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 Open Access
Fixed point results for modified weak and rational αψcontractions in ordered 2metric spaces
 Shahin Fathollahi^{1},
 Nawab Hussain^{2} and
 Liaqat A Khan^{2}Email author
https://doi.org/10.1186/1687181220146
© Fathollahi et al.; licensee Springer. 2014
 Received: 28 July 2013
 Accepted: 5 December 2013
 Published: 6 January 2014
Abstract
We first introduce the concept of a triangular 2αηadmissible mapping which extends the notion of αadmissible mapping with respect to η to 2metric spaces. Next, we introduce the concepts of modified weak and modified rational αψcontractions and establish the existence and uniqueness of fixed points for such mappings in complete 2metric spaces. As an application of the obtained results, we prove some fixed point results in partially ordered 2metric spaces. The presented theorems generalize and improve certain existing results in the literature and provide main results in Dung and Hang (Fixed Point Theory Appl. 2013:161, 2013) as corollaries. Moreover, some examples and an application to integral equations are provided to illustrate the usability of the obtained results.
MSC:46N40, 47H10, 54H25, 46T99.
Keywords
 modified weak αηcontractions
 modified rational αηψcontractions
 triangular 2αηadmissible map
 partially ordered 2metric space
1 Introduction and preliminaries
There exist many generalizations of the concept of metric spaces in the literature ($d(x,y)\in {\mathbb{R}}_{+}$; $d(x,y)\in K$, K a cone in an ordered Banach space; 2metric spaces; probabilistic metric spaces; Gmetric spaces etc.; see, for example, [1–9]). The notion of 2metric was introduced by Gähler in [10]. Note that a 2metric is not a continuous function of its variables, whereas an ordinary metric is. This led Dhage to introduce the notion of Dmetric in [11]. In [12, 13] Mustafa and Sims introduced the notion of Gmetric to overcome flaws of a Dmetric. After that, many fixed point theorems on Gmetric spaces have been proved (see [14] and the references therein). The authors in [15] and [16] noticed that in several situations fixed point results in Gmetric spaces can be in fact deduced from fixed point theorems in metric or quasimetric spaces. It has also been shown by various authors that in several cases the fixed point results in cone metric spaces can be obtained by reducing them to their standard metric counterparts; for example, see [17–19]. It is worth to note that in the above generalizations, a 2metric space was not known to be topologically equivalent to an ordinary metric.
We recollect some essential notations, required definitions and primary results coherent with the literature.
Definition 1.1 [10]
Let X be a nonempty set and let $d:X\times X\times X\to {\mathbb{R}}^{+}$ be a mapping satisfying the following assertions:
(d1) For every pair of distinct points $x,y\in X$, there exists a point $z\in X$ such that $d(x,y,z)\ne 0$;
(d2) If at least two of three points x, y, z are the same, then $d(x,y,z)=0$;
for all $x,y,z\in X$;
(d4) The rectangle inequality: $d(x,y,z)\le d(x,y,t)+d(y,z,t)+d(z,x,t)$ for all $x,y,z,t\in X$.
Then d is called a 2metric on X and $(X,d)$ is called a 2metric space which will be sometimes denoted by X if there is no confusion. Every member $x\in X$ is called a point in X.
Definition 1.2 [10]
is called a 2ball centered at a and b with radius r. The topology generated by the collection of all 2balls as a subbase is called a 2metric topology on X.
Definition 1.3 [20]
Let $\{{x}_{n}\}$ be a sequence in a 2metric space $(X,d)$.

$\{{x}_{n}\}$ is said to be convergent to x in $(X,d)$, written ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$, if for all $a\in X$, ${lim}_{n\to \mathrm{\infty}}d({x}_{n},x,a)=0$;

$\{{x}_{n}\}$ is said to be Cauchy in X if for all $a\in X$, ${lim}_{n,m\to \mathrm{\infty}}d({x}_{n},{x}_{m},a)=0$, that is, for each $\u03f5>0$, there exists ${n}_{0}$ such that $d({x}_{n},{x}_{m},a)<\u03f5$ for all $n,m\ge {n}_{0}$ and $a\in X$;

$(X,d)$ is said to be complete if every Cauchy sequence is a convergent sequence.
Definition 1.4 [20]
A 2metric space $(X,d)$ is said to be compact if every sequence in X has a convergent subsequence.
Lemma 1.1 [20]
Every 2metric space is a ${T}_{1}$space.
Lemma 1.2 [20]
${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$ in a 2metric space $(X,d)$ if and only if ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$ in the 2metric topological space X.
Lemma 1.3 [20]
If $T:X\to Y$ is a continuous map from a 2metric space X to a 2metric space Y, then ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$ in X implies ${lim}_{n\to \mathrm{\infty}}T{x}_{n}=Tx$ in Y.
It is straightforward from Definitions 1.11.3 that every 2metric is nonnegative and every 2metric space contains at least three distinct points. A 2metric $d(x,y,z)$ is sequentially continuous in one argument; moreover, if a 2metric $d(x,y,z)$ is sequentially continuous in two arguments, then it is sequentially continuous in all three arguments (see [21]). A convergent sequence in a 2metric space need not be a Cauchy sequence (see [21]). In a 2metric space $(X,d)$ every convergent sequence is a Cauchy sequence if d is continuous (see [21]). There exists a 2metric space $(X,d)$ such that every convergent sequence is a Cauchy sequence but d is not continuous (see [21]).
Chatterjea in [22] introduced the notion of Ccontraction as follows.
This notion was generalized to a weak Ccontraction by Choudhury in [23].
for all $x,y\in X$.
Samet et al. [24] defined the notion of αadmissible mappings as follows.
In [24] the authors considered the family Ψ of nondecreasing functions $\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ such that ${\sum}_{n=1}^{+\mathrm{\infty}}{\psi}^{n}(t)<+\mathrm{\infty}$ for each $t>0$, where ${\psi}^{n}$ is the n th iterate of ψ, and they gave the following theorem.
 (i)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},T{x}_{0})\ge 1$;
 (ii)
either T is continuous or for any sequence $\{{x}_{n}\}$ in X with $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$.
Then T has a fixed point.
Salimi et al. [25] modified and generalized the notions of αψcontractive and αadmissible mappings as follows.
Definition 1.8 [25]
Note that if we take $\eta (x,y)=1$, then this definition reduces to Definition 1.7. Also, if we take $\alpha (x,y)=1$, then we say that T is an ηsubadmissible mapping.
The following result properly contains Theorem 1.1 and Theorems 2.3 and 2.4 of [26].
Theorem 1.2 [25]
 (i)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},T{x}_{0})\ge 1$;
 (ii)
either T is continuous or for any sequence $\{{x}_{n}\}$ in X with $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$.
Then T has a fixed point.
Recently Karapinar et al. [27] introduced the notion of triangular αadmissible mapping as follows.
Definition 1.9 [27]
Let $T:X\to \phantom{\rule{0.25em}{0ex}}X$ and $\alpha :X\times X\to (\mathrm{\infty},+\mathrm{\infty})$. We say that T is a triangular αadmissible mapping if
(T1) $\alpha (x,y)\ge 1$ implies $\alpha (Tx,Ty)\ge 1$, $x,y\in X$,
(T2) $\{\begin{array}{l}\alpha (x,z)\ge 1,\\ \alpha (z,y)\ge 1\end{array}$ implies $\alpha (x,y)\ge 1$.
Motivated by the abovementioned developments, we first introduce the concepts of 2αηadmissible mappings and weak and rational αηψcontractions and establish the existence and uniqueness of fixed points for such mappings in complete 2metric spaces. As an application of obtained results, we prove some fixed point theorems in partially ordered 2metric spaces. The presented theorems generalize and improve many existing results in the literature. Moreover, some examples and an application to integral equations are provided to illustrate the usability of the proved results.
2 Fixed point results for weak αηCcontraction mappings
Motivated by Karapinar et al. [27] and Salimi et al. [25], we introduce the following notion.
Definition 2.1 Let $(X,d)$ be a 2metric space and $T:X\to X$ and $\alpha ,\eta :X\times X\times X\to [0,+\mathrm{\infty})$ be mappings. We say that T is a triangular 2αηadmissible mapping if for all $a\in X$,
(T1) $\alpha (x,y,a)\ge \eta (x,y,a)$ implies $\alpha (Tx,Ty,a)\ge \eta (Tx,Ty,a)$, $x,y\in X$,
(T2) $\{\begin{array}{l}\alpha (x,z,a)\ge \eta (x,z,a),\\ \alpha (z,y,a)\ge \eta (z,y,a)\end{array}$ implies $\alpha (x,y,a)\ge \eta (x,y,a)$.
If we take $\eta (x,y,a)=1$, then we say that T is a triangular 2αadmissible mapping. Also, if we take $\alpha (x,y,a)=1$, then we say that T is a triangular 2η subadmissible mapping.
Then T is a triangular 2αηadmissible mapping.
then from (T2) we have $\alpha ({x}_{m},{x}_{m+2},a)\ge \eta ({x}_{m},{x}_{m+2},a)$.
then we deduce $\alpha ({x}_{m},{x}_{m+3},a)\ge \eta ({x}_{m},{x}_{m+3},a)$.
By continuing this process, we get $\alpha ({x}_{m},{x}_{n},a)\ge \eta ({x}_{m},{x}_{n},a)$ as required. □
If we take $\eta (x,y,a)=1$, then we say that T is a 2αcontinuous mapping. Also, if we take $\alpha (x,y,a)=1$, then we say that T is a 2ηcontinuous mapping.
Clearly, T is not continuous, but T is 2αηcontinuous on $(X,d)$. Indeed, if ${x}_{n}\to x$ as $n\to \mathrm{\infty}$ and $\alpha ({x}_{n},{x}_{n+1},a)\ge \eta ({x}_{n},{x}_{n+1},a)=1$, then ${x}_{n}\in [0,1]$ and so ${lim}_{n\to \mathrm{\infty}}T{x}_{n}={lim}_{n\to \mathrm{\infty}}{x}_{n}^{2}={x}^{2}=Tx$.
Denote with Ψ the family of continuous functions $\psi :{[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty})$ such that $\psi (s,t)=0$ if and only if $s=t=0$.
We introduce the following notions as a modification of the approach in [25].
Definition 2.3 Let $(X,d)$ be a 2metric space and $T:X\to X$, $\alpha ,\eta :X\times X\times X\to [0,+\mathrm{\infty})$ be three mappings.

We say that T is a weak αηCcontraction mapping if$\begin{array}{r}x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x,y,a)\ge \eta (x,y,a)\\ \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(Tx,Ty,a)\le \frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))\end{array}$(2.2)
for all $a\in X$, where $\psi \in \mathrm{\Psi}$.

We say that T is a modified weak αCcontraction mapping if$\begin{array}{r}x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x,y,a)\ge 1\\ \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(Tx,Ty,a)\le \frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))\end{array}$
for all $a\in X$, where $\psi \in \mathrm{\Psi}$.

We say that T is a modified weak ηCcontraction mapping if$\begin{array}{r}x,y\in X,\phantom{\rule{1em}{0ex}}\eta (x,y,a)\le 1\\ \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(Tx,Ty,a)\le \frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))\end{array}$
for all $a\in X$, where $\psi \in \mathrm{\Psi}$.

We say that T is a weak αCcontraction mapping of type (I) if$\alpha (x,y,a)d(Tx,Ty,a)\le \frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))$
for all $x,y,a\in X$, where $\psi \in \mathrm{\Psi}$.

We say that T is a weak ηCcontraction mapping of type (I) if$d(Tx,Ty,a)\le \frac{\eta (x,y,a)}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))$
for all $x,y,a\in X$, where $\psi \in \mathrm{\Psi}$.

We say that T is a weak αCcontraction mapping of type (II) if${(\alpha (x,y,a)+\ell )}^{d(Tx,Ty,a)}\le {(1+\ell )}^{\frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))}$
for all $x,y,a\in X$, where $\psi \in \mathrm{\Psi}$ and $\ell >0$.

We say that T is a weak ηCcontraction mapping of type (II) if${(1+\ell )}^{d(Tx,Ty,a)}\le {(\eta (x,y,a)+\ell )}^{\frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))}$
for all $x,y,a\in X$, $\ell >0$ and $\psi \in \mathrm{\Psi}$.
Now we are ready to state and prove our first main result of this section.
 (i)
T is a triangular 2αηadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\alpha ({x}_{0},T{x}_{0},a)\ge \eta ({x}_{0},T{x}_{0},a)$ for all $a\in X$;
 (iii)
T is continuous or 2αηcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge \eta ({x}_{n},{x}_{n+1},a)$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge \eta ({x}_{n},x,a)$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
which implies $d({x}^{\ast},T{x}^{\ast},a)=0$, i.e., ${x}^{\ast}=T{x}^{\ast}$. □
By taking $\eta (x,y,a)=1$ in Theorem 2.1, we have the following corollary.
 (i)
T is a triangular 2αadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\alpha ({x}_{0},T{x}_{0},a)\ge 1$ for all $a\in X$;
 (iii)
T is continuous or 2αcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
Now, we prove that all the hypotheses of Corollary 2.1 (Theorem 2.1) are satisfied and hence T has a fixed point.
Proof Let $x,y,a\in X$, if $\alpha (x,y,a)\ge 1$, then $x,y\in [0,1]$. On the other hand, for all $w\in [0,1]$, we have $Tw\le 1$. Hence $\alpha (Tx,Ty,a)\ge 1$ for all $a\in X$. This implies that T is a 2αadmissible mapping. Clearly, $\alpha (0,T0,a)\ge 1$ for all $a\in X$. Now, if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and $a\in X$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, then $\{{x}_{n}\}\subseteq [0,1]$ and hence $x\in [0,1]$. This implies that $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and all $a\in X$.
for all $a\in X$. Hence, T is a modified weak αCcontraction mapping. Then all the hypotheses of Corollary 2.1 (Theorem 2.1) are satisfied and hence T has a fixed point. □
By taking $\alpha (x,y,a)=1$ in Theorem 2.1, we have the following corollary.
 (i)
T is a triangular 2ηsubadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\eta ({x}_{0},T{x}_{0},a)\le 1$ for all $a\in X$;
 (iii)
T is continuous or 2ηcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\eta ({x}_{n},{x}_{n+1},a)\le 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\eta ({x}_{n},x,a)\le 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
Now, we prove that all the hypotheses of Corollary 2.2 (Theorem 2.1) are satisfied and hence T has a fixed point.
Proof Let $x,y,a\in X$, if $\eta (x,y,a)\le 1$, then $x,y\in [0,1]$. On the other hand, for all $w\in [0,1]$, we have $Tw\le 1$. Hence $\eta (Tx,Ty,a)\le 1$ for all $a\in X$. This implies that T is a 2ηadmissible mapping. Clearly, $\eta (0,T0,a)\le 1$ for all $a\in X$.
Now, if $\{{x}_{n}\}$ is a sequence in X such that $\eta ({x}_{n},{x}_{n+1},a)\le 1$ for all $n\in \mathbb{N}\cup \{0\}$ and $a\in X$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, then $\{{x}_{n}\}\subseteq [0,1]$ and hence $x\in [0,1]$. This implies that $\eta ({x}_{n},x,a)\le 1$ for all $n\in \mathbb{N}\cup \{0\}$ and all $a\in X$.
for all $a\in X$. Hence, T is a modified weak ηCcontraction mapping. Then all the hypotheses of Corollary 2.2 (Theorem 2.1) are satisfied and hence T has a fixed point. □
 (i)
T is a triangular 2αadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\alpha ({x}_{0},T{x}_{0},a)\ge 1$ for all $a\in X$;
 (iii)
T is continuous or 2αcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
 (i)
T is a triangular 2ηadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\eta ({x}_{0},T{x}_{0},a)\le 1$ for all $a\in X$;
 (iii)
T is continuous or 2ηcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\eta ({x}_{n},{x}_{n+1},a)\le 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\eta ({x}_{n},x,a)\le 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
 (A)
For all $x,y\in X$, where $\alpha (x,y,a)<\eta (x,y,a)$ and $\alpha (y,x,a)<\eta (y,x,a)$ for all $a\in X$, there exists $z\in X$ such that $\alpha (x,z,a)\ge \eta (x,z,a)$ or $\alpha (z,x,a)\ge \eta (z,x,a)$ and $\alpha (y,z,a)\ge \eta (y,z,a)$ or $\alpha (z,y,a)\ge \eta (z,y,a)$ for all $a\in X$.
Theorem 2.2 Adding condition (A) to the hypotheses of Theorem 2.1 (resp. Corollary 2.1, 2.2, 2.3 and 2.4), we obtain the uniqueness of the fixed point of T.
Proof Assume that ${x}^{\ast}$ and ${y}^{\ast}$ are two fixed points of T. We consider to following cases.
That is, $\psi (d({x}^{\ast},{y}^{\ast},a),d({x}^{\ast},{y}^{\ast},a))=0$ for all $a\in X$. So, $d({x}^{\ast},{y}^{\ast},a)=0$ for all $a\in X$. Hence, ${x}^{\ast}={y}^{\ast}$.
and so $\psi (\ell ,\ell )=0$. Therefore, $\ell =0$. That is, ${lim}_{n\to \mathrm{\infty}}{T}^{n}z={x}^{\ast}$. Similarly, we can deduce ${lim}_{n\to \mathrm{\infty}}{T}^{n}z={y}^{\ast}$. Then by Lemma 1.1 we get ${x}^{\ast}={y}^{\ast}$. □
3 Fixed point results for rational contraction in 2metric spaces
In this section, we prove certain fixed point theorems for a rational contraction mapping via a triangular 2αηadmissible mapping.
Denote with ${\mathrm{\Psi}}_{\phi}$ the family of continuous functions $\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ such that $\phi (t)=0$ if and only if $t=0$.
Definition 3.1 Let $(X,d)$ be a 2metric space and $T:X\to X$, $\alpha ,\eta :X\times X\times X\to [0,+\mathrm{\infty})$ be three mappings.

We say that T is a modified rational αηφcontraction mapping if$\begin{array}{r}x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x,y,a)\ge \eta (x,y,a)\\ \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(Tx,Ty,a)\le M(x,y,a)\phi (M(x,y,a))\end{array}$(3.1)

We say that T is a modified rational αφcontraction mapping if$x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x,y,a)\ge 1\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(Tx,Ty,a)\le M(x,y,a)\phi (M(x,y,a))$
for all $a\in X$, where $\phi \in {\mathrm{\Psi}}_{\phi}$.

We say that T is a modified rational ηφcontraction mapping if$x,y\in X,\phantom{\rule{1em}{0ex}}\eta (x,y,a)\le 1\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(Tx,Ty,a)\le M(x,y,a)\phi (M(x,y,a))$
for all $a\in X$, where $\phi \in {\mathrm{\Psi}}_{\phi}$.

We say that T is a rational αφcontraction mapping if$\alpha (x,y,a)d(Tx,Ty,a)\le M(x,y,a)\phi (M(x,y,a))$
for all $x,y,a\in X$, where $\phi \in {\mathrm{\Psi}}_{\phi}$.

We say that T is a rational ηφcontraction mapping if$d(Tx,Ty,a)\le \eta (x,y,a)M(x,y,a)\phi (M(x,y,a))$
for all $x,y,a\in X$, where $\phi \in {\mathrm{\Psi}}_{\phi}$.
 (i)
T is a triangular 2αηadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\alpha ({x}_{0},T{x}_{0},a)\ge \eta ({x}_{0},T{x}_{0},a)$ for all $a\in X$;
 (iii)
T is continuous or 2αηcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge \eta ({x}_{n},{x}_{n+1},a)$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge \eta ({x}_{n},x,a)$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
Therefore, $M({x}_{n},{x}_{n1},a)=max\{d({x}_{n},{x}_{n1},a),d({x}_{n},{x}_{n+1},a)\}$.
Since ${lim}_{n\to \mathrm{\infty}}M({x}_{n},{x}^{\ast},a)=d({x}^{\ast},T{x}^{\ast},a)$, then $\phi (d({x}^{\ast},T{x}^{\ast},a))=0$, i.e., $d({x}^{\ast},T{x}^{\ast},a)=0$ for all $a\in X$. Thus, ${x}^{\ast}=T{x}^{\ast}$. □
 (i)
T is a triangular 2αadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\alpha ({x}_{0},T{x}_{0},a)\ge 1$ for all $a\in X$;
 (iii)
T is continuous or 2αcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
Now, we prove that all the hypotheses of Corollary 3.1 (Theorem 3.1) are satisfied and hence T has a fixed point.
Proof As in the proof of Example 2.3 we can show that T is a 2αadmissible mapping, $\alpha (0,T0,a)\ge 1$ for all $a\in X$ and if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and $a\in X$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and all $a\in X$.
for all $a\in X$. Hence, T is a modified rational αφcontraction mapping. Then all the conditions of Corollary 3.1 (Theorem 3.1) are satisfied and hence T has a fixed point. □
By taking $\alpha (x,y,a)=1$ in Theorem 3.1, we have the following corollary.
 (i)
T is a triangular 2ηadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\eta ({x}_{0},T{x}_{0},a)\le 1$ for all $a\in X$;
 (iii)
T is continuous or 2ηcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\eta ({x}_{n},{x}_{n+1},a)\le 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\eta ({x}_{n},x,a)\le 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
 (i)
T is a triangular 2αadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\alpha ({x}_{0},T{x}_{0},a)\ge 1$ for all $a\in X$;
 (iii)
T is continuous or 2αcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
 (i)
T is a triangular 2ηadmissible mapping;
 (ii)
there exists ${x}_{0}$ in X such that $\eta ({x}_{0},T{x}_{0},a)\le 1$ for all $a\in X$;
 (iii)
T is continuous or 2ηcontinuous; or
 (iv)
if $\{{x}_{n}\}$ is a sequence in X such that $\eta ({x}_{n},{x}_{n+1},a)\le 1$ for all $a\in X$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\eta ({x}_{n},x,a)\le 1$ for all $n\in \mathbb{N}$ and all $a\in X$.
Then T has a fixed point.
4 Fixed point results in partially ordered 2metric spaces
Recently, there have been so many exciting developments in the field of existence of fixed points in partially ordered sets. This approach was initiated by Ran and Reurings [28] and they also provided some applications to matrix equations. Their results are a hybrid of the two classical theorems: Banach’s fixed point theorem and Tarski’s fixed point theorem. Agarwal et al. [29], Bhaskar and Lakshmikantham [30], Ciric et al. [31] and Hussain et al. [32, 33] presented some new results for nonlinear contractions in partially ordered metric spaces and noted that their theorems can be used to investigate a large class of problems. In this section, as an application of obtained results we prove some fixed point results in partially ordered 2metric spaces. We also note that the recent fixed point results in [34] can be deduced as simple corollaries.
Recall that if $(X,\u2aaf)$ is a partially ordered set and $T:X\to X$ is such that for $x,y\in X$, $x\u2aafy$ implies $Tx\u2aafTy$, then the mapping T is said to be nondecreasing.
Theorem 4.1 (Theorems 2.3 and 2.4 of [34])
 (i)
T is nondecreasing;
 (ii)
there exists ${x}_{0}$ in X such that ${x}_{0}\u2aafT{x}_{0}$;
 (iii)
T is continuous; or
 (iv)
if $\{{x}_{n}\}$ is a nondecreasing sequence in X such that ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then ${x}_{n}\u2aafx$ for all $n\in \mathbb{N}$;
 (v)$d(Tx,Ty,a)\le \frac{1}{2}[d(x,Ty,a)+d(y,Tx,a)]\psi (d(x,Ty,a),d(y,Tx,a))$(4.1)
holds for all $x,y,a\in X$ with $x\u2aafy$ or $y\u2aafx$, where $\psi \in \mathrm{\Psi}$.
Then T has a fixed point.
Again let $x,y,a\in X$ such that $\alpha (x,y,a)\ge 1$. This implies that $x\u2aafy$. As the mapping T is nondecreasing, we deduce that $Tx\u2aafTy$ and hence $\alpha (Tx,Ty,a)\ge 1$ for all $a\in X$. Also, let $\alpha (x,z,a)\ge 1$ and $\alpha (z,y,a)\ge 1$, then $x\u2aafz$ and $z\u2aafy$. So from transitivity we have $x\u2aafy$. That is, $\alpha (x,y,a)\ge 1$ for all $a\in X$. Thus T is a triangular 2αadmissible mapping. The condition (ii) ensures that there exists ${x}_{0}\in X$ such that ${x}_{0}\u2aafT{x}_{0}$. This implies that $\alpha ({x}_{0},T{x}_{0},a)\ge 1$ for all $a\in X$. Let $\{{x}_{n}\}$ be a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $a\in X$ and all $n\in \mathbb{N}$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$. So, ${x}_{n}\u2aaf{x}_{n+1}$ for all $n\in \mathbb{N}$. Then from (iv) we have ${x}_{n}\u2aafx$ for all $n\in \mathbb{N}$. That is, $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}$ and all $a\in X$. Therefore, all the conditions of Corollary 2.1 are satisfied, so T has a fixed point in X. □
(B) For all $x,y\in X$ which are not comparable, there exists $z\in X$ that is comparable to x and y.
Theorem 4.2 (Theorem 2.5 of [34])
Adding condition (B) to the hypotheses of Theorem 4.1, we obtain the uniqueness of the fixed point of T.
Proof Define the mapping $\alpha :X\times X\times X\to {\mathbb{R}}_{+}$ as in the proof of Theorem 4.1. Let $x,y\in X$, where $\alpha (x,y,a)<1$ and $\alpha (y,x,a)<1$ for all $a\in X$. That is, x and y are not comparable. Hence, by condition (B) there exists $z\in X$ that is comparable to x and y, i.e., $z\u2aafx$ or $x\u2aafz$ and $z\u2aafy$ or $y\u2aafx$. That is, $\alpha (z,x,a)\ge 1$ or $\alpha (x,z,a)\ge 1$ and $\alpha (z,y,a)\ge 1$ or $\alpha (y,z,a)\ge 1$ for all $a\in X$. Then the conditions of Theorem 2.2 hold and the fixed point of T is unique. □
 (i)
T is nondecreasing;
 (ii)
there exists ${x}_{0}$ in X such that ${x}_{0}\u2aafT{x}_{0}$;
 (iii)
T is continuous; or
 (iv)
if $\{{x}_{n}\}$ is a nondecreasing sequence in X such that ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then ${x}_{n}\u2aafx$ for all $n\in \mathbb{N}$;
 (v)T is an ordered modified rational φcontraction mapping, that is,$d(Tx,Ty,a)\le M(x,y,a)\phi (M(x,y,a))$(4.2)
Then T has a fixed point.
Again let $x,y,a\in X$ such that $\alpha (x,y,a)\ge 1$. This implies that $x\u2aafy$. As the mapping T is nondecreasing, we deduce that $Tx\u2aafTy$ and hence $\alpha (Tx,Ty,a)\ge 1$ for all $a\in X$. Also, let $\alpha (x,z,a)\ge 1$ and $\alpha (z,y,a)\ge 1$, then $x\u2aafz$ and $z\u2aafy$. So from transitivity we have $x\u2aafy$. That is, $\alpha (x,y,a)\ge 1$ for all $a\in X$. Thus T is a triangular 2αadmissible mapping. The condition (ii) ensures that there exists ${x}_{0}\in X$ such that ${x}_{0}\u2aafT{x}_{0}$. This implies that $\alpha ({x}_{0},T{x}_{0},a)\ge 1$ for all $a\in X$. Let $\{{x}_{n}\}$ be a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge 1$ for all $a\in X$ and all $n\in \mathbb{N}$ and ${x}_{n}\to x$ as $n\to \mathrm{\infty}$. So, ${x}_{n}\u2aaf{x}_{n+1}$ for all $n\in \mathbb{N}$. Then from (iv) we have ${x}_{n}\u2aafx$ for all $n\in \mathbb{N}$. That is, $\alpha ({x}_{n},x,a)\ge 1$ for all $n\in \mathbb{N}$ and all $a\in X$. Therefore, all the conditions of Corollary 3.1 are satisfied, so T has a fixed point in X. □
5 Application to existence of solutions of integral equations
for all $x,y,a\in X$. Then $(X,d)$ is a complete 2metric space.
 (A)
$f:[0,T]\times \mathbb{R}\to \mathbb{R}$ is continuous;
 (B)
$p:[0,T]\to \mathbb{R}$ is continuous;
 (C)
$S:[0,T]\times \mathbb{R}\to [0,+\mathrm{\infty})$ is continuous and ${\int}_{0}^{T}S(t,s)\phantom{\rule{0.2em}{0ex}}ds\le 1$;
 (D)there exist $0\le r<1$ and two functions $\alpha ,\eta :X\times X\times X\to [0,\mathrm{\infty})$ such that for all $s\in [0,T]$,$\begin{array}{r}x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x(s),y(s),a(s))\ge \eta (x(s),y(s),a(s))\\ \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}\alpha (Fx(s),Fy(s),a(s))\ge \eta (Fx(s),Fy(s),a(s))\end{array}$
 (F)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},F({x}_{0}),a)\ge \eta ({x}_{0},F({x}_{0}),a)$;
 (G)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1},a)\ge \eta ({x}_{n},{x}_{n+1},a)$ with ${x}_{n}\to x$ as $n\to \mathrm{\infty}$, then $\alpha ({x}_{n},x,a)\ge \eta ({x}_{n},x,a)$ for all $n\in \mathbb{N}\cup \{0\}$.
Theorem 5.1 Under the assumptions (A)(G), the integral equation (5.1) has a solution in $X=C([0,T],\mathbb{R})$.