THEOREMS HARDY-ROGERS-TYPE IN K-METRIC SPACES. Paunović Lj. University of Pristina, Faculty of Teacher Education in Prizren-Leposavic,Serbia The purpose of the present paper is to establish coincidence point theorem for two mappings and fixed point theorem for one mapping in K-metric spaces which satisfy contractive conditions of Hardy-Rogers type. Results generalize fixed point theorems from K-metric spaces to symmetric spaces and metric spaces.Two examples are given to illustrate the usability of results. Introduction and preliminaries Let E be a real Banach space. A subset P of E is called a cone if and only if (a) P is closed, nonempty and P ; (b) a, b R, a, b 0 , and x, y P imply ax by P; (c) P ( P) {}. Given a cone P E , we define a partial ordering with respect to P by x y if and only if y x P , while x y whill stand for y x int P (interior of P). A cone P E , is called normal if there is a number K 0 such that for all x, y E , x y implies x K y . (1) The least positive number satisfying the above inequality is called the normal constant of P. It is clear that K 1 . Most of ordered Banach spaces used in applications posses a cone with the normal constant K 1 , and if this is the case, proofs of the corresponding results are much alike as in the metric setting. If K 1 , this is not the case. We know that there exists ordered Banach space E with cone P which is not normal but with int P . The cone P is called regular if every increasing sequence in E which is bounded from above is convergent. Every regular cone is normal, but the converse is not true. Definition 1.1. Let X be a nonempty set. Suppose that the mapping d : X X E satisfies (d1) d ( x, y ) for all x, y X and d ( x, y ) if and only if x y ; (d2) 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. Definition 1.2. Let (X,d) be a cone metric space. We say that {x n } is (1) a Cauchy sequence if for every c in E with c , there is an N such that for all n, m N , d ( xn , xm ) c ; (2) a convergent sequence if for every c in E with c , there is an N such that for all n>N, d ( xn , x) c for some fixed x in X. A cone metric space X is said to be complete if every Cauchy sequence in X is convergent in X. For the given cone metric space (X,d), one can construct a symmetric space (X,D) where “symmetric” D : X X R is given by D( x, y) d ( x, y) . In the case when (X,d) is cone metric space with a normal cone P, then (1) implies D( x, y) K D( x, z) D( z, y) , K is a normal constant of P. x, y , z X Definition 1.3. Let (X,d) be a cone metric space and P a cone with nonempty interior. Suppose that the mappings f , g : X X are such that the range of g contains the range of f, and f(X) or g(X) is a complete subspace of X. In this case we will say that the pair (f,g) is Abbas and Jungck’s pair, or shortly AJ’s pair. Definition 1.4. Let f and g be self-maps of a set X. If fx gx for some x in X, then x is called a coincidence point of f and g, and is called a point of coincidence of f and g. Self-maps f and g are said to be weakly compatible if they commute at their coincidence point, that is, if fx gx for x X , then fgx gfx. Proposition 1.1. Let f and g be weakly compatible self-maps of a set X. If f and g have a unique point of coincidence fx gx , then is the unique common fixed point of f and g. Main results Theorem 2.1. Let (X,d) be a complete cone metric space, P a normal cone with normal constant K. Suppose that (f,g) is AJ’s pair, and that there exist nonnegative constants a i , i 1,5 satisfying 3 i 1 ai K (a4 a5 ) 1 such that, for every x, y X , (2) D( fx, fy) a1 D( gx, gy) a2 D( gx, fx) a3 D( gy, fy) a4 D( gx, fy) a5 D( gy, fx) Then, f and g have a unique coincidence point in X. Moreover, if f and g are weakly compatible, f and g have a unique common fixed point. Proof. Let x0 X be arbitrary and let x1 X be chosen such that y0 f ( x0 ) g ( x1 ) . f ( X ) g ( X ) . Let This can be done, since be such that x2 X y1 f ( x1 ) g ( x2 ). Continuing this process, having chosen x n X , we choose x n 1 in X such that y n f ( xn ) g ( xn1 ). We have to show that D( y n , y n1 ) D( y n1 , y n ) for some [0,1), n 1. (3) From D( y n , y n1 ) D( fxn , fxn1 ) a1 D( gxn , gxn1 ) a2 D( gxn , fxn ) a3 D( gxn1 , fxn1 ) a4 D( gxn , fxn1 ) a5 D( gxn1 , fxn ) a1 D( y n1 , y n ) a2 D( y n1 , y n ) a3 D( y n , y n1 ) a4 D( y n1 , y n1 ) a 5 D( y n , y n ) (a1 a2 Ka4 ) D( y n1 , y n ) (a3 Ka4 ) D( y n , y n1 ), and from D( y n1 , y n ) D( fxn1 , fxn ) a1 D( gxn1 , gxn ) a2 D( gxn1 , fxn1 ) a3 D( gxn , fxn ) a4 D( gxn1 , fxn ) a5 D( gxn , fxn1 ) a1 D( y n , y n1 ) a2 D( y n , y n1 ) a3 D( y n1 , y n ) a4 D( y n , y n ) a5 D( y n1 , y n1 ) (a1 a3 Ka5 ) D( y n1 , y n ) (a2 Ka5 ) D( y n , y n1 ), we obtain 2D( y n1 , y n ) 2a1 a2 a3 K (a4 a5 )D( y n , y n1 ) a2 a3 K (a4 a5 )D( y n1 , y n ), that is, 2a a a3 K (a4 a5 ) n 1,2,... D( y n1 , y n ) D( y n , y n1 ), 1 2 1, 2 (a2 a3 K (a4 a5 )) Further, (3) implies that D( y n , y n 1 ) D( y n 1 , y n 2 ) ... n 1 D( y1 , y 0 ). Now we shall show that have (4) { y n } is a Cauchy sequence. By the triangle inequality, for n>m we d ( y n , y m ) d ( y n , y n1 ) d ( y n1 , y n2 ) ... d ( y m1 , y m ) Hence, as P is a normal cone, we have D( yn , ym ) d ( yn , ym ) K d ( yn , yn1 ) d ( yn1 , yn2 ) ... d ( ym1 , ym ) K d ( yn , yn1 d ( yn1 , yn2 ) ... d ( ym1 , ym ) = KD( y n , y n1 ) KD( y n1 , y n2 ) ... KD( y m1 , y m ). Now by (4), it follow that D( y n , y m ) K (n 1 n 2 ... m ) D( y1 , y 0 ) Km D( y1 , y 0 ) 0, as m . 1 From [1] follows that { y n } { fxn } {gxn1} is a Cauchy sequence. Since g(X) is complete, there exists a q in g(X) such that y n q as n . Consequently, we can find p in X such that g(p) = q. We shall show that f(p) = q. Substituting x = p, y = x n in (2), we get D( fp, fxn ) a1 D(q, q) a2 D( gp, fp) a3 D( gxn , fxn ) a4 D( gp, fxn ) a5 D( gxn , fp). According to [1], it follows D( fp, q) a1 D(q, q) a2 D(q, fp) a3 D(q, q) a4 D(q, q) a5 D(q, fp) (a2 a5 ) D( fp, q) D( fp, q), because a2 a5 i 1 ai K (a4 a5 ) 1. Now, if we suppose that fp q , then we have 3 a contradiction. Hence, gp = fp = q. We shall show that f and g have a unique point of coincidence. For this, assume that there exists another point of coincidence q1 q in X such that fp1 gp1 q1 . Now, D(q, q1 ) D( fp, fp1 ) a1 D( gp, gp1 ) a2 D( gp, fp) a3 D( gp1 , fp1 ) a4 D( gp, fp1 ) a5 D( gp1 , fp) a1 D(q, q1 ) a2 D(q, q) a3 D(q1 , q1 ) a4 D(q, q1 ) a5 D(q1 , q) (a1 a4 a5 ) D(q1, q) D(q1, q). As a1 a4 a5 i 1 ai K (a4 a5 ) 1 , we get D(q, q1 ) 0 , that is, q q1 . From the 3 Proposition 1.1, it follows that f and g have a unique common fixed point. Theorem 2.2. Let (X,d) be a sequentially compact cone metric space, P a regular cone and f : X X a continuous mapping such that (5) d ( fx, fy) a1d ( x, y) a2 d ( x, fx) a3 d ( y, fy) a4 d ( x, fy) a5 d ( y, fx) for all x, y X , where ai [0,1) , i 1,5 and x y 5 i 1 ai 1. Then, f has a unique fixed point. In order to prove Theorem 2.2., we shall need the following lemma. Lemma 2.1. Let (X,d) be a cone metric space, f : X X a mapping satisfying (5) for all x, y X , x y where ai [0,1) , i 1,5 and 5 i 1 ai 1. Then d ( f 2 x, fx) d ( x, fx) , for each x X with x fx . Proof. Putting y fx in (5), we have d ( fx, f 2 x) a1 d ( x, fx) a 2 d ( x, fx) a3 d ( fx, f 2 x) a 4 d ( x, f 2 x) a5 d ( fx, fx) (a1 a 2 a 4 )d ( x, fx) (a3 a 4 )d ( fx, f 2 x), that is, a1 a2 a4 (6) d ( x, fx) . 1 a3 a 4 By symmetry in (5), we have (7) d ( fy, fx) a1d ( y, x) a2 d ( y, fy) a3 d ( x, fx) a4 d ( y, fx) a5 d ( x, fy) i.e., putting y fx in (7), we obtain d ( f 2 x, f x) a1 d ( fx, x) a 2 d ( fx, f 2 x) a3 d ( x, f x) a 4 d ( fx, f x) a5 d ( x, f 2 x) d ( fx, f 2 x) (a1 a3 a5 )d ( x, fx) (a 2 a5 )d ( fx, f 2 x), that is, d ( f 2 x, fx) a1 a3 a5 d ( x, fx). 1 a 2 a5 (8) a a 2 a4 a1 a3 a5 k min 1 , [0,1), then 1 a3 a 4 1 a 2 a5 d ( fx, f 2 x) kd( x, fx) d ( x, fx). If (9) Proof of the Theorem 2.2. First, if u and v are two different fixed points of f, according to (5), we have d (u, v) d ( fu, fv) a1d (u, v) a2 d (u, fu) a3 d (v, fv) a4 d (u, fv) a5 d (v, fu) a1d (u, v) a4 d (u, v) a5 d (u, v) (a1 a4 a5 )d (u, v) d (u, v). This is a contradiction. Let x0 X . We define the sequence x n f n x0 , n 0,1,2,... If xn1 xn for some n, then x n is a fixed point of f. Suppose that xn1 xn for each n. Then, we have d ( x n , x n 1 ) d ( f n x0 , f a1 d ( f n 1 a4 d ( f n 1 n 1 x0 ) d ( ff n1 x0 , ff n x0 ) x0 , f n x0 ) a 2 d ( f x0 , f n 1 n 1 x0 , f n x0 ) a3 d ( f n x0 , f n 1 x0 ) x0 ) a5 d ( f x0 , f x0 ) n n a1d ( xn1 , xn ) a2 d ( xn1 , xn ) a3 d ( xn , xn1 ) a4 d ( xn1 , xn ) a 4 d ( xn , xn1 ), that is, d ( x n 1 , x n ) d ( f n 1 x0 , f n x0 ) d ( ff n x0 , ff n 1 x0 ) a1 d ( f n x0 , f n 1 x0 ) a 2 d ( f n x0 , f a 4 d ( f n x0 , f n x0 ) a5 d ( f n 1 x0 , f n 1 x0 ) a3 d ( f n 1 n 1 x0 , f n x0 ) x0 ) a1d ( xn1 , xn ) a2 d ( xn , xn1 ) a3 d ( xn1 , xn ) a5 d ( xn1 , xn ) a5 d ( xn , xn1 ), Now, we obtain 2d ( xn , xn1 ) (2a1 a2 a3 a4 a5 )d ( xn1 , xn ) (a2 a3 a4 a5 )d ( xn , xn1 ), i.e. 2a a2 a3 a4 a5 a 1 d ( xn , xn1 ) 1 d ( xn1 , xn ) 1 d ( xn1 , xn ) d ( xn1 , xn ) . 2 a 2 a3 a 4 a5 a1 1 Hence, the sequence d n d ( xn , xn1 ) is strictly decreasing bounded below by . Since P is regular, there is d * E such that d n d * (n ). From the sequence compactness of (X,d), there are subsequence { x ni } of {x n } and x * X such that xni x * (i ). Since mappings f and f 2 are continuous, we have fxni fx* and f 2 xni f 2 x* . By using [1] , we have d ( fxni , xni ) d ( fx* , x * ) (i ) and d ( f 2 xni , fxni ) d ( f 2 x* , fx* ) It is obvious that (i ) and d ( fxni , xni ) d ni d * d ( fx* , x* ) d ( f 2 xni , fxni ) d ni 1 d * d ( f 2 x * , fx* ) (i ). (i ). (10) Now we shall prove that fx x . If fx x , then d 0. From (10) and according to Lema 2.1., it follows d * lim d ni 1 lim d ( f 2 xni , fxni ) d ( f 2 x * , fx* ) d ( fx* , x * ) d * . * i * * * * i We have a contradiction, so fx* x * . That is, x * is a fixed point of f. This completes the proof of Theorem 2.2. Literature 1. Huang L.G, Zhang X. Cone metric spaces and fixed point theorems of contractive mappings, J. Math. Anal. Appl. 332 (2) (2007) 1468-1476. 2. Abbas M, Jungck G. 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