A fixed point theorem for Meir-Keeler contractions in ordered metric spaces

  • Jackie Harjani1,

    Affiliated with

    • Belén López1 and

      Affiliated with

      • Kishin Sadarangani1Email author

        Affiliated with

        Fixed Point Theory and Applications20112011:83

        DOI: 10.1186/1687-1812-2011-83

        Received: 30 June 2011

        Accepted: 23 November 2011

        Published: 23 November 2011

        Abstract

        The purpose of this paper is to present some fixed point theorems for Meir-Keeler contractions in a complete metric space endowed with a partial order.

        MSC: 47H10.

        Keywords

        fixed point ordered metric spaces Meir-Keeler contraction

        1 Introduction and preliminaries

        The Banach contraction mapping principle is one of the pivotal results of analysis. It is widely considered as the source of metric fixed point theory. Also, its significance lies in its vast applicability in a number of branches of mathematics.

        Generalization of the above principle has been a heavily investigated branch of research. In particular, Meir and Keeler [1] present the following fixed point theorem.

        Theorem 1.1. [1]Let (X,d) be a complete metric space and T: XX an operator. Suppose that for every ε > 0 there exists δ(ε) > 0 such that for x,yX
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equa_HTML.gif

        Then, T admits a unique fixed point ξX and for any xX, the sequence {T n x} converges to ξ.

        The purpose of this article is to present a version of Theorem 1.1 in the context of ordered metric spaces.

        Existence of fixed point in partially ordered sets has been recently studied in [220].

        In the context of ordered metric spaces, the usual contraction is weakened but at the expense that the operator is monotone.

        2 Fixed point results: nondecreasing case

        Our starting point is the following definition.

        Definition 2.1. Let (X, ≤) be a partially ordered set and T: XX a mapping. We say that T is nondecreasing if for x, yX
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equb_HTML.gif

        This definition coincides with the notion of a nondecreasing function in the case where X = ℝ and ≤ represents the usual total order in ℝ.

        Remark 2.2. The contractive condition given by Meir-Keeler:

        For every ε > 0, there exists δ(ε) > 0 such that for x,yX
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equc_HTML.gif

        implies that the operator T: XX is strictly nonexpansive (this means that for x, yX, xy, d(Tx, Ty) < d(x, y)).

        Therefore, any operator T satisfying the Meir-Keeler condition is continuous.

        In what follows, we present the following theorem which is a version of Theorem 1.1 in the context of ordered metric spaces when the operator is nondecreasing.

        Theorem 2.3. Let (X, ≤) be a partially ordered set and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Let T: XX be a continuous and nondecreasing mapping such that for every ε > 0 there exists δ(ε) > 0 satisfying
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ1_HTML.gif
        (1)

        If there exists x0X with x0Tx0, then T has a fixed point.

        Remark 2.4. Condition (??) does not imply that T is continuous as it is proven with the following example.

        Let X = [0, 1] be the unit interval with the usual metric and the order given by R = {(x, x): xX}. Consider the operator T: XX given by Tx = 0 if x ≠ 1 and Tx = 1 if x = 1.

        Obviously, T is not continuous and, as the elements in X are only comparable to themselves, T satisfies (??).

        This shows that the continuity of T in Theorem 2.3 is not redundant (compare with Remark 2.2 and Theorem 1.1).

        Remark 2.5. Condition (1) does not imply that T is strictly nonexpansive but d(Tx, Ty) < d(x, y) is true if x < y as it can be seen using ε = d(x, y) in (1) when x < y.

        Proof of Theorem 2.3. If Tx0 = x0, then the proof is finished.

        Suppose that x 0 < Tx 0 and T is a nondecreasing mapping, we obtain by induction that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equd_HTML.gif

        Put x n+1= T n x 0. Obviously, (x n ) is a nondecreasing sequence.

        For better readability, we divide the proof into several steps.

        Step 1: limn → ∞d(x n , xn + 1) = 0.

        In fact, if the sequence (x n ) is not strictly nondecreasing, then we can find n 0 ∈ ℕ such that http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq1_HTML.gif and the result follows. If (x n ) is strictly nondecreasing, then by Remark 2.5, (d(x n , x n+1)) is strictly decreasing, and hence, it is convergent. Put r = limn → ∞ d(x n , x n+1) (notice that r = inf{d(x n , x n+1): n ∈ ℕ}).

        Now, we will prove that r = 0.

        Suppose that r > 0.

        Applying condition (??) to r > 0 we can find δ(r) > 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Eque_HTML.gif

        Since r = limn→∞ d(x n , x n+1), there exists n 0 ∈ ℕ such that http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq2_HTML.gif and, as http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq3_HTML.gif , we have http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq4_HTML.gif .

        This is a contradiction because r = inf{d(x n , x n+1): n ∈ ℕ}.

        Therefore, r = 0. This means that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ2_HTML.gif
        (2)

        Step 2: (x n ) is a Cauchy sequence.

        In fact, fix ε > 0 and arbitrary.

        By condition (??), there exists δ(ε) > 0 (which can be choosen satisfying δ(ε) ≤ ε) such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ3_HTML.gif
        (3)
        On the other hand, by (??), there exists n 0 ∈ ℕ such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ4_HTML.gif
        (4)
        Fix n > n 0 and in order to prove that {x n } is a Cauchy sequence it is sufficient to show that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ5_HTML.gif
        (5)

        We will use mathematical induction.

        For p = 1 by (??) and the fact that δ(ε) ≤ ε we obtain
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equf_HTML.gif

        Now, we assume that (??) holds for some fixed p.

        Then, using (??) and the inductive hypothesis, we get
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ6_HTML.gif
        (6)

        Now, we consider two cases.

        Case 1:d(xn-1, xn+p) ≥ ε.

        In this case, taking into account (??),
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equg_HTML.gif
        and, since x n-1< x n+p, by (??)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equh_HTML.gif

        This proves that (??) is satisfied by p + 1.

        Case 2:d(xn-1, xn+p) < ε.

        As d(x n-1 , x n+p) > 0 (because {x n } is a nondecreasing sequence and d(x n , x n+1) > 0 for any n = 0,1, 2,...), applying condition (??) for ε 0 = d(x n-1 , x n+p) we can get
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equi_HTML.gif

        (notice that x n-1< x n+p) and this proves that (??) is satisfied by p + 1.

        Therefore, {x n } is a Cauchy sequence.

        Since X is a complete metric space, there exists zX such that limn→∞ x n = z. Finally, the continuity of T implies that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equj_HTML.gif

        and, therefore, z is a fixed point of T.

        This finishes the proof.

        In what follows we prove that Theorem 2.3 is still valid for T not necessarily continuous, assuming the following hypothesis:
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ7_HTML.gif
        (7)

        Theorem 2.6. If in Theorem 2.3 we replace the continuity of T by condition (??) the result is true.

        Proof. Following the proof of Theorem 2.3, we only have to check that Tz = z.

        As (x n ) is a nondecreasing sequence in X with x n z, by condition (??), we can find a subsequence (x n(k)) such that x n(k)z for all k ∈ ℕ.

        If there exists k 0 ∈ ℕ such that http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq5_HTML.gif , then the nondecreasing character of (x n ) gives us that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equk_HTML.gif

        Particularly, http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq6_HTML.gif , and http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq7_HTML.gif is a fixed point of T.

        Suppose that for any k ∈ ℕ, x n(k)< z.

        Applying condition (??) to ε = d(x n(k) , z) for k fixed and arbitrary, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equl_HTML.gif

        As x n(k)z, the last inequality implies that x n(k)+1Tz.

        As (x n(k)+1) is a subsequence of (x n ) and x n z we have x n(k)+1z.

        Now, the uniqueness of the limit in complete metric spaces gives us Tz = z.

        This finishes the proof.

        Now, we present an example where it can be appreciated that assumptions in Theorems 2.3 and 2.6 do not guarantee uniqueness of the fixed point.

        Let X = {(1,0), (0,1)} ⊂ ℝ2 and consider the usual order
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equm_HTML.gif

        Then, (X, ≤) is a partially ordered set whose different elements are not comparable. Besides, (X, d 2) is a complete metric space considering d 2 the Euclidean distance. The identity map T(x, y) = (x, y) is trivially continuous and nondecreasing and condition (??) of Theorem 2 is satisfied since elements in X are only comparable to themselves. Moreover, (1,0) ≤ T(1, 0) = (1, 0) and the operator T has two fixed points.

        In what follows, we present a sufficient condition for the uniqueness of the fixed point in Theorems 2.3 and 2.6. This condition appears in [15] and says:
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ8_HTML.gif
        (8)

        Theorem 2.7. Adding condition (??) to the hypotheses of Theorem 2.3 (or Theorem 2.6) we obtain the uniqueness of the fixed point of T.

        Proof. Suppose that there exist z, yX which are fixed points of T and zy.

        We consider two cases.

        Case 1: Suppose that z and y are comparable.

        Without loss of generality, we suppose z < y.

        Putting ε = d(z, y) and applying condition (??) of Theorem 1, we get
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equn_HTML.gif

        which is a contradiction.

        Case 2: Suppose that z and y are not comparable.

        By condition (??), there exists xX comparable to z and y.

        Suppose z < x (the same argument serves for x < z).

        Monotonicity of T implies that T n z = zT n x for n = 1, 2, 3,...

        We consider two possibilities:

        (a) Suppose that there exists n 0 ∈ ℕ such that http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq8_HTML.gif . Since z is a fixed point of T, T n x = z for nn 0, and, consequently, T n xz.

        (b) Suppose that T n z = z < T n x for any n = 1, 2, 3,.... Applying condition (??) of Theorem 1 for ε = d(T n z, T n x) (where n ∈ ℕ is fixed but arbitrary), we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equo_HTML.gif
        Thus, {d(z, T n x)} is a decreasing sequence of positive real numbers and, consequently, there exists r ≥ 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equp_HTML.gif

        Suppose r > 0.

        Applying condition (??) of Theorem 2.3 for ε = r we find δ(r) > 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equ9_HTML.gif
        (9)
        As limn→∞ d(z, T n x) = r = inf {d(z, T n x): n ∈ ℕ}, there exists n 0 ∈ ℕ such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equq_HTML.gif
        and, since http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq9_HTML.gif , (??) gives us
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equr_HTML.gif

        which contradicts to r = inf{d(z, T n x): n ∈ ℕ}.

        Therefore, z = y.

        This finishes the proof.

        3 Fixed point results: nonincreasing case

        We start this section with the following definition.

        Definition 3.1. Let (X, ≤) be a partially ordered set and T: XX. We say that T is nonincreasing if for x, yX
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equs_HTML.gif

        The main result of this section is the following theorem.

        Theorem 3.2. Let (X, ≤) be a partially ordered set satisfying condition (??) and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Let T: XX be a nonincreasing mapping such that for any ε > 0 there exists δ(ε) > 0 satisfying
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equt_HTML.gif

        (a) If there exists x 0X with x 0Tx 0 or x 0Tx 0, then inf{d(x, Tx): xX} = 0.

        (b) If, in addition, X is compact and T is continuous, then T has a unique fixed point.

        Proof. (a) If Tx0 = x0, then it is obvious that inf{d(x, Tx): xX} = 0. Suppose that x0 < Tx0 (the same argument serves for Tx0 < x0).

        In virtue that T is nonincreasing the consecutive terms of the sequence (T n x 0) are comparable.

        Suppose that there exists n 0 ∈ ℕ such that http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq10_HTML.gif .

        In this case, http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq11_HTML.gif and, consequently, inf{d(x, Tx): xX} = 0 and this finishes the proof.

        Now, we suppose that T n x 0T n+1 x 0 for any n = 1,2,....

        Since T n x 0 and T n+1 x 0 are comparable applying the contractive condition we obtain
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equu_HTML.gif

        and this inequality is satisfied by any n ∈ ℕ.

        Thus, {d(T n x 0, T n+1 x 0)} is a decreasing sequence of positive real numbers and, consequently, limn→∞ d(T n x 0, T n+1 x 0) = r for certain r ≥ 0.

        Using a similar argument that in Theorem 2.3, we prove that r = 0.

        Finally, the fact limn→∞ d(T n x 0, T n+1 x 0) = 0 implies that inf{d(x, Tx): xX} = 0.

        This finishes the proof of (a).

        (b) Suppose that X is compact and T is continuous.

        Taking into account that the mapping
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equv_HTML.gif
        is continuous and the fact that X is compact, we can find zX such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equw_HTML.gif

        By (a), d(z, Tz) = inf{d(x, Tx): xX} = 0 and, therefore, z is a fixed point of T.

        The uniqueness of the fixed point is proved as in Theorem 2.7.

        Remark 3.3. A parallel result in the nonincreasing case cannot be obtained using a similar argument as in Theorem 2.3 because the proof that (x n ) is a Cauchy sequence uses that xn-1 and xn + pare comparable and this can be false when T is a nonincreasing operator (see, Theorem 2.3).

        4 Examples

        In this section, we present some examples which illustrate our results.

        Example 4.1. Let X = {(0,1), (1, 0), (1,1)} ⊂ ℝ2with the Euclidean distance d2. (X, d2) is a complete metric space. We consider the orderin X given by R = {(x, x): xX}.

        Notice that the elements in X are only comparable to themselves. Therefore, condition (??) of Theorem 2.3 is satisfied for any operator T: XX.

        We consider the operator T: XX defined by
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equx_HTML.gif

        Obviously, T is a continuous and nondecreasing operator satisfying (1,1) ≤ T(1, 1) = (1,1).

        Theorem 2.3 gives us the existence of a fixed point for T (which is the point (1,1)).

        On the other hand, the operator T does not satisfy Meir-Keeler condition appearing in Theorem 1.1, because for http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq12_HTML.gif the inequality
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equy_HTML.gif

        fails.

        Therefore, this example cannot be treated by Theorem 1.1.

        Notice that in Example 4.1, we have uniqueness of the fixed point and condition (??) is not satisfied by (X, ≤) (notice that condition (??) fails for the elements (0,1), (1, 0) ∈ X). This proves that condition (??) is not a necessary condition for the uniqueness of the fixed point.

        Example 4.2. Consider the same space X that in Example 4.1 with the Euclidean distance d2and with the order given by
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equz_HTML.gif

        Consider the operator T: XX given by T(0, 1) = (0, 1), T(1, 1) = (0, 1) and T (1, 0) = (1, 0).

        It is easily checked that T is a continuous and nondecreasing operator (notice that (0, 1) ≤ (1, 1) and T(0, 1) = (0, 1) ≤ T(1, 1) = (0,1)).

        Moreover, as the unique pair of elements in X satisfying x < y is ((0,1), (1,1)) and d(T(0, 1), T(1, 1)) = d((0, 1), (0, 1)) = 0, condition (??) of Theorem 2.3 is satisfied.

        As (0, 1) ≤ T(0, 1), Theorem 2.3 says us that T has a fixed point (in this case (0, 1) and (1, 0) are fixed points of T).

        On the other hand, the operator T does not satisfy Meir-Keeler condition appearing in Theorem 1.1 because for http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_IEq12_HTML.gif the inequality
        http://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2011-83/MediaObjects/13663_2011_83_Equaa_HTML.gif

        fails.

        Thus, this example cannot be studied by Meir-Keeler Theorem (Theorem 1.1).

        Declarations

        Acknowledgements

        Partially supported by Ministerio de Ciencia y Tecnología, project MTM 2007-65706.

        Authors’ Affiliations

        (1)
        Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria

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        © Harjani et al; licensee Springer. 2011

        This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.