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Some fixed point results for nondecreasing and mixed monotone mappings with auxiliary functions
Fixed Point Theory and Applicationsvolume 2015, Article number: 209 (2015)
Abstract
The purpose of this paper is to provide sufficient conditions for the existence and uniqueness of fixed points for nondecreasing and mixed monotone mappings with auxiliary functions in the framework of metric space endowed with a partial order. As applications of our results we obtain several interesting corollaries and fixed point theorems in the underlying spaces. In order to illustrate our results, we provide two examples in which other theorems from the literature cannot be applied. In addition, the equivalence property between unidimensional and multidimensional fixed point theorems is investigated. We also present some applications to the existence of solutions of integral and differential equations.
Introduction
One of the newest branches of fixed point theory is devoted to the study of coupled fixed points, introduced by Guo and Lakshmikantham [1] in 1987. Thereafter, Gnana Bhaskar and Lakshmikantham [2] introduced the concept of the mixed monotone property for contractive operators in two variables in the setting of a partially ordered metric space, and they established some coupled fixed point theorems. Their results were extended and generalized by several authors in the last few years; see [3–21] and the references cited therein. Recently, Roldán et al. [22] introduced the notion of coincidence point between mappings in any number of variables, and several special extended notions of socalled coupled, tripled, quadrupled, and multidimensional fixed/coincidence points appeared in the literature; see, for example, [2, 3, 13, 23], respectively.
Harjani and Sadarangani [24] investigated some unidimensional fixed point theorems for generalized contractions in complete partially ordered metric spaces and applications to ordinary differential equations. In [23] and [25] the authors obtained some multidimensional fixed point theorems for mixed monotone mappings, which extended the corresponding coupled, tripled, and quadrupled fixed point results appearing in the literature. In 2014, Wang [26] obtained some multidimensional fixed point theorems for isotone mappings and extended some of the results in coupled, tripled, quadrupled and multidimensional fixed/coincidence points for mixed monotone and nondecreasing mappings in the framework of complete partially ordered metric spaces. She also gave a simple and unified approach to coupled, tripled, quadrupled, and multidimensional fixed point theorems for mixed monotone mappings.
Motivated and inspired by the results of [22–26], we establish some fixed point theorems for nondecreasing and mixed monotone mappings with auxiliary functions in a complete partially ordered metric space. The auxiliary functions used in the paper are more general than the gauge functions appearing in the literature. Our results improve and generalize the wellknown results of Harjani and Sadarangani [24] and Wang [26]. By using the theorems, we also obtain several interesting corollaries and fixed point theorems in the underlying spaces. In order to illustrate our results, we provide two examples in which the theorems of [24] cannot be applied. In addition, the equivalence property between unidimensional and multidimensional fixed point theorems is investigated. Also as applications of our results, we provide two examples for the solution of integral and differential equations.
Basic concepts
In order to fix the framework needed to state our main results, we recall the following notions.
For simplicity, we denote from now on $\underbrace{X\times X\cdot \cdot\cdot X\times X}_{k}$ by $X^{k}$, where $k\in\mathbb{N}$ and X is a nonempty set. If elements x, y of a partially ordered set $(X,\leq)$ are comparable (i.e., $x\leq y$ or $y\leq x$ holds), we write $x\asymp y$. Let $\{A,B\}$ be a partition of the set $\Lambda_{k}=\{1,2,\ldots,k\}$, that is, $A\cup B=\Lambda_{k}$ and $A\cap B=\emptyset$, $\Omega_{A,B}=\{\sigma :\Lambda_{k}\rightarrow\Lambda_{k}:\sigma(A)\subseteq A \mbox{ and } \sigma (B)\subseteq B\} $ and $\Omega^{{\prime }}_{A,B}=\{\sigma:\Lambda_{k}\rightarrow\Lambda_{k}:\sigma(A)\subseteq B \mbox{ and } \sigma(B)\subseteq A\}$. Henceforth, let $\sigma_{1},\sigma _{2},\ldots,\sigma_{k}$ be k mappings from $\Lambda_{k}$ into itself, and ϒ be the ktuple $(\sigma_{1},\sigma_{2},\ldots,\sigma_{k})$. For brevity, $(y_{1},y_{2},\ldots,y_{k}),(v_{1},v_{2},\ldots,v_{k}), (y^{n}_{1},y^{n}_{2},\ldots,y^{n}_{k})$ and $(x^{1}_{0},x^{2}_{0},\ldots,x^{k}_{0})$ will be denoted by Y, V, $Y_{n}$ and $X_{0}$, respectively.
Let $(X,\leq)$ be a partially ordered set and d be a metric on X. We use the next notation from [22]:
The product space $X^{k}$ is endowed with the following natural partial order: for $Y,V\in X^{k}$,
Obviously, $(X^{k},\preceq)$ is a partially ordered set. The mapping $\rho_{k}:X^{k}\times X^{k}\rightarrow[0,\infty)$, defined by
where $Y,V\in X^{k}$, is a metric on $X^{k}$. It is easy to see that, for $Y_{n}$, $Y\in X^{k}$,
Definition 2.1
([7])
Let $(R,\leq)$ be a partially ordered set and d be a metric on R. We say that $(R,d,\leq)$ is regular if the following conditions hold:

(i)
if $\{x_{n}\}$ is a nondecreasing sequence such that $x_{n}\rightarrow x$, then $x_{n}\leq x$ for all n,

(ii)
if $\{y_{n}\}$ is a nonincreasing sequence such that $y_{n}\rightarrow y$, then $y_{n}\geq y$ for all n.
Definition 2.2
([2])
Let $(X,\leq)$ be a partially ordered set and $F:X^{2}\rightarrow X$. A mapping F has the mixed monotone property if F is monotone nondecreasing in its first argument and monotone nonincreasing in its second argument, that is, for any $x,y\in X$,
and
Definition 2.3
([3])
Let $(X,\leq)$ be a partially ordered set and $F:X^{3}\rightarrow X$. A mapping F has the mixed monotone property if $F(x,y,z)$ is monotone nondecreasing in x, monotone nonincreasing in y, and monotone nondecreasing in z, that is, for any $x,y,z\in X$,
and
Definition 2.4
([22])
Let $(X,\leq)$ be a partially ordered set and $F:X^{n}\rightarrow X$. A mapping F has the mixed monotone property if F is monotone nondecreasing in arguments of A and monotone nonincreasing in arguments of B, i.e., for all $x_{1},x_{2},\ldots,x_{n}, y,z\in X$, and all i,
Definition 2.5
Let $F:X^{k}\rightarrow X$ be a mapping. A point $(x_{1},x_{2},\cdot\cdot \cdot,x_{k})\in X^{k}$ is:

(i)
a coupled fixed point [2] if $k=2$, $F(x_{1},x_{2})=x_{1}$, and $F(x_{2},x_{1})=x_{2}$;

(ii)
a tripled fixed point [3] if $k=3$, $F(x_{1},x_{2},x_{3})=x_{1}$, $F(x_{2},x_{1},x_{2})=x_{2}$, and $F(x_{3},x_{2},x_{1})=x_{3}$;

(iii)
a ϒfixed point [25] of F if $F(x_{\sigma_{i}(1)},x_{\sigma_{i}(2)},\ldots,x_{\sigma_{i}(k)})=x_{i}$ for $i\in \Lambda_{k}$.
Definition 2.6
([24])
Let $(X,\leq)$ be a partially ordered set. A mapping $f:X\rightarrow X$ is monotone nondecreasing if $x,y\in X$, $x\leq y\Rightarrow f(x)\leq f(y)$.
Definition 2.7
([26])
An element $Y\in X^{k}$ is called a fixed point of the mapping $T:X^{k}\rightarrow X^{k}$ if $T(Y)=Y$.
Lemma 2.8
([27])
Let $(X,\leq)$ be a partially ordered set and d a metric on X. If $(X,\leq,d)$ is regular, then $(X^{k},\preceq,\rho_{k})$ is regular.
Definition 2.9
A family of functions $f:[0,\infty)^{2}\to \mathbb{R}$ is called Cclass if the following conditions hold:

(a)
$f (s,t)\le s$;

(b)
f is continuous;

(c)
$f (s,t)=s$ implies that either $s=0$ or $t=0$.
For brevity, we denote the Cclass by $\mathcal{C}$.
Example 2.10
([28])
The following functions $f:[0,\infty)^{2}\to\mathbb{R}$ are elements of $\mathcal{C}$. For each $s,t\in[0,\infty)$:

(1)
$f(s,t)=st$, $f (s,t)=s \Rightarrow t=0$;

(2)
$f(s,t)=\frac{st}{1+t}$, $f (s,t)=s \Rightarrow t=0$;

(3)
$f(s,t)=\frac{s}{1+t}$, $f (s,t)=s \Rightarrow s=0$ or $t=0$;

(4)
$f(s,t)=\log_{a}\frac{t+a^{s}}{1+t}$, $a>1$, $f (s,t)=s \Rightarrow s=0$ or $t=0$;

(5)
$f(s,t)=ks$, $0< k<1$, $f(s,t)=s \Rightarrow s=0$;

(6)
$f(s,t)=(s+l)^{\frac{1}{1+t}}l$, $l>1$, $f(s,t)=s \Rightarrow t=0$;

(7)
$f(s,t)=s\log_{a+t}a$, $a>1$, $f(s,t)=s \Rightarrow s=0$ or $t=0$;

(8)
$f(s,t)=s\frac{t}{1+t}$, $f(s,t)=s \Rightarrow t=0$;

(9)
$f(s,t)=s(\frac{1+s}{2+s})t$, $f(s,t)=s \Rightarrow t=0$;

(10)
$f(s,t)=\frac{s}{k+t}$, $k>1$, $f(s,t)=s \Rightarrow s=0$.
Remark 2.11
It is easy to verify that Cclass is a natural generalization of classical Banach contraction principle (see (5) of Example 2.10).
Definition 2.12
([29])
An altering distance function is a function $\psi:[0,\infty )\rightarrow[0,\infty)$ which satisfies:

(a)
ψ is continuous and strictly increasing.

(b)
$\psi(t)=0$ if and only if $t=0$.
Lemma 2.13
([30])
Suppose that $(X,d)$ is a metric space. Let $\{x_{n}\}$ be a sequence in X such that $d(x_{n}, x_{n+1})\rightarrow0$ as $n\rightarrow\infty$. If $\{x_{n}\}$ is not a Cauchy sequence then there exist an $\epsilon>0$ and sequences of positive integers $\{ m(k)\}$ and $\{n(k)\}$ with $m(k)>n(k)>k$ such that $d(x_{m(k)},x_{n(k)})\geq\epsilon$, $d(x_{m(k)1},x_{n(k)})\leq\epsilon$ and:

(i)
$\lim_{k\rightarrow\infty}d(x_{m(k)1},x_{n(k)+1})=\epsilon$;

(ii)
$\lim_{k\rightarrow\infty}d(x_{m(k)},x_{n(k)})=\epsilon$;

(iii)
$\lim_{k\rightarrow\infty}d(x_{m(k)1},x_{n(k)})=\epsilon$.
Remark 2.14
Using a similar argument to the proof of Lemma 2.13, we get
Existence of fixed points
In this section, we state and prove the existence of fixed points for nondecreasing and mixed monotone mappings with auxiliary functions in the setting of complete partially ordered metric spaces. In addition, the equivalence property between unidimensional and multidimensional fixed point theorems is investigated.
We denote Φ the set of all continuous and strictly increasing functions $\varphi:[0,\infty)\rightarrow[0,\infty)$, and Ψ the set of all functions, such that $\lim_{t\rightarrow r}\psi(t)>0$ for every $r>0$ and $\psi (t)=0\Longleftrightarrow t=0$.
Theorem 3.1
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Assume that $T:X\rightarrow X$ is a nondecreasing mapping for which there exist $h \in \mathcal{C}$, $\varphi\in\Phi$, and $\psi\in\Psi$ such that, for all $y, v\in X$ with $y\geq v$,
Suppose that either

(a)
T is continuous, or

(b)
$(X, \leq, d)$ is regular.
If there exists $z_{0}\in X$ such that $z_{0}\asymp T(z_{0})$, then T has a fixed point.
Proof
Starting with $z_{0}$ in X, define the sequence $\{z_{n}\}\subset X$ by $z_{n+1}=T(z_{n})$ for $n\geq0$. Obviously, if $z_{n_{0}+1}=z_{n_{0}}$ for some $n_{0}\geq0$, then $z_{n_{0}}$ is a fixed point of T. So, assume that $z_{n+1}\neq z_{n}$ for every $n\geq0$.
Since $z_{0}\asymp T(z_{0})$, without loss of generality, we may assume that $z_{0}\leq T(z_{0})$ (the case $z_{0}\geq T(z_{0})$ is treated similarly), that is, $z_{0}\leq z_{1}$. Since T is a nondecreasing mapping, it is easy to verify that the sequence $\{z_{n}\}^{\infty}_{n=0}$ is nondecreasing. Taking $y=z_{n}$ and $v=z_{n1}$ in (3), we obtain
for all $n\geq1$.
Since φ is strictly increasing, we have
Hence, the sequence $\{\delta_{n}\}^{\infty}_{n=0}$ given by $\delta_{n}=d(z_{n+1},z_{n})$ is a monotone decreasing and bounded below. Therefore, there exists some $\delta\geq0$ such that $\lim_{n\rightarrow \infty}\delta_{n}=\delta$. Now, we shall prove that $\delta=0$. Assume that $\delta>0$. Using Definition 2.9, we know that when $h(s,t)=s$, then $s=0$ or $t=0$ and $h(s,t)< s$ when $s>0$ and $t>0$. Using the properties of φ and ψ, we have $\varphi(\delta )>\varphi(0)\geq0$ and $\lim_{n\rightarrow\infty}\psi(\delta _{n1})>0$. Therefore, by letting $n\rightarrow\infty$ in (4) and using the properties of h, we have
which is a contradiction. Thus, $\lim_{n\rightarrow\infty}\delta_{n}=0$.
Suppose that $\{z_{n}\}$ is not a Cauchy sequence. Using Lemma 2.13, we see that there exist $\epsilon>0$ and two sequences $\{n(t)\}$ and $\{ m(t)\}$ of positive integers such that, for all $t\in\mathbb{N}$, $m(t)>n(t)\geq t$ and
Since $n(t)< m(t)$, we have $z_{m(t)}\geq z_{n(t)}$. Hence, using (3) with $v=z_{n(t)}$ and $y=z_{m(t)}$, we obtain
Using the properties of φ and ψ, we have $\varphi(\epsilon)>0$ and $\lim_{t\rightarrow\infty}\psi(r_{t})>0$, where $r_{t}=d(z_{n(t)},z_{m(t)})$. Letting $t\rightarrow\infty$ in (6) and using (5), we have
which is a contradiction. Hence, the sequence $\{z_{n}\}^{\infty}_{n=0}$ is a Cauchy sequence in the metric space $(X,d)$. Since $(X,d)$ is complete, there exists $\bar{z}\in X$ such that $\lim_{n\rightarrow\infty}z_{n}=\bar{z}$.
Now suppose that (a) holds. It follows from $z_{n+1}=T(z_{n})$ that z̄ is a fixed point of T, that is, $T(\bar{z})=\bar{z}$.
Suppose that (b) holds. Since $\{z_{n}\}^{\infty}_{n=0}$ is a nondecreasing sequence that converges to z̄, we have $z_{n}\leq\bar{z}$ for all $n\geq0$. From (3) and the fact that $h\in\mathcal{C}$, we obtain
for all $n\geq0$. From (7) and the strict monotonicity of φ, we have
Letting $n\rightarrow\infty$ in (8) and using $\lim_{n\rightarrow \infty}z_{n}=\bar{z}$, we get $d(\bar{z},T(\bar{z}))\leq d(\bar{z},\bar{z})=0$, and so $d(\bar{z},T(\bar{z}))=0$, which implies $\bar {z}=T(\bar{z})$. □
Remark 3.2
In Theorem 3.1, we use $h\in\mathcal{C}$, and $\mathcal{C}$ is a class of more general functions than the gauge function used in Theorems 2.1 and 2.2 of [24]. Indeed, the gauge function, $h(s,t)=st$ in Theorems 2.1 and 2.2 of [24] is an element of $\mathcal{C}$.
Taking $h(s,t)=st$ or $h(s,t)=\frac{s}{1+t}$ in Theorem 3.1, we obtain the following results immediately.
Corollary 3.3
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Assume that $T:X\rightarrow X$ is a nondecreasing mapping for which there exist $\varphi\in\Phi$ and $\psi\in\Psi$ such that, for all $y,v\in X$ with $y\geq v$,
Suppose that either

(a)
T is continuous, or

(b)
$(X,\leq,d)$ is regular.
If there exists $z_{0}\in X$ such that $z_{0}\asymp T(z_{0})$, then T has a fixed point.
Remark 3.4
We note that, if ψ is an altering distance function, then $\psi \in\Psi$. But the reverse is not true in general (see Example 3.5). Therefore, Corollary 3.3 generalizes the wellknown results of Harjani and Sadarangani’s [24] in the framework of partially ordered metric spaces (see Theorems 2.1 and 2.2 in [24]).
Example 3.5
Let $X=\{0,0.5,1,1.5,2\}$ with the usual metric $d(x,y)=xy$, for all $x,y\in X$. We consider the following order relation on X:
Consider $T:X\rightarrow X$ and $\psi,\varphi:[0,\infty)\rightarrow [0,\infty)$ defined by
It is easy to verify the following statements.

(1)
$(X,d)$ is a complete metric space and $(X,\leq,d)$ is regular.

(2)
T is a continuous mapping.

(3)
$\varphi\in\Phi$ and $\psi\in\Psi$.

(4)
Take $z_{0}=1$. Then $1=T(1)$.

(5)
T is a nondecreasing mapping. Indeed, let $x,y\in X$ such that $x\leq y$, (a) if $x=y$, then $T(x)=T(y)$; (b) if $(x,y)=(0,2)$, then $T(x)=T(y)$. Therefore, T is a nondecreasing mapping.

(6)
T satisfies (9). In fact, it is evident that (9) holds when $y=v$. If $(v,y)=(0,2)$, then
$$\varphi\bigl(d\bigl(T(y),T(v)\bigr)\bigr)=\varphi\bigl(d(0,0)\bigr)=0\leq1.5= \varphi(2)\psi (2)=\varphi\bigl(d(y,v)\bigr)\psi\bigl(d(y,v)\bigr). $$
This shows that all the conditions of Corollary 3.3 are satisfied. Therefore by using Corollary 3.3, T has a fixed point. Indeed, 0 and 1 are two fixed points of T. However, Theorems 2.1 and 2.2 of Harjani and Sadarangani’s [24] cannot be applied to this example because ψ is not an altering distance function.
Corollary 3.6
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Assume that $T:X\rightarrow X$ is a nondecreasing mapping for which there exist $\varphi\in\Phi$ and $\psi\in\Psi$ such that, for all $y,v\in X$ with $y\geq v$,
Suppose that either

(a)
T is continuous, or

(b)
$(X,\leq,d)$ is regular.
If there exists $z_{0}\in X$ such that $z_{0}\asymp T(z_{0})$, then T has a fixed point.
Example 3.7
Let $(X,\leq)$ be the partially ordered set with $X=[0,\infty)$ and the natural ordering of the real number as the partial ordering ≤. Consider the metric on X: $d(x,y)=xy$ for all $x,y\in X$. Then $(X,d)$ is complete and $(X,\leq,d)$ is regular.
Let $T:X\rightarrow X$ be defined by $T(x)=\frac{x}{1+x}$ for all $x\in X$. Then T is a nondecreasing and continuous mapping. Consider $\psi,\varphi:[0,\infty)\rightarrow[0,\infty)$ defined by $\psi (t)=\varphi(t)=t$. It is obvious that $\varphi\in\Phi$ and $\psi \in\Psi$.
We now show that T satisfies (10). Consider
This shows that all the conditions of Corollary 3.6 are satisfied and hence T has a fixed point. Indeed, 0 is a unique fixed points of T. However, Theorems 2.1 and 2.2 of Harjani and Sadarangani’s [24] cannot be applied to this example because the T does not satisfy the inequality $\varphi(d(T(x),T(y)))\leq\varphi (d(x,y))\psi(d(x,y))$ for all $x\geq y$. Indeed, if $x\neq y$, then
Now we give the following multidimensional fixed point theorem for mixed monotone mappings.
Theorem 3.8
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Suppose that $\Upsilon=(\sigma_{1},\sigma_{2},\ldots,\sigma_{k})$ is a ktuple of mapping from $\Lambda_{k}$ into itself such that $\sigma_{i}\in\Omega _{A,B}$ if $i\in A$ and $\sigma_{i}\in\Omega^{\prime}_{A,B}$ if $i\in B$ and $F:X^{k}\rightarrow X$ is a mixed monotone mapping and there exist $h \in\mathcal{C}$, $\varphi\in\Phi$, and $\psi\in\Psi$ such that
for which $x_{i}\leq_{i}y_{i}$ for $i\in\Lambda_{k}$. Further assume that either F is continuous or $(X,\leq,d)$ is regular. If there exist $x^{1}_{0},x^{2}_{0},\ldots,x^{k}_{0}\in X$ such that either of the following conditions is satisfied:
or
Then F has one ϒfixed point.
Proof
Let $T:X^{k}\rightarrow X^{k}$ be defined by
for $Y\in X^{k}$. Using similar arguments to the proof of Corollary 3.11 in [26], T is a nondecreasing mapping.
Using the inequalities (1), (11) and (14), we have
where $\rho_{k}$ is defined by (2). Since $(X,\leq,d)$ is a complete partially ordered metric space, so is $(X^{k},\preceq,\rho _{k})$. As F is continuous and using inequality (14), T is continuous. Using Lemma 2.8, we have $(X^{k},\preceq,\rho_{k})$ is regular. From (12) and (13), there exists $X_{0}\in X^{k}$ such that $X_{0}\asymp T(X_{0})$. Therefore, by using Theorem 3.1, T has a fixed point. Hence F has a ϒfixed point. □
Remark 3.9
Note that the multidimensional fixed point theorem (Theorem 3.8) is equivalent to the unidimensional fixed point theorem (Theorem 3.1). Also, we find that Theorem 3.8 is a consequence of Theorem 3.1. Conversely, taking $k=1$, $A=\{1\}$, $B=\emptyset$, and $F=T$ in Theorem 3.8, we obtain Theorem 3.1 immediately.
Using similar arguments to the proof of Theorem 3.8, the following results are immediate consequences of the unidimensional fixed point theorem (Theorem 3.1).
Corollary 3.10
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Assume that $F:X^{2}\rightarrow X$ is a mixed monotone mapping for which there exist $h \in \mathcal{C}$, $\varphi\in\Phi$, and $\psi\in\Psi$ such that, for all $x,y,u,v\in X$ with $x\geq u$, $y\leq v$,
Suppose that either

(a)
F is continuous, or

(b)
$(X,d,\leq)$ is regular.
If there exist $x_{0},y_{0}\in X$ such that
or
then there exist $\bar{x},\bar{y}\in X$ such that $\bar{x}=F(\bar {x},\bar{y}) $ and $\bar{y}=F(\bar{y},\bar{x})$, that is, F has a coupled fixed point.
Corollary 3.11
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Assume that $F:X^{3}\rightarrow X$ is a mixed monotone mapping for which there exist $h \in \mathcal{C}$, $\varphi\in\Phi$, and $\psi\in\Psi$ such that, for all $x,y,z,u,v,w\in X$ with $x\geq u$, $y\leq v$, $z\geq w$,
Suppose that either

(a)
F is continuous, or

(b)
$(X,\leq,d)$ is regular.
If there exist $x_{0},y_{0},z_{0}\in X$ such that
or
then there exist $x,y,z\in X$ such that
that is, F has a tripled fixed point.
Using Theorem 3.1, we obtain the following result on multidimensional fixed points, which generalizes Theorem 3.1 of Wang [26].
Theorem 3.12
Let $(X,\leq)$ be a partially ordered set and suppose that there is a metric d on X such that $(X,d)$ is a complete metric space. Assume that $T:X^{k}\rightarrow X^{k}$ is a nondecreasing mapping for which there exist $h\in\mathcal{C}$, $\varphi\in\Phi$, and $\psi\in\Psi$ such that, for all $Y, V\in X^{k}$ with $Y\succeq V$,
where $\rho_{k}$ is defined by (2). Suppose that either

(a)
T is continuous, or

(b)
$(X, \leq, d)$ is regular.
If there exists $Z_{0}\in X^{k}$ such that $Z_{0}\asymp T(Z_{0})$, then T has a fixed point.
Proof
Using Lemma 2.8, we find that $(X^{k},\preceq,\rho_{k})$ is regular. By our assumptions, all the conditions of Theorem 3.1 are satisfied in the setting of a complete partially ordered metric space $(X^{k},\preceq,\rho_{k})$. Therefore, by using Theorem 3.1, T has a fixed point. □
Remark 3.13
(a) The metric $\rho_{k}$ in Theorem 3.12 can be replaced by some other metrics on $X^{k}$, for example,
where $Y,V\in X^{k}$.
(b) We also find that the multidimensional fixed point theorem (Theorem 3.12) is equivalent to the unidimensional fixed point theorem (Theorem 3.1). In fact, Theorem 3.12 is a consequence of Theorem 3.1. Conversely, taking $k=1$ in Theorem 3.12, we obtain Theorem 3.1 immediately.
Uniqueness of fixed points
Now, we state and prove the uniqueness of fixed points in the setting of a complete partially ordered metric space.
Theorem 4.1
In addition to the hypotheses of Theorem 3.1, suppose that, for all fixed points $\bar{y},y^{*}\in X$ of T, there exists $z\in X$ such that z is comparable to ȳ and to $y^{*}$. Then T has a unique fixed point.
Proof
From Theorem 3.1, the set of fixed points of T is nonempty. Assume that ȳ and $y^{*}$ are two fixed points of T. Put $z_{0}=z$ and $z_{n+1}=T(z_{n})$ for $n\geq0$. Since z is comparable to ȳ, we may assume $z\leq\bar{y}$. Since T is a nondecreasing mapping, we obtain inductively $z_{n}\leq\bar{y}$ for $n\geq0$. Therefore, by using (3), we have
which implies that
Using the inequality (16) and the strict monotonicity of φ, the sequence $\{\Delta_{n}\}$ defined by $\Delta_{n}=d(\bar{y},z_{n})$ is nonincreasing. Hence, there exists $\beta\geq0$ such that $\lim_{n\rightarrow\infty}\Delta_{n}=\beta$. Now we shall prove that $\beta=0$. Suppose, conversely, that $\beta>0$. Using the properties of φ and ψ, we have $\varphi(\beta)>0$ and $\lim_{\Delta_{n}\rightarrow\beta}\psi(\Delta_{n})>0$. Letting $n\rightarrow\infty$ in (15), we get
which is a contradiction. Thus $\beta=0$, that is,
Similarly, we obtain
Combining (17) and (18) yields $y^{*}=\bar{y}$. □
Using similar arguments to the proof of Theorem 3.8, we deduce the following corollaries from Theorem 4.1.
Corollary 4.2
In addition to the hypotheses of Theorem 3.8, suppose that, for all ϒfixed points $\bar{Y},Y^{*}\in X^{k}$ of F, there exists $Z\in X^{k}$ such that Z is comparable to Ȳ and to $Y^{*}$. Then F has a unique ϒfixed point.
Corollary 4.3
In addition to the hypotheses of Corollary 3.10, suppose that, for all coupled fixed points $\bar{Y},Y^{*}\in X^{2}$ of F, there exists $Z\in X^{2}$ such that Z is comparable to Ȳ and to $Y^{*}$. Then F has a unique coupled fixed point.
Corollary 4.4
In addition to the hypotheses of Corollary 3.11, suppose that, for all tripled fixed points $\bar{Y},Y^{*}\in X^{3}$ of F, there exists $Z\in X^{3}$ such that Z is comparable to Ȳ and to $Y^{*}$. Then F has a unique tripled fixed point.
Application to integral equations
In this section, we present two examples where our main results can be applied. Consider the integral equation
where $T>0$. Consider $C[0,T]=\{x:[0,T]\rightarrow\mathbb{R}: x\mbox{ is continuous on }[0,T]\}$ equipped with the metric
It is clear that $(C[0,T],d)$ is a complete metric space. $C[0,T]$ can be equipped with the following partial order:
Clearly, $(C[0,T],\leq)$ satisfies the following condition (see [24]):
Due to [31], we know that $(C[0,T],d,\leq)$ is regular and condition (20) is equivalent to
Theorem 5.1
Assume that the following hypotheses hold:

(i)
$K:[0,T]\times[0,T]\times\mathbb{R}\rightarrow[0,\infty )$ and $g:[0,T]\rightarrow[0,\infty)$ are continuous,

(ii)
there exist $h \in\mathcal{C}$, $\psi\in\Psi$, and a continuous function $\tilde{G}:[0,T]\times[ 0,T]\rightarrow[0,\infty)$ such that
$$ 0\leq K(t,s,y)K(t,s,x)\leq\tilde{G}(t,s)h\bigl(d(x,y),\psi\bigl(d(x,y)\bigr) \bigr) $$for all $s,t\in[0,T]$ and $x,y\in C[0,T]$ with $x\leq y$;

(iii)
$\sup_{t\in[0,T]}\int_{0}^{T}\tilde {G}(t,s)\,ds\leq1$ or $\sup_{t\in[0,T]}(\int_{0}^{T}\tilde{G}(t,s)^{2}\,ds)^{\frac{1}{2}}\leq\frac{1}{\sqrt{T}}$.
Then the integral equation (19) has a unique nonnegative solution $u^{\ast}\in C[0,T]$.
Proof
Let $H:C[0,T]\rightarrow C[0,T]$ be the following function:
Assume that $x\leq y$. From (ii), for all $s,t\in[0,T]$, we have $K(t,s,x(s))\leq K(t,s,y(s))$. So
that is, H is a nondecreasing mapping. Suppose that $\sup_{t\in[0,T]}\int_{0}^{T}\tilde {G}(t,s)\,ds\leq1$. It follows from (ii) that
for all $x\leq y$.
Now suppose that $\sup_{t\in[0,T]}(\int_{0}^{T}\tilde {G}(t,s)^{2}\,ds)^{\frac{1}{2}}\leq\frac{1}{\sqrt{T}}$. From (ii) and CauchySchwarz inequality, we see that
for all $x\leq y$. Now, if choose $\varphi(t)=t$, then we get
for all $x,y\in C[0,T]$ with $x\leq y$. Finally, as K and g are nonnegative functions, we obtain
Therefore by using Theorems 3.1 and 4.1, H has a unique fixed point $u^{\ast}\in C[0,T]$ and $u^{\ast}$ is a unique nonnegative solution of (19). □
Consider the following twopoint boundary value problem of second order differential equation:
It is well known that $x\in C^{2}[0,1]$ being a solution of (21) is equivalent to $x\in C^{2}[0,1]$ being a solution of the integral equation
where $G(t,s)$ is the Green function given by
Corollary 5.2
([24])
Consider problem (21) with $f:[0,1]\times\mathbb {R}\rightarrow[0,\infty)$ continuous and nondecreasing with respect to the second variable and suppose that there exists $0<\alpha\leq8$ such that for $x,y\in\mathbb{R}$ with $y\geq x$
Then our problem (21) has a unique nonnegative solution.
Proof
Take $K(t,s,x(s))=G(t,s)f(s,x(s))$ and $g(t)=0$ for all $s,t\in[0,1]$. It is clear that condition (i) of Theorem 5.1 holds since $f(t,x)$ and $G(t,s)$ are nonnegative continuous functions.
Now we show that condition (ii) holds. As f is nondecreasing with respect to the second variable, for all $s,t\in[0,1]$ and $x,y\in C[0,1]$ with $x\leq y$, we have
It follows from (22) that
for all $s,t\in[0,1]$ and $x,y\in C[0,1]$ with $x\leq y$. If we choose $h(s,t)=\sqrt{\ln(s^{2}+1)}$ and $\tilde{G}(t,s)=\alpha G(t,s)$, then condition (ii) of Theorem 5.1 holds.
It is easy to verify that $\int^{1}_{0}G(t,s)\,ds=\frac{t^{2}}{2}+\frac {t}{2}$ and that
Therefore, all hypotheses of Theorem 5.1 are satisfied. Thus our problem (21) has a unique nonnegative solution. □
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Acknowledgements
The authors are grateful to the anonymous referees for their helpful comments which improved the presentation of the original version of this paper. This work was supported by the Natural Science Foundation of Jiangsu Province under grant (13KJB110028).
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MSC
 47H10
 54H25
Keywords
 nondecreasing mapping
 mixed monotone mapping
 fixed point
 partially ordered metric space