Hacettepe Journal of Mathematics and Statistics Volume 31 (2002), 19 24 A RELATED FIXED POINT THEOREM FOR THREE METRIC SPACES Sarika Jain and Brian Fisher Received 01. 02. 2002 Abstract A related fixed point theorem for set valued mappings on three complete metric spaces is obtained which generalizes a result of [4]. Key Words: Complete metric space, common fixed point. Mathematics Subject Classification: 00 0 00 1. Introduction and Preliminaries Let (X, d) be a complete metric space and let B(X) be the set of all non-empty subsets of X. The function δ(a, B) with A and B in B(X) is defined by δ(a, B) = sup{d(a, b) : a A, b B}. If A consists of a single point a, we write δ(a, B) = δ(a, B) and if B also consists of a single point b, we write δ(a, B) = δ(a, b) = d(a, b). It follows immediately from the definition that δ(a, B) = δ(b, A) 0, δ(a, B) δ(a, C) + δ(c, B) for all A, B, C in B(X). For some preliminary definitions such as the convergence of a sequence of sets in B(X) and the continuity of a mapping F of X into B(X), see [1] or [2]. We need the following lemma in the sequel. 1.1. Lemma. (Fisher [1]). If {A n } and {B n } are sequences of bounded subsets of a complete metric space (X, d) which converge to the bounded subsets A and B respectively, then the sequence {δ(a n, B n )} converges to δ(a, B). Department of Mathematics, Dr H. S. Gour University, Sagar, (M. P.), India. Department of Mathematics and Computer Science, University of Leicester, LE1 7RH and Institute of Simulation Sciences, SERC, De Montfort University, Leicester, LE1 9BH, U.K. Email: fbr@le.ac.uk 19
20 A. Jain, B. Fisher Generalizing a result of Fisher [3] from two metric spaces to three metric spaces, the following related fixed point theorem was proved in [4]. 1.2. Theorem : Let (X, d 1 ), (Y, d 2 ) and (Z, d 3 ) be complete metric spaces. If T is a continuous mapping of X into Y, S is a continuous mapping of Y into Z and R is a mapping of Z into X satisfying the inequalities d 1 (RST x, RST x ) c max{d 1 (x, x ), d 1 (x, RST x), d 1 (x, RST x ), d 2 (T x, T x ), d 3 (ST x, ST x )} d 2 (T RSy, T RSy ) c max{d 2 (y, y ), d 2 (y, T RSy), d 2 (y, T RSy ), d 3 (Sy, Sy ), d 1 (RSy, RSy )} d 3 (ST Rz, ST Rz ) c max{d 3 (z, z ), d 3 (z, ST Rz), d 3 (z, ST Rz ), d 1 (Rz, Rz ), d 2 (T Rz, T Rz )} for all x, x in X, y, y in Y and z, z in Z, where 0 c < 1, then RST has a unique fixed point u in X, T RS has a unique fixed point v in Y and ST R has a unique fixed point w in Z. Further, T u = v,, Sv = w and Rw = u. We now generalize Theorem 1.2 for set valued mappings. 2. Main Result 2.1. Theorem : Let (X, d 1 ), (Y, d 2 ) and (Z, d 3 ) be complete metric spaces. If F is a continuous mapping of X into B(Y ), G is a continuous mapping of Y into B(Z) and H is a mapping of Z into B(X) satisfying the inequalities δ 1 (HGF x, HGF x ) c max{d 1 (x, x ), δ 1 (x, HGF x), δ 1 (x, HGF x ), δ 2 (F x, F x ), δ 3 (GF x, GF x )} (1) δ 2 (F HGy, F HGy ) c max{d 2 (y, y ), δ 2 (y, F HGy), δ 2 (y, F HGy ), δ 3 (Gy, Gy ), δ 1 (HGy, HGy )} (2) δ 3 (GF Hz, GF Hz ) c max{d 3 (z, z ), δ 3 (z, GF Hz), δ 3 (z, GF Hz ), δ 1 (Hz, Hz ), δ 2 (F Hz, F Hz )} (3) for all x, x in X, y, y in Y and z, z in Z, where 0 c < 1, then HGF has a unique fixed point u in X, F HG has a unique fixed point v in Y and GF H has a unique fixed point w in Z. Further, F u = {v}, Gv = {w} and Hw = {u}. Proof. Let x = x 1 be an arbitrary point in X. We define the sequences {x n } in X, {y n } in Y and {z n } in Z inductively as follows. Choose a point y 1 in F x 1, then a point z 1 in Gy 1 and then a point x 2 in Hz 1. In general, having chosen x n in X, y n in Y and z n in Z, choose a point x n+1 in Hz n, then a point y n+1 in F x n+1 and then a point z n+1 in Gy n+1 for n = 1, 2,.... Applying inequality (1), we get d 1 (x n+1, x n+2 ) δ 1 (HGF x n, HGF x n+1 ) c max{d 1 (x n, x n+1 ), δ 1 (x n, HGF x n ), δ 1 (x n+1, HGF x n+1 ), δ 2 (F x n, F x n+1 ), δ 3 (GF x n, GF x n+1 )}
A Fixed Point Theorem 21 c max{δ 1 (HGF x n 1, HGF x n ), δ 1 (HGF x n, HGF x n+1 ), δ 2 (F HGy n 1, F HGy n ), δ 3 (GF Hz n 1, GF Hz n )} c max{δ 1 (HGF x n 1, HGF x n ), δ 2 (F HGy n 1, F HGy n ), δ 3 (GF Hz n 1, GF Hz n )}. (4) Using inequality (2), we get d 2 (y n+1, y n+2 ) δ 2 (F HGy n, F HGy n+1 ) c max{d 2 (y n, y n+1 ), δ 2 (y n, F HGy n ), δ 2 (y n+1, F HGy n+1 ), δ 3 (Gy n, Gy n+1 ), δ 1 (HGy n, HGy n+1 )} c max{δ 2 (F HGy n 1, F HGy n ), δ 2 (F HGy n, F HGy n+1 ), δ 3 (GF Hz n 1, GF Hz n ), δ 1 (HGF x n, HGF x n+1 )} c max{δ 2 (F HGy n 1, F HGy n ), δ 3 (GF Hz n 1, GF Hz n ), δ 1 (HGF x n, HGF x n+1 )} c max{δ 2 (F HGy n 1, F HGy n ), δ 3 (GF Hz n 1, GF Hz n ), δ 1 (HGF x n 1, HGF x n )} (5) on using inequality (4). Further, on using inequality (3), we have d 3 (z n+1, z n+2 ) δ 3 (GF Hz n, GF Hz n+1 ) c max{d 3 (z n, z n+1 ), δ 3 (z n, GF Hz n ), δ 3 (z n+1, GF Hz n+1 ), δ 1 (Hz n, Hz n+1 ), δ 2 (F Hz n, F Hz n+1 )} c max{δ 3 (GF Hz n 1, GF Hz n ), δ 3 (GF Hz n, GF Hz n+1 ), δ 1 (HGF x n, HGF x n+1 ), δ 2 (F HGy n, F HGy n+1 )} c max{δ 3 (GF Hz n 1, GF Hz n ), δ 1 (HGF x n 1, HGF x n ), δ 2 (F HGy n 1, F HGy n )} (6) on using inequalities (4) and (5). It now follows easily by induction on using inequalities (4), (5) and (6) that d 1 (x n+1, x n+2 ) c n 1 max{δ 1 (HGF x 1, HGF x 2 ), δ 2 (F HGy 1, F HGy 2 ), δ 3 (GF Hz 1, GF Hz 2 )} d 2 (y n+1, y n+2 ) c n 1 max{δ 1 (HGF x 1, HGF x 2 ), δ 2 (F HGy 1, F HGy 2 ), δ 3 (GF Hz 1, GF Hz 2 )} d 3 (z n+1, z n+2 ) c n 1 max{δ 1 (HGF x 1, HGF x 2 ), δ 2 (F HGy 1, F HGy 2 ), δ 3 (GF Hz 1, GF Hz 2 )}. Then, for r = 1, 2,... and arbitrary ɛ > 0, we have d 1 (x n+1, x n+r+1 ) δ 1 (HGF x n, HGF x n+r ) δ 1 (HGF x n, HGF x n+1 ) + + δ 1 (HGF x n+r 1, HGF x n+r ) (c n 1 + c n +... + c n+r 2 ) max{δ 1 (HGF x 1, HGF x 2 ), δ 2 (F HGy 1, F HGy 2 ), δ 3 (GF Hz 1, GF Hz 2 )} < ɛ (7)
22 A. Jain, B. Fisher for n greater than some N, since c < 1. The sequence {x n } is therefore a Cauchy sequence in the complete metric space X and so has a limit u in X. Similarly, the sequences {y n } and {z n } are also Cauchy sequences with limits v and w in the complete metric spaces Y and Z respectively. Further, inequality (7) gives δ 1 (u, HGF x n ) d 1 (u, x m+1 ) + δ 1 (x m+1, HGF x n ) d 1 (u, x m+1 ) + δ 1 (HGF x m, HGF x n ) d 1 (u, x m+1 ) + ɛ for m, n N. Letting m tend to infinity, it follows that δ 1 (u, HGF x n ) ɛ, for n > N and so lim HGF x n = {u} = lim HGy n, (8) since ɛ is arbitrary. Similarly, = {v} =, (9) = {w} =. (10) Using the continuity of F and G, we obtain = = F u = {v}, (11) = = Gv = {w} (12) and then we see that GF u = {w}. (13) We now show that u is a fixed point of HGF. Applying inequality (1), we have δ 1 (HGF u, x n ) δ 1 (HGF u, HGF x n 1 ) c max{d 1 (u, x n 1 ), δ 1 (u, HGF u), δ 1 (x n 1, HGF x n 1 ), δ 2 (F u, F x n 1 ), δ 3 (GF u, GF x n 1 )}. Since F and G are continuous, it follows on letting n tend to infinity and using equation (8) that δ 1 (HGF u, u) cδ 1 (u, HGF u). Since c < 1, we have HGF u = {u} and so u is a fixed point of HGF. We now have HGF u = Hw = {u} on using equation (13) and then F HGv = F HGF u = F u = {v}, GF Hw = GF HGv = Gv = {w}
A Fixed Point Theorem 23 on using equations (11) and (12). Hence, v and w are fixed points of F HG and GF H respectively. Further, we see that HGv = {u}, F Hw = {v}. To prove the uniqueness of u, suppose that HGF has a second fixed point u. Then, using inequality (1), we have δ 1 (u, HGF u ) δ 1 (HGF u, HGF u ) Next, using inequality (2), we have c max{d 1 (u, u ), δ 1 (u, HGF u ), δ 2 (F u, F u ), δ 3 (GF u, GF u )} c max{δ 2 (F u, F u ), δ 3 (GF u, GF u )}. (14) δ 2 (F u, F u ) δ 2 (F HGF u, F HGF u ) and using inequality (3), we have c max{d 2 (F u, F u ), δ 2 (F u, F HGF u ), δ 3 (GF u, GF u ), δ 1 (HGF u, HGF u )} c max{δ 3 (GF u, GF u ), δ 1 (HGF u, HGF u )} (15) δ 3 (GF u, GF u ) δ 3 (GF HGF u, GF HGF u ) c max{d 3 (GF u, GF u ), δ 3 (GF u, GF HGF u ), δ 1 (HGF u, HGF u ), δ 2 (F HGF u, F HGF u )} c max{δ 1 (HGF u, HGF u ), δ 2 (F HGF u, F HGF u )}. (16) It now follows easily from inequalities (15) and (16) that and then δ 2 (F HGF u, F HGF u ) cδ 1 (HGF u, HGF u ) (17) δ 3 (GF u, GF u ) cδ 1 (HGF u, HGF u ). (18) Using inequalities (14), (17) and (18), we now have δ 1 (u, HGF u ) δ 1 (HGF u, HGF u ) c 2 δ 1 (HGF u, HF Gu ) and so HGF u is a singleton and HGF u = {u }, since c < 1. It then follows from inequality (18) that GF u is a singleton and from inequality (17) that F u is a singleton. Using inequality (1) again, we have d 1 (u, u ) = δ 1 (HGF u, HGF u ) c max{d 1 (u, u ), d 1 (u, u), d 1 (u, u ), d 2 (F u, F u ), d 3 (GF u, GF u )} = c max{d 2 (F u, F u ), d 3 (GF u, GF u )}. (19)
24 A. Jain, B. Fisher Next, using inequality (2), we have d 2 (F u, F u ) = δ 2 (F HGF u, F HGF u ) and using inequality (3), we have c max{d 2 (F u, F u ), d 2 (F u, F HGF u), d 2 (F u, F HGF u ), d 3 (GF u, GF u ), d 1 (HGF u, HGF u )} = c max{d 3 (GF u, GF u ), d 1 (u, u )} (20) d 3 (GF u, GF u ) = d 3 (GF HGF u, GF HGF u ) c max{d 3 (GF u, GF u ), d 3 (GF u, GF HGF u), d 3 (GF u, GF HGF u ), d 1 (HGF u, HGF u ), d 2 (F HGF u, F HGF u )} = c max{d 1 (u, u ), d 2 (F u, F u )}. (21) It now follows from inequalities (19), (20) and (21) that d 1 (u, u ) = 0, proving the uniqueness of u. The uniqueness of v and w follow similarly. 2.2. Remark. If we let F be a single valued mapping T of X into Y, G be a single valued mapping S of Y into Z and H be a single valued mapping R of Z into X, we obtain Theorem 1.2. References [1] Fisher, B. Common fixed points of mappings and set valued mappings, Rostock. Math. Kolloq. 18, 69 77, 1981. [2] Fisher, B. Set valued mappings on metric spaces, Fund. Math., 112, 141 145, 1981. [3] Fisher, B. Related fixed points on two metric spaces, Math. Sem. Notes, Kobe Univ., 10, 17 26, 1982. [4] Jain, R. K., Sahu, H. K. and Fisher, B. Related fixed point theorems for three metric spaces, Novi Sad J. Math., 26, 11 17, 1996.