Common Fixed Point Theorems for Ćirić-Berinde Type Hybrid Contractions

International Journal of Mathematical Analysis Vol. 9, 2015, no. 31, 1545-1561 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ijma.2015.54125 Common Fixed Point Theorems for Ćirić-Berinde Type Hybrid Contractions Ki-Hwan Kim and Seong-Hoon Cho Department of Mathematics Hanseo University, Seosan Chungnam, 356-706, South Korea Corresponding author Copyright c 2015 Ki-Hwan Kim and Seong-Hoon Cho. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Ćirić-Berinde type hybrid contraction for single valued and multivalued mappings is introduced, and the existence of common fixed points for such contractions is proved. Mathematics Subject Classification: 47H10, 54H25 Keywords: Fixed point, Contractive multivalued mapping, Metric space 1 Introduction and preliminaries The authors of [13] gave generalizations of fixed point results in ordered metric spaces by introducing α-ψ-contractions and proving coresponding fixed point theorems. Since then, the authors of [2, 3, 6, 10, 12] obtained fixed point theorems in the field. Recently, the author of [5] obtained fixed point results for Ćirić-Berinde type contractive multivalued mappings, which are generalizations of the results of [1, 2, 4, 7]. The purpose of this paper is to introduce the notion of Ćirić-Berinde type hybrid contractions and to establish fixed point theorems for such contractions.

1546 Kim and Cho Let (X, d) be a metric space. We denote by CB(X) the class of nonempty closed and bounded subsets of X and by CL(X) the class of nonempty closed subsets of X. Let H(, ) be the generalized Hausdorff distance on CL(X), i.e., for all A, B CL(X), { max{sup H(A, B) = a A d(a, B), sup b B d(b, A)}, if the maximum exists,, otherwise, where d(a, B) = inf{d(a, b) : b B} is the distance from the point a to the subset B. We denote by Ξ the class of all functions ξ : [0, ) [0, ) such that (1) ξ is continuous; (2) ξ is nondecreasing on [0, ); (3) ξ(t) = 0 if and only if t = 0; (4) ξ is subadditive. Also, we denote by Ψ the family of all strictly increasing functions ψ : [0, ) [0, ) such that n=1 ψn (t) < for each t > 0, where ψ n is the n-th iteration of ψ. Note that if ψ Ψ, then ψ(0) = 0 and 0 < ψ(t) < t for all t > 0. Let (X, d) be a metric space, and let α : X X [0, ) be a function. The following are found in [5]: (1) for any sequence {x n } in X with α(x n, x n+1 ) 1 for all n N and lim n x n = x, we have α(x n, x) 1 for all n N; (2) for any sequence {x n } in X with α(x n, x n+1 ) 1 for all n N and a cluster point x of {x n }, we have lim inf α(x n, x) 1; n (3) for any sequence {x n } in X with α(x n, x n+1 ) 1 for all n N and a cluster point x of {x n }, there exists a subsequence {x n(k) } of {x n } such that α(x n(k), x) 1 for all k N. Remark 1.1. [5] (1) implies (2), and (2) implies (3).

Common fixed point theorems 1547 Note that if (X, d) is a metric space and ξ Ξ, then (X, ξ d) is a metric space. Let (X, d) be a metric space, and let α : X X [0, ) be a function and T : X CL(X) be a multivalued mapping. Then, we say that (1) T is called α -admissible [3] if α(x, y) 1 implies α (T x, T y) 1, where α (T x, T y) = inf{α(a, b) : a T x, b T y}; (2) T is called α-admissible [11] if, for each x X and y T x with α(x, y) 1, we have α(y, z) 1 for all z T y. We give generalizations of concept of the above definitions: Let f : X X be a mapping such that T (X) f(x). Then, we say that (3) T is called α -admissible with respect to f if α(fx, fy) 1 implies α (T x, T y) 1, where α (T x, T y) = inf{α(a, b) : a T x, b T y}; (4) T is called α-admissible with respect to f if, for each x X and fy T x with α(fx, fy) 1, we have α(fy, z) 1 for all z T y. Lemma 1.1. Let (X, d) be a metric space, and let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x). If T is α -admissible with respect to f, then it is α-admissible with respect to f. Proof. Suppose that T is an α -admissible mapping with respect to f. Let x X and fy T x be such that α(fx, fy) 1. Let z T y be given. Then, α(fy, z) α (T x, T y). Because T is α -admissible with respect to f, α (T x, T y) 1. Hence α(fy, z) 1, and hence T is α-admissible with respect to f. Corollary 1.1. Let (X, d) be a metric space, and let T : X CL(X) be a multivalued mapping. If T is α -admissible, then it is α-admissible. Lemma 1.2. [5] Let (X, d) be a metric space, and let ξ Ξ and B CL(X). If a X and ξ(d(a, B)) < c, then there exists b B such that ξ(d(a, b)) < c.

1548 Kim and Cho Let (X, d) be a metric space. A function g : X [0, ) is called upper semi-continuous if, for each x X and {x n } X with lim n x n = x, we have lim n g(x n ) g(x). A function g : X [0, ) is called lower semi-continuous if, for each x X and {x n } X with lim n x n = x, we have g(x) lim n g(x n ). Let T : X CL(X) be a multivalued mapping and f : X X be a mapping. A point x X is called fixed point of T if x T x. A point x X is called coincidence point (resp. common fixed point) of f and T if fx T x (resp. x = fx T x). Let f T : X [0, ) be a function defined by f T (x) = d(fx, T x). The graph G(T, f) of T with respect to f is defined as G(T, f) = {(fx, fy) : fy T x}. The graph G(T, f) of T with respect to f is called (T, f)-orbitally closed if, for any sequence {fx n }, we have (fx, fx) G(T, f) whenever (fx n, fx n+1 ) G(T, f) and lim n fx n = fx. Definition 1.1. Let (X, d) be a metric space. Mappings f : X X and T : X CL(X) are said to be compatible [9] if ft x CL(X) and H(fT x, T fx) < for all x X and lim n H(fT x n, T fx n ) = 0 whenever {x n } is a sequence in X such that lim n T x n = A CL(X) and lim n fx n = t A. Definition 1.2. Let (X, d) be a metric space. Mappings f : X X and T : X CL(X) are said to be weakly compatible [7] if ft x = T fx whenever fx T x. Definition 1.3. Let (X, d) be a metric space. Mappings f : X X and T : X CL(X) are said to be (IT)-commuting at x X [14] if ft x T fx. Definition 1.4. Let (X, d) be a metric space. Mappings f : X X and T : X CL(X) are said to be R-weakly commuting [15] if ft x CL(X) and H(fT x, T fx) < for all x X and there exists some positive real number R such that H(fT x, T fx) Rd(fx, T x). Definition 1.5. Let (X, d) be a metric space, and let f : X X be a mapping and T : X CL(X) be a multivalued mapping. Then, f is called T -weakly commuting at x X [8] if ffx T fx. Note that compatiblity implies weak compatibilitye but the converse is not true (see [7]). Also, note that (IT)-commuting at coincidence points implies f is T -weakly commuting but the converse is not true in general (see [8]).

Common fixed point theorems 1549 From now on, we denote M(x, y, f, T ) = max{d(fx, fy), d(fx, T x), d(fy, T y), 1 {d(fx, T y)+d(fy, T x)}} 2 for all x, y X. 2 Fixed point theorems Let (X, d) be a metric space, and let T : X CL(X) be a multivalued mapping and f : X X be a mapping. To obtain common fixed points of f and T, we consider the following conditions: (C1) f and T are weakly compatible and lim n f n z = u for some u X. (C2) f is T -weakly commuting at z and fz = ffz for any coincidence point z of f and T. Theorem 2.1. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α-admissible with respect to f. Assume that there exist L 0 and ψ Ψ such that, for all x, y X with α(fx, fy) 1, ξ(h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) (2.1) where ξ Ξ. Also, suppose that the following are satisfied: (1) there exist x 0 X and y 1 T x 0 such that α(fx 0, y 1 ) 1; (2) one of the following is satisfied: (i) f is bijective continuous and T is continuous; (ii) f is bijective continuous and f T is lower semi- continuous; (iii) G(T, f) is (T, f)-orbitally closed. Then f and T have a coincidence point in X. Moreover, assume that one of the following is satisfied: (iv) conditions (i) and (C1) are satisfied;

1550 Kim and Cho (v) conditions (ii) and (C2) are satisfied; (vi) conditions (iii) and (C2) are satisfied. Then f and T have a common fixed point in X. Proof. Let x 0 X and y 1 T x 0 be such that α(fx 0, y 1 ) 1. Since T x 0 f(x), y 1 f(x) and so there exists x 1 X such that y 1 = f(x 1 ). Hence, fx 1 T x 0 and α(fx 0, fx 1 ) 1 (2.2) Let c be a real number with ξ(d(fx 0, fx 1 )) < ξ(c). If x 0 = x 1, then x 1 is a coincidence point of f and T. Let x 0 x 1. If fx 1 T x 1, then x 1 is a coincidence point of f and T. Let fx 1 / T x 1. Then d(fx 1, T x 1 ) > 0. From (2.1) we obtain 0 <ξ(d(fx 1, T x 1 )) ξ(h(t x 0, T x 1 )) ψ(ξ(max{d(fx 0, fx 1 ), d(fx 0, T x 0 ), d(fx 1, T x 1 ), 1 2 {d(fx 0, T x 1 ) + d(fx 1, T x 0 )}})) + Lξ(d(fx 1, T x 0 ) ψ(ξ(max{d(fx 0, fx 1 ), d(fx 0, fx 1 ), d(fx 1, T x 1 ), 1 2 {d(fx 0, T x 1 ) + d(fx 1, x 1 )}})) + Lξ(d(fx 1, fx 1 )) ψ(ξ(max{d(fx 0, fx 1 ), d(fx 0, fx 1 ), d(fx 1, T x 1 ), 1 2 {d(fx 0, fx 1 ) + d(fx 1, T x 1 )}})) =ψ(ξ(max{d(fx 0, fx 1 ), d(fx 1, T x 1 )})). If max{d(fx 0, fx 1 ), d(fx 1, T x 1 )} = d(fx 1, T x 1 ), then we have 0 < ξ(d(fx 1, T x 1 )) ψ(ξ(d(fx 1, T x 1 ))) < ξ(d(fx 1, T x 1 )), which is a contradiction. Thus, max{d(fx 0, fx 1 ), d(fx 1, T x 1 )} = d(fx 0, fx 1 ), and hence we have 0 < ξ(d(fx 1, T x 1 )) ψ(ξ(d(fx 0, fx 1 ))) < ψ(ξ(c)). Hence, there exists y 2 T x 1 such that ξ(d(fx 1, y 2 )) < ψ(ξ(c)). Since y 2 T x 1 f(x), there exists x 2 X such that y 2 = f(x 2 ). Hence, ξ(d(fx 1, fx 2 )) < ψ(ξ(c)). (2.3)

Common fixed point theorems 1551 Since T is α-admissible with respect to f, from (2.2) and fx 2 T x 1, we have α(fx 1, fx 2 ) 1. If fx 2 T x 2, then x 2 is a coincidence point. Let fx 2 T x 2. Then d(fx 2, T x 2 ) > 0, and so ξ(d(fx 2, T x 2 )) > 0. From (2.1) we obtain 0 <ξ(d(fx 2, T x 2 )) ξ(h(t x 1, T x 2 )) ψ(ξ(max{d(fx 1, fx 2 ), d(fx 1, T x 1 ), d(fx 2, T x 2 ), 1 2 {d(fx 1, T x 2 ) + d(fx 2, T x 1 )}})) + Lξ(d(fx 2, T x 1 )) ψ(ξ(max{d(fx 1, fx 2 ), d(fx 1, fx 2 ), d(fx 2, T x 2 ), 1 2 {d(fx 1, T x 2 ) + d(fx 2, x 2 )}})) + Lξ(d(fx 2, fx 2 )) ψ(ξ(max{d(fx 1, fx 2 ), d(fx 1, fx 2 ), d(fx 2, T x 2 ), 1 2 {d(fx 1, fx 2 ) + d(fx 2, T x 2 )}})) =ψ(ξ(max{d(fx 1, fx 2 ), d(fx 2, T x 2 )})). (2.4) If max{d(fx 1, fx 2 ), d(fx 2, T x 2 )} = d(fx 2, T x 2 ), then we have ξ(d(fx 2, T x 2 )) ψ(ξ(d(fx 2, T x 2 ))) < ξ(d(fx 2, T x 2 )), which is a contradiction. Thus, max{d(fx 1, fx 2 ), d(fx 2, T x 2 )} = d(fx 1, fx 2 ). From (2.3) and (2.4) we have ξ(d(fx 2, T x 2 )) ψ(ξ(d(fx 1, fx 2 ))) < ψ 2 (ξ(c)), because ψ is strictly increasing. Hence, there exists y 3 T x 2 such that So, there exists x 3 X such that ξ(d(fx 2, y 3 )) < ψ 2 (ξ(c)). ξ(d(fx 2, fx 3 )) < ψ 2 (ξ(c)). Since T is α-admissible with respect to f, from fx 3 T x 2 and α(fx 1, fx 2 ) 1, we have α(fx 2, fx 3 ) 1. By induction, we obtain a sequence {fx n } X such that, for all n N {0},

1552 Kim and Cho α(fx n, fx n+1 ) 1, fx n+1 T x n, fx n fx n+1, ξ(d(fx n, fx n+1 )) < ψ n (ξ(c)) and ξ(h(t x n, T x n+1 )) ψ(ξ(d(fx n, fx n+1 ))). Let ɛ > 0 be given. Since n=0 ψn (ξ(c)) <, there exists N N such that ψ n (ξ(c)) < ξ(ɛ). n N For all m > n N, we have ξ(d(fx n, fx m )) < n N ψ n (ξ(c)) < ξ(ɛ) m 1 k=n ψ k (ξ(c)) (2.5) which implies d(fx n, fx m ) < ɛ for all m > n N. Hence, {fx n } is a Cauchy sequence in X. It follows from the completeness of f(x) that there exists z f(x) such that z = lim fx n. n So, there exists x X such that z = f(x ), and so lim fx n = fx. n Suppose that f is bijective continuous and T is continuous. Then, f 1 is continuous, and so lim x n = x. n Thus, We have lim T x n = T x. n d(fx, T x ) d(fx, fx n+1 ) + d(fx n+1, T x ) d(fx, fx n+1 ) + H(T x n, T x ). By letting n in the above inequality, we obtain d(fx, T x ) = 0, and so fx T x.

Common fixed point theorems 1553 Assume that f is bijective continuous and f T is lower semi-continuous. Then, since lim n x n = x, we have f T (x ) lim n f T (x n ). Hence, d(fx, T x ) lim n d(fx n, T x n ) lim n d(fx n, fx n+1 ) = 0. Thus, fx T x. Assume that G(T, f) is (T, f)-orbitally closed. Since (fx n, fx n+1 ) G(T, f) for all n N {0} and lim n fx n = fx, we have (fx, fx ) G(T, f) by the (T, f)-orbitally closedness. Hence, fx T x. Suppose that conditions (i) and (C1) are satisfied. From (i) there exists x X such that fx T x. From (C1) we have lim n f n x = u for some u X. Hence, fu = u by continuity of f. Since f and T are weakly compatible, Also, we have f 2 x = ffx ft x = T fx. f 3 x = ff 2 x ft fx = T f 2 x. Inductivly, we obtain f n x T f n 1 x. Thus we have d(u, T u) d(u, f n x ) + d(f n x, T u) d(u, f n x ) + H(T f n 1 x, T u). Letting n in the above inequality and by using continuity of f and T, we have d(u, T u) = 0. Thus, u = fu T u Suppose that conditions (ii) and (C2) are satisfied. From (ii) there exists x X such that fx T x. Since (C2) is satisfied, fx = ffx T fx and so fx is a common fixed point of f and T. Suppose that (iii) and (C2) are satisfied. From (iii) there exists x X such that fx T x. From condition (C2) fx is a common fixed point of f and T. Remark 2.1. If we have f = id, where id is the identity map of X, then Theorem 2.1 reduce to Theorem 2.1 of [5]. Corollary 2.2. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α-admissible with respect to f.

1554 Kim and Cho Assume that there exist L 0 and ψ Ψ such that, for all x, y X, ξ(α(fx, fy)h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) where ξ Ξ. Also, suppose that conditions (1) and (2) of Theorem 2.1 are satisfied. Then, f and T have a coincidence point in X. Moreover, if one of (iv), (v) and (vi) of Theorem 2.1 is satisfied, then f and T have a common fixed point in X. Remark 2.2. (1) If we have f = id in Corollary 2.2, then Corollary 2.2 become Corollary 2.2 of [5]. (2)Corollary 2.2 is a generalization of Theorem 3.4 of [11]. If we have ξ(t) = t for all t 0, L = 0, f = id and T is continuous, then Corollary 2.2 reduce to Theorem 3.4 of [11]. From Theorem 2.1 we obtain the following result. Corollary 2.3. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α -admissible with respect to f. Assume that there exist L 0 and ψ Ψ such that, for all x, y X with α(fx, fy) 1, ξ(h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) where ξ Ξ. Also, suppose that conditions (1) and (2) of Theorem 2.1 are satisfied. Then, f and T have a coincidence point in X. Moreover, if one of (iv), (v) and (vi) of Theorem 2.1 is satisfied, then f and T have a common fixed point in X. Remark 2.3. (1) Corollary 2.3 is a generalization of Corollary 2.4 of [5]. (2) Corollary 2.3 is a generalization of Theorem 2.5 of [2]. In fact, if we have L = 0 and f = id in Corollary 2.3, then Corollary 2.3 reduce to Theorem 2.5 of [2]. Corollary 2.4. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α -admissible with respect to f. Assume that there exist L 0 and ψ Ψ such that, for all x, y X, ξ(α(fx, fy)h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x))

Common fixed point theorems 1555 where ξ Ξ. Also, suppose that conditions (1) and (2) of Theorem 2.1 are satisfied. Then, f and T have a coincidence point in X. Moreover, if one of (iv), (v) and (vi) of Theorem 2.1 is satisfied, then f and T have a common fixed point in X. Remark 2.4. In Corollary 2.4, let ξ(t) = t for all t 0 and α(x, y) = 1 for all x, y X and ψ(t) = kt for all t 0, where k [0, 1). If f = id and T is single valued map, then Corollary 2.4 reduce to Theorem 2.2 of [4]. Theorem 2.5. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α-admissible with respect to f. Assume that there exist L 0 and ψ Ψ such that, for all x, y X with α(fx, fy) 1, ξ(h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) where ξ Ξ. Also, suppose that the following are satisfied: (1) there exist x 0 X and y 1 T x 0 such that α(fx 0, y 1 ) 1; (2) for a sequence {x n } in X with α(fx n, fx n+1 ) 1 for all n N {0} and a cluster point x of {fx n }, there exists a subsequence {fx n(k) } of {fx n } such that, for all k N {0}, α(fx n(k), x) 1. If ψ is upper semi-continuous, then f and T have a coincidence point in X. Moreover, if (C2) is satisfied, then f and T have a common fixed point in X. Proof. Following the proof of Theorem 2.1, we obtain a sequence {x n } X with lim n fx n = fx X such that, for all n N {0}, fx n+1 T x n, fx n fx n+1 and α(fx n, fx n+1 ) 1. From (2) there exists a subsequence {fx n(k) } of {fx n } such that α(fx n(k), fx ) 1.

1556 Kim and Cho Thus, we have ξ(d(x n(k)+1, T x )) =ξ(h(t x n(k), T x )) ψ(ξ(m(x n(k), x, f, T ))) + Lξ(d(fx, fx n(k)+1 )) ψ(ξ(m(x n(k), x, f, T ))) + Lξ(d(fx, fx n(k)+1 )) (2.8) where m(x n(k), x f, T ) := max{d(fx n(k), fx ), d(fx n(k), fx n(k)+1 ), d(fx, T x ), 1 2 {d(fx n(k), T x ) + d(fx, fx n(k)+1 )}}. and so We have lim m(x n(k), x, f, T ) = d(fx, T x ), k lim ξ(m(x n(k), x )) = ξ(d(fx, T x )). k Suppose that d(fx, T x ) 0. Since ψ is upper semi-continuous, lim ψ(ξ(m(x n(k), x, f, T ))) ψ(ξ(d(fx, T x ))). k Letting k in the inequality (2.8), and using continuity of ξ, we obtain 0 < ξ(d(fx, T x )) lim ψ(ξ(m(x n(k), x, f, T ))) + lim Lξ(d(fx, fx n(k)+1 )) k k ψ(ξ(d(fx, T x ))) <ξ(d(fx, T x )) which is a contradiction. Hence, d(fx, T x ) = 0, and hence fx T x. Assume that (C2) is satisfied. Then, fx = ffx T fx and so fx is a common fixed point of f and T. Remark 2.5. If we have f = id, where id is the identity map of X, then Theorem 2.5 reduce to Theorem 2.6 of [5]. Note that upper semi-continuity of ψ can not be dropped in Theorem 2.6 (see Example 2.1 of [5] with f = id). Corollary 2.6. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α-admissible with respect to f.

Common fixed point theorems 1557 Assume that there exist L 0 and ψ Ψ such that, for all x, y X with α(fx, fy) 1, ξ(α(fx, fy)h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) where ξ Ξ. Also, suppose that the following are satisfied: (1) there exist x 0 X and y 1 T x 0 such that α(fx 0, y 1 ) 1; (2) for a sequence {x n } in X with α(fx n, fx n+1 ) 1 for all n N {0} and a cluster point x of {fx n }, there exists a subsequence {fx n(k) } of {fx n } such that, for all k N {0}, α(fx n(k), x) 1. If ψ is upper semi-continuous, then f and T have a coincidence point in X. Moreover, if (C2) is satisfied, then f and T have a common fixed point in X. Corollary 2.7. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α -admissible with respect to f. Assume that there exist L 0 and ψ Ψ such that, for all x, y X with α(fx, fy) 1, ξ(h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) where ξ Ξ. Also, suppose that the following are satisfied: (1) there exist x 0 X and y 1 T x 0 such that α(fx 0, y 1 ) 1; (2) for a sequence {x n } in X with α(fx n, fx n+1 ) 1 for all n N {0} and a cluster point x of {fx n }, there exists a subsequence {fx n(k) } of {fx n } such that, for all k N {0}, α(fx n(k), x) 1. If ψ is upper semi-continuous, then f and T have a coincidence point in X. Moreover, if (C2) is satisfied, then f and T have a common fixed point in X.

1558 Kim and Cho Remark 2.6. By taking L = 0 and f = id in Corollary 2.7 and by applying Remark 1.1, Corollary 2.7 reduces to Theorem 2.6 of [2]. Corollary 2.8. Let (X, d) be a complete metric space, and let α : X X [0, ) be a function. Let T : X CL(X) be a multivalued mapping and f : X X be a mapping such that T (X) f(x) and f(x) is closed. Suppose that T is α -admissible with respect to f. Assume that there exist L 0 and ψ Ψ such that ψ, for all x, y X, ξ(α(fx, fy)h(t x, T y)) ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)) where ξ Ξ. Also, suppose that the following are satisfied: (1) there exist x 0 X and y 1 T x 0 such that α(fx 0, y 1 ) 1; (2) for a sequence {x n } in X with α(fx n, fx n+1 ) 1 for all n N {0} and a cluster point x of {fx n }, there exists a subsequence {fx n(k) } of {fx n } such that, for all k N {0}, α(fx n(k), x) 1. If ψ is upper semi-continuous, then f and T have a coincidence point in X. Moreover, if (C2) is satisfied, then f and T have a common fixed point in X. We give an example to illustrate Theorem 2.1. Example. Let X = [0, ), and let d(x, y) = x y for all x, y 0. Define mappings f : X X by fx = 1 2 x and T : X CL(X) by {0, 1} (x = 0), T x = { 1 x} (0 < x 2), 2, 1} (x > 2). { 1 2 Then, (X, d) is complete, T (X) f(x) and f(x) is closed. Let L = 2. Let ξ(t) = t for all t 0, and let ψ(t) = 1 t for all t 0. 2 Then, ξ Ξ and ψ Ψ. Let α : X X [0, ) be defined by

Common fixed point theorems 1559 { 2 (0 x, y 1), α(x, y) = 0 otherwise. { 0 (0 x 2), We have f T (x) = 1 x 1 otherwise. 2 So, condition (ii) of (2) in Theorem 2.1 is satisfied. Condition (1) in Theorem 2.1 is satisfied with x 0 = 1 and y 4 1 = 1. 4 We show that T is α-admissible with respect to f. Case 1: Let x = 0 and fy T x be such that α(fx, fy) 1. Then, we have either y = 0 or y = 2. If y = 0, then α(fy, z) = α(0, z) 1 for z T 0. If y = 2, then α(fy, z) = α(1, z) 1 for z T 2. Case 2: Let 0 < x 2 and fy T x be such that α(fx, fy) 1. Then, we have y = x. Hence, α(fy, z) = α( 1 x, z) 1 for z T y. 2 Case 3: Let x > 2 and fy T x be such that α(fx, fy) 1. Then, we have either y = 1 or y = 2. If y = 1, then α(fy, z) = α( 1, z) 1 for z T 1. 2 If y = 2, then α(fy, z) = α(1, z) 1 for z T 2. Thus, T is α-admissible with respect to f. We now show that (2.1) is satisfied. We have α(fx, fy) 1 for all x, y [0, 2]. Hence, for all x, y X with α(fx, fy) 1, T x = 1x and T y = 1 y. Thus 2 2 we have, for all x, y X with α(fx, fy) 1 ξ(h(t x, T y)) = 1 x y 2 L 1 x y = Lξ(d(fy, T x)) 2 ψ(ξ(m(x, y, f, T ))) + Lξ(d(fy, T x)). Hence, (2.1) is satisfied. Thus, all hypothesis of Theorem 2.1 are satisfied, and f0 T 0. Moreover, f0 = ff0 and ff0 T f0, and so (C2) is satisfied and 0 = f0 T 0. Acknowledgment This research (Ki-Hwan Kim) was supported by Hanseo University.

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