A Generalized Contraction Principle Under w-Distance for a Partially Ordered Metric Space

Abstract

Fixed point results for generalized weak contractions under w-distance are proved using discontinuous control functions that are more relaxed than functions used in related work for metric spaces. Later, we apply our theory to coupled coincidence point problems and existence of solution of Fredholm type integral equation. We present examples to justify our claims.

1 Introduction and Preliminaries

Applications of fixed point theorems are very important in diverse disciplines of mathematics, statistics, chemistry, biology, computer science, engineering, and economics in dealing with problems arising in approximation theory, potential theory, game theory, mathematical economics, theory of differential equations, theory of integral equations, theory of matrix equations, etc. For example, fixed point theorems are extremely useful when it comes to prove the existence of various types of Nash equilibria in economics. One of the very popular tools of fixed point theory is the Banach contraction principle which first appeared in 1922. It states that if (X, d) is a complete metric space and $T:X\rightarrow X$ is a contraction mapping, that is, $d(Tx,Ty)\le kd(x,y)$ for all $x,y\in X$, where k is a non-negative number such that $k<1$, then T has a unique fixed point. Several mathematicians have been dedicated to improvement and generalization of this principle. Fixed point theory in partially ordered metric spaces is of relatively recent origin. An early result in this direction is due to Turinici [18], in which fixed point problems were studied in partially ordered uniform spaces. Later, in 2004, Ran and Reurings [17] showed the existence of fixed points of nonlinear contraction mappings in metric spaces endowed with a partial ordering and presented applications of their results to matrix equations. Subsequently, Nieto and Rodríguez-López [16] extended the result of Ran and Reurings [17] for nondecreasing mappings and applied their results to get a unique solution for a first-order differential equation. Bhaskar and Lakshmikantham [5] introduced the notion of a coupled fixed point of a mapping F from $X\times X$ into X. They established some coupled fixed point results and applied their results to the study of existence and uniqueness of solution for a periodic boundary value problem. Lakshmikantham and Ćirić [15] introduced the concept of coupled coincidence points and proved coupled coincidence and coupled common fixed point results for mappings F from $X \times X$ into X and g from X into X satisfying nonlinear contraction in ordered metric space. The studies of asymmetric structures and their applications in mathematics are important. One of the types of asymmetric structures on a metric space was introduced by Kada et al. [13] in 1996 known as a w-distance, and he proved some fixed point theorems using it. Since then, many fixed point results have been developed by different authors using w-distance on metric spaces or a generalized w-distance such as c-distance on cone metric spaces. For more study in this area, one may refer to [1–4, 6]. Weak contraction was studied in partially ordered metric spaces by Harjani et al. [10]. In a recent result by Choudhury et al. [8], a generalization of the above result to a coincidence point theorem has been done using three control functions. Later, in 2013, Choudhury et al. [9] proved coincidence point results by assuming a weak contraction inequality with three control functions, two of which are not continuous. The results were obtained under two sets of additional conditions.

By ${\mathbb {R}}$, ${\mathbb {Q}}$, ${\mathbb {N}}$, and $\upchi _A$, we shall mean the set of real numbers, the set of rational numbers , the set of natural numbers, and the characteristic function of a set A defined by $\upchi (x)=1$ if $x\in A$ and $\upchi (x)=0$ otherwise, respectively. Following basic definitions can be referred in [5, 7, 9, 11, 12, 14, 15]. Let T and S be any two self maps on a metric space (X, d). T and S are said to be compatible if $\lim _{n\rightarrow \infty }d(TSx_n,STx_n)=0$ whenever $\{x_n\}$ is a sequence in X such that $\lim _{n\rightarrow \infty }Tx_n=\lim _{n\rightarrow \infty }Sx_n=t$ for some $t\in X$ and weakly compatible if T and S commute at their coincidence points. Let $\sqsubseteq$ be a partial order relation on X and g be a self mapping on X. Then T is said to be g-nondecreasing if $gx\sqsubseteq gy\Rightarrow Tx\sqsubseteq Ty$ for all $x,y\in X$ and g-nonincreasing if $gx\sqsupseteq gy\Rightarrow Tx\sqsubseteq Ty$ for all $x,y\in X$. Let $F:X\times X\rightarrow X$ be any mapping. F is said to have mixed monotone property if for any $x, y\in X$ we have $x_1, x_2\in X, x_1\sqsubseteq x_2 \Rightarrow F(x_1, y)\sqsubseteq F(x_2, y)$ and $y_1, y_2\in X, y_1\sqsubseteq y_2\Rightarrow F(x, y_1)\sqsupseteq F (x, y_2)$ and mixed g-monotone property if for any $x, y\in X$ we have $x_1, x_2 \in X,gx_1\sqsubseteq gx_2\Rightarrow F(x_1, y)\sqsubseteq F(x_2, y)$ and $y_1, y_2 \in X,gy_1\sqsubseteq gy_2\Rightarrow F(x, y_1)\sqsupseteq F(x, y_2).$ An element $(x, y)\in X \times X$ is called a coupled fixed point of F if $F(x, y) = x$ and $F(y, x) = y$ and a coupled coincidence point of F and g if $F(x, y) = gx$ and $F(y, x) = gy$. The mappings F and g are said to commute if $gF(x, y) = F(gx, gy)$ for all $x, y\in X$ , compatible if $\lim _{n\rightarrow \infty }d(F(gx_n,gy_n),gF(x_n,y_n))=0$, $\lim _{n\rightarrow \infty }d(F(gy_n,gx_n),gF(y_n,x_n))=0$ whenever $\{x_n\}$ and $\{y_n\}$ are sequences in X such that $\lim _{n\rightarrow \infty }F(x_n,y_n)=\lim _{n\rightarrow \infty }gx_n=x$ and $\lim _{n\rightarrow \infty }F(y_n,x_n)=\lim _{n\rightarrow \infty }gy_n=y$ for some $x,y\in X$ and weakly compatible if they commute at their coupled coincidence points. It is simple to observe that compatibility implies weakly compatibility. X is called regular if for any nonincreasing sequence $\{x_n\}$ in X such that $x_n\rightarrow x\in X$ we have $x\sqsubseteq x_n$ for all n and for any nondecreasing sequence $\{x_n\}$ in X such that $x_n\rightarrow x\in X$ we have $x_n\sqsubseteq x$ for all n. Next we give the notation of w-distance of Kada et al. [13] with some properties.

Definition 1.1

([13]) Let (X, d) be a metric space. A function $p:X\times X\rightarrow [0,\infty )$ is called a w-distance on X if the following conditions hold:

(w1):

$p(x,z)\le p(x,y)+p(y,z)$ for all $x,y,z\in X$,

(w2):

p(x, . ) is lower semi-continuous for all $x\in X$. That is, if $x\in X$ and $y_n\rightarrow y$ in X then $p(x,y)\le \liminf p(x,y_n)$.

(w3):

For all $\epsilon >0$, there exists $\delta >0$ such that $p(z,x)\le \delta$ and $p(z,y)\le \delta$ imply $d(x,y)\le \epsilon$.

Example 1.2

• (i) Every metric d is a w-distance on (X, d) as $d(x,y)\ge 0$ for all $x,y\in X$, (w1) holds because it is nothing but the triangular inequality of the metric d on X, (w2) follows by taking $z=y_n$ in the inequality $|d(x,y)-d(x,z)|\le d(y,z)$ that holds for all $x,y,z\in X$, and for (w3), one can use triangular inequality and take $\delta =\epsilon /2$.

• (ii) For any positive real number $\lambda$, $p(x,y)=\lambda d(x,y)$ is a w-distance on a metric space (X, d) and argument is similar to what is applied to (i).

• (iii) Consider $X=[0,\infty )$ together with the Euclidean metric d on X. Let $f:X\rightarrow X$ be any given mapping. Define a mapping $p:X\times X\rightarrow {\mathbb {R}}$ by $p(x,y) = f(x)+y$ for all $x,y\in X$. Then p is a w-distance on X as $p:X\times X\rightarrow [0,\infty )$, (w1) holds because for any $x,y,z\in X$ we have $p(x,y)+p(y,z)-p(x,z)=f(x)+y+f(y)+z-f(x)-z=y+f(y)\ge 0$, (w2) holds since for any sequence $y_n\rightarrow y$ in X we have for any $x\in X$, $f(x)+y_n\rightarrow f(x)+y$ and hence $p(x,y)=f(x)+y=\lim _{n\rightarrow \infty } (f(x)+y_n)=\liminf (f(x)+y_n)=\liminf p(x,y_n)$. Further, for any $\epsilon >0$, there exists $\delta =\epsilon /2>0$ such that $p(z,x)=f(z)+x\le \delta$ and $p(z,y)=f(z)+y\le \delta$ together imply $d(x,y)=|x-y|=|(f(z)+x)-(f(z)+y)|\le |f(z)+x|+|f(z)+y|=p(z,x)+p(z,y)\le 2\delta =\epsilon$. So (w3) holds.

• (iv) Consider $X=[0,\infty )$ together with the Euclidean metric d on X. Define a mapping $p:X\times X\rightarrow {\mathbb {R}}$ by $p(x,y) = y$ for all $x,y\in X$. Then p is a w-distance on X as it is a particular case of w-distance defined in (iii) for $f=0$.

• (v) $p(x,y)=d(u,y)$ is a w-distance on a metric space (X, d) where u is a fixed point in X. Clearly $p(x,y)=d(u,y)\ge 0$ for all $x,y\in X$. Further for any $x,y,z\in X$, we have $p(x,y)=d(u,y)\le d(u,z)+d(u,y)=p(x,z)+p(z,y)$. So (w1) holds. (w2) holds because of continuity of the function d(u, .). Further, for any $\epsilon >0$, there exists $\delta =\epsilon /2>0$ such that $p(z,x)=d(u,x)\le \delta$ and $p(z,y)=d(u,y)\le \delta$ together imply $d(x,y)=d(x,u)+d(u,y)\le 2\delta =\epsilon$. So (w3) holds.

Lemma 1.3

([13]) Let (X, d) be a metric space and p be a w-distance on X. Let $\{x_n\}$ and $\{y_n\}$ be sequences in X and $x,y,z\in X$. Suppose that $u_n$ and $v_n$ are sequences in $[0,\infty )$ converging to 0. Then the following hold:

• (i) If $p(x_n,y)\le u_n$ and $p(x_n,z)\le v_n$, then $y=z$. In particular if $p(x,y)=0$ and $p(x,z)=0$ then $y=z$.

• (ii) If $p(x_n,y_n)\le u_n$ and $p(x_n,z)\le v_n$, then $y_n$ converges to z.

• (iii) If $p(x_n,x_m)\le u_n$ for $m>n$, then $\{x_n\}$ is a Cauchy sequence in X.

• (iv) If $p(y,x_n)\le u_n$, then $\{x_n\}$ is a Cauchy sequence in X.

Lemma 1.4

([13]) Let p be a w-distance on metric space (X, d) and $\{x_n\}$ be a sequence in X such that for each $\epsilon > 0$ there exist $N_{\epsilon }\in {\mathbb {N}}$ such that $m > n > N_{\epsilon }$ implies $p(x_n,x_m) <\epsilon$ (or $\lim _{m,n} p(x_n, x_m) = 0$ ) then $\{x_n\}$ is a Cauchy sequence.

In this paper, we relax some conditions on control functions used by Choudhury et al. in [9] and use them to prove coincidence point results using w-distance in weak contraction inequalities. Since every metric is a w-distance, therefore, our results may be applied even when we use metric in weak contraction inequalities. Thus all results of Choudhury et al. in [9] hold even after ignoring some conditions on the control function used by them and are implied by our results for such control functions.

2 Main Results

We denote by $\Psi$, the set of all functions $\psi :[0,\infty )\rightarrow [0,\infty )$ such that $\psi$ is continuous and nondecreasing and $\psi (t)=0\Rightarrow t=0$. Further, by $\Theta$ we denote the set of all functions $\alpha :[0,\infty )\rightarrow [0,\infty )$ such that $\alpha$ is bounded on any bounded interval in $[0,\infty )$.

Theorem 2.1

Let $(X, \sqsubseteq ,d)$ be a partially ordered metric space equipped with a w-distance p and T, g be two self maps on X satisfying the following conditions:

• (i) $T(X)\subseteq g(X)$ ,T is g-nondecreasing, there exists an $x_0 \in X$ with $gx_0\sqsubseteq Tx_0$ or $Tx_0\sqsubseteq gx_0$ and either T and g are continuous, compatible with (X, d) complete or g(X) is complete and (X, d) is regular.

• (ii) There exist $\psi \in \Psi$ and $\varphi ,\theta \in \Theta$ such that for all $s,t\in [0,\infty )$, we have $\psi (s)\le \varphi (t)\Rightarrow s\le t.$ Further, for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, we have $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$ and for all $x,y\in X$ with $gx\sqsubseteq gy$ or $gx\sqsupseteq gy$, $\psi (p(Tx,Ty))\le \varphi (p(gx,gy))-\theta (p(gx,gy))$.

Then T and g have a coincidence point in X.

Proof

Let us first assume that (X, d) is complete, T and g are continuous and the pair (T, g) is compatible. Also assume that $Tx_0\sqsubseteq gx_0$. Since $T(X)\subseteq g(X)$, one can find $x_1\in X$ such that $Tx_0=gx_1$. Again, we can choose $x_2\in X$ such that $Tx_1=gx_2$. Continuing like this we can find $x_n\in X$ such that $Tx_n=gx_{n+1}$ for all $n\ge 0.$ Further, $gx_1\sqsubseteq gx_0$ and T is g-nondecreasing implies $Tx_1\sqsubseteq Tx_0$ or $gx_2\sqsubseteq gx_1$ and hence $Tx_2\sqsubseteq Tx_1$. Continuing this process, we obtain $gx_n\sqsupseteq gx_{n+1}$ and $Tx_n\sqsupseteq Tx_{n+1}$ for all $n\ge 0$. Let $Q_n=p(gx_n,gx_{n+1})$ and $R_n=p(gx_{n+1},gx_n)$ for all $n\ge 0$. Then

$$\begin{aligned} \psi (p(gx_{n+1},gx_{n+2}))=\psi (p(Tx_n,Tx_{n+1}))\le \varphi (p(gx_n,gx_{n+1}))-\theta (p(gx_n,gx_{n+1})),\\ \psi (p(gx_{n+2},gx_{n+1}))=\psi (p(Tx_{n+1},Tx_n)) \le \varphi (p(gx_{n+1},gx_n))-\theta (p(gx_{n+1},gx_n)). \end{aligned}$$

That is, for all $n\ge 0$, $\psi (Q_{n+1})\le \varphi (Q_n)-\theta (Q_n)$ and $\psi (R_{n+1})\le \varphi (R_n)-\theta (R_n)$. This implies that $\psi (Q_{n+1})\le \varphi (Q_n)$ and $\psi (R_{n+1})\le \varphi (R_n)$ and hence $Q_{n+1}\le Q_n$, $R_{n+1}\le R_n$ for all $n\ge 0$. So, there exist $q\ge 0$ and $r\ge 0$ such that $Q_n=p(gx_n,gx_{n+1})\rightarrow q$ and $R_n=p(gx_{n+1},gx_n)\rightarrow r$ as $n\rightarrow \infty .$ Then for all $n\ge 0$, we have $\psi (q)\le \limsup \varphi (Q_n)+\limsup (-\theta (Q_n))$ and $\psi (r)\le \limsup \varphi (R_n)+\limsup (-\theta (R_n)).$ That is, $\psi (q)\le \limsup \varphi (Q_n)-\liminf \theta (Q_n)$ and $\psi (r)\le \limsup \varphi (R_n)-\liminf \theta (R_n)$. or $\psi (q)-\limsup \varphi (Q_n)+\liminf \theta (Q_n)\le 0$ and $\psi (r)-\limsup \varphi (R_n)+\liminf \theta (R_n)\le 0.$ This gives a contradiction unless $q=0$ and $r=0$. Thus $Q_n=p(gx_n,gx_{n+1})\rightarrow 0$ and $R_n=p(gx_{n+1},gx_n)\rightarrow 0$ as $n\rightarrow \infty .$ Next, we show that

$$\begin{aligned} \lim _{m,n} p(gx_m,gx_n)=0. \end{aligned}$$

(2.1)

Suppose (2.1) does not hold. Then there exists an $\epsilon >0$ such that for all positive integers p, we get two positive integers $n_p,m_p$ with $n_p>m_p>p$ and $p(gx_{m_p},gx_{n_p})\ge \epsilon .$ Also, we can find a positive integer k such that $p(gx_n,gx_{n+1})<\epsilon$ for all $n\ge k$. For the choice of $p=k$ and $n=m_k$, we get two positive integers $n_k,m_k$ with $n_k>m_k>k$,

$$\begin{aligned} p(gx_{m_k},gx_{n_k})\ge \epsilon \end{aligned}$$

(2.2)

and $p(gx_{m_k},gx_{m_k+1})<\epsilon$. Clearly $n_k\ne m_k+1$. We can assume that $n_k$ is the smallest positive integer such that (2.2) holds. Then $p(gx_{m_k},gx_r)< \epsilon$ for $r\in \{m_k+1,m_k+2,\ldots n_k-1\}.$ In particular, $p(gx_{m_k},gx_{n_k-1})<\epsilon$. Now

$$\begin{aligned} \epsilon \le p(gx_{m_k},gx_{n_k})\le p(gx_{m_k},gx_{n_k-1})+p(gx_{n_k-1},gx_{n_k})<\epsilon +p(gx_{n_k-1},gx_{n_k}). \end{aligned}$$

Letting $k\rightarrow \infty$, we get $\lim _{k\rightarrow \infty }p(gx_{m_k},gx_{n_k}) =\epsilon .$ Again

$$\begin{aligned}&p(gx_{m_k},gx_{n_k})\le p(gx_{m_k},gx_{m_k+1})+p(gx_{m_k+1},gx_{n_k+1})+p(gx_{n_k+1},gx_{n_k})\text { and }\\&p(gx_{m_k+1},gx_{n_k+1})\le p(gx_{m_k+1},gx_{m_k})+p(gx_{m_k},gx_{n_k})+p(gx_{n_k},gx_{n_k+1}). \end{aligned}$$

Letting $k\rightarrow \infty$ in above inequalities, we get $\lim _{k\rightarrow \infty }p(gx_{m_k+1},gx_{n_k+1})=\epsilon$. Since $n_k>m_k$, we have $gx_{n_k}\sqsubseteq gx_{m_k}$. So we get

$$\begin{aligned} \psi (p(gx_{m_k+1},gx_{n_k+1}))=\psi (p(Tx_{m_k},Tx_{n_k}))\le \varphi (p(gx_{m_k},gx_{n_k}))-\theta (p(gx_{m_k},gx_{n_k})) \end{aligned}$$

This gives

$$\begin{aligned} \psi (\epsilon )&\le \limsup \varphi (p(gx_{m_k},gx_{n_k}))+\limsup (-\theta (p(gx_{m_k},gx_{n_k})))\\&=\limsup \varphi (p(gx_{m_k},gx_{n_k}))-\liminf \theta (p(gx_{m_k},gx_{n_k})). \end{aligned}$$

Hence $\psi (\epsilon )- \limsup \varphi (p(gx_{m_k},gx_{n_k}))+\liminf \theta (p(gx_{m_k},gx_{n_k}))\le 0.$ Thus we get a contradiction. So (2.1) must be true. By Lemma 1.4, the sequence $\{gx_n\}$ is a Cauchy sequence in X. Since X is a complete metric space, therefore, there exists an $x\in X$ such that $\lim _{n\rightarrow \infty }Tx_n=\lim _{n\rightarrow \infty }gx_n=x.$ Since the pair (T, g) is compatible, we have $\lim _{n\rightarrow \infty }d(Tgx_n,gTx_n)=0$. By triangular inequality,

$$\begin{aligned} d(gx,Tgx_n)\le d(gx,gTx_n)+d(gTx_n,Tgx_n) \end{aligned}$$

Taking limit $n\rightarrow \infty$ in above inequality, and using continuities of T and g, we get $d(gx,Tx)=0$. That is $Tx=gx$. Thus T and g have a coincidence point at x.

Now, we consider the case that (X, d) is regular and g(X) is a complete subspace of X. As in the first case, choose $x_0\in X$ satisfying $Tx_0\sqsubseteq gx_0$ and by using it, construct a sequence $\{x_n\}$ in X satisfying $Tx_n=gx_{n+1}$ such that $\{gx_n\}$ is a nonincreasing Cauchy sequence satisfying (2.1). Since g(X) is complete, therefore, there exists $x\in X$ such that $gx_n\rightarrow gx$ as $n\rightarrow \infty$. By regularity of X we have $gx_n\ge gx$ for all $n\ge 0$. This implies that, for all $n\ge 1$

$$\begin{aligned} \psi (p(gx_n,Tx))&=\psi (p(Tx_{n-1},Tx))\le \varphi (p(gx_{n-1},gx))-\theta (p(gx_{n-1},gx))\\&\le \varphi (p(gx_{n-1},gx)) \end{aligned}$$

and hence $p(gx_n,Tx)\le p(gx_{n-1},gx).$ Since $\{gx_n\}$ satisfies (2.1), for any $k\in {{\mathbb {N}}}$, there exists a positive integer $m_k$ such that $p(gx_{m_k},gx_n)<1/k$ for all $n>m_k$. This implies that for all integers $k\ge 1$,

$$\begin{aligned} p(gx_{m_k+1},Tx)\le p(gx_{m_k},gx)\le \liminf p(gx_{m_k},gx_n)\le 1/k=\alpha _k \end{aligned}$$

and also

$$\begin{aligned} p(gx_{m_k},Tx)\le p(gx_{m_k},gx_{m_k+1})+p(gx_{m_k+1},Tx)\le p(gx_{m_k},gx_{m_k+1})+\alpha _k=\beta _k \end{aligned}$$

Now, using Lemma 1.3(i) and the fact that $\alpha _k\rightarrow 0$ and $\beta _k\rightarrow 0$ as $k\rightarrow \infty$, we get $Tx=gx$. That is, T and g have coincidence point at x. $\square$

Corollary 2.2

In the hypothesis of Theorem 2.1, if we replace (ii) by any one of the following:

($ii_a$):

There exists $\varphi \in \Theta$ such that $\varphi (0)=0$ and for all $x,y\in X$ with $gx\sqsubseteq gy$ or $gx\sqsupseteq gy$, we have $p(Tx,Ty)\le \varphi (p(gx,gy))$. Further, assume that for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, we have $\limsup \varphi (x_n)<t$.

($ii_b$):

There exist $\psi \in \Psi$ and $k\in [0,1)$ such that for all $x,y\in X$ with $gx\sqsubseteq gy$ or $gx\sqsupseteq gy$, we have $\psi (p(Tx,Ty))\le k\,\psi (p(gx,gy))$.

($ii_c$):

There exists an increasing function $\psi \in \Psi$ and a function $\theta \in \Theta$ such that for all $x,y\in X$ with $gx\sqsubseteq gy$ or $gx\sqsupseteq gy$, we have $\psi (p(Tx,Ty))\le \psi (p(gx,gy))-\theta (p(gx,gy))$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, we have $\liminf \theta (x_n)>0.$

($ii_d$):

There exists $\theta \in \Theta$ such that for all $x,y\in X$ with $gx\sqsubseteq gy$ or $gx\sqsupseteq gy$, we have $p(Tx,Ty)\le p(gx,gy))-\theta (p(gx,gy))$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, we have $\liminf \theta (x_n)>0.$

($ii_e$):

There exists $k\in [0,1)$ such that for all $x,y\in X$ with $gx\sqsubseteq gy$ or $gx\sqsupseteq gy$, we have $p(Tx,Ty)\le$ kp(gx, gy).

Then T and g have a coincidence point in X.

Proof

• ($ii_{a}$) Take $\psi$ as an identity mapping on $[0,\infty )$ and $\theta =0$ in Theorem 2.1. Let $\psi (t)\le \varphi (s)$ for $t,s\ge 0$. This implies $t\le \varphi (s)\text { for }t,s\ge 0.$ If $s>0$ then take $x_n=s$. So we have $\varphi (s)<s$ whenever $s>0$. Since $\varphi (0)=0$ we get $t\le \varphi (s)\le s$ for all $s\ge 0$. Conclusion follows now by Theorem 2.1.

• ($ii_{b}$) Take $\theta =0$ and $\varphi =k\,\psi$ in Theorem 2.1.

• ($ii_{c}$) Take $\varphi =\psi$ in Theorem 2.1.

• ($ii_{d}$) Take $\varphi =\psi$ as an identity mapping on $[0,\infty )$ in Theorem 2.1.

• ($ii_{e}$) Take $\theta =0$, $\varphi (t)=kt$ for all $t\ge 0$ and $\psi$ as an identity mapping on $[0,\infty )$, in Theorem 2.1. $\square$

Remark 2.3

In Theorem 2.1, compatibility of T and g can be substituted with commutativity because commutativity of T and g implies compatibility of the pair (T, g).

Theorem 2.4

In addition to the hypotheses of Theorem 2.1 and Corollary 2.2 as stated above, suppose that for any $x,y\in X$ there exists $u\in X$ such that Tu is comparable to both Tx and Ty, then, there is a unique point of coincidence. Further, if the pair (T, g) is weakly compatible, then, T and g have a unique common fixed point.

Proof

Let x and y be coincidence points of T and g. That is $gx=Tx$ and $gy=Ty$. By assumption, there exists $u\in X$ such that Tu is comparable with Tx and Ty. Let $Tu\sqsubseteq Tx$ (proof is similar for the other case and can be outlined with a little adjustment). Let $u_0=u$ and since $T(X)\subseteq g(X)$, there exists $u_1\in X$ such that $Tu_0=gu_1$. Clearly $gu_1\sqsubseteq gx$. Since T is g-increasing, we have $Tu_1\sqsubseteq Tx=gx$. Let $Tu_1=gu_2$ for some $u_2\in X$. Then $gu_2\sqsubseteq gx$. Inductively, one can find a sequence $\{u_n\}$ such that $gu_{n+1}=Tu_n$ for all $n\ge 0$ and $gu_n\sqsubseteq gx$ for all $n\ge 1$. Let $R_n=p(gx,gu_n),\,n\ge 1$. Since gx and $gu_n$ are comparable for all $n\ge 1$, we have for all $n\ge 1$, $\psi (R_{n+1})=\psi (p(gx,gu_{n+1}))=\psi (p(Tx,Tu_n))\le \varphi (p(gx,gu_n))-\theta (p(gx,gu_n))=\varphi (R_n)-\theta (R_n)\le \varphi (R_n$). This implies that $R_{n+1}\le R_n$ for all $n\ge 1$. So $\{R_n\}$ is a nonincreasing sequence of non-negative real numbers. Then as in the proof of Theorem 2.1, we can show that $\lim _{n\rightarrow \infty }R_n=0$. That is, $\lim _{n\rightarrow \infty }p(gx,gu_n)=0$. Similarly $\lim _{n\rightarrow \infty }p(gu_n,gx)=0$, $\lim _{n\rightarrow \infty }p(gy,gu_n)=0$ and $\lim _{n\rightarrow \infty }p(gu_n,gy)=0$. By triangular inequality, $p(gx,gx)\le p(gx,gu_n)+p(gu_n,gx) \text { for all } n\ge 1.$ Letting $n\rightarrow \infty$, we get $p(gx,gx)=0$. Again, $p(gx,gy)\le p(gx,gu_n)+p(gu_n,gy)$ for all $n\ge 1$. Letting $n\rightarrow \infty$ we get $p(gx,gy)=0$. Lemma 1.3(i) now implies that $gx=gy$. Thus, there is a unique point of coincidence. Further, assume that the pair (T, g) is weakly compatible. Since $gx=Tx$, therefore, $ggx=gTx=Tgx.$ Let $gx=z$. Then we have $gz=Tz$. Due to uniqueness of point of coincidence we get $gx=gz$. It follows that $z=gz=Tz$. Thus, z is a common fixed point of T and g. To prove uniqueness, let w be a common fixed point of T and g. Then $Tw=gw=w$. By uniqueness of point of coincidence, we have $z=gz=gw=w$. Hence, T and g have a unique common fixed point in X. $\square$

Example 2.5

Consider $X=[0,\infty )$, together with the Euclidean metric d and let $p(x,y)=y$ for all $x,y\in X$. Then $(X,d,\le )$ is a partially ordered complete metric space with w-distance p as proved in Example 1.2(iv). Define self maps T and g on X by $Tx=x^2/3$ and $gx=x^2$ for all $x\in X.$ Then $T(X)=g(X)=[0,\infty )$. Further, for any x, $y\in X$ we have $gx\le gy\Rightarrow x^2\le y^2\Rightarrow (1/3)x^2\le (1/3)y^2\Rightarrow Tx\le Ty$. So T is g-nondecreasing. Also there exists $0\in X$ with $T0=g0=0$. T and g both are continuous being polynomial functions. T and g are compatible as for any sequence $\{x_n\}$ in X with $\lim _{n\rightarrow \infty }Tx_n=\lim _{n\rightarrow \infty }gx_n= t\in [0,\infty )$, we have $\lim _{n\rightarrow \infty }(x_n^2/3)=\lim _{n\rightarrow \infty }x_n^2=t\Rightarrow \lim _{n\rightarrow \infty }x_n^2=0\Rightarrow \lim _{n\rightarrow \infty }d(Tgx_n,gTx_n)=\lim _{n\rightarrow \infty }|(1/3)x_n^4-(1/9)x_n^4|=\lim _{n\rightarrow \infty }(2/9)x_n^4=0$. Let $\psi (t)=t^2$, $\varphi (t)=4.5\upchi _{(3,4)}(t)+(4/9)t^2\upchi _{[0,3]\cup [4,\infty )}(t)$ for all $t\in [0,\infty )$ and $\theta =\upchi _{(3,4)}$. Then $\psi \in \Psi$ as it is continuous being a polynomial function, nondecreasing for having non-negative first derivative and $\psi (t)=t^2=0 \Rightarrow t=0$. Also $\varphi ,\theta \in \Theta$ as let $I=[a,b]$ be any bounded interval in $[0,\infty )$. For any $t\in [a,b],$ $\varphi (t)$ is equal to either $(4/9)t^2$ or 4.5. So it lies between $\min \{(4/9)a^2,4.5\}$ and $\max \{(4/9)b^2,4.5\}$. Thus $\varphi$ is bounded on [a, b]. $\theta$ taking just two values 0 and 1 is bounded on any interval. Now let $\psi (s)\le \varphi (t)$ for some $s,t\in [0,\infty )$. Then $s^2\le (4/9)t^2$ if $t\in [0,3]\cup [4,\infty )$ and $s^2\le 4.5$ if $t\in (3,4)$. In the former case, $s^2\le t^2$ and in the later case, $s^2\le 4.5<9<t^2$. Thus in both cases, we have $s\le t$. Consider now a sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$. If $t\notin \{3,4\}$, then both $\varphi$ and $\theta$ are continuous at t and hence $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)=\psi (t)-\lim _{n\rightarrow \infty } \varphi (x_n)+\lim _{n\rightarrow \infty } \theta (x_n)=\psi (t)-\varphi (t)+\theta (t)$ which is equal to $t^2-(4/9)t^2=(5/9)t^2>0$ if $t\in [0,3)\cup (4,\infty )$ and to $t^2-4.5+1=t^2-3.5>0$ if $t\in (3,4)$. Now consider the case when $t=3$. Since $x_n\rightarrow 3$, we can choose a positive integer m such that $2.5<x_n<3.5$ for all $n\ge m$. If $x_n\in (3,3.5)$ for all $n\ge m$ then $A_m=\sup \{\varphi (x_m),\varphi (x_{m+1}),\ldots \}=\sup \{4.5\}=4.5.$ If $x_n\in (2.5,3]$ for all $n\ge m$ then $A_m=\sup \{\varphi (x_m),\varphi (x_{m+1}),\ldots \}= \sup \{(4/9)x_n^2: n\ge m\}=4$. In case, if both the sets $\{n\ge m:x_n\in (3,3.5)\}$ and $\{n\ge m:x_n\in (2.5,3]\}$ are not empty, then we have $A_m=\sup \{\varphi (x_m),\varphi (x_{m+1}),\ldots \}=\max \{\sup \{ \varphi (x_n): n\ge m,x_n\in (3,3.5)\},\sup \{\varphi (x_n): n\ge m,x_n\in (2.5,3]\}\}=\max \{\sup \{(4/9)x_n^2: n\ge m,x_n\in (2.5,3]\},4.5\}=4.5.$ Thus $A_m\le 4.5$. This implies that $\limsup \varphi (x_n)= \inf _{n\ge 1}\sup _{k\ge n}\varphi (x_k)\le \sup _{k\ge m}\varphi (x_k)=A_m\le 4.5$. Thus $\psi (3)-\limsup \varphi (x_n)+\liminf \theta (x_n)\ge 9-\limsup \varphi (x_n)\ge 9-4.5=4.5>0.$ Consider the case when $t=4$. Since $x_n\rightarrow 4$, we can choose a positive integer m such that $3.5<x_n<4.5$ for all $n\ge m$. If $x_n\in (3.5,4)$ for all $n\ge m$ then $A_m=\sup \{\varphi (x_m),\varphi (x_{m+1}),\ldots \}=\sup \{4.5\}=4.5.$ If $x_n\in [4,4.5)$ for all $n\ge m$, then $A_m=\sup \{\varphi (x_m),\varphi (x_{m+1}),\ldots \}= \sup \{(4/9)x_n^2: n\ge m\}\le (4/9)(4.5)^2=9$. In case, if both the sets $\{n\ge m:x_n\in (3.5,4)\}$ and $\{n\ge m:x_n\in [4,4.5)\}$ are not empty, then we have $A_m=\sup \{\varphi (x_m),\varphi (x_{m+1}),\ldots \}=\max \{\sup \{ \varphi (x_n): n\ge m,x_n\in (3.5,4)\},\sup \{\varphi (x_n): n\ge m,x_n\in [4,4.5)\}\}=\max \{4.5,\sup \{(4/9)x_n^2: n\ge m,x_n\in [4,4.5)\}\}=\sup \{(4/9)x_n^2: n\ge m,x_n\in [4,4.5)\}\le (4/9)(4.5)^2=9$. Thus $A_m\le 9$. This implies that $\limsup \varphi (x_n)= \inf _{n\ge 1}\sup _{k\ge n}\varphi (x_k)\le \sup _{k\ge m}\varphi (x_k)=A_m\le 9$. Thus $\psi (4)-\limsup \varphi (x_n)+\liminf \theta (x_n)\ge 16-\limsup \varphi (x_n)\ge 16-9=7>0.$ Now we show that $\psi (p(Tx,Ty))\le \varphi (p(gx,gy))-\theta (p(gx,gy))$ for all $x,y\in X$. Let us first assume that $3<y^2<4$. Then $\varphi (p(gx,gy))-\theta (p(gx,gy))=\varphi (gy)-\theta (gy)=\varphi (y^2)-\theta (y^2)=4.5-1=3.5>(16/9)>(1/9)y^4=\psi (y^2/3)=\psi (Ty)=\psi (p(Tx,Ty))$. Now assume that $y^2\in [0,3]\cup [4,\infty )$. Then $\varphi (p(gx,gy))-\theta (p(gx,gy))=\varphi (gy)-\theta (gy)=\varphi (y^2)-\theta (y^2)=(4/9)y^4\ge (1/9)y^4=\psi (y^2/3)=\psi (Ty)=\psi (p(Tx,Ty))$. Thus $T,g,\psi ,\varphi$, and $\theta$ satisfy all the conditions in Theorem 2.1. We observe that T and g have a coincidence point at 0. Further, because of comparability of any two real numbers, and compatibility (and hence weakly compatibility) of T and g all additional conditions as mentioned in Theorem 2.4 is also satisfied, and we see that 0 is a unique common fixed point of T and g.

Example 2.6

Consider $X=[0,\infty )$ together with the Euclidean metric d and let $p(x,y)=\lambda d(x,y)$ for all $x,y\in X$ where $\lambda$ is any positive real number. Then, by Example 1.2(ii), p is a w-distance on (X, d). Further $(X,d,\le )$ is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define self maps T and g on X by $T=\upchi _X$ and $g=\upchi _{{\mathbb {Q}}}+(1/2)\upchi _{{\mathbb {X}}-{\mathbb {Q}}}$. Then $T(X)=\{1\}\subseteq \{1/2,1\}=g(X)$. T is g-nondecreasing being a constant function. Further for $0\in X, T0=g0=1.$ $g(X)=\{1/2,1\}$, being finite is a closed subset of a complete space (X, d) and hence complete. Consider the functions $\psi ,\varphi$, and $\theta$ as defined in Example 2.5 where we have already established that these functions satisfy the conditions $\psi \in \Psi$, $\varphi ,\theta \in \Theta$, for all $s,t\in [0,\infty )$, $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$. Now we show that $\psi (p(Tx,Ty))\le \varphi (p(gx,gy))-\theta (p(gx,gy))$ for all $x,y\in X$. If x, y both are rational or both are irrational, then $gx=gy$. So, $p(gx,gy)=\lambda d(gx,gy)=0$. This implies that $\psi (p(Tx,Ty))=\psi (p(1,1))=\psi (0)=0 =\varphi (0)-\theta (0)=\varphi (p(gx,gy))-\theta (p(gx,gy))$. Now assume that exactly one of the two numbers x and y is rational. Then $d(gx,gy)=|1-(1/2)|=1/2$. So $\psi (p(Tx,Ty))=\psi (p(1,1))=\psi (0)=0$ $\varphi (p(gx,gy))-\theta (p(gx,gy))=\varphi (\lambda /2)-\theta (\lambda /2)$ which is equal to $(4/9)(\lambda /2)^2=\lambda ^2/9$ if $\lambda \notin (6,8)$ and to $4.5-1=3.5$ otherwise. In both, cases $\varphi (p(gx,gy))-\theta (p(gx,gy))\ge 0=\psi (p(Tx,Ty))$. So $T,g,\psi ,\varphi$ and $\theta$ satisfy all the conditions in Theorem 2.1. We observe that every rational number is a coincidence point of T and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of T and g, all additional conditions as mentioned in Theorem 2.4 are also satisfied and we see that 1 is a unique common fixed point of T and g.

Example 2.7

Consider $X=[0,\infty )$ together with the Euclidean metric d and let $p(x,y)=f(x)+y$ for all $x,y\in X$ where $f:X\rightarrow X$ is any arbitrary function such that $f(0)=0$. Then, by Example 1.2(iii), p is a w-distance on (X, d). Further $(X,d,\le )$ is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define self map T on X by $T=\upchi _{\phi }$ ($\phi$ denotes an empty set) and take g as in Example 2.5. Then $T(X)=\{0\}\subseteq [0,\infty )=g(X)$. T is g-nondecreasing being a constant function. Further for $0\in X, T0=g0=0.$ $g(X)=X$ is complete. Consider the functions $\psi ,\varphi$, and $\theta$ as defined in Example 2.5 where we have already established that these functions satisfy the conditions $\psi \in \Psi$, $\varphi ,\theta \in \Theta$, for all $s,t\in [0,\infty )$, $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$. Now we show that $\psi (p(Tx,Ty))\le \varphi (p(gx,gy))-\theta (p(gx,gy))$ for all $x,y\in X$. We note that $\varphi (t)-\theta (t)$ is equal to $(4/9)t^2$ if $t\in [0,3]\cup [4,\infty )$ and to $4.5-1=3.5$ if $t\in (3,4)$. In both cases, $\varphi (t)-\theta (t)\ge 0$. So, $\varphi (p(gx,gy))-\theta (p(gx,gy))\ge 0=\psi (0)=\psi (p(0,0))=\psi (p(Tx,Ty))$ for all $x,y\in X$. Thus, $T,g,\psi ,\varphi$, and $\theta$ satisfy all the conditions in Theorem 2.1. We observe that 0 is a coincidence point of T and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of T and g, all additional conditions as mentioned in Theorem 2.4 are also satisfied and we see that 0 is a unique common fixed point of T and g.

3 Applications

3.1 Coupled Coincidence Point Theory

In this section, we establish some coupled coincidence point results in partially ordered metric spaces by using results in the previous section. These results extend some of the existing results. Let $(X,d,\sqsubseteq )$ be a partially ordered metric space with a partial order $\sqsubseteq$ and a metric d. Define a relation “$\preceq$” on $X\times X$ by $(x,y)\preceq (u,v)$ if and only if $x\sqsubseteq u$ and $y\sqsupseteq v.$ Further, define a function $d_1$ by $d_1((x,y),(u,v))=d(x,u)+d(y,v)$ for all $(x,y),(u,v)\in X\times X.$ Then $X\times X$ becomes a partially ordered metric space with a partial order $\preceq$ and a metric $d_1$. Now, consider two function $F:X\times X\rightarrow X$ and $g:X\rightarrow X$. Define $T:X\times X\rightarrow X\times X$ and $G:X\times X\rightarrow X\times X$ by $T(x,y)=(F(x,y),F(y,x))$ for all $(x,y)\in X\times X$ and $G(x,y)=(gx,gy)$ for all $(x,y)\in X\times X.$ Now, following results have been settled in [9, Lemma 4.1, Lemma 4.2, Lemma 4.3 , and Lemma 4.4, respectively].

• (i) If F satisfies mixed g-monotone property, then T is G-nondecreasing.

• (ii) If F and g are commutative, then so are T and G.

• (iii) If F and g are compatible, then so are T and G.

• (iv) If F and g are weakly compatible, then so are T and G.

Theorem 3.1

Let $(X, \sqsubseteq ,d)$ be a partially ordered metric space equipped with a w-distance p and suppose $F:X\times X \rightarrow X$ and $g:X\rightarrow X$ be two functions with $F(X\times X) \subseteq g(X)$ and F satisfying mixed g-monotone property. Suppose there exist $x_0,y_0\in X$ with $gx_0\sqsubseteq F(x_0,y_0)$ and $gy_0\sqsupseteq F(y_0,x_0)$ or $gx_0\sqsupseteq F(x_0,y_0)$ and $gy_0\sqsubseteq F(y_0,x_0)$. Further assume that either F and g are continuous, (F, g) compatible with (X, d) as a complete metric space or (g(X), d) is a complete subspace of a regular metric space (X, d). Let $\psi ,\varphi$, and $\theta$ be functions having the same properties as in Theorem 2.1 except the contractive condition which is replaced by the following condition. For all $x,y,u,v\in X$ with $gx\sqsubseteq gu$ and $gy\sqsupseteq gv$ or $gx\sqsupseteq gu$ and $gy\sqsubseteq gv$ we have $\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))\le \varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))$. Then F and g have a coupled coincidence point in X. In addition to above, suppose for any pair of elements (x, y), $(x^{*},y^{*})\in X\times X$, there exists $(u,v)\in X\times X$ such that (F(u, v), F(v, u)) is comparable with (F(x, y), F(y, x)) and $(F(x^{*},y^{*}),F(y^{*},x^{*}))$ and also the pair (F, g) is weakly compatible. Then there exists a unique coupled common fixed point of F and g.

Proof

Let $S=X\times X$ and $d_1$ , a relation $\preceq$ on S and the self maps T and G on S be as defined just before Theorem 3.1. Consider $p_1((x,y),(u,v))=p(x,u)+p(y,v)$ for all $(x,y),(u,v)\in S$. Then $(S,d_1,\preceq )$ is a partially ordered metric space equipped with a w-distance $p_1$, T is G-nondecreasing and $T(S)\subseteq G(S)$. Let $p_0=(x_0,y_0)$. Then $gx_0\sqsubseteq F(x_0,y_0)$ and $gy_0\sqsupseteq F(y_0,x_0)$ or $gx_0\sqsupseteq F(x_0,y_0)$ and $gy_0\sqsubseteq F(y_0,x_0)$ imply that $Gp_0\sqsubseteq Tp_0$ or $Gp_0\sqsupseteq Tp_0$ for some point $p_0\in S$. Also the pair (T, G) is compatible and T and G are continuous on S if so are F and g. Also regularity of (X, d) implies regularity of $(S,d_1)$ and completeness of $(G(S),d_1)$ is implied by completeness of (g(X), d). Let $p=(x,y)$ and $q=(u,v)$. Then $gx\sqsubseteq gu$ and $gy\sqsupseteq gv$ or $gx\sqsupseteq gu$ and $gy\sqsubseteq gv$ means $Gp\preceq Gq$ or $Gp\succeq Gq$. So $\psi (p_1(Tp,Tq))\le \varphi (p_1(Gp,Gq))-\theta (p_1(Gp,Gq))$ for all $p,q\in S$ with $Gp\preceq Gq$ or $Gp\succeq Gq$. Thus all the conditions of Theorem 2.1 are satisfied. As a result, there exists a point $s=(x,y)\in S$ such that $Ts=Gs$. That is $(F(x,y),F(y,x))=(gx,gy)$ or $F(x,y)=gx$ and $F(y,x)=gy$. Hence (x, y) is a coupled coincidence point of F and g. Since weak compatibility of T and G follows by weak compatibility of F and g, the last part of the theorem follows by Theorem 2.4. $\square$

Example 3.2

Consider $X=[0,\infty )$ together with the Euclidean metric d and $p(x,y)=y$ for all $x,y\in X$. Then $(X,d,\le )$ is a partially ordered complete metric space and as proved in Example 1.2(iv), p is a w-distance on (X, d). Consider now the function defined by $F(x, y)=(1/3)(x^{2}-y^{2})$ if $x\ge y$, $F(x,y)= 0$ if $x<y$ and take g as in Example 2.5. Let $x_1,x_2\in X$ with $gx_1\le gx_2$ and $y\in X$. Then $x_1^2\le x_2^2\Rightarrow x_1\le x_2$. If $y\le x_1\le x_2$ then $F(x_1,y)=(1/3)(x_1^2-y^2)\le (1/3)(x_2^2-y^2)=F(x_2,y)$. If $x_1\le y\le x_2$ then $F(x_1,y)=0\le (1/3)(x_2^2-y^2)=F(x_2,y)$. Further, if $x_1\le x_2\le y$ then $F(x_1,y)=0=F(x_2,y)$. Thus we always have $F(x_1,y)\le F(x_2,y)$. Now, let $y_1,y_2\in X$ with $gy_1\le gy_2$ and $x\in X$. Then $y_1^2\le y_2^2\Rightarrow y_1\le y_2$. If $y_1\le y_2\le x$ then $F(x,y_1)=(1/3)(x^2-y_1^2)\ge (1/3)(x^2-y_2^2)=F(x,y_2)$. If $y_1\le x\le y_2$ then $F(x,y_1)=(1/3)(x^2-y_1^2)\ge 0=F(x,y_2)$. Further, if $x\le y_1\le y_2$ then $F(x,y_1)=0=F(x,y_2)$. Thus we always have $F(x,y_1)\ge F(x,y_2)$ . So F satisfies mixed g-monotone property. Further, $F(X\times X)= g(X)=[0,\infty )$ and g is continuous on X being a polynomial function. Next we show that F is continuous on $X\times X$. Let $(x_n,y_n)\rightarrow (x,y)$ in $X\times X$. This implies that $x_n\rightarrow x$ and $y_n\rightarrow y$ in X. If $x> y$ then choose $r\le (x-y)/\sqrt{2}$ which is nothing but the perpendicular distance of the point (x, y) from the line $y=x$. Choose now a positive integer m such that for all $n\ge m$, $(x_n,y_n)$ belongs to the circular open neighborhood D centered at (x, y) having radius r. Then we have $x_n>y_n$ for all $n\ge m$. This implies that, for all $n\ge m$, $|F(x_n,y_n)-F(x,y)|=(1/3)|(x_n^2-y_n^2)-(x^2-y^2)|<|(x_n^2-x^2)-(y_n^2-y^2)|\le |x_n^2-x^2|+|y_n^2-y^2|\rightarrow 0$ as $n\rightarrow \infty$. If $x<y$, by the same argument as above we can find a positive integer N such that $x_n<y_n$ for all $n\ge N$. So, $|F(x_n,y_n)-F(x,y)|=0$ for all $n\ge N$. Now assume that $x=y$. Then $|F(x_n,y_n)-F(x,y)|= |F(x_n,y_n)|\le |(1/3)(x_n^2-y_n^2)|\rightarrow (1/3)|x^2-y^2|=0$ as $n\rightarrow \infty$. Thus $F(x_n,y_n)\rightarrow F(x,y)$ whenever $(x_n,y_n)\rightarrow (x,y)$ in $X\times X$. So F is a continuous function. Consider the functions $\psi ,\varphi$, and $\theta$ as defined in Example 2.5 where we have already established that these functions satisfy the conditions $\psi \in \Psi$, $\varphi ,\theta \in \Theta$, for all $s,t\in [0,\infty )$, $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$. In [9] (Example 4.1), it is proved that (F, g) is a compatible pair of mappings. Further, if $x_0=0$ and $y_0=c>0$, then $F(x_0,y_0)=F(0,c)=0=gx_0$ and $F(y_0,x_0)=F(c,0)=c^2/3\le c^{2}=g(y_0)$. For all $x,y,u,v\in X$ we have $\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))=\psi (F(u,v)+F(v,u))=\psi (|u^{2}-v^{2}|/3)=(u^{2}-v^{2})^{2}/9$ and $\varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))=\varphi (gu+gv)-\theta (gu+gv) =\varphi (u^{2}+v^{2})-\theta (u^{2}+v^{2})=3.5$ if $3<u^{2}+v^{2}<4$ and $(4/9)(u^{2}+v^{2})^{2}$ otherwise. It is easy to observe that $\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))\le \varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))$ for all $x,y,u,v\in X$. So all the conditions of Theorem 3.1 are satisfied. We observe that F and g have a coupled coincidence point at (0, 0). Further, all additional conditions as mentioned in Theorem 3.1 are also satisfied and we see that (0, 0) is a unique coupled common fixed point of F and g.

Example 3.3

Consider $X=[0,\infty )$ together with the Euclidean metric d and let $p(x,y)=f(x)+y$ for all $x,y\in X$ where $f:X\rightarrow X$ is any arbitrary function such that f(0)=0. Then, by Example 1.2(iii), p is a w-distance on (X, d). Further $(X,d,\le )$ is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define $F:X\times X\rightarrow X$ by $F=\upchi _{\phi }$ ($\phi$ denotes an empty set) and take g as in Example 2.5. Then $F(X\times X)=\{0\}\subseteq [0,\infty )=g(X)$. F is mixed g-monotonic being a constant function. Further for $(0,0)\in X\times X$, $F(0,0)=g0=0.$ $g(X)=X$ is complete. Consider the functions $\psi ,\varphi$ and $\theta$ as defined in Example 2.5 where we have already established that these functions satisfy the conditions $\psi \in \Psi$, $\varphi ,\theta \in \Theta$, for all $s,t\in [0,\infty )$, $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$. Now we show that $\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))\le \varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))$ for all $x,y,u,v\in X$. As proved in Example 2.7, $\varphi (t)-\theta (t)\ge 0$ for all $t\ge 0$. So, $\varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))\ge 0=\psi (0)=\psi (p(0,0)+p(0,0))=\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))$ for all $x,y,u,v\in X$. Thus, $F,g,\psi ,\varphi$, and $\theta$ satisfy all the conditions in Theorem 3.1. We observe that 0 is a coupled coincidence point of F and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of F and g, all additional conditions as mentioned in Theorem 3.1 are also satisfied and we see that 0 is a unique coupled common fixed point of F and g.

Example 3.4

Consider $X=[0,\infty )$ together with the Euclidean metric d and let $p(x,y)=\lambda d(x,y)$ for all $x,y\in X$ where $\lambda$ is any positive real number. Then, by Example 1.2(ii), p is a w-distance on (X, d). Further $(X,d,\le )$ is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define $F:X\times X\rightarrow X$ by $F=\upchi _{X\times X}$ and consider g as defined in Example 2.6. Then $F(X\times X)=\{1\}\subseteq \{1/2,1\}=g(X)$. F is mixed g-monotonic being a constant function. Further for $(0,0)\in X\times X, F(0,0)=g0=1.$ Now, $g(X)=\{1/2,1\}$ being finite is a closed subset of a complete space (X, d) and hence complete. Consider the functions $\psi ,\varphi$ and $\theta$ as defined in Example 2.5 where we have already established that these functions satisfy the conditions $\psi \in \Psi$, $\varphi ,\theta \in \Theta$, for all $s,t\in [0,\infty )$, $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$. Now we show that $\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))\le \varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))$ for all $x,y,u,v\in X$. As proved in Example 2.7, $\varphi (t)-\theta (t)\ge 0$ for all $t\ge 0$. So, $\varphi (p(gx,gu)+p(gy,gv))-\theta (p(gx,gu)+p(gy,gv))\ge 0=\psi (0)=\psi (p(1,1)+p(1,1))=\psi (p(F(x,y),F(u,v))+p(F(y,x),F(v,u)))$ for all $x,y,u,v\in X$. So $F,g,\psi ,\varphi$, and $\theta$ satisfy all the conditions in Theorem 3.1. We observe that every point (x, y) of the set ${\mathbb {Q}}\times {\mathbb {Q}}$ is a coupled coincidence point of F and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of F and g, all additional conditions as mentioned in Theorem 3.1 are also satisfied and we see that (1, 1) is a unique coupled common fixed point of F and g.

3.2 Integral Equations

Let $I=[a,b]$ be any interval of the real line. Consider the componentwise partial order on ${\mathbb {R}}^n$. Let $X=C(I,{\mathbb {R}}^n)$ be the space of all continuous functions defined on I with values in ${\mathbb {R}}^n$ and pointwise partial order. Let $d:X\times X\rightarrow {\mathbb {R}}$ defined by $d(x,y)=||x-y||_{\infty }$ for any $x,y\in X$, be a complete metric on X. By Example 1.2(v), $p:X\times X\rightarrow {\mathbb {R}}$ defined by $p(x,y)=d(0,y)=||y||_{\infty }$ for any $x,y\in X$, is a w-distance on X. Let $h:I\times I\times {\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ be continuous. Consider the Fredholm type integral equation

$$\begin{aligned} x(t)=\int _Ih(t,s,x(s))\mathrm{d}s, \end{aligned}$$

(3.1)

for $t\in I$.

Definition 3.5

A function $\alpha \in C(I,{\mathbb {R}}^n)$ is said to be a lower solution of the integral equation (3.1) if $\alpha (t)\le \int _Ih(t,s,\alpha (s))\mathrm{d}s$ for all $t\in I$ and an upper solution if $\alpha (t)\ge \int _Ih(t,s,\alpha (s))\mathrm{d}s$ for all $t\in I$.

Now we consider the following assumptions:

• (a) $h(t,s,.):{\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ is nondecreasing for each $t,s\in I$,

• (b) There exist $\psi \in \Psi$ and $\varphi ,\theta \in \Theta$ such that for all $s,t\in [0,\infty )$ we have $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and $\psi (t)\le t$ for all $t\ge 0$. Further, for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, suppose that $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$ and for all $x\in X,t,s\in I$, $|h(t,s,x(s))|\le (1/(b-a))(\varphi (||x||_{\infty })-\theta (||x||_{\infty }))$.

Next we give an existence theorem for the solution of the integral equation (3.1).

Theorem 3.6

Suppose that assumptions (a) and (b) hold. Then the existence of a lower or an upper solution for the integral equation (3.1) provides the existence of a solution of the equation (3.1).

Proof

Define a mapping $T:X\rightarrow X$ by $T(x)(t):=\int _Ih(t,s,x(s))ds$ for all $t\in I$. Clearly T is a well-defined mapping. Take g as the identity function on X. By our assumption (a), we observe that T is nondecreasing. Now for any $x,y\in X$ and $t\in I$, we have, by assumption (b)

$$\begin{aligned} \psi (|T(y)(t)|)&= \psi (|\int _Ih(t,s,y(s))\mathrm{d}s|)\le \psi (\int _I|h(t,s,y(s))|\mathrm{d}s)\\&\le \int _I|h(t,s,y(s))|\mathrm{d}s\le (1/(b-a))(\varphi (||y||_{\infty })-\theta (||y||_{\infty }))\int _{I}\mathrm{d}s\\&=\varphi (||y||_{\infty })-\theta (||y||_{\infty }). \end{aligned}$$

So for all $x,y\in X$, $\psi (p(T(x),T(y)))=\psi (||T(y)||_\infty )=\psi (\max _{t\in I}|T(y)(t)|)=\max _{t\in I}\psi (|T(y)(t)|)\le \varphi (||y||_{\infty })-\theta (||y||_{\infty })=\varphi (p(gx,gy))-\theta (p(gx,gy))$. From the existence of a lower or an upper solution of the equation (3.1), there exists an $x_0\in X$ such that $x_0\le Tx_0$ or $x_0\ge Tx_0$. Further, X is regular as ${\mathbb {R}}^n$ is so. The conclusion follows by Theorem 2.1. $\square$

By Example 1.2(i), $q:X\times X\rightarrow {\mathbb {R}}$ defined by $q(x,y)=d(x,y)$ for any $x,y\in X$, is a w-distance on X. Let $k:I\times I\times {\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ be continuous. For $t\in I$, consider the Fredholm type integral equations

$$\begin{aligned} x(t)&=\int _Ik(t,x(s),y(s))\mathrm{d}s,\nonumber \\ y(t)&=\int _Ik(t,y(s),x(s))\mathrm{d}s. \end{aligned}$$

(3.2)

Definition 3.7

An element $(\alpha ,\beta )\in C(I,{\mathbb {R}}^n)\times C(I,{\mathbb {R}}^n)$ is said to be a coupled lower and upper solution of the integral equation (3.2) if $\alpha (t)\le \beta (t)$, $\alpha (t)\le \int _Ik(t,\alpha (s),\beta (s))ds$ for all $t\in I$ and $\beta (t)\ge \int _Ik(t,\beta (s),\alpha (s))ds$ for all $t\in I$.

Now we consider the following assumptions:

• (a) For any $x,y\in X$ and $t\in I$, we have $x_1,x_2\in X,$ $x_1\le x_2\Rightarrow k(t,x_1,y)\le k(t,x_2,y)$ and $y_1,y_2\in X$, $y_1\le y_2\Rightarrow k(t,x,y_1)\ge k(t,x,y_2)$

• (b) There exist $\psi \in \Psi$ and $\varphi ,\theta \in \Theta$ such that for all $s,t\in [0,\infty )$, we have $\psi (s)\le \varphi (t)\Rightarrow s\le t$ and $\psi (t)\le t$ for all $t\ge 0$. Further, for any sequence $\{x_n\}$ in $[0,\infty )$ with $x_n\rightarrow t>0$, suppose that $\psi (t)-\limsup \varphi (x_n)+\liminf \theta (x_n)>0$ and for all $x,y,u,v\in X$ with $x\le u$, $y\ge v$ and $t,s\in I$, $|k(t,x(s),y(s))-k(t,u(s),v(s))|\le (1/2)(1/(b-a))(\varphi (||x-u||_{\infty }+||y-v||_{\infty })-\theta (||x-u||_{\infty }+||y-v||_{\infty }))$.

Next we give an existence theorem for the solution of the integral equation (3.2).

Theorem 3.8

Suppose that assumptions (a) and (b) hold. Then the existence of a lower and upper solution for the integral equation (3.2) provides the existence of a solution of the equation (3.2).

Proof

Define a mapping $F:X\times X\rightarrow X$ by $F(x,y)(t):=\int _Ik(t,x(s),y(s))\mathrm{d}s$ for all $t\in I$. Clearly F is a well-defined mapping. Take g as the identity function on X. Let $x,y\in X$ with $x_1,x_2\in X,$ $x_1\le x_2$ and $y_1,y_2\in X$, $y_1\le y_2$. By our assumption (a), $k(t,x_1,y)\le k(t,x_2,y)$ and $k(t,x,y_1)\ge k(t,x,y_2)$ for all $t\in I$. This implies that $\int _{I} k(t,x_1(s),y(s))\mathrm{d}s\le \int _{I}k(t,x_2(s),y(s))\mathrm{d}s$ and $\int _{I} k(t,x(s),y_1(s))\mathrm{d}s\ge \int _{I}k(t,x(s),y_2(s))\mathrm{d}s$. That is $F(x_1,y)\le F(x_2,y)$ and $F(x,y_1)\ge F(x,y_2)$. So, F is mixed g-monotonic. Now for any $x,y,u,v\in X$ with $x\le u$, $y\ge v$ and $t\in I$, we have, by assumption (b),

$$\begin{aligned}&|F(x,y)(t)-F(u,v)(t)|\\&\quad =\left| \int _Ik(t,x(s),y(s))\mathrm{d}s-\int _Ik(t,u(s),v(s))\mathrm{d}s\right| \\&\quad \le \int _I|k(t,x(s),y(s))-k(t,u(s),v(s))|\mathrm{d}s\\&\quad \le (1/2)(1/(b-a))(\varphi (||x-u||_{\infty }+||y-v||_{\infty })\\&\qquad -\theta (||x-u||_{\infty }+||y-v||_{\infty }))\int _{I}\mathrm{d}s\\&\quad =(1/2)(\varphi (\mathrm{d}(x,u)+\mathrm{d}(y,v))-\theta (\mathrm{d}(x,u)+\mathrm{d}(y,v)))\\&\quad =(1/2)(\varphi (q(gx,gu)+q(gy,gv))-\theta (q(gx,gu)+q(gy,gv))). \end{aligned}$$

This implies that for all $x,y,u,v\in X$ with $x\le u$, $y\ge v$,

$$\begin{aligned}&q(F(x,y),F(u,v))\nonumber \\&\quad =d(F(x,y),F(u,v))\nonumber \\&\quad =\sup _{t\in I}|F(x,y)(t)-F(u,v)(t)|\nonumber \\&\quad \le (1/2)(\varphi (q(gx,gu)+q(gy,gv))-\theta (q(gx,gu)+q(gy,gv))) \end{aligned}$$

(3.3)

Since $x\le u$, $y\ge v$ is equivalent to $v\le y$, $u\ge x$, we also have $q(F(v,u),F(y,x))\le (1/2)(\varphi (q(gv,gy)+q(gu,gx))-\theta (q(gv,gy)+q(gu,gx)))$. That is

$$\begin{aligned} q(F(y,x),F(v,u))\le (1/2)(\varphi (q(gx,gu)+q(gv,gy))-\theta (q(gx,gu)+q(gv,gy))) \end{aligned}$$

(3.4)

Adding (3.3) and (3.4) we get for all $x,y,u,v\in X$ with $x\le u$, $y\ge v$,

$$\begin{aligned}&q(F(x,y),F(u,v))+q(F(y,x),F(v,u))\\&\quad \le \varphi (q(gx,gu)+q(gv,gy))-\theta (q(gx,gu)+q(gv,gy)). \end{aligned}$$

Now for all $x,y,u,v\in X$ with $gx\le gu$, $gy\ge gv$,

$$\begin{aligned}&\psi (q(F(x,y),F(u,v))+q(F(y,x),F(v,u)))\\&\quad \le q(F(x,y),F(u,v))+q(F(y,x),F(v,u))\\&\quad \le \varphi (q(gx,gu)+q(gv,gy))-\theta (q(gx,gu)+q(gv,gy)). \end{aligned}$$

From the existence of a lower and upper solution of the equation (3.2), there exists a point $(x_0,y_0)\in X\times X$ such that $x_0\le F(x_0,y_0)$ and $y_0\ge F(y_0,x_0)$. Further, X is regular as ${\mathbb {R}}^n$ is so. The conclusion follows by Theorem 3.1. $\square$