Multi-valued contraction mappings in complete metric
Transcript
Multi-valued contraction mappings in complete metric
R ENDICONTI del S EMINARIO M ATEMATICO della U NIVERSITÀ DI PADOVA K IYOSHI I SEKI Multi-valued contraction mappings in complete metric spaces Rendiconti del Seminario Matematico della Università di Padova, tome 53 (1975), p. 15-19. <http://www.numdam.org/item?id=RSMUP_1975__53__15_0> © Rendiconti del Seminario Matematico della Università di Padova, 1975, tous droits réservés. L’accès aux archives de la revue « Rendiconti del Seminario Matematico della Università di Padova » (http://rendiconti.math.unipd.it/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/legal. php). Toute utilisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright. Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques http://www.numdam.org/ Multi-Valued Contraction in Complete Metric KIYOSHI ISEKI Let of ~, be define (x, d) we a metric space. Mappings Spaces. (*) For any B nonempty subsets A, Let CB(X) be the set of all nonempty closed and bounded subsets of X. The space CB(X) is a metric space with respect to the above defined distance H (see K. Kuratowski [1] p. 214). Then we have the following theorem which is a generalization of S. Reich result [2] (or see I. Rus [3]). THEOREM]. Let (X, d) be a complete metric space, and let CB(X ) be a nlulti-valued mapping with the following every x, y X - for E X, where 0153, (3, y are point, i. e. there is (*) f: non-negative and 2a --~- 2~ -fa point x such that x E I(x). Indirizzo dell’A.: Kobe University, y 1. Then f has a Rokko Nada Kobe, Giappone. 16 PROOF. Let xo be a point in X, and xl E f (xo). If since H is a metric on CB(X). Therefore then we have 0. This contains the proof of the case a == f3 = y we have Next we suppose 0 2a + 2~3 + y and = Let h By = .H( f (xo), f(x1))(p, the definition of Therefore there is Hence Hence we have a H, we then we have have point x, of such that 17 Therefore If If have the first case, then x2 a similar such that we H( f (xl), !(X2)) > 0, by In general, if all i (i 0, 1, = H( f (xi), ... ), H( f (x2), = 0 for > 0, which completes the proof. there is a point X3 of method, i, then xi E f (xi). there is a point Xi+2 If, some for E satisfying Hence, for n > m, is a Cauchy sequence. The completeness of X This shows that Let x’ be the limit of the existence of the limit of then implies Hence, by the assumption, we have 18 Let n - From 00, then (1 ) implies 12013x2013~>0y we the following relation. have which means This completes the proof. Let BN(X) be the set of all nonempty bounded subset of X. Then we have a fixed point theorem. THEOREM 2. Let (X, d) be a complete mctric space. is ac functions which satisfies for every x, y then f has a If f : X -- BN(X ) in X, where oc, f3, y are non-negative and 2~ -f- 4f3 -funique f ixed point, i. e. for some x’, f (x’ ) Theorem 2 is is due to an idea y 1, - generalization of S. Reich by S. Reich [2]. a result [2]. The proof PROOF. If oc = ~ 0, then the result is trivial. We suppose y 0 2a + 4fl + y. Now put p (2oc + 4fl + 1 )~. Then we have p 1. Hence there is a single-valued function g : X - X such that g(x) is a which satisfies point y in = == = For such a function g, 19 Hence we The 1. For the + y have assumption 2a -f- 4~ -f- y C 1 implies By a well known theorem, g has a fixed point x’, i. e. point x’, Hence If and .H(z, f (z) ) > 0, which is impossible. Hence To show z consider we = x’. then have f(z) = ~z~. = Hence we have z x’, which shows that The proof is complete. f = has a unique fixed point. REFERENCES [1] K. KURATOWSKI, Topology, 1, PWN, Warszawa (1966). [2] S. REICH, Fixed points of contractive functions, Boll. Un. Mat. Ital., (4) 5 (1972), pp. 26-42. [3] I. A. Rus, Teoria punctului fix II, Univ. « Babes-Bolyai », Cluj (1973). Manoscritto pervenuto in redazione il 19 marzo 1974.