positive integer and φ : [0,∞) → [0,∞) is continuous,decreasing and φ(t) > 0∀t > 0.Proof. Let ψ1(t1, t2, t3, t4) = ats1 + (1 − a)max{ts2, ts3}, ψ2(t1, t2, t3, t4) =ψ1(t1, t2, t3, t4) − φ(ψ1(t1, t2, t3, t4)). Then φ1(t) = ψ1(t, t, t, t) = ts. Now thecorollary follows from Theorem 2.1. Note: Corollary 2.6 with S = T = I is a different version of theorem of Sessaand Fisher [13].Finally we give the following theorem for compatible of type(β) pairs of map-pings.Theorem 2.7. Theorem 2.1 with condition (iii) is replaced by(iii)1 (f, S) and (g, T ) are compatible of type (β) pairs.Proof. As in Theorem 2.1, the sequence {yn} converges to z ∈ X .Case: Suppose S is continuous. Then Sfx2n → Sz, S2x2n → Sz. Since (f, S) iscompatible of type (β), we have f2x2n → Sz. Putting x = fx2n, y = x2n+1 in (i)and letting n → ∞ , we get Sz = z. Putting x = z, y = x2n+1 in (i) and lettingn→ ∞ , we get fz = z. As in Steps III, IV in Theorem 2.1, we have gz = z = Tz.Case: Suppose f is continuous. Then fSx2n → fz, f2x2n → fz . Since (f, S) iscompatible pair of type(β), we have S2x2n → fz.Putting x = Sx2n, y = x2n+1 in (i) and letting n → ∞, we getfz = z. Sincez = fz ∈ f(X) ⊆ T (X), there exists w ∈ Xsuch that z = Tw.Putting x = x2n, y = w in (i) and letting n → ∞ , we get z = gw so thatgw = Tw. Since (g, T ) is compatible pair of type (β), we have gz = Tz. Puttingx = x2n, y = z in (i) and letting n → ∞ , we get gz = z so that Tz = z.Since z = gz ∈ g(X) ⊆ S(X), there exists v ∈ X such that z = Sv. Puttingx = v, y = x2n+1 in (i) and letting n → ∞ , we get fv = z so that z = fv = Sv.Since (f, S) is compatible pair of type (β), we have fz = Sz so that Sz = fz = z.Similarly we can show that z is a common fixed point of f, g, S and T when T or gis continuous. Uniqueness of z follows easily from (i). References[1] B. S. Choudhury, P.N. Dutta, A unified fixed point result in metric spacesinvolving a two variable function, Filomat 14(2000), 43-48.Common fixed point theorems 73[2] B. S. Choudhury, A common unique fixed point result in metric spaces involv-ing generalized altering distances, Mathematical Communications 10(2005),105-110.[3] B. Singh, S. Jain, Semi compatibility, compatibility and fixed point theoremsin fuzzy metric spaces, J. Chungcheong Math. Soc. 18(2005), 1-22.[4] G.V .R. Babu, Generalization of fixed point theorems relating to the diameterof orbits by using a control function, Tamkang J. Math. 35(2004), 159-168.[5] G.V .R. Babu, S. Ismail, A fixed point theorem by altering distances, Bull.Cal. Math. Soc. 93(2001), 393-398.[6] H.K. Pathak, Y. J. Cho, S.M. Kang, B. S. Lee, Fixed point theorems forcompatible mappings of type (p) and application to dynamic programming, LeMatematiche (Fasc.I) 50(1995), 15-33.[7] K.P. R.Sastry, S.V.R. Naidu, Fixed point theorems for generalized con-traction mappings, Yokahama Math. J. XXVIII, No.1& 2,(1980), 15-29.[8] K.P. R. Sastry, G.V.R. Babu, Fixed point theorems in metric spaces byaltering distances, Bull. Cal. Math. Soc. 90(1998), 175-182.[9] K.P. R. Sastry, G.V.R. Babu, Some fixed point theorems by altering dis-tances between the points, Indian J. Pure. Appl. Math. 30(1999), 641-647.[10] K.P. R. Sastry, S. V.R. Naidu, G.V.R. Babu, G.A. Naidu, General-ization of common fixed point theorems for weakly commuting maps by alteringdistances, Tamkang J. Math. 31(2000), 243-250.[11] M. Jr. Gregus, A fixed point theorem in Banach space, Boll. Un. Math. ItalA(5) 17(1980), 193-198.[12] M.S. Khan, M. Swaleh, S. Sessa, Fixed point theorem by altering distancesbetween the points, Bull. Austral. Math. Soc. 30(1984), 1-9.[13] S. Sessa, B. Fisher, Common fixed points of two mappings in Banach spaces,J. Math. Phy. Sci. 18(1984), 353-360.[14] Y. J. Cho, B.K. Sharma, D.R. Sahu, Semi compatibility and fixed points,Math Japonica 42(1995), 91-98.

Common fixed point theorems through generalized altering distance functions

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