Fixed point theorem for generalized weak contractions satisfying rational expressions in ordered metric spaces

RESEARCH Open Access Fixed point theorem for generalized weak contractions satisfying rational expressions in ordered metric spaces Nguyen Van Luong * and Nguyen Xuan Thuan * Correspondence: luonghdu@gmail.com Department of Natural Sciences, Hong Duc University, Thanh Hoa, Vietnam Abstract In this paper, we prove a fixed point theorem for generalized weak contractions satisfying rational expressions in partially ordered metric spaces. The result is a generalization of a recent result of Harjani et al. (Abstr. Appl. Anal, Vol.200, -8, 200. An example is also given to show that our result is a proper generalization of the existing one. 2000 Mathematics Subject Classification: 47H0, 54H25. Keywords: fixed point, generalized weak contraction, rational type, ordered metric spaces Introduction and preinaries It is well known that the Banach contraction mapping principle is one of the pivotal results of analysis. Generalizations of this principle have been obtained in several directions. The following is an example of such generalizations. Jaggi in [] proved the following theorem satisfying a contractive condition of rational type Theorem.. ([] Let T be a continuous self-map defined on a complete metric space (X, d. Suppose that T satisfies the following condition: d ( Tx, Ty α d (x, Tx.d ( y, Ty d ( x, y + βd ( x, y for all x, y Î X, x yandforsomea, b 0 with a + b <,then T has a unique fixed point in X. Another generalization of the contraction principle was suggested by Alber and Guerre-Delabriere [2] in Hilbert spaces. Rhoades [3] has shown that their result is still valid in complete metric spaces. Definition.2. ([3]Let (X, d be a metric space. A mapping T : X X is said to be -weak contraction if d ( Tx, Ty d ( x, y ϕ ( d ( x, y for all x, y Î X, where :[0, [0, is a continuous and non-decreasing function with (t =0if and only if t =0. 20 Luong and Thuan; licensee Springer. 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.

Page 2 of 0 Theorem.3. ([3] Let (X, d be a complete metric space and T be a -weak contraction on X. Then, T has a unique fixed point. In fact, while Alber and Guerre-Delabriere assumed an additional assumption t (t = on, but Rhoades proved Theorem.3 without this particular condition. A number of extensions of Theorem.3 were presented in [4-9] and references therein. Some of these results were presented without the continuity and monotonicity of. Recently, existence of fixed points in partially ordered sets has been considered, and first results were obtain by Ran and Reurings [0] and then by Nieto and Lopez []. The following fixed point theorem is the version of theorems, which were proved in those papers. Theorem.4. ([0,] Let (X, be a partially ordered set, and suppose that there is a metric d such that (X, d be a complete metric space. Let T : X Xbeanondecreasing mapping satisfying the following inequality d(tx, Ty kd(x, y, for all x, y X with x y, where k Î (0,. Also, assume either (i T is continuous or (ii X has the property: If anon decreasing sequence {x n } in X converges to x X thenx n x for all n ( If there exists x 0 Î X such that x 0 Tx 0, then T has a fixed point. Besides, applications to matrix equations and ordinary differential equations were presented in [0,]. Afterward, coupled fixed point and common fixed point theorems and their applications to periodic boundary value problems and integral equations were given in [5-7,2-9]. In particular, Harjani and Sadarangani [5] proved some fixed point theorems in the context of ordered metric spaces as the extensions of Theorem.3. We state one of their results. Theorem.5. ([5] Let (X, be a partially ordered set and suppose that there is a metric d such that (X, d be a complete metric space. Let T : X X be a non-decreasing mapping satisfying the following inequality d ( Tx, Ty d ( x, y ϕ ( d ( x, y, for all x, y X with x y, where : [0, [0, is a continuous and non-decreasing function with (t =0if and only if t =0.Also, assume either (i T is continuous or (ii X has the property (. If there exists x 0 Î X such that x 0 Tx 0, then T has a fixed point. In addition, Harjani et al. in [2] proved the following theorem as a version of Theorem. in partially ordered metric spaces where they replaced the condition ( by a stronger condition, that is If{x n } is a non - decreasing sequence in X such that x n x then x =sup{x n }. (2

Page 3 of 0 Theorem.6. ([2]Let (X, be a partially ordered set and suppose that there is a metric d such that (X, d be a complete metric space. Let T : X X be a nondecreasing mapping such that d ( Tx, Ty α d (x, Tx.d ( y, Ty d ( x, y + βd ( x, y, for all x, y Xwithx y, x y, (3 where 0 a, b and a + b <.Also, assume either (i T is continuous or (ii X has the property (2. If there exists x 0 Î X such that x 0 Tx 0, then T has a fixed point. In this paper, we prove a fixed point theorem for generalized weak contractions satisfying rational expressions in partially metric spaces, which is a generalization of the result of Harjani et al. [2]. We also give an example to show that our result is a proper extension of the result in [2]. 2 Main theorem Theorem 2.. Let (X, be a partially ordered set, and suppose that there is a metric d such that (X, d be a complete metric space. Let T : X X be a non-decreasing mapping satisfying the following inequality d ( Tx, Ty M ( x, y ϕ ( M ( x, y, for all x, y X with x y, x y, (4 where :[0, [0, is a lower semi-continuous function with (t =0if and only if t =0,and { ( d (x, Tx.d y, Ty M(x, y =max d ( x, y, d ( x, y }. Also, assume either (i T is continuous or (ii X has the property (2. If there exists x 0 Î X such that x 0 Tx 0, then T has a fixed point. Proof. Letx 0 Î X be such that x 0 Tx 0, we construct the sequence {x n }inx as follows x n+ = Tx n, n =0,,2,... (5 Since T is a non-decreasing mapping, by induction, we can show that x 0 x x 2 x n x n+ (6 If there exists n 0 such that x n0 = x n0 +, thenx n0 = x n0 + = Tx n0.thismeansthatx n0 is a fixed point of T and the proof is finished. Thus, we can suppose that x n x n+ for all n.

Page 4 of 0 Since x n >x n- for all n, from (4, we have d(x n+, x n =d(tx n, Tx n { d(xn, Tx n. d(x n, Tx n max d (x n, x n ( { d(xn, Tx n. d(x n, Tx n ϕ max d(x n, x n =max{d(x n+, x n, d(x n, x n } ϕ(max{d(x n+, x n, d(x n, x n } }, d(x n, x n }, d(x n, x n (7 Suppose that there exists m 0 such that d ( x m0 +, x m0 > d ( xm0, x m0,from(7,we have d(x m0 +, x m0 max{d(x m0 +, x m0, d(x m0, x m0 } ϕ(max{d(x m0 +, x m0, d(x m0, x m0 } = d(x m0 +, x m0 ϕ(d(x m0 +, x m0 < d(x m0 +, x m0 which is a contradiction. Hence, d (x n+, x n d (x n, x n- for all n. Since {d(x n+, x n } is a non-increasing sequence of positive real numbers, there exists δ 0 such that d (x n+, x n = δ We shall show that δ = 0. Assume, to the contray, that δ >0. Taking the upper it as n in (7 and using the properties of the function, we get δ δ inf ϕ (max {d (x n+, x n, d (x n, x n } δ ϕ (δ < δ which is a contradiction. Therefore, δ = 0, that is, d (x n+, x n = 0 (8 In what follows, we shall prove that {x n } is a Cauchy sequence. Suppose, to the contrary, that {x n } is not a Cauchy sequence. Then, there exists ε >0 such that we can find subsequences {x m(k }, {x n(k }of{x n } with n(k >m(k k satisfying d(x m(k, x n(k ε (9 Further, corresponding to m(k, we can choose n(k in such way that it is the smallest integer with n(k > m(k k satisfying (9. Hence, We have d(x m(k, x n(k < ε (0 ε d(x m(k, x n(k d(x m(k, x n(k +d(x n(k, x n(k <ε+ d(x n(k, x n(k Taking k and using (8, we get k d ( x m(k, x n(k = ε ( By the triangle inequality, d(x m(k, x n(k d(x m(k, x m(k +d(x m(k, x n(k +d(x n(k, x n(k, d(x m(k, x n(k d(x m(k, x m(k +d(x m(k, x n(k +d(x n(k, x n(k

Page 5 of 0 Taking k in the above inequalities and using (7, (, we obtain k d ( x m(k, x n(k = ε (2 Since m(k <n(k, x n(k- >x m(k-, from (4, we have d ( ( x n(k, x m(k = d Txn(k, Tx m(k { ( ( d xn(k, Tx n(k d xm(k, Tx m(k max d (, d ( } x n(k, x m(k x n(k, x m(k ( { ( ( d xn(k, Tx n(k.d xm(k, Tx m(k ϕ max d (, d ( } x n(k, x m(k x n(k, x m(k { ( ( d xn(k, x n(k.d xm(k, x m(k max d (, d ( } x n(k, x m(k x n(k, x m(k ( { ( ( d xn(k, x n(k.d xm(k, x m(k ϕ max d (, d ( } x n(k, x m(k x n(k, x m(k (3 Taking upper it as k in (3 and using (7, (, (2 and the properties of the function, we have ε max {0, ε} ϕ (max {0, ε} = ε ϕ (ε < ε which is a contradiction. Therefore, {x n } is a Cauchy sequence. Since X is a complete metric space, there exists x Î X such that n x n = x. Now, suppose that the assumption (a holds. The continuity of T implies x = x n = Tx n = T ( x n = Tx and this proved that x is a fixed point of T. Finally, suppose that the assumption (b holds. Since {x n } is a non-decreasing sequence and x n x, thenx =sup{x n }. Particularly, x n x for all n. SinceT is nondecreasing, Tx n Tx for all n, thatis,x n+ Tx for all n. Moreover, as x n x n+ Tx for all n and x =sup{x n }, we obtain x Tx. Consider the sequence {y n }thatisconstructed as follows y 0 = x, y n+ = Ty n, n =0,,2,... Since y 0 Ty 0, arguing like above part, we obtain that {y n } is a non-decreasing sequence and y n = y for certain y Î X. By the assumption (b, we have y = sup{y n }. Since x n <x = y 0 Tx = Ty 0 y n y for all n, suppose that x y, from (4, we have d(y n+, x n+ =d ( Tx n, Ty n { } d(yn, Ty n. d(x n, Tx n max, d(y n, x n d(y n, x n ( { } d(yn, Ty n. d(x n, Tx n ϕ max, d(y n, x n d(y n, x n { } d(yn, y n+. d(x n, x n+ =max, d(y n, x n d(y n, x n ( { } d(yn, y n+. d(x n, x n+ ϕ max, d(y n, x n d(y n, x n

Page 6 of 0 Taking upper it as n in the above inequality, we have d(y, x max{0, d(y, x} ϕ(max{0, d(y, x} < d(y, x which is a contradiction. Hence, x = y. Wehavex Tx x, thereforetx = x. That is, x is a fixed point of T. The proof is complete. Corollary 2.2. Let (X, be a partially ordered set, and suppose that there is a metric d such that (X, d be a complete metric space. Let T : X X be a non-decreasing mapping such that { } d(x, Tx. d(y, Ty d(tx, Ty k max, d(x, y, (4 d(x, y for all x, y Î X with x y, x y, where k Î (0,. Also, assume either (i T is continuous or (ii X has the property (2. If there exists x 0 Î X such that x 0 Tx 0, then T has a fixed point. Proof. In Theorem 2., taking (t = ( - kt, for all t Î [0,, we get Corollary 2.2. Remark 2.3. For a, b >0, a + b <and for all x, y Î X, x y, we have d(x, Tx. d(y, Ty d(tx, Ty α + βd(x, y d(x, y { } d(x, Tx. d(y, Ty (α + βmax, d(x, y d(x, y { } d(x, Tx. d(y, Ty = k max, d(x, y d(x, y where k = a + b Î (0,. Therefore, Corollary 2.2 is a generalization of Theorem.6, so is Theorem 2.. Now, we shall prove the uniqueness of the fixed point. Theorem 2.4. In addition to the hypotheses of Theorem 2., suppose that for every x, y X, there exists z X thatiscomparable to x and y, (5 then T has a unique fixed point. Proof. From Theorem 2., the set of fixed points of T is non-empty. Suppose that x, y Î X are two fixed points of T. Bytheassumption,thereexistsz Î X that is comparable to x and y. We define the sequence {z n } as follows z 0 = z, z n+ = Tz n, n =0,,2,... Since z is comparable with x, we may assume that z x. Using the mathematical induction, it is easy to show that z n x for all n. Suppose that there exists n 0 such that z n0 = x, thenz n = Tz n- = Tx = x for all n n 0 -. Hence, z n x as n.

Page 7 of 0 On the other hand, if z n x for all n, from (4, we have d(x, z n =d(tx, Tz n { d(x, Tx. d(zn, Tz n max d(x, z n ( { d(x, Tx. d(zn, Tz n ϕ max d(x, z n = d(x, z n ϕ(d(x, z n }, d(x, z n }, d(x, z n (6 It implies that d (x, z n <d (x, z n- foralln, that is, {d(x, z n } is a decreasing sequence of positive real numbers. Therefore, there is an a 0 such that d(x, z n a. We shall show that a = 0. Suppose, to the contrary, that a >0. Taking the upper it as n in (6 and using the properties of, we have α = d(x, z n α inf ϕ(d(x, z n α ϕ(α < α which is a contradiction. Hence, a =0,thatis,z n x as n. Therefore, in both cases, we have z n = x (7 Similarly, we have z n = y (8 From (7 and (8, we get x = y. Example 2.5. Let X = [ 0, 2] with the usual metric d (x, y = x - y, x, y Î X. Obviously, (X, d is a complete metric space. We consider the ordered relation in X as follows ( { } x, y X, x y x = yor x, y {0} n : n =2,3,... and x y where be the usual ordering. Let T : X X be given by 0, if x =0, Tx = /(n +, if x =/n, n =2,3,... 2/2, otherwise It is easy to see that T is non-decreasing and X has the property (2. Also, there is x 0 =0inX such that x 0 =0 0=Tx 0. Clearly, T has a fixed point that is 0. However, we cannot apply Theorem.6 because the condition (3 is not true. Indeed, suppose that the condition (3 holds. Taking y = 0 and x =/n, n = 2, 3, 4,... in (3, we have d (T n, T0 α d ( n, T ( n.d (0, T0 d ( n,0 + βd n,0, n =2,3,4,... This implies n + β, n =2,3,4,... n

Page 8 of 0 or n β, n =2,3,4,... n + Taking n in the last inequality, we have b and we obtain a contradiction. We now show that T satisfies (4 with : [0, [0, which is given by ϕ(t = t 3, t [0,. We have x, y Î X, x y, x y if x =/n, y =0orx =/n, y =/m, m>n 2. So, we have two possible cases. Case. x =/n, n 2 and y = 0, we have M(x, y ϕ ( M(x, y = n n 3 n n (n + = n + = d ( Tx, Ty Case 2. x =/n, y =/m, m>n 2, we have M(x, y =max { For m>n 2, we have n n+ m m+ n m is equivalent to n n+ m m+ n m n m (m n2 (n +(m + mn, n } m or mn (m n2 (n +(m + The last inequality holds since mn < (m n2 (n +(m + Therefore, M(x, y = n m We have d ( Tx, Ty M(x, y ϕ ( M(x, y, m > n 2 (9 is equivalent to n + m + n m n 3 m, m > n 2

Page 9 of 0 or m n (n +(m + m n mn 3 (m n (mn 3, m > n 2 or 2 (m n (mn 3 mn (n +(m + = m + n + mn (n +(m +, m > n 2 or ( n 2 m + n +, m > n 2 (20 m (n +(m + We have ( n 2 < m n 2 < n + + n (n +(m + = m + n + (n +(m +, m > n 2 Thus, the inequality (20 holds, so does the inequality (9. Therefore, all the conditions of Theorem 2. are satisfied. Applying Theorem 2., we conclude that T has a fixed point in X. Notice that since T is not continuous, this example cannot apply to Theorem.. Moreover, since the condition (5 is not satisfied, the uniqueness of fixed point of T does not guarantee. In fact, T has two fixed points that are 0 and 2/2. Acknowledgements The authors express their sincere thanks to the reviewers for their valuable suggestions in improving the paper. Authors contributions All authors contribute equally and significantly in this research work. All authors read and approved the final manuscript. Competing interests The authors declare that they have no competing interests. Received: March 20 Accepted: 5 September 20 Published: 5 September 20 References. Jaggi, DS: Some unique fixed point theorems. Indian J Pure Appl Math. 8, 223 230 (977 2. Ya, I, Alber, S: Guerre-Delabriere, Principles of weakly contractive maps in Hilbert spaces, new results in operator theory. In: Gohberg I, Lyubich Yu (eds. Advances and Application, vol. 98, pp. 7 22. Birkhauser Verlag, Basel (997 3. Rhoades, BE: Some theorems on weakly contractive maps. Nonlinear Anal. 47, 2683 2693 (200. doi:0.06/s0362-546x(000388-4. Doric, D: Common fixed point for generalized (j, ψ-weak contractions. Appl Math Lett. 22, 896 900 (2009. doi:0.06/j.aml.2009.08.00 5. Harjani, J, Sadarangani, K: Fixed point theorems for weakly contractive mappings in partially ordered sets. Nonlinear Anal. 7, 3403 340 (2009. doi:0.06/j.na.2009.0.240 6. Harjani, J, Sadarangani, K: Generalized contractions in partially ordered metric spaces and applications to ordinary differential equations. Nonlinear Anal. 72, 88 97 (200. doi:0.06/j.na.2009.08.003 7. Radenovic, S, Kadelburg, Z: Generalized weak contractions in partially ordered metric spaces. Comput Math Appl. 60, 776 783 (200. doi:0.06/j.camwa.200.07.008 8. Popescu, O: Fixed points for j-weak contractions. Appl Math Lett. 24, 4 (20. doi:0.06/j.aml.200.06.024 9. Zhang, Q, Song, Y: Fixed point theory for generalized j-weak contractions. Appl Math Lett. 22, 75 78 (2009. doi:0.06/j.aml.2008.02.007 0. Ran, ACM, Reurings, MCB: A fixed point theorem in partially ordered sets and some applications to matrix equations. Proc Am Math Soc. 32, 435 443 (2004. doi:0.090/s0002-9939-03-07220-4. Nieto, JJ, Rodrguez-Lopez, R: Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations. Order. 22, 223 239 (2005. doi:0.007/s083-005-908-5 2. Harjani, J, Lopez, B, Sadarangani, K: A fixed point theorem for mappings satisfying a contractive condition of rational type on a partially ordered metric space. Abstr Appl Anal. 200, 8 (200

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