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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) doi:10.1186/1029242X2014219
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Keywords
 ordered set
 metric space
 fixed point