International Journal of Analysis and Applications ISSN 2291-8639 Volume 6, Number 1 (2014), 97-112 http://www.etamaths.com COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS ANIMESH GUPTA1,∗ AND NEELESH PANDEY2 Abstract. In this paper, we obtain some common fixed point theorems for four fuzzy mappings in complete ordered metric linear spaces. These mappings are assumed to satisfy certain contractive inequality involving functions which are generalizations of altering distance functions. We also note that this fuzzy fixed point result is derivable from a multi-valued fixed point result. 1. Introduction In 1965, the theory of fuzzy sets was investigated by Zadeh [23]. In 1981, Heilpern [11] first introduced the concept of fuzzy contractive mappings and proved a fixed point theorem for these mappings in metric linear spaces. Estruch and Vidal [10] proved a fixed point theorem for fuzzy contraction mappings in a complete metric spaces which in turn generalized Heilpern fixed point theorem. Afterwards a number of works appeared in which fixed points of fuzzy mappings satisfying contractive inequalities have been studied (see [9]) A new category of contractive fixed point problems was addressed by M.S. Khan et. al [13]. There they introduced Altering Distance Function, which is a control function that alters distance between two points in a metric space. Afterwards a number of works have appeared in which altering distances have been used. In references [20] and [21] for example, fixed points of single valued mappings and in [6] fixed points of set valued mappings have been obtained by using altering distance functions. Altering distances have been generalized to func- tions with more than one argument. In [7] a generalization of such functions to a two-variable function and in [8] a generalization to a three-variable function were introduced and applied for obtaining fixed point results in metric spaces. In this paper we introduce a contractive inequality for four fuzzy mappings through a 4-variable generalization of altering distance function and then prove that the two fuzzy mappings defined on a complete ordered metric linear space satisfying such inequality have a common fixed point. We have discussed some spe- cific results, which are obtainable under special choices of the generalized altering distance function. We also show that a more general result in the fixed point theory of multi-valued mappings can be established and the result we obtained for fuzzy mappings can be deduced from the general theorem. 2010 Mathematics Subject Classification. 46S40, 47H10, 54H25. Key words and phrases. annihilator, weak annihilator, dominating maps, fuzzy mapping, com- mon fixed point. c©2014 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 97 98 GUPTA AND PANDEY 2. Preliminaries Throughout the rest of the paper unless otherwise stated (X,d) stands for a complete metric space. A fuzzy set in X is a function with domain X and values in [0, 1]. If A is a fuzzy set on X and x ∈ X then the functional value Ax is called the grade of membership of x in A. The α−level set of A, denoted by Aα, is defined by Aα = {x : Ax ≥ α}, if α ∈ (0, 1], A0 = {x : Ax ≥ 0}, where B denoted the closure of the set B. For any two subsets A and B of X we denote by H(A,B) the Hausdroff distance. For any two subsets A and B of X we write δ(A,B) = supα∈A,β∈B d(a,b). Definition 2.1. A function ψ : [0, +∞) → [0, +∞) is called an altering distance function if and only if (i) ψ is continuous, (ii) ψ is non-decreasing, (iii) ψ(t) = 0 ⇐⇒ t = 0. Choudhury [9] introduced the concept of a generalized altering distance function for three variables. Definition 2.2. A function ψ : [0, +∞) × [0, +∞) × [0, +∞) → [0, +∞) is called an altering distance function if and only if (i) ψ is continuous, (ii) ψ is non-decreasing in all three variables, (iii) ψ(x,y,z) = 0 ⇐⇒ x = y = z = 0. Rao et al. [18] introduced the concept of a generalized altering distance function for four variables. Definition 2.3. A function ψ : [0, +∞)× [0, +∞)× [0, +∞)× [0, +∞) → [0, +∞) is called an altering distance function if and only if (i) ψ is continuous, (ii) ψ is non-decreasing in all three variables, (iii) ψ(x,y,z,w) = 0 ⇐⇒ x = y = z = w = 0. Definition 2.4. Let (X,d) be a metric space and f,g : X → X. If w = fx = gx, for some x ∈ X, then x is called a coincidence point of f and g, and w is called a coincidence point of f and g. If x = w, then x is a common fixed point of f and g. The pair {f,g} is said to be comparable if and only if limn→+∞d(fgxn,gfxn) = 0, whenever {xn} is a sequence in X such that limn→+∞fxn = limn→+∞gxn = t for some t ∈ X. Definition 2.5. Let f and g be two self mappings defined on a set X. Then f and g are said to be weakly comparable if they commute at every coincidence point. Definition 2.6. Let X be a nonempty set. Then (X,d,�) is called an ordered metric linear space iff (i) (X,d) is a metric linear space, (ii) (X,�) is a partial order. Definition 2.7. Let (X,�) be a partially ordered set. Then x,y ∈ X are compa- rable if x � y or y � x holds. Definition 2.8. Let (X,�) be a partially ordered set. A pair (f,g) of self maps of X is said to be weakly increasing if gx � gfx and gx � fgx for all x ∈ X. COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 99 The notion of partially weakly increasing of pair of mappings is introduced by Abbas et al [1]. Definition 2.9. Let (X,�) be a partially ordered set and f and g be two self maps on X. An ordered pair (f,g) is said to be partially weakly increasing if gx � gfx and gx � fgx for all x ∈ X. Note that a pair (f,g) is weakly increasing if and only if ordered pair (f,g) and (g,f) are partially weakly increasing. In the following, an example of an ordered pair (f,g) of self-maps f and g which is partially weakly increasing but not weakly increasing. Example 2.10. Let X = [0, 1] be endowed with a usual ordering and f,g : X → X be defined by fx = x2 and gx = √ x. Clearly, (f,g) is partially weakly increasing but (g,f) is not partially weakly increasing. Definition 2.11. Let (X,�) be a partially ordered set. A mapping f is called weak annihilator of g if fgx � x for all x ∈ X. Example 2.12. Let X = [0, 1] be endowed with a usual ordering and f,g : X → X be defined by fx = x2 and gx = x3. Thus f is a weak annihilator of g. Definition 2.13. Let (X,�) be a partially ordered set. A mapping f is called domination if x � fx for each x ∈ X. Example 2.14. Let X = [0, 1] be endowed with a usual ordering and f : X → X be defined by fx = n √ x. Thus f is domination for each x ∈ X. Definition 2.15. A subset K of a partially ordered set X is called totally ordered when every two elements of K are comparable. 3. Main results Now, we proof our main results of this section. Theorem 3.1. Let (X,d,�) be an ordered complete metric linear space. Let T,S,I,J : X → W(X) be four fuzzy mappings satisfying for every pair (x,y) ∈ X ×X such that x and y are comparable, φ1(δ1(Sx,Ty)) ≤ ψ1 (M(Ix,Sx)) −ψ2 (M(Ix,Sx))(3.1) where M(Ix,Sx) = {d(Ix,Jy),D1(Ix,Sx),D1(Jy,Ty), 1 2 [D1(Ix,Ty) + D1(Jy,Sx)]} and ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and φ1(x) = ψ1(x,x,x,x). Suppose that (i) (I,T) and (J,S) be partially weakly increasing, (ii) T(X) ⊆ I(X) and S(X) ⊆ J(X), (iii) S and T are dominating maps, (iv) T is weak annihilator of I and S is weak annihilator of J, (v) if for a nondecreasing sequence {xn} with xn � yn for all n and yn → u implies that xn � u. Assume either 100 GUPTA AND PANDEY (a) {S,I} are comparable, S or I is continuous and {T,J} are weakly compa- rable or (b) {T,J} are comparable, T or J is continuous and {S,I} are weakly compa- rable . Then S,T,I and J have a common fixed point. Moreover, the set of common fixed points of S,T,I and J is totally ordered if and only if S,T,I and J have one and only one common fixed point. Proof. Let x0 ∈ X be an arbitrary point in X. Since T(X) ⊆ I(X) and S(X) ⊆ J(X), we can define the sequences {xn} and {yn} in X by {y2n−1} = Jx2n−1 ⊂ Sx2n−2, {y2n} = Ix2n ⊂ Tx2n−1,(3.2) for all n ∈ N. By given assumptions x2n−2 � Sx2n−2 = Jx2n−1 � SJx2n−1 � x2n−1 and x2n−1 � Tx2n−1 = Ix2n � TIx2n � x2n. Thus for all n ≥ 1, we have xn � xn+1.(3.3) Without loss of generality, we may assume that d(y2n,y2n+1) > 0 ∀n ∈ N.(3.4) If not, then y2n = y2n+1, for some n. Putting x = x2n+1 and y = x2n, form (3.3) and the considered contraction (3.1), we have COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 101 φ1(d(y2n+2,y2n+1)) = φ1(δ1(Sx2n+1,Tx2n)) ≤ ψ1 (d(Ix2n+1,Jx2n),D1(Ix2n+1,Sx2n+1),D1(Jx2n,Tx2n), 1 2 [D1(Ix2n+1,Tx2n) + D1(Jx2n,Sx2n+1)] ) −ψ2 (d(Ix2n+1,Jx2n),D1(Ix2n+1,Sx2n+1),D1(Jx2n,Tx2n), 1 2 [D1(Ix2n+1,Tx2n) + D1(Jx2n,Sx2n+1)] ) ≤ ψ1 (d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 [d(y2n+1,y2n+1) + d(y2n,y2n+2)] ) −ψ2 (d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 [d(y2n+1,y2n+1) + d(y2n,y2n+2)] ) (3.5) ≤ ψ1 ( 0,d(y2n+1,y2n+2), 0, 1 2 d(y2n,y2n+2) ) −ψ2 ( 0,d(y2n+1,y2n+2), 0, 1 2 d(y2n,y2n+2) ) .(3.6) Using a triangular inequality, we have 1 2 d(y2n,y2n+2) ≤ 1 2 [d(y2n,y2n+1) + d(y2n+1,y2n+2)] ≤ 1 2 d(y2n+1,y2n+2). Using this together with a property of the generalized altering function ψ1, we get ψ1 ( 0,d(y2n+1,y2n+2), 0, 1 2 d(y2n,y2n+2) ) ≤ φ1(d(y2n+1,y2n+2). Hence, we obtain φ1(d(y2n+1,y2n+2) ≤ φ1(d(y2n+1,y2n+2) −ψ2 ( 0,d(y2n+1,y2n+2), 0, 1 2 d(y2n,y2n+2) ) . This implies that ψ2 ( 0,d(y2n+1,y2n+2), 0, 1 2 d(y2n,y2n+2) ) = 0 which yields that d(y2n,y2n+1) = 0. Following the similar arguments, we obtain y2n+2 = y2n+3 and so on. Thus {yn} becomes a constant sequence and {y2n} is the common fixed point of I,J,S and T. Take for each n, d(y2n,y2n+1) > 0. We claim that 102 GUPTA AND PANDEY lim n→+∞ d(y2n,y2n+1) = 0.(3.7) By (3.6), we have φ1(d(y2n+2,y2n+1)) = φ1(δ1(Sx2n+1,Tx2n)) ≤ ψ1 (M(y2n+1,y2n)) −ψ2 (M(y2n+1,y2n))(3.8) where M(y2n+1,y2n) = {d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 d(y2n,y2n+2)}. Suppose for some n ∈ N, that d(y2n+2,y2n+1) > d(y2n,y2n+1).(3.9) Using (3.9) and a triangular inequality, we have 1 2 d(y2n,y2n+2) ≤ 1 2 [d(y2n,y2n+1) + d(y2n+1,y2n+2)] < d(y2n+1,y2n+2). Using this and (3.9) together with a property of the generalized altering distance function ψ1, we get ψ1 (M(y2n+1,y2n)) ≤ φ1(d(y2n+1,y2n+2). where M(y2n+1,y2n) = {d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 d(y2n,y2n+2)}. Hence, we obtain φ1(d(y2n+2,y2n+1)) ≤ φ1(d(y2n+2,y2n+1)) −ψ2 (M(y2n+1,y2n)) . where M(y2n+1,y2n) = {d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 d(y2n,y2n+2)}. This implies that ψ2 ( d(y2n+1,y2n),d(y2n+2,y2n+1),d(y2n+1,y2n), 1 2 d(y2n,y2n+2) ) = 0 which yields that d(y2n+1,y2n) = 0. Hence, we obtain a contradiction with (3.4). We deduce that d(y2n+1,y2n+2) ≤ d(y2n,y2n+1), ∀n ∈ N.(3.10) Similarly, putting x = x2n+1 and y = x2n+2, form (3.3) and the considered contraction (3.1), we have COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 103 φ1(d(y2n+2,y2n+3)) = φ1(δ1(Sx2n+1,Tx2n+2)) ≤ ψ1 (d(y2n+1,y2n+2),d(y2n+2,y2n+3),d(y2n+1,y2n+2), 1 2 d(y2n+1,y2n+3) ) −ψ2 (d(y2n+1,y2n+2),d(y2n+2,y2n+3),d(y2n+1,y2n+2), 1 2 d(y2n+1,y2n+3) ) .(3.11) Suppose, for some n ∈ N, that d(y2n+2,y2n+3) > d(y2n+1,y2n+2).(3.12) Then, by a triangular inequality, we have 1 2 d(y2n+1,y2n+3) ≤ 1 2 [d(y2n+1,y2n+2) + d(y2n+2,y2n+3)] < d(y2n+2,y2n+3). Hence, from this, (3.11) and (3.12), we obtain φ1(d(y2n+1,y2n+3)) ≤ φ1(d(y2n+2,y2n+3)) −ψ2 (d(y2n+1,y2n+2),d(y2n+2,y2n+3),d(y2n+1,y2n+2), 1 2 d(y2n+1,y2n+3) ) . This implies that ψ2 ( d(y2n+1,y2n+2),d(y2n+1,y2n+2),d(y2n+2,y2n+3), 1 2 d(y2n+1,y2n+3) ) = 0 which yields that d(y2n+1,y2n+2) = 0. Hence, we obtain a contradiction with (3.4). We deduce that d(y2n+1,y2n+2) ≥ d(y2n+2,y2n+3), ∀n ∈ N.(3.13) Combining (3.10) and (3.13), we obtain d(y2n,y2n+1) > d(y2n+2,y2n+3), ∀n ∈ N.(3.14) Then, {d(y2n+1,y2n+2)} is a nonincreasing sequence of positive real numbers. This implies that there exists r ≥ 0 such that lim n→+∞ d(y2n+1,y2n+2) = r.(3.15) 104 GUPTA AND PANDEY By (3.8), we have φ1(d(y2n+2,y2n+1)) = φ1(δ1(Sx2n+1,Tx2n)) ≤ ψ1 (d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 d(y2n,y2n+2) ) −ψ2 (d(y2n+1,y2n),d(y2n+1,y2n+2),d(y2n,y2n+1), 1 2 d(y2n,y2n+2) ) ≤ φ(d(y2n+1,y2n)) −ψ2 (d(y2n+1,y2n),d(y2n+1,y2n+2), d(y2n,y2n+1), 0) .(3.16) Letting n → +∞ in (3.16) and using the continuities of φ1 and ψ2, we obtain φ1(r) ≤ φ1(r) −ψ2(r,r,r, 0), which implies that ψ2(r,r,r, 0) = 0 so r = 0. Hence lim n→+∞ d(y2n+1,y2n+2) = 0. Hence, (3.7) is proved. Next, we claim that {yn} is a Cauchy sequence. From (3.7), it will be sufficient to prove that {y2n} is a Cauchy sequence. We proceed by negation and suppose that {y2n} is not a Cauchy sequence. Then, there exists ε > 0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that for all positive integer i, n(i) > m(i) > i, d(ym(i),yn(i)) ≥ ε, d(ym(i),yn(i)−2) < ε.(3.17) From (3.17) and using a triangular inequality, we get ε ≤ d(ym(i),yn(i)) ≤ d(ym(i),yn(i)−2) + d(yn(i)−2,yn(i)−1) + d(yn(i)−1,yn(i)) ≤ ε + d(yn(i)−2,yn(i)−1) + d(yn(i)−1,yn(i)). Letting i → +∞ in the above inequality and using (3.7), we obtain lim i→+∞ d(ym(i),yn(i)) = ε.(3.18) Again, a triangular inequality gives us |d(yn(i),ym(i)−1) −d(yn(i),ym(i))| ≤ d(ym(i)−1,ym(i)).(3.19) Letting i → +∞ in the above inequality and using (3.7) and (3.18), we get lim i→+∞ d(yn(i),ym(i)−1) = ε.(3.20) COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 105 Similarly, we have lim i→+∞ d(yn(i)+1,ym(i)−1) = ε.(3.21) On the other hand, we have d(yn(i),ym(i)) ≤ d(yn(i),yn(i)+1) + d(yn(i)+1,ym(i)) = d(yn(i),yn(i)+1) + d(Txn(i),Sxm(i)−1). Then, from (3.7), (3.18) and the continuity of φ1, we get by letting i → +∞ in the above inequality φ1(ε) ≤ lim i→+∞ d(Txn(i),Sxm(i)−1).(3.22) Now, using the considered contractive condition (3.1) for x = x2m(i)−1 and y = x2n(i), we have φ1(δ1(Sx2m(i)−1,Tx2n(i))) ≤ ψ1 ( M(x2m(i)−1,Tx2n(i)) ) −ψ2 ( M(x2m(i)−1,Tx2n(i)) ) M(x2m(i)−1,Tx2n(i)) = {d(Ix2m(i)−1,Jx2n(i)),D1(Ix2m(i)−1,Sx2m(i)−1), D1(Jx2n(i),Tx2n(i)), 1 2 [D1(Ix2m(i)−1,Tx2n(i)) + D1(Jx2n(i),Sx2m(i)−1)]} φ1(δ1(y2m(i)−1,y2n(i))) ≤ ψ1 ( d(y2m(i)−1,y2n(i)),d(y2m(i)−1,y2m(i)),d(y2n(i),y2n(i)+1), 1 2 [d(y2m(i)−1,y2n(i)+1) + d(y2n(i),y2m(i))] ) −ψ2 ( d(y2m(i)−1,y2n(i)),d(y2m(i)−1,y2m(i)),d(y2n(i),y2n(i)+1), 1 2 [d(y2m(i)−1,y2n(i)+1) + d(y2n(i),y2m(i))] ) . Then, from (3.7), (3.20), (3.21) and the continuity of ψ1 and ψ2, we get by letting i → +∞ in the above inequality lim i→+∞ φ1(δ1(Sx2m(i)−1,Tx2n(i))) ≤ ψ1(ε, 0, 0,ε) −ψ2(ε, 0, 0,ε) ≤ φ1(ε) −ψ2(ε, 0, 0,ε). Now, combining (3.1) with the above inequality, we get φ1(ε) ≤ φ1(ε) −ψ2(ε, 0, 0,ε), which implies that ψ2(ε, 0, 0,ε) = 0, that is a contradiction since ε > 0. We deduce that {yn} is a Cauchy sequence. 106 GUPTA AND PANDEY Finally, we prove existence of a common fixed point of the four mappings I,J,S and T . Since {yn} is a Cauchy sequence in complete metric linear space (X,d), there exists a point z ∈ X, such that {y2n} converges to z. Therefore, {y2n+1} = Jx2n+1 ⊂ Sx2n → z, as n → +∞(3.23) and {y2n+2} = Ix2n+2 ⊂ Tx2n+1 → z, as n → +∞.(3.24) Suppose that (a) holds. Since {S,I} are comparable, we have lim n→+∞ SIx2n+2 = lim n→+∞ SIx2n+2 = Iz. Also, x2n+1 � Tx2n+1 = Ix2n+2. Now φ1(δ1(SIx2n+2,Tx2n+1)) ≤ ψ1 (d(IIx2n+2,Jx2n+1),D1(IIx2n+2,SIx2n+2), D1(Jx2n+1,Tx2n+1), 1 2 [D1(IIx2n+2,Tx2n+1) + D1(Jx2n+1,SIx2n+2)] ) −ψ2 (d(IIx2n+2,Jx2n+1),D1(IIx2n+2,SIx2n+2), D1(Jx2n+1,Tx2n+1), 1 2 [D1(IIx2n+2,Tx2n+1) + D1(Jx2n+1,SIx2n+2)] ) Assume that I is continuous. On passing limit as n → +∞, we obtain φ1(d(Iz,z)) ≤ ψ1 (d(Iz,z), 0, 0,d(Iz,z)) −ψ2 (d(Iz,z), 0, 0,d(Iz,z)) ≤ φ1(d(Iz,z)) −ψ2 (d(Iz,z), 0, 0,d(Iz,z)) , so ψ2 (d(Iz,z), 0, 0,d(Iz,z)) = 0, which implies that Iz = z.(3.25) Now, x2n+1 � Tx2n+1 and Tx2n+1 → z as n → +∞, so by assumption we have x2n+1 � z and (3.1) becomes φ1(δ1(Sz,Tx2n+1)) ≤ ψ1 (M(z,x2n+1)) −ψ2 (M(z,x2n+1)) . where M(z,x2n+1) = (d(Iz,Jx2n+1),D1(Iz,Sz),D1(Jx2n+1,Tx2n+1), 1 2 [D1(Iz,Tx2n+1) + D1(Jx2n+1,Sz)] ) Passing to the limit n → +∞ in the above inequality and using (3.25), COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 107 φ1(δ1(Sz,z)) ≤ ψ1 ( 0,D1(z,Sz), 0, 1 2 D1(z,Sz) ) −ψ2 ( 0,D1(z,Sz), 0, 1 2 D1(z,Sz) ) which holds unless ψ2 ( 0,d(z,Sz), 0, 1 2 D1(z,Sz) ) = 0, so Sz = z.(3.26) Since S(X) ⊆ J(X), there exists a point w ∈ X such that Sz = Jw. Suppose that Tw 6= Jw. Since z � Sz = Jw � SJw � w implies z � w. From (3.1), we obtain φ1(δ1(Sz,Tw)) ≤ ψ1 (d(Iz,Jw),D1(Iz,Sz),D1(Jw,Tw), 1 2 [D1(Iz,Tw) + D1(Jw,Sz)] ) −ψ2 (d(Iz,Jw),D1(Iz,Sz),D1(Jw,Tw), 1 2 [D1(Iz,Tw) + D1(Jw,Sz)] ) ≤ ψ1 ( 0, 0,D1(Jw,Tw), 1 2 D1(Jw,Tw) ) −ψ2 ( 0, 0,D1(Jw,Tw), 1 2 D1(Jw,Tw) ) Hence Jw = Tw.(3.27) Since T and J are weakly compatible, Tz = TSz = TJw = JTw = JSz = Jz. Thus z is a coincidence point of T and J. Now, since x2n � Sx2n and Sx2n → z as n → +∞, implies that x2n � z, from (3.1) φ1(δ1(Sx2n,Tz)) ≤ ψ1 (M((x2n,z))) −ψ2 (M((x2n,z))) where M((x2n,z) = (d(Ix2n,Jz),D1(Ix2n,Sx2n),D1(Jz,Tz), 1 2 [D1(Ix2n,Tz) + D1(Jz,Sx2n)]) Passing to the limit n → +∞ in the above inequality, we have φ1(δ1(z,Tz)) ≤ ψ1 (d(z,Tz), 0, 0,d(z,Tz)) −ψ2 (d(z,Tz), 0, 0,d(z,Tz)) which gives that 108 GUPTA AND PANDEY z = Tz.(3.28) Therefore, Sz = Tz = Iz = Jz = z, so z is a common fixed point of I,J,S and T. The proof is similar when S is continuous. Similarly, the result follows when (b) holds. Now, suppose that set of common fixed points of I,J,S and T is totally ordered. We claim that there is a unique common fixed point of I,J,S and T. Assume on contrary that, Su = Tu = Iu = Ju = u and Sv = Tv = Iv = Jv = v but u 6= v. By supposition, we can replace x = u and y = v in (3.1) to obtain φ1(d(u,v)) ≤ φ1(δ1(Su,Ty)) ≤ ψ1 (d(Iu,Jv),D1(Iu,Su),D1(Jv,Tv), 1 2 [D1(Iu,Tv) + D1(Jv,Su)] ) −ψ2 (d(Iu,Jv),D1(Iu,Su),D1(Jv,Tv), 1 2 [D1(Iu,Tv) + D1(Jv,Su)] ) ≤ ψ1 (d(u,v), 0, 0,d(u,v)) −ψ2 (d(u,v), 0, 0,d(u,v)) ≤ φ1(d(u,v)) a contraction, so u = v. Conversely, if I,J,S and T have only one common fixed point, then the set of common fixed point of I,J,S and T being singleton is totally ordered. � Corollary 3.2. Let (X,d,�) be an ordered complete metric linear space. Let T,S,I,J : X → W(X) be four fuzzy mappings satisfying for every pair (x,y) ∈ X ×X such that x and y are comparable and there exists a positive Lebesgue inte- grable function u on R+ such that ∫ � 0 u(t)dt > 0 for each � > 0 and that , ∫ φ1(δ1(Sx,Ty)) 0 u(t)dt ≤ ∫ ψ1(M(x,y)) 0 u(t)dt− ∫ ψ2(M(x,y)) 0 u(t)dt(3.29) where ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and φ1(x) = ψ1(x,x,x,x) also M(x,y) = {d(Ix,Jy),D1(Ix,Sx),D1(Jy,Ty), 1 2 [D1(Ix,Ty) + D1(Jy,Sx)]} Suppose that (i) (I,T) and (J,S) be partially weakly increasing, (ii) T(X) ⊆ I(X) and S(X) ⊆ J(X), (iii) S and T are dominating maps, (iv) T is weak annihilator of I and S is weak annihilator of J, (v) if for a nondecreasing sequence {xn} with xn � yn for all n and yn → u implies that xn � u. Assume either (a) {S,I} are comparable, S or I is continuous and {T,J} are weakly compa- rable or COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 109 (b) {T,J} are comparable, T or J is continuous and {S,I} are weakly compa- rable . Then S,T,I and J have a common fixed point. Moreover, the set of common fixed points of S,T,I and J is totally ordered if and only if S,T,I and J have one and only one common fixed point. Remark 3.3. If we take ψ1(t1, t2, t3, t4) = max{t1, t2, t3, t4} and ψ2(t1, t2, t3, t4) = (1 − k) max{t1, t2, t3, t4}, for k ∈ (0, 1) then φ1(t) = t for all t1, t2, t3, t4 ≥ 0 then the we get following result. Corollary 3.4. Let (X,d,�) be an ordered complete metric linear space. Let T,S,I,J : X → W(X) be four fuzzy mappings satisfying for every pair (x,y) ∈ X ×X such that x and y are comparable, δ1(Sx,Ty) ≤ k max{d(Ix,Jy),D1(Ix,Sx),D1(Jy,Ty), 1 2 [D1(Ix,Ty) + D1(Jy,Sx)]}(3.30) where k ∈ (0, 1) Suppose that (i) (I,T) and (J,S) be partially weakly increasing, (ii) T(X) ⊆ I(X) and S(X) ⊆ J(X), (iii) S and T are dominating maps, (iv) T is weak annihilator of I and S is weak annihilator of J, (v) if for a nondecreasing sequence {xn} with xn � yn for all n and yn → u implies that xn � u. Assume either (a) {S,I} are comparable, S or I is continuous and {T,J} are weakly compa- rable or (b) {T,J} are comparable, T or J is continuous and {S,I} are weakly compa- rable . Then S,T,I and J have a common fixed point. Moreover, the set of common fixed points of S,T,I and J is totally ordered if and only if S,T,I and J have one and only one common fixed point. Remark 3.5. Other results could be derived for other choices of ψ1 and ψ2. Corollary 3.6. Let (X,d,�) be an ordered complete metric linear space. Let T,S,I : X → W(X) be three fuzzy mappings satisfying for every pair (x,y) ∈ X×X such that x and y are comparable, φ1(δ1(Sx,Ty)) ≤ ψ1 (d(Ix,Iy),D1(Ix,Sx),D1(Iy,Ty), 1 2 [D1(Ix,Ty) + D1(Iy,Sx)] ) −ψ2 (d(Ix,Jy),D1(Ix,Sx),D1(Iy,Ty), 1 2 [D1(Ix,Ty) + D1(Iy,Sx)] ) (3.31) where ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and φ1(x) = ψ1(x,x,x,x). Suppose that 110 GUPTA AND PANDEY (i) (I,T) and (I,S) be partially weakly increasing, (ii) T(X) ⊆ I(X) and S(X) ⊆ I(X), (iii) S and T are dominating maps, (iv) T is weak annihilator of I and S is weak annihilator of I, (v) if for a nondecreasing sequence {xn} with xn � yn for all n and yn → u implies that xn � u. Assume either (a) {S,I} are comparable, S or I is continuous and {T,I} are weakly compa- rable or (b) {T,I} are comparable, T or I is continuous and {S,I} are weakly compa- rable . Then S,T,I have a common fixed point. Moreover, the set of common fixed points of S,T,I is totally ordered if and only if S,T,I have one and only one common fixed point. Proof. It follows by taking J = I in Theorem 3.1. � Corollary 3.7. Let (X,d,�) be an ordered complete metric linear space. Let T,I,J : X → W(X) be three fuzzy mappings satisfying for every pair (x,y) ∈ X×X such that x and y are comparable, φ1(δ1(Tx,Ty)) ≤ ψ1 (d(Ix,Jy),D1(Ix,Tx),D1(Jy,Ty), 1 2 [D1(Ix,Ty) + D1(Jy,Tx)] ) −ψ2 (d(Ix,Jy),D1(Ix,Tx),D1(Jy,Ty), 1 2 [D1(Ix,Ty) + D1(Jy,Tx)] ) (3.32) where ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and φ1(x) = ψ1(x,x,x,x). Suppose that (i) (I,T) and (J,T) be partially weakly increasing, (ii) T(X) ⊆ I(X) and T(X) ⊆ J(X), (iii) T is dominating maps, (iv) T is weak annihilator of I and J, (v) if for a nondecreasing sequence {xn} with xn � yn for all n and yn → u implies that xn � u. Assume either (a) {T,I} are comparable, T or I is continuous and {T,J} are weakly compa- rable or (b) {T,J} are comparable, T or J is continuous and {T,I} are weakly compa- rable . Then T,I and J have a common fixed point. Moreover, the set of common fixed points of T,I and J is totally ordered if and only if T,I and J have one and only one common fixed point. Proof. It follows by taking S = T in Theorem 3.1. � COMMON FIXED POINT THEOREMS FOR FOUR FUZZY MAPPINGS 111 Corollary 3.8. Let (X,d,�) be an ordered complete metric linear space. Let T,I : X → W(X) be three fuzzy mappings satisfying for every pair (x,y) ∈ X ×X such that x and y are comparable, φ1(δ1(Tx,Ty)) ≤ ψ1 (d(Ix,Iy),D1(Ix,Tx),D1(Iy,Ty), 1 2 [D1(Ix,Ty) + D1(Iy,Tx)] ) −ψ2 (d(Ix,Iy),D1(Ix,Tx),D1(Iy,Ty), 1 2 [D1(Ix,Ty) + D1(Iy,Tx)] ) (3.33) where ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and φ1(x) = ψ1(x,x,x,x). Suppose that (i) (I,T) be partially weakly increasing, (ii) T(X) ⊆ I(X), (iii) T is dominating maps, (iv) T is weak annihilator of I, (v) if for a nondecreasing sequence {xn} with xn � yn for all n and yn → u implies that xn � u, (vi) {T,I} are comparable, T or I is continuous. Then T,I have a common fixed point. Moreover, the set of common fixed points of T,I is totally ordered if and only if T,I have one and only one common fixed point. Proof. It follows by taking S = T and J = I in Theorem 3.1. � 4. acknowledgements The authors are grateful referees for their critical reading of the original manu- script and for their valuable suggestions which improved the quality of this work also we express our sincere thanks to the members of editorial board for consideration this article for publication. References [1] M. Abbas, N. Talat, S. Radenovic, Common fixed points of four maps in partially ordered metric spaces, Appl. Math. Lett, 24(2011), 1520-1526. [2] H.M. 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