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A discussion on generalized almost contractions via rational expressions in partially ordered metric spaces
Journal of Inequalities and Applications volume 2014, Article number: 219 (2014)
Abstract
The main purpose of this paper is to give some fixed point results for mappings involving generalized (\varphi ,\psi )contractions in partially ordered metric spaces. Our results generalize, extend, and unify several wellknown comparable results in the literature (Jaggi in Indian J. Pure Appl. Math. 8(2):223230, 1977, Harjani et al. in Nonlinear Anal. 71:34033410, 2009, Luong and Thuan in Fixed Point Theory Appl. 2011:46, 2011). The presented results are supported by three illustrative examples.
MSC: 46N40, 47H10, 54H25, 46T99.
1 Introduction and preliminaries
The Banach contraction mapping principle [1] 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 application in a vast number of branches of mathematics. Generalizations of this principle have been investigated heavily (see Jaggi [2], Harjani et al. [3], Luong and Thuan [4]). In particular, in 1977, Jaggi [2] proved the following theorem satisfying a contractive condition of a rational type.
Theorem 1 Let (X,d) be a complete metric space. Let T:X\to X be a continuous mapping such that
for all distinct points x,y\in X where \alpha ,\beta \in [0,1) with \alpha +\beta <1. Then T has a unique fixed point.
Existence of fixed point in partially ordered sets has been recently studied in [3–53].
Recently, Harjani et al. [3] proved the ordered version of Theorem 1. Very recently, Luong and Thuan [4] generalized the results of [3] and proved the following.
Theorem 2 Let (X,\le ) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a metric space. Let T:X\to X be a nondecreasing mapping such that
for all distinct points x,y\in X with y\le x where \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a lower semicontinuous function with the property that \psi (t)=0 if and only if t=0, and
Also, assume either

(i)
T is continuous or

(ii)
if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}.
If there exists {x}_{0}\in X such that {x}_{0}\le T{x}_{0}, then T has a fixed point.
Set \mathrm{\Phi}=\{\varphi \mid \varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})\text{is continuous and nondecreasing with}\varphi (t)=0\text{if and}\text{only if}t=0\} and \mathrm{\Psi}=\{\psi \mid \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})\text{is lower semi continuous},\psi (t)0\text{for all}t0,\text{and}\psi (0)=0\}. For some work on the class of Φ or the class of Ψ, we refer the reader to [21, 51, 54].
In 2004, Berinde [55] introduced an almost contraction, a new class of contractive type mappings which exhibits totally different features more than the one of the particular results incorporated [1, 16, 39, 50], i.e., an almost contraction generally does not have a unique fixed point; see Example 1 in [55]. Thereafter, many authors presented several interesting and useful facts about almost contractions; see [42, 56–59].
The purpose of this article is to generalize the above results for a mapping T:X\to X involving a generalized (\varphi ,\psi )almost contraction. Some examples are also presented to show that our results are effective.
2 Main result
Our essential result is given as follows.
Theorem 3 Let (X,\le ) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping which satisfies the inequality
for all distinct points x,y\in X with y\le x where \varphi \in \mathrm{\Phi}, \psi \in \mathrm{\Psi}, L\ge 0 and
Also, assume either

(i)
T is continuous or

(ii)
if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}.
If there exists {x}_{0}\in X such that {x}_{0}\le T{x}_{0}, then T has a fixed point.
Proof Let {x}_{0}\in X such that {x}_{0}\le T{x}_{0}. We define a sequence \{{x}_{n}\} in X as follows:
Since T is a nondecreasing mapping together with (2.2), we have {x}_{2}=T{x}_{1}. Inductively, we obtain
Assume that there exists {n}_{0} such that {x}_{{n}_{0}}={x}_{{n}_{0}+1}. Since {x}_{{n}_{0}}={x}_{{n}_{0}+1}=T{x}_{{n}_{0}}, then T has a fixed point. Suppose that {x}_{n}\ne {x}_{n+1} for all n\in \mathbb{N}. Thus, by (2.3) we have
Regarding (2.4), the condition (2.1) implies that
where
Suppose that M({x}_{n1},{x}_{n})=d({x}_{n},{x}_{n+1}) for some n\ge 1. Then the inequality (2.5) turns into
Regarding (2.4) and the property of ψ, this is a contradiction. Thus, M({x}_{n1},{x}_{n})=d({x}_{n1},{x}_{n}) for all n\ge 1. Therefore, the inequality (2.5) yields
Since ϕ is nondecreasing, we have d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n}). Consequently, \{d({x}_{n1},{x}_{n})\} is a decreasing sequence of positive real numbers which is bounded below. So, there exists \alpha \ge 0 such that {lim}_{n\to \mathrm{\infty}}d({x}_{n1},{x}_{n})=\alpha. We claim that \alpha =0. Suppose, to the contrary, that \alpha >0. By taking the limit of the supremum in the relation \varphi (d({x}_{n},{x}_{n+1}))\le \varphi (d({x}_{n1},{x}_{n}))\psi (d({x}_{n1},{x}_{n})), as n\to \mathrm{\infty}, we get
which is a contradiction. Hence, we conclude that \alpha =0, that is,
We prove that the sequence \{{x}_{n}\} is Cauchy in X. Suppose, to the contrary, that \{{x}_{n}\} is not a Cauchy sequence. So, there exists \epsilon >0 such that
where \{{x}_{m(k)}\} and \{{x}_{n(k)}\} are subsequences of \{{x}_{n}\} with
Moreover, n(k) is chosen to be the smallest integer satisfying (2.8). Thus, we have
By the triangle inequality, we get
Keeping (2.7) in mind and letting n\to \mathrm{\infty} in the above inequality, we get
Due to the triangle inequality, we have
and
By using (2.7), (2.11), and letting n\to \mathrm{\infty} in (2.12) and (2.13), we get
Analogously, we derive
Since m(k)<n(k) we have {x}_{m(k)1}<{x}_{n(k)1}. By (2.1) we have
where
Letting n\to \mathrm{\infty} in (2.16) (and hence in (2.17)), and taking (2.7), (2.11), (2.14), and (2.15) into account, we obtain
which is a contradiction. Thus, \{{x}_{n}\} is a Cauchy sequence in X. Since X is a complete metric space, there exists z\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=z.
We will show that z is a fixed point of T. Assume that (i) holds. Then by the continuity of T, we have
Suppose that (ii) holds. Since \{{x}_{n}\} is a nondecreasing sequence and {lim}_{n\to \mathrm{\infty}}{x}_{n}=z then z=sup\{{x}_{n}\}. Hence, {x}_{n}\le z for all n\in \mathbb{N}. Since T is a nondecreasing mapping, we conclude that T{x}_{n}\le Tz, or equivalently,
Then z=sup\{{x}_{n}\}, and we get z\le Tz.
To this end, we construct a new sequence \{{y}_{n}\} as follows:
Since z\le Tz, we have {y}_{0}\le T{y}_{0}={y}_{1}. Hence we find that \{{y}_{n}\} is a nondecreasing sequence. By repeating the discussion above, one can conclude that \{{y}_{n}\} is Cauchy. Thus there exists y\in X such that {lim}_{n\to \mathrm{\infty}}{y}_{n}=y. By (ii), we have y=sup\{{y}_{n}\} and so we have {y}_{n}\le y. From (2.19), we get
If z=y then the proof is finished. Suppose that z\ne y. On account of (2.20), the expression (2.1) implies that
where
Letting n\to \mathrm{\infty} in (2.21) (and hence (2.22)), we obtain
which is a contradiction. So y=z and we have z\le Tz\le z, then Tz=z. □
If we take L=0 in Theorem 3 we get the following result.
Theorem 4 Let (X,\le ) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping which satisfies the inequality
for all distinct x,y\in X with y\le x where \varphi \in \mathrm{\Phi}, \psi \in \mathrm{\Psi} and
Also, assume either

(i)
T is continuous or

(ii)
if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}.
If there exists {x}_{0}\in X such that {x}_{0}\le T{x}_{0}, then T has a fixed point.
Other corollaries could be derived.
Corollary 5 Let (X,\le ) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping such that
for all distinct x,y\in X with y\le x where \psi \in \mathrm{\Psi}, L\ge 0 and
Also, assume either

(i)
T is continuous or

(ii)
if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}.
If there exists {x}_{0}\in X such that {x}_{0}\le T{x}_{0}, then T has a fixed point.
Proof Take \varphi (t)=t in Theorem 3. □
Corollary 6 Let (X,\le ) be a partially ordered set. Suppose there exists a metric d X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping such that
for all distinct x,y\in X with y\le x where L\ge 0 and
Also, assume either

(i)
T is continuous or

(ii)
if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}.
If there exists {x}_{0}\in X such that {x}_{0}\le T{x}_{0}, then T has a fixed point.
Proof Take \psi (t)=(1k)\psi (t) for all t\in [0,\mathrm{\infty}) in Corollary 5. □
Corollary 7 Let (X,\le ) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping such that
for all distinct x,y\in X with y\le x where \alpha ,\beta \in [0,1) with \alpha +\beta <1. Also, assume either

(i)
T is continuous or

(ii)
if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}.
If there exists {x}_{0}\in X such that {x}_{0}\le T{x}_{0}, then T has a fixed point.
Proof Take L=0 and k=\alpha +\beta for all t\in [0,\mathrm{\infty}) in Corollary 6. Indeed,
□
Theorem 8 In addition to the hypotheses of Theorem 3, assume that
then T has a unique fixed point.
Proof Suppose, to the contrary, that x and y are fixed points of T where x\ne y. By (2.28), there exists a point z\in X which is comparable with x and y. Without loss of generality, we choose z\le x. We construct a sequence \{{z}_{n}\} as follows:
Since T is nondecreasing, z\le x implies Tz\le Tx=x. By induction, we get {z}_{n}\le x.
If x={z}_{{N}_{0}} for some {N}_{0}\ge 1 then {z}_{n}=T{z}_{n1}=Tx=x for all n\ge {N}_{0}1. So {lim}_{n\to \mathrm{\infty}}{z}_{n}=x. Analogously, we get {lim}_{n\to \mathrm{\infty}}{z}_{n}=y, which completes the proof.
Consider the other case, that is, x\ne {z}_{n} for all n=0,1,2,\dots . Then, by (2.1), we observe that
for all distinct x,y\in X with y\le x where \varphi \in \mathrm{\Phi}, \psi \in \mathrm{\Psi} and
Thus,
which is a contradiction. This ends the proof. □
Remark

Corollary 5 is a generalization of Theorem 2.1 of Luong and Thuan [4].

Corollary 7 (with L=0) corresponds to Theorem 2.2 and Theorem 2.3 of Harjani, López and Sadarangani [3].

Theorem 2.28 generalizes Theorem 2.4 of Luong and Thuan [4].
Now, we give some examples illustrating our results.
Example 9 Let X=\{4,5,6\} be endowed with the usual metric d(x,y)=xy for all x,y\in X, and \u2aaf:=\{(4,4),(5,5),(6,6),(6,4)\}. Consider the mapping
We define the functions \varphi ,\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) by \varphi (t)=2t and \psi (t)=\frac{3}{2}t. Now, we will check that all the hypotheses required by Theorem 4 (Theorem 3 with L=0) are satisfied.
First, X has the property: if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then x=sup\{{x}_{n}\}. Indeed, let \{{z}_{n}\} be a nondecreasing sequence in X with respect to ⪯ such that {z}_{n}\to z\in X as n\to +\mathrm{\infty}. We have {z}_{n}\u2aaf{z}_{n+1} for all n\in \mathbb{N}.

If {z}_{0}=4, then {z}_{0}=4\u2aaf{z}_{1}. From the definition of ⪯, we have {z}_{1}=4. By induction, we get {z}_{n}=4 for all n\in \mathbb{N} and z=4. Then {z}_{n}\u2aafz for all n\in \mathbb{N} and z=sup\{{z}_{n}\}.

If {z}_{0}=5, then {z}_{0}=5\u2aaf{z}_{1}. From the definition of ⪯, we have {z}_{1}=5. By induction, we get {z}_{n}=5 for all n\in \mathbb{N} and z=5. Then {z}_{n}\u2aafz for all n\in \mathbb{N} and z=sup\{{z}_{n}\}.

If {z}_{0}=6, then {z}_{0}=6\u2aaf{z}_{1}. From the definition of ⪯, we have {z}_{1}\in \{6,4\}. By induction, we get {z}_{n}\in \{6,4\} for all n\in \mathbb{N}. Suppose that there exists p\ge 1 such that {z}_{p}=4. From the definition of ⪯, we get {z}_{n}={z}_{p}=4 for all n\ge p. Thus, we have z=4 and {z}_{n}\u2aafz for all n\in \mathbb{N}. Now, suppose that {z}_{n}=6 for all n\in \mathbb{N}. In this case, we get z=6 and {z}_{n}\u2aafz for all n\in \mathbb{N} and z=sup\{{z}_{n}\}.
Thus, we proved that in all cases, we have z=sup\{{z}_{n}\}.
Let x,y\in X such that x\u2aafy and x\ne y, so we have only x=6 and y=4. In particular
so (2.23) holds easily. On the other hand, it is obvious that T is a nondecreasing mapping with respect to ⪯ and there exists {x}_{0}=6 such that {x}_{0}\u2aafT{x}_{0}. All the hypotheses of Theorem 4 are verified and u=4 is a fixed point of T.
Note that Theorem 1 is not applicable. Indeed, taking x=4 and y=5
for any \alpha ,\beta \ge 0 such that \alpha +\beta <1. Also, we could not apply Theorem 2 in this example. Indeed, for x=6 and y=4 (that is, x\ne y and x\u2aafy), we have
Example 10 Let X=[0,\mathrm{\infty}) be endowed with the Euclidean metric and the order ⪯ given as follows:
Define T:X\to X by Tx=x if 0\le x<1 and Tx=0 if x\ge 1. Define the functions \varphi ,\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) by \varphi (t)=4t and \psi (t)=3t.
Take x\u2aafy and x\ne y. It means that 1\le x<y. In particular, d(Tx,Ty)=0 and M(x,y)=yx. This implies that (2.23) holds. It is easy that X satisfies the property: if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x, then {x}_{n}\u2aafx for all n\in \mathbb{N}. Also, the other conditions of Theorem 4 are satisfied and u=0 is a fixed point of T.
Notice that we cannot apply Theorem 1 (since T is not continuous) nor Theorem 2 to this example. Indeed, letting x\u2aafy and x\ne y (that is, 1\le x<y), we have
Example 11 Let X=\{(0,1),(1,0),(1,1)\}\subset {\mathbb{R}}^{2} with the Euclidean distance {d}_{2}. (X,{d}_{2}) is, obviously, a complete metric space. Moreover, we consider the order ≤ in X given by R=\{(x,x),x\in X\}\cup \{((0,1),(1,1))\}. We also consider T:X\to X given by T((0,1))=(0,1), T((1,0))=(1,0) and T((1,1))=(0,1). Take \varphi (t)=3t and \psi (t)=2t. Obviously, T is a continuous and nondecreasing mapping since (0,1)\le (1,1) and T(0,1)=(0,1)\le T(1,1)=(0,1). Let x\le y and x\ne y, then necessarily x=(0,1) and y=(1,1). Then
so (2.23) holds. Also, (0,1)\le T((0,1)), therefore all conditions in Theorem 4 hold and there are two fixed points which are (0,1) and (1,0). The nonuniqueness follows from the fact that the partial order ≤ is not total.
Note that Theorem 1 is not applicable. Indeed, taking x=(0,1) and y=(1,0)
for any \alpha ,\beta \ge 0 such that \alpha +\beta <1. Also, we could not apply Theorem 2 in this example. Indeed, for x=(0,1) and y=(1,1) we have
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Mustafa, Z., Karapınar, E. & Aydi, H. A discussion on generalized almost contractions via rational expressions in partially ordered metric spaces. J Inequal Appl 2014, 219 (2014). https://doi.org/10.1186/1029242X2014219
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DOI: https://doi.org/10.1186/1029242X2014219
Keywords
 ordered set
 metric space
 fixed point