Cirics fixed point theorem in a cone metric space

´ CIRI ´ C’S FIXED POINT THEOREM IN A CONE METRIC SP A CE BESSEM SAMET Abstract. In this pap er, we extend a fixed p oin t theorem due to ´ Ciri´ c to a cone metric space. 1. Introduction and preliminaries Many generaliza tions of the Ba nach contraction pr inciple [4] hav e b een consid- ered in the literature (see [1]-[3], [5]-[17]). Huang and Zhang [12] recently hav e introduced the conc e pt of cone metric spac e , where the set o f real n umbers is replaced b y an o rdered Banach space, and they hav e established some fixed p o in t theorems for contractive type mapping s in a no rmal cone metric spa ce. The study of fixed p oint theorems in such spaces is followed by some other mathematicians (see [1]- [3], [5], [13], [14], [16]). In this pap er, we extend a fixed p oint theorem due to ´ Ciri´ c ([8]-Theorem 2.5) to a co ne metr ic space. Before presenting our result, we start by r ecalling some definitions. Let E be a r eal Bana ch space a nd P a subset of E . P is ca lle d a co ne if and only if: (i) P is closed, nonempty , and P 6 = { 0 } . (ii) a, b ∈ R , a, b ≥ 0, x, y ∈ P ⇒ ax + b y ∈ P . (iii) x ∈ P and − x ∈ P ⇒ x = 0. Given a cone P ⊂ E , we define a partial order ing ≤ with resp ect to P b y: x ≤ y ⇔ y − x ∈ P. W e s hall write x < y to indicate tha t x ≤ y but x 6 = y , while x ≪ y will sta nd for y − x ∈ intP , where intP denotes the interior of P . The cone P is called nor mal if there is a n umber k > 0 such that for all x, y ∈ E , 0 ≤ x ≤ y ⇒ k x k ≤ k k y k , where k · k is the no rm in E . In this case, the num be r k is called the normal constant of P . Reza pour and Hamlbarani [16] prov ed that there are no no rmal cones with normal constant k < 1 and for each c > 1 there ar e cones with normal constant k > c . F or this re ason, in a ll this pap e r, we tak e k ≥ 1 . In the following we always suppo se E is a Banach s pa ce, P is a cone in E with int P 6 = ∅ and ≤ is partial ordering with respect to P . As it has been defined in [12], a function d : X × X → E is called a co ne metric on X if it s atisfies the following co nditions: (a) 0 < d ( x, y ) for all x, y ∈ X , x 6 = y and d ( x, y ) = 0 if and o nly if x = y . 2000 Mathematics Subje ct Classific ation. 54H25, 47H10, 34B15. Key wor ds and phr ases. ´ Ciri´ c’s theo rem; Cone metric space; Fixed p oin t. 1 2 BESSEM SAMET (b) d ( x, y ) = d ( y , x ) for a ll x, y ∈ X . (c) d ( x, y ) ≤ d ( x, z ) + d ( z , y ) for all x, y , z ∈ X . Then ( X , d ) is called a cone metric space. Let ( x n ) b e a sequence in X and x ∈ X . • If for every c ∈ E , c ≫ 0 there is N such that for all n > N , d ( x n , x ) ≪ c , then ( x n ) is said to be conv ergent to x and x is the limit of ( x n ). W e denote this by x n → x as n → + ∞ . • If for an y c ∈ E with 0 ≪ c , there is N such that for a ll n, m > N , d ( x n , x m ) ≪ c , then ( x n ) is called a Cauchy sequence in X . Let ( X, d ) b e a cone metric space. If e v ery Cauc hy sequence is conv ergent in X , then X is called a complete c one metric space. The following lemmas will be useful later. Lemma 1. (Huang and Zhang [12] ) L et ( X , d ) b e a c one metric sp ac e, P b e a normal c one. L et ( x n ) b e a se quenc e in X . Then ( x n ) c onver ges to x if and only if k d ( x n , x ) k → 0 as n → + ∞ . Lemma 2. (Huang and Zhang [1 2] ) L et ( X , d ) b e a c one metric sp ac e, ( x n ) b e a se quenc e in X . If ( x n ) is c onver gent, then it is a Cauchy se quenc e, t o o. Lemma 3. (Huang and Zhang [12] ) L et ( X , d ) b e a c one metric sp ac e, P b e a normal c one. L et ( x n ) b e a se quenc e in X . Then, ( x n ) is a Cauchy se quenc e if and only if k d ( x n , x m ) k → 0 as n, m → + ∞ . W e denote L ( E ) the s e t o f linea r b ounded opera tors on E , endo wed with the following no rm: k S k = sup x ∈ E ,x 6 =0 k S x k k x k , ∀ S ∈ L ( E ) . It is clear that if S ∈ L ( E ), we hav e: k S x k ≤ k S k k x k , ∀ x ∈ E . W e denote by I : E → E the identit y op erator, i.e ., I x = x, ∀ x ∈ X . If S ∈ L ( E ), we deno te b y S − 1 ∈ L ( E ) (if s uc h op erator exists) the op erator defined by: S − 1 S x = S S − 1 x = x, ∀ x ∈ E . 2. Fixed point theorem The ma in result of this pap er is the following. Theorem 1. L et ( X , d ) b e a c omplete c one metric sp ac e, P b e a normal c one with normal c onstant k ( k ≥ 1 ). Supp ose the mapping T : X → X satisfies the fol lowing c ontr active c ondition: d ( T x, T y ) ≤ A 1 ( x, y ) d ( x, y ) + A 2 ( x, y ) d ( x, T x ) + A 3 ( x, y ) d ( y , T y ) (2.1) + A 4 ( x, y ) d ( x, T y ) + A 4 ( x, y ) d ( y , T x ) , ´ CIRI ´ C’S FIXED P O I NT THEOREM IN A CONE M ETRIC SP ACE 3 for al l x, y ∈ X , wher e A i : X × X → L ( E ) , i = 1 , · · · , 4 . F urt her, assume that for al l x, y ∈ X , we have: ∃ α ∈ [0 , 1 / k ) | 4 X i =1 k A i ( x, y ) k + k A 4 ( x, y ) k ≤ α (2.2) ∃ β ∈ [0 , 1) | k S ( x, y ) k ≤ β (2.3) ( A 1 ( x, y ) + A 2 ( x, y ))( P ) ⊆ P (2.4) A 2 ( x, y )( P ) ⊆ P (2.5) A 4 ( x, y )( P ) ⊆ P (2.6) ( I − A 3 ( x, y ) − A 4 ( x, y )) − 1 ( P ) ⊆ P . (2.7) Her e, S : X × X → L ( E ) is given by: S ( x, y ) = ( I − A 3 ( x, y ) − A 4 ( x, y )) − 1 ( A 1 ( x, y ) + A 2 ( x, y ) + A 4 ( x, y )) , ∀ x, y ∈ X . Then, T has a un ique fix e d p oint. Pr o of. L e t x ∈ X b e ar bitrary and define the sequence ( x n ) n ∈ N ⊂ X by: x 0 = x, x 1 = T x 0 , · · · , x n = T x n − 1 = T n x 0 , · · · By (2.1), we get: d ( x n , x n +1 ) = d ( T x n − 1 , T x n ) ≤ A 1 ( x n − 1 , x n ) d ( x n − 1 , x n ) + A 2 ( x n − 1 , x n ) d ( x n − 1 , x n ) + A 3 ( x n − 1 , x n ) d ( x n , x n +1 ) + A 4 ( x n − 1 , x n ) d ( x n − 1 , x n +1 ) + A 4 ( x n − 1 , x n ) d ( x n , x n ) = ( A 1 ( x n − 1 , x n ) + A 2 ( x n − 1 , x n )) d ( x n − 1 , x n ) + A 3 ( x n − 1 , x n ) d ( x n , x n +1 ) + A 4 ( x n − 1 , x n ) d ( x n − 1 , x n +1 ) . Using the tria ngular inequalit y , w e g et: d ( x n − 1 , x n +1 ) ≤ d ( x n − 1 , x n ) + d ( x n , x n +1 ) , i.e., d ( x n − 1 , x n ) + d ( x n , x n +1 ) − d ( x n − 1 , x n +1 ) ∈ P. F rom (2.6), it follows that: A 4 ( x n − 1 , x n )[ d ( x n − 1 , x n ) + d ( x n , x n +1 ) − d ( x n − 1 , x n +1 )] ∈ P, i.e., A 4 ( x n − 1 , x n ) d ( x n − 1 , x n +1 ) ≤ A 4 ( x n − 1 , x n ) d ( x n − 1 , x n ) + A 4 ( x n − 1 , x n ) d ( x n , x n +1 ) . Then, we hav e: d ( x n , x n +1 ) ≤ ( A 1 ( x n − 1 , x n ) + A 2 ( x n − 1 , x n ) + A 4 ( x n − 1 , x n )) d ( x n − 1 , x n ) +( A 3 ( x n − 1 , x n ) + A 4 ( x n − 1 , x n )) d ( x n , x n +1 ) . Hence, ( I − A 3 ( x n − 1 , x n ) − A 4 ( x n − 1 , x n )) d ( x n , x n +1 ) ≤ ( A 1 ( x n − 1 , x n ) + A 2 ( x n − 1 , x n ) + A 4 ( x n − 1 , x n )) d ( x n − 1 , x n ) . Using (2.7), we get: (2.8) d ( x n , x n +1 ) ≤ S ( x n − 1 , x n ) d ( x n − 1 , x n ) . 4 BESSEM SAMET It is not difficult to see tha t under hypo theses (2.4), (2 .6 ) and (2 .7), we hav e: S ( x, y )( P ) ⊆ P , ∀ x, y ∈ X . Using this remar k , (2 .8) and pro ceeding by iterations, w e g et: d ( x n , x n +1 ) ≤ S ( x n − 1 , x n ) S ( x n − 2 , x n − 1 ) · · · S ( x 0 , x 1 ) d ( x 0 , x 1 ) , which implies by (2.3) that: k d ( x n , x n +1 ) k ≤ k k S ( x n − 1 , x n ) kk S ( x n − 2 , x n − 1 ) k · · · k S ( x 0 , x 1 ) kk d ( x 0 , x 1 ) k ≤ k β n k d ( x 0 , x 1 ) k . F or any positive integer p , we ha ve: d ( x n , x n + p ) ≤ p X i =1 d ( x n + i − 1 , x n + i ) , which implies that: k d ( x n , x n + p ) k ≤ k p X i =1 k d ( x n + i − 1 , x n + i ) k ≤ k 2 p X i =1 β n + i − 1 k d ( x 0 , x 1 ) k ≤ k 2 β n 1 − β k d ( x 0 , x 1 ) k . (2.9) Since β ∈ [0 , 1), β n → 0 as n → + ∞ . So from (2.9) it follows that the se q uence ( x n ) n ∈ N is Ca uc hy . Since ( X , d ) is co mplete, there is a p oint u ∈ X s uch that: (2.10) lim n → + ∞ d ( T x n , u ) = lim n → + ∞ d ( x n , u ) = lim n → + ∞ d ( x n , x n +1 ) = 0 . Now, us ing the co n tractive condition (2.1), we get: d ( T u , T x n ) ≤ A 1 ( u, x n ) d ( u, x n ) + A 2 ( u, x n ) d ( u, T u ) + A 3 ( u, x n ) d ( x n , x n +1 ) + A 4 ( u, x n ) d ( u, x n +1 ) + A 4 ( u, x n ) d ( x n , T u ) . By the triangular inequality , we hav e: d ( u, T u ) ≤ d ( u, x n +1 ) + d ( x n +1 , T u ) d ( x n , T u ) ≤ d ( x n , T x n ) + d ( T x n , T u ) . By (2.5) and (2.6), we get: A 2 ( u, x n ) d ( u, T u ) ≤ A 2 ( u, x n )( d ( u, x n +1 ) + d ( x n +1 , T u )) A 4 ( u, x n ) d ( x n , T u ) ≤ A 4 ( u, x n ) d ( x n , T x n ) + A 4 ( u, x n ) d ( T x n , T u ) . Hence, d ( T u , T x n ) ≤ A 1 ( u, x n ) d ( u, x n ) + ( A 2 ( u, x n ) + A 4 ( u, x n )) d ( u, x n +1 ) +( A 2 ( u, x n ) + A 4 ( u, x n )) d ( x n +1 , T u ) +( A 3 ( u, x n ) + A 4 ( u, x n )) d ( x n , x n +1 ) . Using (2.2), this ineq ua lit y implies that: k d ( T u, T x n ) k ≤ k α 1 − k α ( k d ( u, x n ) k + k d ( u, x n +1 ) k + k d ( x n , x n +1 ) k ) . ´ CIRI ´ C’S FIXED P O I NT THEOREM IN A CONE M ETRIC SP ACE 5 F rom (2.10), it follows immediately that: (2.11) lim n → + ∞ d ( T u , T x n ) = 0 . Then, (2.1 0 ), (2.11) a nd the uniqueness o f the limit imply that u = T u , i.e., u is a fixed p oint of T . So we prov ed that T has lea s t one fixed p oin t u ∈ X . Now, if v ∈ X is a no ther fixed p oint of T , by (2.1), we get: d ( u, v ) = d ( T u , T v ) ≤ A 1 ( u, v ) d ( u, v ) + 2 A 4 ( u, v ) d ( u, v ) , which implies that: k d ( u, v ) k ≤ k ( k A 1 ( u, v ) k + 2 k A 4 ( u, v ) k ) k d ( u, v ) k ≤ k α k d ( u, v ) k , i.e., (1 − k α ) k d ( u, v ) k ≤ 0 . Since 0 ≤ α < 1 / k , we get d ( u, v ) = 0, i.e., u = v . So the pro of of the theorem is complete.  Now, we will show that The o rem 2.5 of ´ Ciri´ c [8] is a par ticular ca se of Theorem 1. Corollary 1. L et ( X , d ) b e a c omple te met ric sp ac e and T : X → X b e a mapping satisfying the fol lowing c ontr active c ondition: d ( T x, T y ) ≤ a 1 ( x, y ) d ( x, y ) + a 2 ( x, y ) d ( x, T x ) + a 3 ( x, y ) d ( y , T y ) (2.12) + a 4 ( x, y )( d ( x, T y ) + d ( y , T x )) , for al l x, y ∈ X , wher e a i : X × X → [0 , + ∞ ) , i = 1 , · · · , 4 and 4 X i =1 α i ( x, y ) + α 4 ( x, y ) ≤ α for e ach x, y ∈ X and some α ∈ [0 , 1) . Then, T has a unique fixe d p oint. Pr o of. W e take E = R (with the usual norm) and P = [0 , + ∞ ). Then, ( X, d ) is a complete cone metric space and P is a normal cone with normal constant k = 1. F or each i = 1 , · · · , 4, we define A i : X × X → L ( E ) by: A i ( x, y ) : t ∈ R 7→ a i ( x, y ) t, for all x, y ∈ X . let us check now that all the required hypotheses of Theo rem 1 are satisfied. • Condition (2.12) implies that: d ( T x, T y ) ≤ A 1 ( x, y ) d ( x, y ) + A 2 ( x, y ) d ( x, T x ) + A 3 ( x, y ) d ( y , T y ) + A 4 ( x, y ) d ( x, T y ) + A 4 ( x, y ) d ( y , T x ) , for all x, y ∈ X . Then, condition (2.1) o f Theore m 1 is satisfied. • F or a ll i = 1 , · · · , 4, we hav e: k A i ( x, y ) k = a i ( x, y ) , ∀ x, y ∈ X . Then, 4 X i =1 k A i ( x, y ) k + k A 4 ( x, y ) k ≤ α, ∀ x, y ∈ X and c ondition (2.2) o f Theor em 1 is s atisfied. 6 BESSEM SAMET • F or a ll x, y ∈ X , we ha ve: S ( x, y ) t = a 1 ( x, y ) + a 2 ( x, y ) + a 4 ( x, y ) 1 − a 3 ( x, y ) − a 4 ( x, y ) t, ∀ t ∈ R . Then, for all x, y ∈ X , we ha ve: k S ( x, y ) k = a 1 ( x, y ) + a 2 ( x, y ) + a 4 ( x, y ) 1 − a 3 ( x, y ) − a 4 ( x, y ) . Since α ∈ [0 , 1), we hav e: a 1 ( x, y ) + a 2 ( x, y ) + a 4 ( x, y ) + αa 3 ( x, y ) + αa 4 ( x, y ) ≤ α, ∀ x, y ∈ X . Then, k S ( x, y ) k ≤ α, ∀ x, y ∈ X and c ondition (2.3) o f Theor em 1 holds with β = α . • Conditions (2.4), (2.5) and (2.6) are ea sy to check. • F or a ll x, y ∈ X , we ha ve: ( I − A 3 ( x, y ) − A 4 ( x, y )) − 1 s = s 1 − a 3 ( x, y ) − a 4 ( x, y ) , ∀ s ∈ R . Since a 3 ( x, y ) + a 4 ( x, y ) < 1 fo r all x, y ∈ X , then s ≥ 0 ⇒ ( I − A 3 ( x, y ) − A 4 ( x, y )) − 1 s ≥ 0 . Hence, co ndition (2.7) of Theore m 1 is satisfied. Now, we are able to apply Theorem 1 and then, T has a unique fixed p oint.  3. Open problem W e present the fo llowing o pen pr oblem. In hypothesis (2.2), w e ass umed that α ∈ [0 , 1 /k ), where k is the no rmal constant of the cone P . What can we s a y a bout the cas e when α ∈ [1 /k , 1) with k > 1? References [1] M . A bbas and G. 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