Revisiting of some outstanding metric fixed point theorems via E-contraction

Transcription

1 DOI: /auom An. Şt. Univ. Ovidius Constanţa Vol. 6(3),018, Revisiting of some outstanding metric fixed point theorems via E-contraction Andreea Fulga and Erdal Karapınar Abstract In this paper, we introduce the notion of α-ψ-contractive mapping of type E, to remedy of the weakness of the existing contraction mappings. We investigate the existence and uniqueness of a fixed point of such mappings. We also list some examples to illustrate our results that unify and generalize the several well-known results including the famous Banach contraction mapping principle. 1 Introduction and Preliminaries Throughout the manuscript, we denote N 0 := N {0} where N is the positive integers. Further, R represent the real numbers and R + 0 := [0, ). Let Ψ be the family of nondecreasing functions ψ : [0, ) [0, ) satisfying the following conditions: (Ψ 1 ) ψ(t) < t, for any t > 0. (Ψ ) ϕ is continuous at 0; In the literatures, there are well-known examples for such functions, such as, comparison functions [3, 5] and (c)-comparison functions [3, 5]. Popescu [] suggested the concept of α-orbital admissible as a refinement of the alpha-admissible notion, defined in [9, 11]. Key Words: α-ψ-contractive mapping of type, Fixed point, metric space. 010 Mathematics Subject Classification: Primary 46T99, 47H10; Secondary 54H5. Received: Accepted:

2 E-CONTRACTION 74 Definition 1.1. [] Let T : X X be a mapping and α : X X [0, ) be a function. We say that T is an α-orbital admissible if If the additional condition, α(x, T x) 1 α(t x, T x) 1. α(x, y) 1 and α(y, T y) 1 α(x, T y) 1, is fulfilled, then the α-admissible mapping T is called triangular α-orbital admissible. Notice that each α-admissible mapping is an α-orbital admissible. For more details and counter examples, see e.g. [1, 4, 5, 11, 1, 16, ]. In this manuscript, we define a new notion, α-ψ-contractive mapping of type E, and derive some existence and uniqueness fixed point theorems for such mappings. Our results remove some weakness of the existing fixed point theorems including the initial metric fixed point theorem, Banach contraction mapping principle. We list some immediate consequence and examples that illustrates our results. Main results Definition.1 (cf. [8]). Let (X, d) be a metric space and T : X X be given mapping. We say that T is an α-ψ-contractive mapping of type E if there exist two functions α : X X [0, ) and ψ Ψ such that where α(x, y)d(t x, T y) ψ(e(x, y)), for all x, y X, (1) E(x, y) = d(x, y) + d(x, T x) d(y, T y) () The following lemma is a standard argument to prove that the given sequence is Cauchy Lemma.1. (See e.g. [3]) Let (X, d) be a metric space and let {x n } be a sequence in X such that d(x n+1, x n ) is nonincreasing and lim n d(x n+1, x n ) = 0. If {x n } is not a Cauchy sequence, then there exist an ε > 0 and two sequences {m k } and {n k } of positive integers such that the following four sequences tend to ε when k : d(x mk, x nk ), d(x mk +1, x nk +1), d(x mk 1, x nk ), d(x mk, x nk 1), d(x mk 1, x nk 1) Theorem.1. Let (X, d) be a complete metric space. Let T : X X be α-ψ-contractive mapping of type E satisfying the following conditions:

3 E-CONTRACTION 75 (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, T x 0 ) 1 ; (iii) T is continuous. Then, there exists a fixed point x such that T x = x. Proof. By assumption (ii), there exists a point x 0 X such that α(x 0, T x 0 ) 1. We construct an iterative sequence {x n } such that x n = T x n 1 for all n N. Owing to the fact that T is α orbital admissible, we derive α(x 0, x 1 ) = α(x 0, T x 0 ) 1 α(t x 0, T x 1 ) = α(x 1, x ) 1. Recursively, we have α(x n, x n+1 ) 1, for all n N 0. (3) Since T is triangular α orbital admissible, we find from (3) that α(x n, x n+1 ) 1 and α(x n+1, x n+ ) 1 α(x n, x n+ ), for any n N. Inductively, we conclude that α(x n, x n+j ) 1, for all n, j N. (4) If x n0 = x n0+1 = T x n0 for some n 0 N 0, then x = x n0 forms a fixed point for T that the proof finishes. Hence, from now on, we assume that x n x n+1 for all n N 0. (5) We shall prove that the sequence {d(x n, x n+1 )} is monotone. By taking x = x n and y = x n+1 in the inequality (1) and by regarding (3) and (5), we obtain d(x n, x n+1 ) = d(t x n 1, T x n ) α(x n 1, x n )d(t x n 1, T x n ) ψ(e(x n 1, x n )) = ψ(d(x n 1, x n ) + d(x n 1, T x n 1 ) d(x n, Sx n ) ) (6) ψ(d(x n 1, x n ) + d(x n 1, x n ) d(x n, x n+1 ) ) Suppose that d(x n, x n+1 ) d(x n 1, x n ). In this case, the inequality (6), becomes

4 E-CONTRACTION 76 d(x n, x n+1 ) ψ(d(x n 1, x n ) d(x n 1, x n ) + d(x n, x n+1 )) = ψ(d(x n, x n+1 )) < d(x n, x n+1 ), a contradiction. Hence, we deduce that d(x n, x n+1 ) < d(x n 1, x n ) for each n. Moreover, we can derive an estimation from (6) that d(x n, x n+1 ) ψ(e(x n 1, x n )) = ψ(d(x n 1, x n ) d(x n, x n+1 )) ψ(d(x n 1, x n )) < d(x n 1, x n ), for each n. Since the sequence {d(x n, x n+1 )} decreasing and bounded from below, we conclude that it converges to some nonnegative number d 0, that is, lim d(x n, x n+1 ) = d. n Notice also that lim E(x n, x n+1 ) = d. n We claim that d = 0. Suppose, on the contrary, that d > 0. Taking lim sup of the inequality (7), and taking the basic condition (Ψ 1 ) into account, we get that d ψ(d) < d, a contradiction. Hence, we conclude that (7) lim d(x n, x n+1 ) = 0. (8) n We shall proved that the sequence {x n } is Cauchy. Suppose, on the contrary, that there exist ε > 0 and sequences {x n (k)},{x m (k)} of positive integers such that n(k) > m(k) > k and {d(x n(k), x m(k) ) ε, d(x n(k) 1, x m(k) ) < ε, for all k N. Due to Lemma.1, we have lim d(x n(k), x m(k) ) = lim d(x n(k) 1, x m(k) 1 ) = ε (9) k k On the other hand, from (4), we have α(x n(k) 1, x m(k) 1 ) 1. Thus, from (1) and the assumption (ii), we deduce that ε d(x n(k), x m(k) ) = d(t x n(k) 1, T x m(k) 1 ) αd(x n(k) 1, x m(k) 1 )d(t x n(k) 1, T x m(k) 1 ) (10) ψ(e(x n(k) 1, x m(k) 1 ))

5 E-CONTRACTION 77 Since E(x n(k) 1, x m(k) 1 ) = d(x n(k) 1, x m(k) 1 )+ d(xn(k) 1, x T n(k) 1 ) d(x m(k) 1, x T m(k) 1 ) using (9) and (8) we obtain that = d(x n(k) 1, x m(k) 1 )+ d(x n(k) 1, x n(k) ) d(x m(k) 1, x m(k) ) lim E(x n(k) 1, x m(k) 1 ) = ε. (11) k Finally, letting k in (10) and using (9) and (11), we get On account of (Ψ 1 ), the inequality (1) turns into ε ψ(ε). (1) ε ψ(ε) < ε, which is a contradiction. Hence, we find that ε = 0. Therefore, x n is a Cauchy sequences. By completeness of (X, d), the sequence x n converges to some point x X as n. From the continuity of T, it follows that x n+1 = T x n T x ) as n. By the uniqueness of the limit, we get x = T x, that is, x is a fixed point of T. We say that a complete metric space (X, d) is regular if {x n } is a sequence in X such that α(x n, x n+1 ) 1 for all n and x n x X as n, then there exists a subsequence x n(k) of x n such that α(x n(k), x) 1 for all k. As it is known well, the continuity condition is a very heavy condition. Like in [9], we realize that we can replace the continuity of the operator T by a regularity condition on a complete metric space (X, d). Theorem.. Let (X, d) be a complete metric space. Let T : X X be α-ψ-contractive mapping of type E satisfying the following conditions: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, T x 0 ) 1 ; (iii) (X, d) is regular. Then, there exists a fixed point x such that T x = x.

6 E-CONTRACTION 78 Proof. Following the proof of Theorem.1, we know that {x n } is a Cauchy sequence in the complete metric space (X, d). Then, there exists x X such that x n x as n. By the hypothesis (iii) we deduce that there exists a subsequence x n(k) of x n such that α(x n(k), x ) 1 for all k. Therefore, we have where d(x n(k), T x ) = d(t x n(k) 1, T x ) α(x n(k) 1, x )d(t x n(k) 1, T x ) ψ(e(x n(k) 1, x )) E(x n(k) 1, x ) = d(x n(k) 1, x ) + d(xn(k) 1, T x n(k) 1 ) d(x, T x ) (13) Now, we shall show that T x = x. Suppose, on the contrary, that T x x that is, d(x, T x ) > 0. Letting k in the above inequality, and taking (Ψ 1 ) into account, we find that 0 < d(x, T x ) ψ(d(x, x )) < d(x, x ), which yields d(x, T x ) = 0, a contradiction. Therefore T x = x. The assure the uniqueness of the fixed point, we will consider the following hypothesis (U) for all x y X, there exists z X such that α(x, z) 1, α(y, z) 1 and α(z, T z) 1 Theorem.3. The fixed point x of T, in Theorem.1 (resp. Theorem.), is unique, if assume an additional condition (U). Proof. If we consider that z = x 0 we obtain that α(x 0, T x 0 ) 1, so even hypotheses (ii), from Theorem.1 (resp. Theorem.) so we obtain that x is a point fixed of T, where x = lim n x n = lim n T n x. Suppose that x and y are two fixed points of T such that x = y. Then, from (U) there exists z X such that α(x, z) 1, α(y, z) 1 and α(z, T z) 1. Since T is a triangular α-orbital admissible, we get that α(x, T n z) 1 and α(y, T n z) 1. Thus, from (1) we have d(x, T n+1 z) = d(t x, T (T n z)) α(x, T n z)d(t x, T (T n z)) ψ(e(x, T n z)) < E(x, T n z) (14)

7 E-CONTRACTION 79 This imply that d(x, T n+1 z) < d(x, T n z) + d(x, T x ) d(t n x, T n+1 x) By Theorem. we deduce that the sequence T n z converges to a fixed point z of T. Letting n in the above inequality, we get d(x, z ) < d(x, z ). This implies that d(x, z ) = 0 so x = z. Similarly, we get y = z. Hence, x = y, which is a contradiction..1 Examples Now, we shall consider some examples that illustrate and support our main results. Example.1. Let X = R. Consider the self mapping T : X X such that 3x 8 if x [0, 1], T x = x otherwise Define a metric d : X X R + x y 0 as d(x, y) = Define α : X X R + 0 such that { 4 if (x, y) [0, 1] α(x, y) = 0 otherwise. Notice that the self-mapping T is continuous. Further, several well-known contraction types does not hold for (x, y) [0, 1], that is, d(t x, T y) = 3 x y 16 and hence α(x, y)d(t x, T y) = 3 x y, 4 but d(x, y) = x y. Hence, there is no k [0, 1) or ψ Ψ such that the conditions of α-ψ-contractive mapping or Banach contraction mapping principle are fulfilled. On the other hand, for all x, y [0, 1], one can easily derive that E(x, y) = x y + 1 x 3x 8 y 3y 8 = 13 x y, for all x, y [0, 1], 16 which yields that T : X X is an α-ψ-contractive mapping of type E with ψ(t) = at with a , for example a = 6. Indeed, we have for x, y [0, 1] α(x, y)d(t x, T y) = 3 x y 4 5 x y 3 = 5E(x, y) 6 = ψ(e(x, y)).

8 E-CONTRACTION 80 Note that if x, y R\[0, 1], then the result is provided easily from the fact that α(x, y) = 0. Let us check that T is α-orbital admissible: α(x, T x) = α(x, x 4 ) 1 α(t x, T x) = α( x 4, x 16 ) 1. As a second step, let us prove that T is triangular α-orbital admissible: α(x, y) 1 and α(y, T y) = α(y, y 4 ) 1 α(x, T y) = α(x, y 4 ) 1, Thus, the first condition (i) of Theorem.1 is satisfied. The second condition (ii) of Theorem.1 is also fulfilled. Indeed, for x 0 = 0, we have α(0, T 0) = α(0, 0) = 5 1. Thus, all conditions of Theorem.1 are satisfied. Here, T 0 = 0 is the fixed point of T. Example.. Let X = A R + 0 T : X X such that T x = where A = {a, b, c}. Consider the self mapping x + 1 if x R + 0, c if x = a, b if x {b, c}. Define a metric d : X X R + 0 as x y if x, y R + 0, 1 if (x, y) {(a, b), (b, a)} 3 if (x, y) {(a, c), (c, a)}, d(x, y) = if (x, y) {(b, c), (c, b)}, 0 if (x, y) {(a, a), (b, b), (c, c)}, 0 if (x, y) A R + 0 R+ 0 A. Define α : X X R + 0 such that { 1 if (x, y) {(a, c), (a, b), (b, b), (c, b)} α(x, y) = 0 otherwise. It is easy to get that E(a, b) = = 4, E(a, c) = = 4, E(b, c) = + 0 = 4. From the above calculations, we can easily conclude that T : X X is an α- ψ-contractive mapping of type E, for ψ(t) = t. Indeed, we have the following possibilities: α(a, b)d(t a, T b) = E(a,b) =, α(a, c)d(t a, T c) = E(a,c) = 4 =, α(b, c)d(t b, T c) = 0 E(b,c) = 4 =.

9 E-CONTRACTION 81 Thus, the condition (1) is satisfied for all x, y A. Notice that the condition (1) is fulfilled trivially for x, y R + 0, since α(x, y) = 0 for x, y R+ 0. Let us check that T is α-orbital admissible: α(a, T a) = α(a, c) 1 α(t a, T a) = α(c, b) 1 α(b, T b) = α(b, b) 1 α(t b, T b) = α(b, b) 1 α(c, T c) = α(c, b) 1 α(t c, T c) = α(b, b) 1 As a second step, let us prove that T is triangular α-orbital admissible: α(a, b) 1 and α(b, T b) = α(b, b) 1 α(a, T b) = α(a, b) 1, α(a, c) 1 and α(c, T c) = α(c, b) 1 α(a, T c) = α(a, b) 1, α(b, b) 1 and α(b, T b) = α(b, b) 1 α(b, T b) = α(b, b) 1, α(c, b) 1 and α(b, T b) = α(b, b) 1 α(c, T b) = α(b, b) 1. Thus, the first condition (i) of Theorem. is satisfied. The second condition (ii) of Theorem. is also fulfilled. Indeed, for any x 0 A, we have α(x 0, T x 0 ) 1. It is also easy to see that (X, d) is regular. Indeed, whatever the initial point x 0 A is chosen, the sequence {x n } tends to b. By definition of the auxiliary function α, we have α(a, b) 1, α(b, b) 1 and α(c, b) 1. Thus, all conditions of Theorem. are provided. Notice that T b = b is the fixed point of T. Example.3. Let X = [ 0, 3] [1, 4] equipped with a metric d : X X R + 0 such that d(x, y) = x y. Consider the self mapping T : X X such that T x = We define α : X X R + 0 α(x, y) = x 3 if x [ 0, 3], 5 x 4 if x [1, 4]. such that 4 if x, y [ 0, 3 1 if x, y [1, 3] if (x, y) {(0, 4), (1, 4)}, 0 otherwise. First of all, we remark that, for x, y [ 0, 3], d(x, y) = x y 3 x y =, and d(t x, T y) = 1 x 6 3 y = 3 ] x y 6

10 E-CONTRACTION 8 Thus, we have 4 x y α(x, y)d(t x, T y) = 6 Hence, there is no ψ Ψ to provide that T is α ψ contraction. We have consider the following two cases: (i) for all x, y [ ] 0, 3 E(x, y) = x y + 1 x x 3 y y 3 5 x y =. 6 Consequently, we conclude that T is α ψ contraction type of E, with ψ(t) = at, where a > 4 5. In particular, for the choice of ψ(t) = 9 10t, we have α(x, y)d(t x, T y) = 4 x y 6 ψ( 9 9 x y ) = E(x, y) = ψ(e(x, y)) which yields that T : X X is an α-ψ-contractive mapping of type E with ψ(t) = 9 10 t. (ii) for x, y [1, 3], we have d(t x, T y) = 1 x 4 y = 4 x y 8 and d(x, y) = and α(x, y)d(t x, T y) = x y 8 Moreover, we find that E(x, y) = Hence, we conclude that α(x, y)d(t x, T y) = x y x y 8 x y + 1 x x 4 y y 4 7 x y = (iii) for x = 0 and y = 4, we have ( ) 7 x y 8 = 9 E(x, y) = ψ(e(x, y)). 10 d(t 0, T 4) = = 3 4 and d(0, 4) = 0 4 = and α(0, 4)d(T 0, T 4) = 6 4. Moreover, we find that E(0, 4) = = = 13 4.

11 E-CONTRACTION 83 Hence, we deduce that α(0, 4)d(T 0, T 4) = = 9 E(0, 4) = ψ(e(0, 4)). 10 (iv) for x = 1 and y = 4, we have d(t 1, T 4) = = 3 8 and d(1, 4) = 1 4 = 3 4 and α(1, 4)d(T 1, T 4) = 6 8. Moreover, we find that 1 4 E(1, 4) = = = Hence, we deduce that α(1, 4)d(T 1, T 4) = = 9 E(1, 4) = ψ(e(1, 4)). 10 Notice that for any other possibilities, the result is provided easily from the fact that α(x, y) = 0. Let us check that T is α-orbital admissible: (i) if x [ ] 0, 3 α(x, T x) = α(x, x 3 ) 1 α(t x, T x) = α( x 3, x 9 ) 1. (ii) if x [1, 3] then T x [1, 3] and α(x, T x) = 1 α(t x, T x) 1. As a second step, let us prove that T is triangular α-orbital admissible: (i) if x [ ] 0, 3 α(x, y) 1 and α(y, T y) = α(y, y 3 ) 1 α(x, T y) = α(x, y 3 ) 1, (ii) if x, y [1, 3] then T y [1, 3] and α(x, y) 1 and α(y, T y) = α(y, 5 y 4 ) 1 α(x, T y) = α(x, 5 y 4 ) 1, Thus, the first condition (i) of Theorem. is satisfied. The second condition (ii) of Theorem. is also fulfilled. Indeed, for x 0 = 0, we have α(0, T 0) = α(0, 0) = 1 1. It is also easy to see that (X, d) is regular. Indeed let x n be a sequence in X such that for all n and x n x as n. Since α(x n, x n+1 ) 1 for all n, by the definition of α, we have x n [0, 1] for all n and x [0, 1]. Then, α(x n, x) = 1.

12 E-CONTRACTION 84 Thus, all conditions of Theorem. are provided and so we derive that T has a fixed point. We shall show that the condition (U) are not satisfied. Thus, we could not guarantee the uniqueness of a fixed point. α(0, 4) 1, α(1, 4) 1 and α(4, T 4)α(4, 3 ) = 0, Notice that T 0 = 0 and T = are the fixed points of T. Example.4. Let X = R equipped with a metric d : X X R + 0 d(x, y) = x y. Consider the self mapping T : X X such that { x T x = 4 if x [0, 1], 0 otherwise such that We define α : X X R + 0 such that α(x, y) = { 5 if (x, y) [0, 1] 0 otherwise. Notice that the self-mapping T is not continuous at x = 1. Further, several well-known contraction types does not hold for (x, y) [0, 1], that is, d(t x, T y) = x y 8 and hence α(x, y)d(t x, T y) = 5 x y 8 but d(x, y) = x y. Hence, there is no ψ Ψ such that the conditions of α-ψ-contractive mapping is fulfilled. On the other hand, for all x, y [0, 1], one can easily derive that E(x, y) = x y + 1 x x 4 y y 4 7 x y =, for all x, y [0, 1], 8 which yields that T : X X is an α-ψ-contractive mapping of type E with ψ(t) = at with a , for example a = 14. Indeed, we have for x, y [0, 1] α(x, y)d(t x, T y) = 5 x y 8 13 x y 16 = 13E(x, y) 14 = ψ(e(x, y)). Note that if x, y R\[0, 1], then the result is provided easily from the fact that α(x, y) = 0. Let us check that T is α-orbital admissible: α(x, T x) = α(x, x 4 ) 1 α(t x, T x) = α( x 4, x 16 ) 1.

13 E-CONTRACTION 85 As a second step, let us prove that T is triangular α-orbital admissible: α(a, y) 1 and α(y, T y) = α(y, y 4 ) 1 α(x, T y) = α(x, y 4 ) 1, Thus, the first condition (i) of Theorem. is satisfied. The second condition (ii) of Theorem. is also fulfilled. Indeed, for x 0 = 0, we have α(0, T 0) = α(0, 0) = 5 1. Finally, it is also easy to see that (X, d) is regular. Indeed let x n be a sequence in X such that for all n and x n x as n. Since α(x n, x n+1 ) 1 for all n, by the definition of α, we have x n [0, 1] for all n and x [0, 1]. Then, α(x n, x) = 5 1. Thus, all conditions of Theorem. are provided. Notice that T 0 = 0 is the fixed point of T. Moreover, because for any z [0, 1], T z = z 4 [0, 1] we deduce that for all x, y [0, 1], x y there exist z [0, 1] such that α(x, z) 1, α(y, z) 1 and α(z, T z) 1, so condition (U) is satisfied. Thus, Theorem.3 guarantee the uniqueness of the fixed point. 3 Outcomes of the main results In this section, we shall list some basic consequences of the main results. 3.1 Outcomes of the main results in the setting of standard metric structure. Theorem 3.1. Let (X, d) be a complete metric space and ψ Ψ. Suppose that a continuous self-mapping T : X X satisfies where d(t x, T y) ψ(e(x, y)), for all x, y X, (15) E(x, y) = d(x, y) + d(x, T x) d(y, T y). Then, there exists a fixed point x such that T x = x. Proof. It is sufficient to take α(x, y) = 1 for all x, y X in Theorem.3. Theorem 3.. Let (X, d) be a complete metric space and k [0, 1). Suppose that a continuous self-mapping T : X X satisfies d(t x, T y) ke(x, y), for all x, y X, (16)

14 E-CONTRACTION 86 where E(x, y) = d(x, y) + d(x, T x) d(y, T y). Then, there exists a fixed point x such that T x = x. Proof. It is sufficient to take ψ(t) = kt in Theorem 3.1. In the following theorems, the continuity condition of the self-mapping is not necessary, since the contraction condition (17) and (18) imply the continuity of the mentioned self-mapping. Theorem 3.3. Let (X, d) be a complete metric space and ψ Ψ. Suppose that a self-mapping T : X X satisfies d(t x, T y) ψ(d(x, y)), for all x, y X. (17) Then, there exists a fixed point x such that T x = x. Proof. It follows from Theorem 3.1, by regarding the monotonicity of the function ψ, that is, d(t x, T y) ψ(d(x, y)) ψ(e(x, y)), for all x, y X. Theorem 3.4. Let (X, d) be a complete metric space and k [0, 1). Suppose that a self-mapping T : X X satisfies d(t x, T y) kd(x, y), for all x, y X. (18) Then, there exists a fixed point x such that T x = x. Proof. It follows from Theorem 3.3, by letting ψ(t) = kt. 3. Outcomes of the main results in the setting of metric spaces endowed with a partial order In the last decades, one of exciting research topics for the metric fixed point theory researchers is to investigate the existence and uniqueness of a fixed point of certain mapping in context of metric spaces endowed with partial orders, see e.g.[4, 18] We shall show that Theorem.3 infer various existing fixed point results on a metric space endowed with a partial order. For this purpose, we, first, recollect some basic concepts. Definition 3.1. For a partially ordered non-empty set (X, ), the self-mapping T : X X T is called nondecreasing with respect to if x, y X, x y = T x T y.

15 E-CONTRACTION 87 Definition 3.. A sequence {x n } in a partially ordered set (X, ) is called nondecreasing with respect to, if x n x n+1 for all n. Definition 3.3. Let (X, ) be a partially ordered set and d be a metric on X. We say that (X,, d) is regular if for every nondecreasing sequence {x n } X such that x n x X as n, there exists a subsequence {x n(k) } of {x n } such that x n(k) x for all k. Suppose that (X, ) is a partially ordered set and d be a metric on X. We say that (X, ) have a property of (S) if it fulfills the following condition (S) for all x, y X there exists z X such that x z and y z, For the simplicity, we shall use the notation (X, d, ) to represent the partially ordered set (X, ) equipped with a metric d. The triple (X, d, ) is called metric spaces endowed with a partial order. Theorem 3.5. Let (X, d, ) be a metric spaces endowed with a partial order, where (X, d) is complete. Let T : X X be a nondecreasing mapping with respect to. Suppose that there exists a function ψ Ψ such that for all x, y X with x y, where d(t x, T y) ψ(e(x, y), (19) E(x, y) = d(x, y) + d(x, T x) d(y, T y). Suppose also that the following conditions hold: (i) there exists x 0 X such that x 0 T x 0 ; (ii) T is continuous or (X,, d) is regular. Then, T has a fixed point. Moreover, if (X, ) have a property of (S), the observed fixed point is unique. Proof. Consider the mapping α : X X [0, ) as { 1 if x y or x y, α(x, y) = 0 otherwise. It is obvious that T T : X X is an α-ψ-contractive mapping of type E, that is, α(x, y)d(t x, T y) ψ(e T (x, y)), for all x, y X. From condition (i), the definition of α yields that α(x 0, T x 0 ) 1.

16 E-CONTRACTION 88 Moreover, for all x, y X, from the monotone property of T, we have α(x, y) 1 = x y or x y = T x T y or T x T y = α(t x, T y) 1. Consequently, T is α orbital admissible. For a last step, we examine the following cases: If T is continuous, the existence of a fixed point follows from Theorem.1. Suppose now that (X,, d) is regular. Let {x n } be a sequence in X such that α(x n, x n+1 ) 1 for all n and x n x X as n. Due to regularity, there is a subsequence {x n(k) } of {x n } such that x n(k) x for all k. Hence, we have α(x n(k), x) 1 for all k. So, the existence of a fixed point follows from Theorem.. For the uniqueness, let x, y X. By assumption of the theorem, there exists z X such that x z and y z, which yields that α(x, z) 1 and α(y, z) 1. Consequently, we conclude the uniqueness of the fixed point by Theorem.3. Theorem 3.6. Let (X, d, ) be a metric spaces endowed with a partial order, where (X, d) is complete. Let T : X X be a nondecreasing mapping with respect to. Suppose that there exists a k [0, 1) such that for all x, y X with x y, where d(t x, T y) ke(x, y), E(x, y) = d(x, y) + d(x, T x) d(y, T y). Suppose also that the following conditions hold: (i) there exists x 0 X such that x 0 T x 0 ; (ii) T is continuous or (X,, d) is regular. Then, T has a fixed point. Moreover, if (X, ) have a property of (S), the observed fixed point is unique. Theorem 3.7. Let (X, d, ) be a metric spaces endowed with a partial order, where (X, d) is complete. Let T : X X be a nondecreasing mapping with respect to. Suppose that there exists a function ψ Ψ such that d(t x, T y) ψ(d(x, y), for all x, y X with x y. Suppose also that the following conditions hold: (i) there exists x 0 X such that x 0 T x 0 ;

17 E-CONTRACTION 89 (ii) T is continuous or (X,, d) is regular. Then, T has a fixed point. Moreover, if (X, ) have a property of (S), the observed fixed point is unique. Theorem 3.8. Let (X, d, ) be a metric spaces endowed with a partial order, where (X, d) is complete. Let T : X X be a nondecreasing mapping with respect to. Suppose that there exists a k [0, 1) such that d(t x, T y) kd(x, y), for all x, y X with x y. Suppose also that the following conditions hold: (i) there exists x 0 X such that x 0 T x 0 ; (ii) T is continuous or (X,, d) is regular. Then, T has a fixed point. Moreover, if (X, ) have a property of (S), the observed fixed point is unique. 3.3 Outcomes of the main results in the setting of the cyclic contractive mappings Investigation of the existence and uniqueness of a fixed point of certain cyclic contractive mappings was initiated by Kirk, Srinivasan and Veeramani [17]. Following this pioneer paper [17], this trend has been appreciated by a number of authors (see e.g. [6, 13, 14] and the related references therein). Here, we shall indicate that our main result, Theorem.3, infer a fixed point theorems for cyclic contractive mappings. Theorem 3.9. Suppose that {A i } i=1 are nonempty closed subsets of a complete metric space (X, d) and T : Y Y is a given mapping with Y = A 1 A. If the the following conditions are fulfilled (I) T (A 1 ) A and T (A ) A 1 ; (II) there exists a function ψ Ψ such that d(t x, T y) ψ(e(x, y)), for all (x, y) A 1 A, where E(x, y) is defined as in (), then, T has a unique fixed point that belongs to A 1 A.

18 E-CONTRACTION 90 Proof. The proof consists of several steps. Step 1. the pair (Y, d) forms a complete metric space since A 1 and A are closed subsets of (X, d). Step. We shall indicate that T is an α ψ contractive mapping type of E. For this purpose, we specify the mapping α : Y Y [0, ) as { 1 if (x, y) (A1 A α(x, y) = ) (A A 1 ), 0 otherwise. Regarding (II) and α, we are able to write α(x, y)d(t x, T y) ψ(m(x, y)), for all x, y Y. In other words, T is an α ψ contractive mapping type of E. Step 3. We shall show that T is α admissible. Suppose that (x, y) Y Y with α(x, y) 1. For the case (x, y) A 1 A, from (I), (T x, T y) A A 1, which yields that α(t x, T y) 1. For the other case, (x, y) A A 1, again from (I), (T x, T y) A 1 A, which implies that α(t x, T y) 1. So, we find that α(t x, T y) 1 whenever α(x, y) 1. Step 4. We notice that for any a A 1, from (I), we get (a, T a) A 1 A, and hence α(a, T a) 1. Step 5. We shall show that (X, d) is regular. Suppose that {x n } is a sequence in X such that α(x n, x n+1 ) 1 for all n and x n x X as n. On account of the definition of the α mapping, we find (x n, x n+1 ) (A 1 A ) (A A 1 ), for all n. Since (A 1 A ) (A A 1 ) is a closed set with respect to the Euclidean metric, we derive that (x, x) (A 1 A ) (A A 1 ), which yields that x A 1 A. Consequently, we have α(x n, x) 1 for all n. Finally, assume that x, y Fix(T ). From (I), we find that x, y A 1 A. As a result, we conclude that α(x, z) 1 and α(y, z) 1, for any z Y. Thus, condition (U) is satisfied. Thus, all the hypotheses of Theorem.3 are fulfilled that guarantees the existence and uniqueness of a fixed point of T in A 1 A (from (I)). As an immediate outcome of Theorem 3.9 is the following: Theorem Suppose that {A i } i=1 are nonempty closed subsets of a complete metric space (X, d) and T : Y Y is a given mapping with Y = A 1 A. If the the following conditions are fulfilled (I) T (A 1 ) A and T (A ) A 1 ;

19 E-CONTRACTION 91 (II) there exists a function ψ Ψ such that d(t x, T y) ψ(d(x, y)), for all (x, y) A 1 A, where d(x, y) is defined as in (), then, T has a unique fixed point that belongs to A 1 A. Theorem Suppose that {A i } i=1 are nonempty closed subsets of a complete metric space (X, d) and T : Y Y is a given mapping with Y = A 1 A. If the the following conditions are fulfilled (I) T (A 1 ) A and T (A ) A 1 ; (II) there exists a k [0, 1) such that d(t x, T y) ke(x, y), for all (x, y) A 1 A, where E(x, y) is defined as in (), then, T has a unique fixed point that belongs to A 1 A. The following is the main result of [17]. Theorem 3.1. Suppose that {A i } i=1 are nonempty closed subsets of a complete metric space (X, d) and T : Y Y is a given mapping with Y = A 1 A. If the the following conditions are fulfilled (I) T (A 1 ) A and T (A ) A 1 ; (II) there exists a k [0, 1) such that d(t x, T y) kd(x, y), for all (x, y) A 1 A, then, T has a unique fixed point that belongs to A 1 A. 4 An Applications to the solutions of periodic boundary value problems of first order In this section we examine the existence and uniqueness of solutions of periodic boundary value problems of first order. Although these problems have been investigated under different conditions in [19], [0]-[], we suggest weaker condition for the existence and uniqueness conditions.

20 E-CONTRACTION 9 We consider X = C[0, T ] with the partial ordering x y x(t) y(t) for all t [0, T ] d(x, y) = sup{ x(t) y(t), t [0, T ]}. (0) equipped with the metric d(x, y) = sup{ x(t) y(t), t [0, T ]}. (1) The space (X, d, ) satisfies the condition (S). Indeed, it is obvious that for every pair x(t), y(t) in X, we have x(t) max{x(t), y(t)} and y(t) max{x(t), y(t)}. We shall discuss the following first order periodic boundary value problem { x (t) = f(t, x(t)), t [0, T ] () x(0) = x(t ). Definition 4.1. A lower solution of the problem () is a function x(t) C[0, T ] fulfilling { x (t) f(t, x(t)), t [0, T ] (3) x(0) x(t ). An upper solution the problem () is a function x(t) C[0, T ] R satisfying { x (t) f(t, x(t)), t [0, T ] (4) x(0) x(t ). Note that the problem () can be written as { x (t) + λx(t) = f(t, x(t)) + λx(t), t [0, T ] x(0) = x(t ). (5) This problem is equivalent to the integral equation x(t) = T 0 G(t, s)[f(s, x(s) + λx(s)]ds, (6) where G(t, s) is the Green function defined by λ(t +s t) e G(t, s) = e λt 1, 0 s < t T e λ(s t) e λt 1, 0 t < s T. (7) Now, we state a theorem for the existence and uniqueness of a solution of the problem (3).

21 E-CONTRACTION 93 Theorem 4.1. We take the periodic boundary value problem () into account. Suppose that f is continuous and that there exists λ > 0 such that for all x, y C[0, T ] satisfying x y, the following condition holds: 0 f(t, y(t)) + λy(t) f(t, x(t)) λx(t) ψ((y x)), (8) for some ψ Ψ, λ [0, ), such that 0 < ψ(u) < u < λ, for all u R +. If the problem () has a lower solution, then it has a unique solution. Proof. We set the map F : C[0, T ] R as follows F x(t) = T 0 G(t, s)[f(s, x(s) + λx(s)]ds, (9) where G(t, s) is the Green function given in (7). Then the solution of the problem () is the fixed point of F. Suppose that x y are functions in C[0, T ] R satisfying (8). Furthermore, since f fulfills (8), we get F x(t) = T 0 T 0 G(t, s)[f(s, x(s)) + λx(s)]ds G(t, s)[f(s, y(s)) + λy(s)]ds = F y(t), that is, F is nondecreasing. Let us examine T d(f y, F x) = sup G(t, s)[f(s, y(s)) + λy(s) f(s, x(s)) λx(s)]ds 0 (30) sup T 0 G(t, s)ψ( y(s) x(s) )ds ψ(d(x, y)) T 0 G(t, s)ds = ψ(d(x, y)) λ ψ(e(x, y)) λ where E(x, y) = d(x, y) + d(x, T x) d(y, T y). By selecting λ in a way that 0 < ψ(e(x, y)) < λ we conclude that the nondecreasing map F fulfills the condition (19) of Corollary 3.5. As a next step, we shall indicate that x 0 F x 0 for some x 0 X. Since the problem () has a lower solution, then, there exists x 0 X provides (3). Hence, we get that x 0(t) + λx 0 (t) f(t, x 0 (t)) + λx 0 (t), t [0, T ] x ( 0) x 0 (T ). Multiplying both sides by e λt and then integrating from 0 to t we derive x 0 (t)e λt x 0 (0) + t 0 (31) e λs [x 0 (s) + f(s, x 0 (s))]ds. (3)

22 E-CONTRACTION 94 By using the inequality x 0 (0) x 0 (T ) we get T x 0 (0)e λt x 0 (T )e λt x 0 (0) + e λs [x 0 (s) + f(s, x 0 (s))]ds, 0 or equivalently, x 0 (0) T Combining (3) and (33) we get 0 e λs e λt 1 [x 0(s) + f(s, x 0 (s))]ds. (33) x 0 (t) = + T 0 t 0 T 0 e λ(s t) e λt 1 [x 0(s) + f(s, x 0 (s))]ds e λ(s t) [x 0 (s) + f(s, x 0 (s))]ds G(t, s)[x 0 (s) + f(s, x 0 (s))]ds (34) where G(s, t) is the Green s function given in (8). So, we have x 0 (t) F x 0 (t) for the lower solution x 0 (t) of (). Then, by the Corollary 3.5 the map F has a unique fixed point. Accordingly, the boundary value problem () has a unique solution. Example 4.1. Let X = C[0, 1] with a partial order x y and a metric Consider the BVP d(x, y) = sup{ x(t) y(t), t [0, 1]}. { x (t) = x e t + 1, t [0, 1] x(0) = x(1). For this specific example the function fulfils the condition f(t, x) = x e t + x f(t, y(t)) + λy(t) f(t, x(t)) λx(t) k(y(t) x(t))

23 E-CONTRACTION 95 for all 0 < x y. Indeed, f(t, y(t)) + λy(t) f(t, x(t)) λx(t) = y e t λy ( x e t + 1) λx = [λ (y + x)](y x) [λ M](y x) ψ(y x) for ψ(t) = (λ M)t, where M = min t [0,1] {(x + y)et }. Observe that x 0 (t) = 0 is a lower solution of the BVP. Clearly, and x 0(t) = 0 x e t + 1, t [0, 1], x 0 (0) = 0 = x 0 (1). By the Theorem 4.1, the BVP has a unique solution. References [1] H.Aydi, M. Jellali, E. Karapınar, Common fixed points for generalized α-implicit contractions in partial metric spaces: Consequences and application, RACSAM - Revista de la Real Academia de Ciencias Exactas, Fisicas y Naturales. Serie A. Matemiticas, September 015, Volume 109, Issue, pp [] A. Amini-Harandi and H. Emami, A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations, Nonlinear Anal., 7, 38 4, (010). [3] V. Berinde, Contracţii generalizate şi aplicaţii, Editura Club Press, Baia Mare, [4] N. Bilgili, E. Karapınar, A note on common fixed points for (ψ, α, β)- weakly contractive mappings in generalized metric spaces, Fixed Point Theory Appl., 013 (013), Article ID 87. [5] SH. Cho, JS. Bae, E. Karapinar, Fixed point theorems for α-geraghty contraction type maps in metric spaces, Fixed Point Theory Appl. 013, Article ID 39 (013). [6] B.S. Choudhury, N. Metiya, Fixed point and common fixed point results in ordered cone metric spaces, Analele Universitatii Ovidius Constanta- Seria Matematica, Volume 0, Issue 1, Pages 557, ISSN (Online) , DOI: 01.

24 E-CONTRACTION 96 [7] J. Cringanu, D. Paşca, A fixed point method for a class of nonlinear equations involving a duality mapping defined on product spaces, Electronic Journal of Differential Equations, ISSN , 013. [8] A. Fulga and A. Proca, A new Generalization of Wardowski Fixed Point Theorem in Complete Metric Spaces, Advances in the Theory of Nonlinear Analysis and its Applications, Volume 1, Issue 1, Pages: 57-63, Year: 017, Article Id: 017:5 PDF [9] T. Gnana Bhaskar and V. Lakshmikantham, Fixed point theorems in partially ordered metric spaces and applications, Nonlinear Anal., 65, , (006). [10] J. Harjani and K. Sadarangani, Fixed point theorems for weakly contractive mappings in partially ordered sets, Nonlinear Anal., 71, , (009). [11] E. Karapınar, P. Kuman, P. Salimi, On α ψ-meri-keeler contractive mappings, Fixed Point Theory Appl. (013), 013:94. [1] E. Karapınar and B. Samet. Generalized (α ψ)-contractive Type Mappings and Related Fixed Point Theorems with Applications Abstract and Applied Analysis Volume 01 (01), Article ID , 17 pages [13] E. Karapınar, Fixed point theory for cyclic weak φ-contraction, Appl. Math. Lett. 4 (6) (011) [14] E. Karapınar, K. Sadaranagni, Fixed point theory for cyclic (φ-ψ)- contractions, Fixed Point Theory Appl. 011, 011:69. [15] E. Karapinar, I. M, Erhan and U. Aksoy Weak -contractions on partially ordered metric spaces and applications to boundary value problems Boundary Value Problems, 014, 014:149 [16] E. Karapınar, H.H. Alsulami and M. Noorwali, Some extensions for Geraghty type contractive mappings Journal of Inequalities and Applications 015, 015:303 (6 September 015) [17] W.A. Kirk, P.S. Srinivasan, P. Veeramani, Fixed points for mappings satisfying cyclical contractive conditions, Fixed Point Theory. 4(1) (003) [18] J.J. Nieto, R. Rodríguez-López, Contractive Mapping Theorems in Partially Ordered Sets and Applications to Ordinary Differential Equations. Order. (005) 3 39.

25 E-CONTRACTION 97 [19] J. J. Nieto and R. R. Lopez, Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations, Order,, 3 39, (005). [0] J. J. Nieto, R. L. Pouso, and R. Rodrguez-Lpez, Fixed point theorems in ordered abstract spaces, Proc. of the American Math. Soc., 135, , (007). [1] T. P. Petru, A. Petruşel and J.-C. Yao, Ulam-Hyers stability for operatorial equations and inclusions via nonself operators, Taiwanese Journal of Mathematics, Vol. 15, No. 5, pp , October 011. [] O. Popescu, Some new fixed point theorems for α-geraghty contractive type maps in metric spaces, Fixed Point Theory Appl. 014, 014:190 [3] S. Radenović, Z. Kadelburg, D. Jandrlić and A. Jandrlić, Some results on weak contraction maps, Bulletin of the Iranian Mathematical Society Vol. 38 No. 3 (01), pp [4] A.C.M. Ran, M.C.B. Reurings, A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc. 13 (003) [5] I. A. Rus, Generalized contractions and applications, Cluj University Press, Cluj-Napoca, 001. [6] I.A. Rus, Cyclic representations and fixed points, Ann. T. Popoviciu, Seminar Funct. Eq. Approx. Convexity 3(005) [7] I. A. Rus, The theory of a metrical fixed point theorem: theoretical and applicative relevances, Fixed Point Theory, 9(008), No., [8] I. A. Rus, Remarks on Ulam stability of the operatorial equations, Fixed Point Theory, 10(009), No., [9] B. Samet, C. Vetro, P. Vetro, Fixed point theorems for α-ψ-contractive type mappings, Nonlinear Analysis 75 (01), Andreea Fulga, Department of Mathematics and Computer Sciences Universitatea Transilvania Brasov, Str. Iuliu Maniu 50, , Brasov, Romania. afulga@unitbv.ro Erdal Karapınar, Atilim University, Department of Mathematics,06836, İncek, Ankara, Turkey erdalkarapinar@yahoo.com; erdal.karapinar@atilim.edu.tr

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