In this paper we work out a unique common fixed point result for two self-mappings defined on a complete metric space. These mappings are assumed to satisfy a contractive inequality which involves two generalised altering distances

B. S. Choudhury

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Mathematical Communications 10(2005), 105-110 105A common unique fixed point result in metricspaces involving generalised altering distancesBinayak S. Choudhury∗Abstract. In this paper we work out a unique common fixed pointresult for two self-mappings defined on a complete metric space. Thesemappings are assumed to satisfy a contractive inequality which involvestwo generalised altering distances.Key words: generalised altering distances, metric space, fixed pointAMS subject classifications: 54H25Received December 18, 2003 Accepted May 17, 20041. IntroductionFixed point theory in spaces has a vast literature. In particular, there has been anumber of works on fixed points involving Altering Distance Functions. These arecontrol functions which alter the distance between two points in a metric space.Such functions were introduced by M. S. Khan et al. in [4] and used in the samepaper for defining and solving a new category of fixed points problems in metricspaces.Definition 1 [see [4]]. An altering distance function is a function ψ : [0,∞)→[0,∞) which is(i) monotone increasing and continuous and(ii) ψ(t) = 0 if and only if t = 0.Afterwards a number of works has appeared in which altering distances hasbeen used. In references [5], [6] and [7], for example, fixed points of single valuedmappings and in [1] fixed points of multi-mappings have been obtained by usingaltering distance functions. Altering distances have been generalised to a two-variable function and in [3] a generalisation to a three-variable function has beenintroduced and applied for obtaining fixed point results in metric spaces.In this paper we propose a generalisation of altering distances to a three-variablefunction and with the help of such function we derive a unique common fixed pointresult for two self-mappings in a complete metric space. We note that specific fixed∗Department of Mathematics, Bengal Engineering and Science University, P.O. - B. Garden,Shibpur, 711 103 Howrah, West Bengal, India, e-mail: binayak12@yahoo.co.in106 B. S.Choudhurypoint results follow by specific choices of the functional form of ψ. We propose thefollowing definition.Definition 2. A function ψ : [0,∞)3 → [0,∞] is said to be a generalised alteringdistance function if(i) ψ(x, y, z) is continuous,(ii) ψ is monotone increasing in all the three variables and(iii) ψ(x, y, z) = 0 only if x = y = z = 0.We define φ (x) = ψ(x, x, x) for x ∈ [0,∞). Clearly, φ(x) = 0 if and only ifx = 0. Examples of ψ areψ(a, b, c) = kmax{a, b, c}, for k > 0,ψ(a, b, c) = ap + bq + cr, p, q, r ≥ 1,ψ(a, b, c) = (a+ α bq)r + cs, where p, q, r, s ≥ 1 and α > 0Other examples may also be constructed.2. Fixed point theoremTheorem 1. Let(X,d) be a complete metric space and S and T two self - mappingssuch that the following inequality is satisfied:φ1(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)) (1)where ψ1 and ψ2 are generalised altering distance functions and φ1(x) = ψ1(x, x, x).Then S and T have a common fixed point.Proof. Let x0 ∈ X be an arbitrary point. For n = 0, 1, 2, . . .x2n+1 = Sx2nandx2n+2 = Tx2n+1letan = d(xn, xn+1). (2)Putting x = x2n and y = x2n+1 in (1), for all n = 0, 1, 2, . . . we getφ1(d(Sx2n, Tx2n+1)) = φ1(d(x2n+1, x2n+2))≤ ψ1(d(x2n, x2n+1), d(x2n, Sx2n), d(x2n+1, Tx2n+1))−ψ2(d(x2n, x2n+1), d(x2n, Sx2n), d(x2n+1, Tx2n+1))= ψ1(d(x2n, x2n+1), d(x2n, x2n+1), d(x2n+1, x2n+2)−ψ2(d(x2n, x2n+1), d(x2n, x2n+1), d(x2n+1, x2n+2))or by(2), for all n = 0, 1, 2, . . .φ1(a2n+1) ≤ ψ1(a2n, a2n, a2n+1)− ψ2(a2n, a2n, a2n+1). (3)A common unique fixed point result in metric spaces 107If a2n+1 > a2n, thenφ1(a2n+1) < ψ1(a2n+1, a2n+1, a2n+1). = φ(a2n+1) (4)This is due to the fact that ψ1 is monotone increasing in all variables and ψ2(a2n, a2n, a2n+1) 	=0 whenever a2n+1 	= 0. Thus we arrive at a contradiction, so thata2n+1 ≤ a2n, n = 0, 1, 2, . . . (5)Putting x = x2n and y = x2n−1 in (1) we obtainφ1(a2n) ≤ ψ1(a2n−1, a2n−1, a2n)− ψ2(a2n−1, a2n−1, a2n) (6)By an identical argument we obtaina2n+2 ≤ a2n+1, n = 0, 1, 2 . . . (7)From (5) and (7), we obtain for all n = 0, 1, 2, . . .an+1 ≤ an, n = 0, 1, 2 . . . (8)Then from (3) and (6), for all n = 0, 1, 2, . . . we obtainφ1(an+1) ≤ φ1(an)− φ2(an+1), where φ2(x) = ψ2(x, x, x)or equivalentlyφ2(an+1) ≤ φ1(an)− φ1(an+1).Summing up in (8) we obtain∞∑n=0φ2(an+1) ≤ φ1(a0) <∞which impliesφ2(an)→ 0 as n→∞. (9)Again from (8) {an} is convergent and let an → a (say) as n → ∞. Since φ iscontinuous, from (9) we obtain φ2(a) = 0 which implies that a = 0, that isan = d(xn+1, xn)→ 0 as n→∞. (10)We next prove that {xn} is a Cauchy sequence. In view of (10) it is sufficient toprove that {x2r}∞r=1 ⊂ {xn}is a Cauchy sequence. If {x2r}∞r=1 is not a Cauchysequence, then given ∈> 0 we can find monotone increasing sequences of naturalnumbers {2m(k)} and {2n(k)} such thatn(k) > m(k), d(x2m(k), x2n(k)) >∈andd(x2m(k), x2n(k)−1) <∈ (11)108 B. S.ChoudhuryThen by (11)∈ < d(x2m(k), x2n(k)) ≤ d(x2m(k), x2n(k)−1) + d(x2n(k)−1, x2n(k))< ∈ +d(x2n(k)−1, x2n(k))Making k →∞ in the above inequality by virtue of (10) we obtainlimk→∞d(x2m(k), x2n(k)) =∈ (12)For all k = 1, 2, . . .d(x2n(k)+1, x2m(k)) ≤ d(x2n(k)+1, x2n(k)) + d(x2n(k), x2m(k)). (13)Also for all k = 1, 2, . . .d(x2n(k), x2m(k)) ≤ d(x2n(k), x2n(k)+1) + d(x2n(k)+1, x2m(k)). (14)Making k →∞ in (13) and (14) respectively, by using f (10) and (12) we havelimk→∞d(x2n(k)+1, x2m(k)) ≤∈and∈≤ limk→∞d(x2n(k)+1, x2m(k))that islimk→∞d(x2n(k)+1, x2m(k)) =∈ . (15)For all k = 1, 2, . . .d(x2n(k), x2m(k)−1) ≤ d(x2n(k), x2m(k)) + d(x2m(k), x2m(k), x2m(k)−1)d(x2n(k), x2m(k)) ≤ d(x2n(k), x2m(k)−1) + d(x2m(k)−1, x2m(k)).Making k →∞ in the above two inequalities and using (10) and (12) we obtainlimk→∞d(x2n(k), x2m(k)−1) =∈ . (16)Putting x = x2n(k) and y = x2m(k)−1 in (1), for all k = 1, 2, . . . we obtainφ1d(x2n(k)+1, x2m(k))≤ ψ1(d(x2n(k), x2m(k)−1), d(x2n(k), x2n(k)+1), d(x2m(k)−1, x2m(k)))−ψ2(d(x2n(k), x2m(k)−1), d(x2n(k), x2n(k)+1), d(x2m(k)−1, x2m(k)))Making k → ∞ in the above inequality and taking into account the continuity ofψ1 and ψ2, by virtue of (10), (15) and (16) we haveφ1(∈) ≤ ψ1(∈, 0, 0)− ψ2(∈, 0, 0) < φ1(∈)This is due to the fact that ψ1 is monotone increasing in its variables and by propertyof ψ2 that ψ(x, y, z) = 0 if and only if x = y = z = 0.A common unique fixed point result in metric spaces 109The above inequality gives a contradiction so that ∈= 0. This establishes thefact {x2n}∞n=1is a Cauchy sequence and hence in view of (10) {xn} is also a Cauchysequence and hence convergent in (X,d).Letxn → z as n→∞. (17)Putting x = x2n and y = z in (1), for all n = 1, 2, . . . we obtainφ1(d(x2n+1, T z)) ≤ ψ1(d(x2n, z), d(x2n, x2n+1), d(z, T z))−ψ2(d(x2n, z), d(x2n, x2n+1), d(z, T z)).Making n → ∞ in the above inequality, by using (10) and (17) and continuity ofψ1 and ψ2 we obtainφ1(d(z, T z)) ≤ ψ1(0, 0, d(z, T z))− ψ2(0, 0, d(z, T z)).If d(z, T z) 	= 0, then using the property that ψ1 and ψ2 are monotone increasingand ψ2(x, y, z) = 0 if and only if x = y = z = 0, we obtainφ1(d(z, T z)) < φ1(d(z, T z))which is a contradiction. Hence, we obtaind(z, T z) = 0, or z = Tz. (18)In an exactly similar way we provez = Sz. (19)Equations (18) and (19) show that z is a common fixed point of S and T . ✷Let z1 and z2 be two common fixed points of S and T and z1 	= z2. Thend(z1, z2) 	= 0. From (1) we obtainφ1(d(z1, z2)) ≤ ψ1(d(z1, z2), 0, 0)− ψ2(d(z1, z2), 0, 0) < φ1(d(z1, z2)).This is again due to the fact that ψ1 is monotonic increasing in all its variables andψ(x, y, z) ≤ 0 if at least one of x, y, z is non-zero.The above inequality is a contradiction which shows that z1 = z2. This estab-lishes the uniqueness property of the fixed point.A number of fixed point results may be obtained by assuming different forms forthe functions ψ1 and ψ2. In particular, fixed point results under various contrac-tive conditions are obtainable from the above theorems. Contractive mappings areimportant in fixed point theory. A comprehensive survey of various types of con-tractive mappings and related fixed point theorems may be obtained in [8]. Here,for example, we derive the following corollary of our theorem.Corollary 1. Let S, T : X → Xwhere (X, d) is a complete metric spacesatisfying[d(Sx, T y]s ≤ k1[d(x, y)]s + k2[d(x, Tx)]s + k3[d(y, T y)]s (20)110 B. S.Choudhurywhere 0 < k1 + k2 + k3 < 1 and s > 0. Then S and T have a common fixed point.Proof. We make particular choices of ψ1 and ψ2 as the follows:ψ1(a, b, c) = k1as + k2bs + k3csψ2(a, b, c) = (1− k)[k1as + k2bs + k3cs]with k = k1 + k2 + k3. Then (20) is implied by (1). The corollary then follows byan applying of Theorem 1. ✷Other fixed point results may also be obtained under specific choices of ψ1 andψ2. As a final remark we observe that there is no continuity assumption on thefunctions S and T .References[1] B.S.Choudhury, A.Upadhyay, Bulletin of Pure and Applied Science 19E(2000), 529-533.[2] B.S.Choudhury, P.N.Dutta, A unified fixed point result in metric spacesinvolving a two variable functuion, Filomat 14(2000), 43 - 48.[3] B.S.Choudhury, P.N.Dutta, A unified approach to fixed points of self-mappings in metric spaces, preprint.[4] M.S.Khan, M. Swaleh, S. Sessa, Fixed point theorem by altering distancesbetween the points, Bull.Austral. Math.Soc. 30(1984), 1-9.[5] K.P.R. Sastry, G.V.R.Babu, Fixed point theorems in metric spaces byaltering distances, Bull.Calcutta Math.Soc. 90(1998), 175 - 182.[6] K.P.R. Sastry, G.V.R.Babu, Fixed point theorems in metric spaces byaltering distances between the points, Indian J.pure Appl.Math. 30(1999), 641- 647.[7] K.P.R. Sastry, S. V.R.Naidu, G.V.R.Babu, G.A.Naidu, Generaliza-tion of common fixed point theorems for weakly commuting maps by alteringdistance,Tmkang J.Math. 31(2000), 243 - 250.[8] M.R. Taskovic, Some new principles in fixed point theory, Math. Joponica35(1990), 645 - 666.

A common unique fixed point result in metric spaces involving generalised altering distances

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