A discussion on generalized almost contractions via rational expressions in partially ordered metric spaces
 Zead Mustafa^{1},
 Erdal Karapınar^{2, 3}Email author and
 Hassen Aydi^{4, 5}
https://doi.org/10.1186/1029242X2014219
© Mustafa et al.; licensee Springer. 2014
Received: 9 October 2013
Accepted: 15 May 2014
Published: 2 June 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.
Keywords
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.
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.
 (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.
 (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.
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 ${lim}_{n\to \mathrm{\infty}}{x}_{n}=z$.
Then $z=sup\{{x}_{n}\}$, and we get $z\le Tz$.
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.
 (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.
 (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. □
 (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. □
 (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.
□
then T has a unique fixed point.
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.
which is a contradiction. This ends the proof. □
Remark
Now, we give some examples illustrating our results.
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 $z=sup\{{z}_{n}\}$.
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.
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.
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.
Notes
Declarations
Acknowledgements
The authors express their gratitude to the referees for constructive and useful remarks and suggestions.
Authors’ Affiliations
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