Common fixed points of generalized contractive multivalued mappings in cone metric spaces

Transcription

1 MATHEMATICAL COMMUNICATIONS 365 Math. Commun., Vol. 14, No., pp (009) Common fixed points of generalized contractive multivalued mappings in cone metric spaces Mujahid Abbas 1,, B. E. Rhoades and Talat Nazir 1 1 Department of Mathematics, Lahore University of Management Sciences, Lahore-54 79, Pakistan. Department of Mathematics, Indiana University, Bloomington, Indiana , U. S. A. Received July 1, 009; accepted September 0, 009 Abstract. In this paper we obtain some sufficient conditions for the existence of common fixed points of multivalued mappings satisfying generalized contractive conditions in non normal cone metric spaces. These results establish some of the most general common fixed point theorems for two multivalued maps in cone metric spaces. AMS subject classifications: 47H10, 54H5, 54C60, 46B40 Key words: common fixed point, multivalued mapping, cone metric spaces, non normal cone 1. Introduction and premilinaries The study of fixed points for multivalued contractions and nonexpansive maps using the Hausdorff metric was initiated by Markin [9]. Later, an interesting and rich fixed point theory for such maps was developed. The theory of multivalued maps has application in control theory, convex optimization, differential equations and economics. Recently, Huang and Zhang [4] introduced the concept of a cone metric space, replacing the set of positive real numbers by an ordered Banach space. They obtained some fixed point theorems in cone metric spaces using the normality of the cone, which induces an order in Banach spaces (see also, [1], []). Rezapour and Hamlbarani Haghi [10] showed the existence of a non normal cone metric space and also obtained some fixed point theorems in cone metric spaces (see also [5], [7], [11], [1], [13] and [16]). Wardowski [15] introduced the concept of multivalued contractions in cone metric spaces and, using the notion of normal cones, obtained fixed point theorems for such mappings. The aim of this paper is to prove some common fixed points results for multivalued mappings taking closed values in cone metric spaces. It is worth mentioning that our results do not require the assumption of a normal cone. Our results extend and unify various comparable results in the literature ([6], [8] and [14]). Let E be a topological vector space. A subset P of E is called a cone if and only if: Corresponding author. addresses: mujahid@lums.edu.pk (M. Abbas), rhoades@indiana.edu (B. E. Rhoades), talat@lums.edu.pk (T. Nazir) c 009 Department of Mathematics, University of Osijek

2 366 M. Abbas, B. E. Rhoades and T. Nazir (a) P is closed, nonempty and P {0}; (b) a, b R, a, b 0, x, y P imply that ax + by P ; (c) P ( P ) = {0}. Given a cone P E, we define a partial ordering with respect to P by x y if and only if y x P. A cone P is said to be normal in a normed space E if there is a number K > 0 such that for all x, y E, 0 x y implies x K y. The least positive number satisfying the above inequality is called the normal constant of P, while x y stands for y x intp (interior of P ). Rezapour and Hamlbarani Haghi [10] proved that there is no normal cone with normal constant K < 1 and for each k > 1 there is a cone with normal constant K > k. Definition 1. Let X be a nonempty set. Suppose that the mapping d : X X E satisfies: (d1) 0 d(x, y) for all x, y X and d(x, y) = 0 if and only if x = y; (d) d(x, y) = d(y, x) for all x, y X; (d3) d(x, y) d(x, z) + d(z, y) for all x, y, z X. Then d is called a cone metric on X and (X, d) is called a cone metric space. The concept of a cone metric space is more general than that of a metric space. Definition. Let (X, d) be a cone metric space, {x n } a sequence in X and x X. For every c E with 0 c, we say that {x n } is (i) a Cauchy sequence if there is a natural number N such that for all n, m > N, d(x n, x m ) c. (ii) a convergent sequence if there is a natural number N such that for all n > N, d(x n, x) c for some x in X. Remark 1. If a ha, for some a P and h (0, 1), then a = 0. A cone metric space X is said to be complete if every Cauchy sequence in X is convergent in X. It is known that {x n } converges to x X if and only if d(x n, x) 0 as n. A set A in a cone metric space X is closed if for every sequence {x n } in A which converges to some x in X implies that x A. Let X be a cone metric space. We denote by P (X) the family of all nonempty subsets of X, and by P cl (X) the family of all nonempty closed subsets of X. A point x in X is called a fixed point of a multivalued mapping T : X P cl (X) provided x T x. The collection of all fixed points of T is denoted by F (T ).

3 Multivalued mappings in cone metric spaces 367. Common fixed point results Kannan [6] proved a fixed point theorem for a single valued self mapping T of a metric space X satisfying the property d(t x, T y) h{d(x, T x) + d(y, T y)} for all x, y in X and for a fixed h [0, 1 ). Latif and Beg [8] introduced the notion of a K multivalued mapping, which is the extension of Kannan mappings to multivalued mappings. Recently, Rus [14] coined the term R multivalued mapping, which is a generalization of a K multivalued mapping. In this section we obtain common fixed point theorems for two multivalued mappings on a cone metric space without using the condition of a normal cone. Given the fact that in a cone one has only a partial ordering, it is doubtful whether the following theorem can be further generalized. Theorem 1. Let (X, d) be a complete cone metric space and T 1, T : X P cl (X) two multivalued mappings such that for i, j {1, } with i j and for each x, y X, u x T i (x), there exists u y T j (y) such that where h (0, 1) is a constant and d(u x, u y ) hu(x, y; u x, u y ), (1) u(x, y; u x, u y ) {d(x, y), d(x, u x ), d(y, u y ), d(x, u x) + d(y, u y ) Then F (T 1 ) = F (T ). Also F (T 1 ) = F (T ) P cl (X). Proof. Let x T 1 (x ). Then there exists an x T (x ) such that where d(x, x) hu(x, x ; x, x) u(x, x ; x, x) {d(x, x ), d(x, x ), d(x, x ), d(x, x ) + d(x, x) = {0, d(x, x ), d(x, x) }., d(x, x) + d(x, x ) }, d (x, u y) + d (y, u x ) }. Now u(x, x ; x, x) = 0 implies that x = x and, u(x, x ; x, x) = d(x, x) gives d(x, x) hd(x, x), which by Remark 1 implies that x = x. Similarly, for u(x, x ; x, x) = d(x, x)/, we obtain x = x. Thus F (T 1 ) F (T ). Also, F (T ) F (T 1 ) and therefore F (T 1 ) = F (T ). Suppose that x 0 is an arbitrary point of X. For i, j {1, } with i j and x 1 T i (x 0 ), there exists x T j (x 1 ) such that d(x 1, x ) hu(x 0, x 1 ; x 1, x ),

4 368 M. Abbas, B. E. Rhoades and T. Nazir where u(x 0, x 1 ; x 1, x ) {d(x 0, x 1 ), d(x 0, x 1 ), d(x 1, x ), d(x 0, x 1 ) + d(x 1, x ), d(x 0, x ) + d(x 1, x 1 ) } = {d(x 0, x 1 ), d(x 1, x ), d(x 0, x 1 ) + d(x 1, x ) Now, u(x 0, x 1 ; x 1, x ) = d(x 0, x 1 ) implies that d(x 1, x ) hd(x 0, x 1 ). If u(x 0, x 1 ; x 1, x ) = d(x 1, x ), d(x 0, x ) }. then d(x 1, x ) hd(x 1, x ), which by Remark 1, implies that x 1 = x. Also, if then we obtain u(x 0, x 1 ; x 1, x ) = d(x 0, x 1 ) + d(x 1, x ), d(x 1, x ) h d(x 0, x 1 ) + h d(x 1, x ) h d(x 0, x 1 ) + 1 d(x 1, x ) and d(x 1, x ) hd(x 0, x 1 ). Finally, for u(x 0, x 1 ; x 1, x ) = d(x 0, x ) we get d(x 1, x ) h d(x 0, x ) h d(x 0, x 1 ) + h d(x 1, x ) h d(x 0, x 1 ) + 1 d(x 1, x ), which also implies that d(x 1, x ) hd(x 0, x 1 ). Continuing this process, for x n T j (x n 1 ), there exists x n+1 T i (x n ) such that where d(x n, x n+1 ) hu(x n 1, x n ; x n, x n+1 ), u(x n 1, x n ; x n, x n+1 ) {d(x n 1, x n ), d(x n 1, x n ), d(x n, x n+1 ), d(x n 1, x n ) + d(x n, x n+1 ), d(x n 1, x n+1 ) + d(x n, x n ) } = {d(x n 1, x n ), d(x n, x n+1 ), d(x n 1, x n ) + d(x n, x n+1 ), d(x n 1, x n+1 ) }. If u(x n 1, x n ; x n, x n+1 ) = d(x n 1, x n ), then d(x n, x n+1 ) h d(x n 1, x n ). For u(x n 1, x n ; x n, x n+1 ) = d(x n, x n+1 ), d(x n, x n+1 ) h d(x n, x n+1 ), which by Remark 1 gives x n = x n+1. When u(x n 1, x n ; x n, x n+1 ) = d(x n 1, x n ) + d(x n, x n+1 ),

5 Multivalued mappings in cone metric spaces 369 we obtain and d(x n, x n+1 ) h [d(x n 1, x n ) + d(x n, x n+1 )] h d(x n 1, x n ) + 1 d(x n, x n+1 ) d(x n, x n+1 ) hd(x n 1, x n ). Finally, u(x n 1, x n ; x n, x n+1 ) = d(x n 1, x n+1 )/ gives that and d(x n, x n+1 ) h d(x n 1, x n+1 ) h [d(x n 1, x n ) + d(x n, x n+1 )] h d(x n 1, x n ) + 1 d(x n, x n+1 ) d(x n, x n+1 ) hd(x n 1, x n ). In a similar manner, for x n+1 T i (x n ), there exists x n+ T i (x n+1 ) so that Therefore d(x n+1, x n+ ) hd(x n, x n+1 ). d(x n, x n+1 ) hd(x n 1, x n ) h d(x n, x n )... h n d(x 0, x 1 ) for all n 1 for m > n we have d(x n, x m ) h n d(x 0, x 1 )/(1 h). Let 0 c be given. Choose a symmetric open neighborhood V of 0 such that c + V P. Also, choose a natural number N 1 such that h n d(x 0, x 1 )/(1 h) V for all n N 1 which implies that h n d(x 0, x 1 )/(1 h) c for all n > N 1. Hence d(x n, x m ) c for all n, m > N 1. Therefore {x n } is a Cauchy sequence in X. Since X is complete, there exists an element x X such that x n x as n. Let 0 c be given and 0 < δ < min{ 1 4, 1 h}. Choose a natural number N such that d(x m, x ) δc for all m N. Let n N be given. Then, for x n T j (x n 1 ), there exists u n T i (x ) such that where d(x n, u n ) hu(x n 1, x ; x n, u n ), u(x n, x ; x n+1, u n ) {d(x n 1, x ), d(x n 1, x n ), d(x, u n ), Note that d(x n 1, x n ) + d(x, u n ) d(u n, x ) d(u n, x n ) + d(x n, x ) hu(x n, x ; x n+1, u n ) + d(x n, x )., d(x n 1, u n ) + d(x, x n ) }.

6 370 M. Abbas, B. E. Rhoades and T. Nazir Now, u(x n, x ; x n+1, u n ) = d(x n 1, x ) implies that d(u n, x ) hd(x n 1, x ) + d(x n, x ) h( c h ) + c = c. If u(x n, x ; x n+1, u n ) = d(x n 1, x n ), then d(u n, x ) hd(x n 1, x n ) + d(x n, x ) In case u(x n, x ; x n+1, u n ) = d(x, u n ), then hd(x n 1, x ) + hd(x, x n ) + d(x n, x ) h( c 3h ) + h( c 3h ) + c 3 = c. d(u n, x ) hd(x, u n ) + d(x n, x ) d(u n, x ) 1 1 h d(x n, x ) c. If u(x n, x ; x n+1, u n ) = d(x n 1, x n ) + d(x, u n ), we get d(u n, x ) h [d(x n 1, x n ) + d(x, u n )] + d(x n, x ) h [d(x n 1, x ) + d(x, x n )] + 1 d(x, u n ) + d(x n, x ) d(u n, x ) h[d(x n 1, x ) + d(x, x n )] + d(x, x n ) c. Finally, if u(x n, x ; x n+1, u n ) = d(x n 1, u n ) + d(x, x n ), then d(u n, x ) h [d(x n 1, u n ) + d(x, x n )] + d(x n, x ) h [d(x n 1, x ) + d(x, u n )] + h d(x, x n ) + d(x n, x ) 1 d(x n 1, x ) + 1 d(x, u n ) + 3 d(x n, x ) d(u n, x ) d(x n 1, x ) + 3d(x n, x ) c. Thus u n x as n. Since T i (x ) is closed, x F (T j ) = F (T i ).

7 Multivalued mappings in cone metric spaces 371 Now, we prove that F (T i ) is closed. Let {p n } be a sequence in F (T j ) = F (T i ) such that p n p as n. Since p n T i (p n ), there exists q n T j (p) such that where d(p n, q n ) hu(p n, p; p n, q n ), u(p n, p; p n, q n ) {d(p n, p), d(p n, p n ), d(p, q n ), d(p n, p n ) + d(p, q n ), d(p n, q n ) + d(p, p n ) } = {d(p n, p), 0, d(p, q n ), d(p, q n), d(p n, q n ) + d(p, p n ) }. Now we show that q n p as n. Let 0 c be given and 0 < δ < min{ 1 4, 1 h}. Choose a natural number N such that d(p m, p) δc for all m N is given. If u(p n, p; p n, q n ) = d(p n, p) for some n, then d(q n, p) d(q n, p n ) + d(p n, p) hd(p n, p) + d(p n, p) c. If u(p n, p; p n, q n )=0, then our claim follows immediately. If u(p n, p; p n, q n )= d(p, q n ), then d(q n, p) d(q n, p n ) + d(p n, p) hd(p, q n ) + d(p n, p) d(q n, p) 1 1 h d(p n, p) c. If u(p n, p; p n, q n ) = d(p, q n )/, then d(q n, p) d(q n, p n ) + d(p n, p) h d(p, q n) + d(p n, p) d(p, q n) + d(p n, p) d(q n, p) d(p n, p) c. If u(p n, p; p n, q n ) = [d(p n, q n ) + d(p, p n )]/, then d(q n, p) d(q n, p n ) + d(p n, p) h [d(p n, q n ) + d(p, p n )] + d(p n, p) h [d(p n, p) + d(p, q n ) + d(p, p n )] + d(p n, p) 1 d(p, q n) + d(p, p n )

8 37 M. Abbas, B. E. Rhoades and T. Nazir d(q n, p) 4d(p n, p) c. Thus q n p as n. Since q n T j (p) for each n 1 and T j (p) is closed, p T j (p). Therefore, F (T j ) = F (T i ) P cl (X). The following theorem generalizes [14, Theorem 3.4] to cone metric spaces. Theorem. Let (X, d) be a complete cone metric space and T 1, T : X P cl (X) two multivalued mappings such that for i, j {1, } with i j and for each x, y X, u x T i (x), there exists u y T j (y) such that d(u x, u y ) αd(x, y) + βd(x, u x ) + γd(y, u y ), () where α, β, γ 0 and α+β +γ < 1. Then F (T 1 ) = F (T ) φ and F (T 1 ) = F (T ) P cl (X). Proof. Suppose that x 0 is an arbitrary point of X. For i, j {1, } with i j, take x 1 T i (x 0 ). Then there exists x T j (x 1 ) such that which implies that d(x 1, x ) αd(x 0, x 1 ) + βd(x 0, x 1 ) + γd(x 1, x ), d(x 1, x ) kd(x 0, x 1 ), where 0 < k = (α + β)/(1 γ) < 1. Now for x T j (x 1 ) there exists x 3 T i (x ) such that d(x, x 3 ) kd(x 1, x ). Continuing this process we obtain a sequence {x n } in X with x n 1 T i (x n ), x n T j (x n 1 ) such that d(x n, x n+1 ) kd(x n 1, x n ) which further implies that d(x n, x n+1 ) k n d(x 0, x 1 ) for all n 1 Then, for m > n, d(x n, x m ) d(x n, x n+1 ) + d(x n+1, x n+ ) d(x m+1, x m ) [k n + k n k m 1 ]d(x 1, x 0 ) k n d(x 0, x 1 )/ (1 k). Let 0 c be given. Choose a symmetric open neighborhood V of 0 such that c + V P. Also, choose a natural number N 1 such that k n d(x 0, x 1 )/(1 k) V for all n N 1 which implies that k n d(x 0, x 1 )/(1 k) c for all n > N 1 and hence d(x n, x m ) c for all n, m > N 1. Therefore {x n } is a Cauchy sequence in X. Since X is complete, there exists an element x X such that x n x as n. Let 0 c be given. Now for x n T j (x n 1 ), there exists u n T i (x ) such that which further gives d(x n, u n ) αd(x n 1, x ) + βd(x n 1, x n ) + γd(x, u n ), d(x, u n ) d(x, x n ) + αd(x n 1, x ) + βd(x n 1, x n ) + γd(x, u n ), d(x, u n ) 1 1 γ [d(x, x n ) + αd(x n 1, x ) + βd(x n 1, x n )] c

9 Multivalued mappings in cone metric spaces 373 for a sufficiently large n which shows that u n x as n. Since T i (x ) is closed, x F (T i ) F (T i ) φ. Let x X be a fixed point of T 1. Then, by hypothesis, there exists x T x such that d(x, x) αd(x, x ) + βd(x, x ) + γd(x, x ) = γd(x, x ), which by using Remark1, implies that d(x, x) = 0, x = x. Thus, F (T 1 ) F (T ). Similarly, F (T ) F (T 1 ). Now we prove that F (T i ) is closed. Let {x n } be a sequence in F (T j ) = F (T i ) such that x n x as n. Since x n T i (x n ), there exists v n T j (x) such that Thus, d(x n, v n ) αd(x n, x) + βd(x n, x n ) + γd(x, v n ), d(x, v n ) d(x, x n ) + αd(x n, x) + γd(x, v n ). d(x, v n ) 1 + α 1 γ d(x, x n) c for a sufficiently large n. Thus d(x, v n ) 0 as n. Since v n T j (x) for each n N and T j (x) is closed, x T j (x). Hence, x F (T j ) = F (T i ). Example 1. Let X = [0, 1], E = R and P = {(x, y) E : x, y 0}. Let d : X X E be defined by d(x, y) = ( x y, h x y ), where h 0. Define T 1, T : X P cl (X) by T 1 x = [0, x 4 ] and T x = [0, x 3 ]. Note that for x = y = 0, () is satisfied as u x = u y = 0. For x = y 0 and u x T 1 x, take u y = 0. Then d(u x, u y ) = (u x, hu x ) ( x 4, hx 4 ) 1 6 (0, 0) + 6 (3x 4, h3x 4 ) + (x, hx) (x y, h(x y)) + 6 (x u x, h(x u x )) + (x, hx) 6 = αd(x, y) + βd(x, u x ) + γd(y, u y ), () is satisfied with α = 1/6, β = γ = /6. Now when x = 0, y 0, then () is satisfied for u y = 0. If x 0, y = 0, then since u y = 0, for any u x T 1 x we have d(u x, u y ) ( x 4, hx 4 ) αd(x, y) + βd(x, u x ) + γd(y, u y ).

10 374 M. Abbas, B. E. Rhoades and T. Nazir Since d(x, y) is symmetric in x and y and β = γ, it is sufficient to consider 0 < x < y, u x T 1 x. Take u y = 0. Then d(u x, u y ) = (u x, hu x ) ( x 4, hx 4 ) 1 6 (0, 0) + 6 (3x 4, h3x 4 ) + (y, hy) 6 αd(x, y) + βd(x, u x ) + γd(y, u y ). Now we show that for x, y X, u x T x, there exists u y T 1 y such that () is satisfied. For x = y = 0, () is satisfied as u x = u y = 0. For x = y 0, u x T x, take u y = 0. Then we have d(u x, u y ) = (u x, hu x ) ( x 3, hx 3 ) 1 6 (0, 0) + 6 (x 3, hx 3 ) + (x, hx) 6 αd(x, y) + βd(x, u x ) + γd(y, u y ). Now when x = 0, y 0, then () is satisfied for u y = 0. If x 0, y = 0, then for any u x T x, we have u y = 0, and d(u x, u y ) = d(u x, hu x ) ( x 3, hx 3 ) αd(x, y) + βd(x, u x ) + γd(y, u y ). For 0 < x < y, u x T x, take u y = 0. Then we have d(u x, u y ) = (u x, hu x ) ( x 3, hx 3 ) 1 6 (0, 0) + 6 (x 3, hx 3 ) + (y, hy) 6 αd(x, y) + βd(x, u x ) + γd(y, u y ), with α + β + γ = 5/6. Note that F (T 1 ) = F (T ) φ. Moreover, F (T 1 ) = F (T ) P cl (X). Now we present another example with a different topological vector space as a range. Example. Let E = C R [0, ), P = {f E : f(x) 0, x [0, )}, X = [0, 1] with a usual metric and with a cone metric d : X X E defined by d(x, y) = f x,y, where f x,y (t) = t x y ([3]). Define T 1, T : X P cl (X) as T 1 x = [ x 7, x 5 ] and T x = [ x 6, x 3 ]. Now if x = y, then () is satisfied with u x = u y, α = β = 4/10 and γ = 1/10. Also the following cases arise:

11 Multivalued mappings in cone metric spaces 375 Case (i): x = 0, y > 0. Then u x = 0 T 1 x, choosing u y = y/6 T y we have [d(u x, u y )] (t) = f ux,u y (t) = tu y = t y 6 t[ 4 10 (y) (0) (5 6 y)] t[ 4 10 (y x) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t), with α = β = 4/10, γ = 1/10 and for all t [0, ). Case (ii): x > 0, y = 0. Then, for u x T 1 x, choosing u y = 0 T y we have [d(u x, u y )] (t) = f ux,u y (t) = tu x t x 5 t[ 4 10 (x) (4x 5 ) (0)] t[ 4 10 (x y) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t). Case (iii): 0 < y < x. Then for u x T 1 x, taking u y = y/6, we have [d(u x, u y )] t = f ux,u y (t) t( x 5 y 6 ) t[ 4 10 (0) (4x 5 ) (5 6 y)] t[ 4 4 (x y) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )]t. Case (iv): 0 < x < y. Then, for u x T 1 x, taking u y = y/5, yields [d(u x, u y )] (t) = f ux,u y (t) t( y 5 x 7 ) t[ 4 10 (y x) (4x 5 ) (4 5 y)] t[ 4 10 (y x) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t). Now we show that for x, y X, u x T x, there exists u y T 1 y such that () is satisfied. We consider the following cases. Case (i): x = 0, y > 0. Then with u x = 0 T x, u y = y/7 T 1 y we have [d(u x, u y )] (t) = f ux,u y (t) = tu y = t y 7 t[ 4 10 (y) (0) (6 7 y)] t[ 4 10 (y x) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t).

12 376 M. Abbas, B. E. Rhoades and T. Nazir Case (ii): x > 0, y = 0. Then, for u x T x, u y = 0 T 1 y we have [d(u x, u y )] (t) = f ux,u y (t) = tu x t x 3 t[ 4 10 (x) (x 3 ) (0)] t[ 4 4 (x y) (x u x) (0)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t). Case (iii): 0 < y < x. Then, for u x T x, take u y = 8y 50 to obtain [d(u x, u y )] (t) = f ux,u y (t) = t(u x 8y 50 ) t(x 3 8y 50 ) t[ 4 10 (x y) (x 3 ) (4 50 y)] t[ 4 10 (x y) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t). Case (iv): 0 < x < y. Then, for u x T x, taking u y = y/6 yields [d(u x, u y )] (t) = f ux,u y (t) t x 3 y 6 t[ 4 10 (y x) (x 3 ) (5 6 y)] t[ 4 10 (y x) (x u x) (y u y)] = [αd(x, y) + βd(x, u x ) + γd(y, u y )](t). Also note that F (T 1 ) = F (T ) φ. Moreover, F (T 1 ) = F (T ) P cl (X). Other examples in the support of Theorem can be constructed by taking E = l p ( p > 0), P = {{x n } n 1 E x n 0}, and d(x, y) = {(ρ(x, y)/ n ) 1/p } n 1, where ρ is a metric on any nonempty set X. The following corollary extends Theorem 4.1 of [8] to the case of two mappings on cone metric spaces. Corollary 1. Let (X, d) be a complete cone metric space and P a non normal cone. If T 1, T : X P cl (X) are two multivalued mappings such that for i, j {1, } with i j, x, y X and u x T i (x), there exists u y T j (y) such that d(u x, u y ) h[d(x, u x ) + d(y, u y )], where α, β, γ 0 are fixed constants with α + β + γ < 1. Then F (T 1 ) = F (T ) φ and F (T 1 ) = F (T ) P cl (X). The following corollary extends Theorem 4.1 of [8] to cone metric spaces.

13 Multivalued mappings in cone metric spaces 377 Corollary. Let (X, d) be a complete cone metric space and P a non normal cone. If T : X P cl (X) is a multivalued mapping such that for each x, y X and u x T (x), there exists u y T (y) such that d(u x, u y ) h[d(x, u x ) + d(y, u y )], where 0 h < 1. Then F (T ) φ and F (T ) P cl (X). Corollary 3. Let (X, d) be a complete cone metric space and P a non normal cone. If T : X P cl (X) is a multivalued mapping such that for each x, y X and u x T (x), there exists u y T (y) such that d(u x, u y ) αd(x, y), where 0 α < 1. Then F (T ) φ and F (T ) P cl (X). Proof. Take β = γ = 0, and T 1 = T = T in Corollary 1. Corollary 4. Let (X, d) be a complete cone metric space and P a non normal cone. If T : X P cl (X) is a multivalued mapping such that for each x, y X and u x T (x), there exists u y T (y) such that d(u x, u y ) αd(x, y) + βd(x, u x ) + γd(y, u y ), where α, β, γ 0 are fixed constants with α + β + γ < 1, then T has a fixed point. The above corollary is an extension of Theorem 3.1 of [14] to cone metric spaces. Acknowledgement The authors are grateful to the referees for the critical remarks which improved the work. References [1] M. Abbas, B. E. Rhoades, Fixed and periodic point results in cone metric spaces, Applied Mathematics Letters (009), [] M. Abbas, G. Jungck, Common fixed point results of noncommuting mappings without continuity in cone metric spaces, J. Math. Anal. Appl. 341(008), [3] R. H. Haghi, Sh. Rezapour, Fixed points of multifunctions on regular cone metric spaces, Expositiones Mathematicae, to appear. [4] L. G. Huang, X. Zhang, Cone metric spaces and fixed point theorems of contractive mappings, J. Math. Anal. Appl. 33(007), [5] D. Ilić, V. Rakočević, Common fixed points for maps on cone metric space, J. Math. Anal. Appl. 341(008), [6] R. Kannan, Some results on fixed points, Bull. Calcutta. Math. Soc. 60(1968), [7] D. Kilm, D. Wardowski, Dynamic processes and fixed points of set-valued nonlinear contractions in cone metric spaces, Nonlinear. Anal. 71(009), [8] A. Latif, I. Beg, Geometric fixed points for single and multivalued mappings, Demonstratio Math. 30(1997),

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