COMMON FIXED POINT THEOREMS FOR MULTIVALUED OPERATORS ON COMPLETE METRIC SPACES

Transcription

1 STUDIA UNIV. BABEŞ BOLYAI MATHEMATICA Volume XLVII Number 1 March 00 COMMON FIXED POINT THEOREMS FOR MULTIVALUED OPERATORS ON COMPLETE METRIC SPACES 1. Introduction The purpose of this paper is to prove a common fixed point theorem for multivalued operators defined on a complete metric space. Then as consequences we obtain some generalizations of several results proved in [6] for singlevalued operators. For other results of this type see [1] [] [3] and [5]. The metric conditions which appears in Theorem 3.1 generalize some conditions given in [6].. Preliminaries Let X be a nonempty set. We denote: P X := {A X A } and P cl X := {A P X A = Ā}. If X d is a metric space B P X and a A then Da B := inf{da b b B}. Definition.1. If T : X X is a multivalued operator then an element x X is a fixed point of T iff x T x. We denote by F T := {x X x T x} the fixed points set of T. Definition.. Let T n n N be a sequence of multivalued operators T n : X P X n N. Then we denote by ComT := {x X x T n x n N } = the common fixed points set of the sequence T n n N. n N F Tn 000 Mathematics Subject Classification. 7H10 5H5. Key words and phrases. fixed point multivalued operator common fixed point. 73

2 Lemma.3. I.A.Rus []. Let ϕ : R k + R + k N be a function and denote by ψ : R + R + the mapping given by ψt = ϕt t... t t R +. Suppose that the following conditions are satisfied: i r s r s R k + ϕr ϕs; ii ϕ is upper semi-continuous; iii ψt < t for each t > 0. Then lim ψn t = 0 for each t 0. In [6] T.Veerapandi and S.A.Kumar gave the following result: Theorem.. Let X be a Hilbert space Y P cl X and T n : Y Y for n N be a sequence of mappings. We suppose that at least one of the following conditions is satisfied: i there exist real numbers a b c satisfying 0 a b c < 1 and a + b + c < 1 such that for each x y Y and x y T i x T j y a x y + b x T i x + y T j y + + c x T j y + y T i x for i j; ii there exist a real number h satisfying 0 h < 1 such that for all x y Y and x y T i x T j y h max { x y 1 x T i x + y T j y } 1 x T j y + y T i x for i j. Then T n n N has a unique common fixed point. 3. The main results The first result of this section improve and generalize Theorem. in the multivoque case. Theorem 3.1. Let X d be a complete metric space and S T : X P cl X multivalued operators. We suppose that there exists a function ϕ : R 3 + R + such that: i r s r s R 3 + ϕr ϕs; ii ϕt t t < t for each t > 0; 7

3 COMMON FIXED POINT THEOREMS so that we have iii ϕ is continuous; iv for each x X any u x Sx and for all y X there exists u y T y d u x u y ϕ such that d x y d x u x + d y u y In these conditions F S = F T = {x }. d x u y + d y u x. Proof. Let x 0 X arbitrarily. Then we can construct a sequence x n X x n+1 Sx n x n+ T x n+1 Denote by d n := dx n x n+1 n N. n N. We have several steps in our proof. Step I. Let us prove that the sequence d n is monotone decreasing. Indeed we have successively: d n+1 = d x n+1 x n+ ϕ d x n x n+1 d x n x n+1 + d x n+1 x n+ d x n x n+ + d x n+1 x n+1 { < max d n ϕ d n d n + d n+1 d n + d n+1 d n + d n+1 } = d n from where it follows d n+1 < d n. By an analogous method we have d n+ < d n+1. Step II. We prove that lim d n = 0. For this purpose let us define ψ : R + R + by ψt = ϕt t t. Obviously ψ is monotone increasing and ψt < t t > 0. By induction we can prove that d n ψ n d 0 n 1. Indeed we have d 1 ϕ d 0 d 1 + d 0 d 0 + d 1 ϕd 0 d 0 d 0 = ψd 0. If inequality d n ψ n d 0 is true then we get successively: d n+1 ϕ d n d n + d n+1 d n + d n+1 ϕd n d n d n = ψd n ψψ n d 0 = ψ n+1 d 0. By passing to limit as n if d 0 > 0 it follows lim d n lim ψn d 0 = 0 and hence lim d n = 0. 75

4 lim d n = 0. For d 0 = 0 the sequence d n being decreasing it is obviously that Step III. We ll prove that the sequence x n is Cauchy in X i.e. for each ε > 0 there exists k N such that for each m n k dx m x n < ε. Suppose by contradiction that x n is not Cauchy sequence. Then there exists ε > 0 such that for each k N there exist m k n k N m k > n k k with the property dx mk x nk > ε. In what follows let us suppose the numbers mk and nk as follows: mk := inf{m k N m k > n k k dx nk x mk ε dx nk x mk > ε} and nk := n k. Then k N we have: ε < dx nk x mk dx nk x mk + dx mk x mk 1 + and Using step II we deduce that +dx mk 1 x mk. From the triangle inquality we get: lim dx nk x mk = ε. 1 k dx nk x mk 1 dx nk x mk dx mk 1 x mk dx nk+1 x mk 1 dx nk x mk dx mk 1 x mk + dx nk x nk+1. Using again step III and the relation 1 it follows lim dx nk x mk 1 = ε k Then we have successively: lim dx nk+1 x mk 1 = ε. k 76 dx nk x mk dx nk x nk+1 + dx nk+1 x mk dx nk x nk [ ϕd x nk x mk 1 d x nk x nk+1 + d x mk 1 x mk

5 COMMON FIXED POINT THEOREMS d x nk x mk + d x mk 1 x nk+1 ] 1. Because ϕ is continuous passing to the limit as k we have: ε [ ϕ ε 0 ε Step IV. We prove that F T. ] 1 [ ψε ] 1 < ε a contradiction. Because x n is Cauchy sequence in the complete metric space X d we obtain that there exists x X such that lim x n = x. From x n+1 Sx n we have that there exists u n T x such that: d x n+1 u n ϕ d x n x d x n x n+1 + d x u n d x n u n + d x x n+1 { < max d x n x d x n x n+1 + d x u n d x n u n + d x x n+1 from where i.e. := M. Consequently we have the following situations: a. Case M = d x n x. In this case we have d x n+1 u n d x n x lim dx n+1 u n lim dx n x = 0 lim dx n+1 u n = 0. < b. Case M = d x n x n+1 + d x u n. We deduce successively: d x n+1 u n d x n x n+1 + d x u n d x n x n+1 + [dx x n+1 + dx n+1 u n ] i.e. d x n+1 u n dx x n+1 dx n+1 u n [d x n x n+1 +d x x n+1 ] 0 therefore dx n+1 u n dx x n+1 + d x x n+1 + d x n x n+1. } 77

6 c. Case M = d x n u n + d x x n+1. In this case from the inequal- ity we have again Passing to the limit in this inequality as n we obtain lim dx n+1 u n = 0. d x n+1 u n d x n u n + d x x n+1 lim dx n+1 u n = 0. Passing to the limit as n in inequality dx u n dx x n+1 + dx n+1 u n on the basis of the limit lim dx n+1 u n = 0 we obtain dx u n 0 as n 0. Since u n T x n N and T x is a closed set it follows that x T x i.e. x F T. Step V. that F S F T. that We ll obtain now the conclusion of our theorem. We first prove Let x F S. From x Sx we have that there exists u T x d x u ϕ d x x d x x + d x u d x u + d x x. If we suppose that dx u > 0 then we obtain d x d x u d x u u ϕ 0 < d x u such a contradiction. Thus dx u = 0 which means that u = x. It follows that x T x and so F S F T. and T. we have 78 We shall prove now the equality F S = F T betwen the fixed points set for S If we assume that there exists y F T such that y x F S then d x y ϕ d x y = ϕ d x y 0 d x y + d y x = d x x + d y y d x y ψd x y < d x y

7 COMMON FIXED POINT THEOREMS a contradiction proving the fact that F S = F T P X. In fact we have obtained even more namely that F S = F T = {x }. Corollary 3.. Let X d be a complete metric space and S T : X P cl X multivalued operators. We suppose that there exist a b c R + a + b + c < 1 such that for each x X each u x Sx and for all y X have there exists u y T y so that we d u x u y a d x y + b [d x u x + d y u y ] + c [d x u y + d y u x ]. Then F S = F T = {x }. Proof. Applying Theorem 3.1 for the function ϕ : R 3 + R + ϕt 1 t t 3 = at 1 + bt + ct 3 which satisfies the conditions i ii and iii of this theorem we obtain the conclusion. Remark 3.3. If T and S are singlevalued operators then Corollary 3. is Theorem 3 from [6]. Corollary 3.. Let X d be a complete metric space and P cl X multivalued operators. We suppose that there exists h ]0 1[ S T : X such that for each x X any u x Sx and for all y X there exists u y T y so that we have { d u x u y h max d x y d x u x + d y u y In these conditions F S = F T = {x }. d x u y + d y u x Proof. We apply Theorem 3.1 for the function ϕ : R 3 + R + ϕt 1 t t 3 = h max{t 1 t t 3 } which satisfies the conditions i ii and iii of this theorem. Remark 3.5. Corollary 3. is a generalization for multivalued operators of Theorem from [6] theorem proved for singlevalued operators in Hilbert spaces. Remark 3.6. Let X d be a complete metric space and T n n N be a sequence of multivalued operators T n : X P cl X n N. If each pair of multivalued operators T 0 T n for n N satisfies similar conditions as in Theorem 3.1 then F Tn = F T0 = {x } for all n N. We next give a generalization of Theorem 1 of N.Negoescu []. }. 79

8 Theorem 3.7. Let X d be a compact metric space S T : X P cl X and ϕ : R 3 + R +. Suppose that the following conditions are satisfies: i r s; r s R 3 + ϕr ϕs; ii ϕt t t < t t > 0; iii iv S or T be continuous; d u x u y < ϕ d x y dx u x dy u y dx u y dy u x for all x y X x y and for all u x u y Sx T x. fixed point. In these conditions: a. S or T has a strict fixed point; b. if both S and T have such fixed points then the pair S T has a common Proof. a. Let S be continuous and we consider the function fx := Dx Sx. Because f is continuous on X it follows that f takes its minimum value i.e. there exists x 0 X such that fx 0 = inf{fx x X}. point of T. We prove that x 0 is a fixed point of S or some x 1 Sx 0 is a fixed Indeed we choose: x 1 Sx 0 be such that dx 0 x 1 = Dx 0 Sx 0 ; x T x 1 be such that dx 1 x = Dx 1 T x 1 ; x 3 Sx be such that dx x 3 = Dx Sx. We shall prove that Dx 0 Sx 0 = 0 or Dx 1 T x 1 = 0 i.e. x 0 Sx 0 or x 1 T x 1. We suppose that Dx 0 Sx 0 > 0 and Dx 1 T x 1 > 0. Hence using the inequality iv we have: d x 1 x < ϕ d x 0 x 1 dx 0 x 1 dx 1 x dx 0 x dx 1 x 1 { } max d x 0 x 1 dx 0 x 1 dx 1 x := M. Consequently we distinguish the following situations: I. Case M = d x 0 x 1. In this case we deduce dx 1 x < dx 0 x 1. II. Case M = dx 0 x 1 dx 1 x. In this case we have d x 1 x < dx 0 x 1 dx 1 x. Since dx 1 x = Dx 1 T x 1 > 0 it follows that dx 1 x < 80

9 COMMON FIXED POINT THEOREMS dx 0 x 1. Now d x 3 x < ϕ d x x 1 dx x 3 dx 1 x dx x dx 1 x 3 { } max d x 1 x dx x 3 dx 1 x. Analogously it follows that d x x 3 < d x 1 x or d x x 3 < dx x 3 dx 1 x. In the second situations if dx x 3 = 0 we obtain a contradiction. Thus it follows that dx x 3 < dx 1 x. Similarly we deduce successively: Dx Sx = dx x 3 < dx 1 x < dx 0 x 1 = fx 0 which contradict the minimality of fx 0. Therefore Dx 0 Sx 0 = 0 or Dx 1 T x 1 = 0. So x 0 Sx 0 or x 1 T x 1. b. We assume that there exist u Su and v T v such that u v. Then using the hypothesis iv we get again a contradiction: d u v < ϕ d u v du u dv v d u v d u v. So u=v meaning that u is a common fixed point of S and T. Remark 3.8. If ϕ : R 3 + R + ϕt 1 t t 3 = max{t 1 t t 3 } from Theorem 3.7 we get a result of Negoescu [ Theorem 1]. References Remark 3.9. We note that Theorem 3.7 is true for S = T : X P cl X. [1] I. Kubiaczyk Fixed points for contractive correspondences Demonstratio Math [] N. Negoescu Observations sur des paires d applications multivoques d un certain type de contractivité Bul. Instit. Politehnic Iaşi fasc [3] V. Popa Common fixed points of sequence of multifunctions Babes-Bolyai Univ. Seminar on Fixed Point Theory Preprint nr [] I.A. Rus Generalized contractions Babes-Bolyai Univ Seminar on Fixed Point Theory Preprint nr [5] I.A. Rus A. Petruşel A. Sîntămărian Data dependence of the fixed points set of c- multivalued weakly Picard operators Mathematica Cluj-Napoca to appear. [6] T. Veerapandi S.A. Kumar Common fixed point theorems of a sequence of mappings in Hilbert space Bull. Cal. Math. Soc Carol I High School Sibiu 00 Sibiu ROMANIA address: aurelmuntean@yahoo.com 81

10 Received: