Remarks on contractive mappings via Ω-distance

Abstract

Very recently, some authors discovered that some fixed point results in thecontext of a G-metric space can be derived from the fixed point resultsin the context of a quasi-metric space and hence the usual metric space. In thisarticle, we investigate some fixed point results in the framework of aG-metric space via Ω-distance that cannot be obtained by theusual fixed point results in the literature. We also add an application toillustrate our results.

MSC: 47H10, 54H25, 46J10, 46J15.

1 Introduction and preliminaries

Very recently, Jleli and Samet [1] and Samet et al.[2] proved that some fixed point results in the setting of G-metricspaces, introduced by Sims and Mustafa [3], are consequences of the well-known fixed point theorem in the context ofthe usual metric space. Indeed, authors in [1, 2] noticed that $G(x,y,y)=q(x,y)$ is a quasi-metric and obtained that the results arejust a characterization of existence results in the framework of a quasi-metric. Onthe other hand, a G-metric was introduced as a generalization of the(usual) metric. Basically, G-metrics claim the geometry of three pointsinstead of two points. Consequently, Jleli and Samet [1] and Samet et al.[2] concluded that if the expression in the fixed point theorem can bereduced to two points, then it can be written as a consequence of the relatedexistence result in the literature.

Recently, Saadati et al.[4] introduced the concept of Ω-distance on a complete G-metricspace as a generalized notion of ω-distance due to Kada etal.[5]. In these papers, the authors investigate the existence/uniqueness of afixed point of certain operators in this setting. In this paper, we revise somepublished papers (see, e.g., [6, 7]) and improve the statements in a way that cannot be manipulated by thetechniques used in [1, 2] (see also [8–10]).

We first recall some necessary definitions and basic results on the topics in theliterature.

Definition 1 ([3])

Let X be a non-empty set. A function $G:X\times X\times X\to [0,\mathrm{\infty })$ is called a G-metric if the followingconditions are satisfied:

• (i) $G(x,y,z)=0$ if $x=y=z$ (coincidence),

• (ii) $G(x,x,y)>0$ for all $x,y\in X$, where $x\ne y$,

• (iii) $G(x,x,z)\le G(x,y,z)$ for all $x,y,z\in X$, with $z\ne y$,

• (iv) $G(x,y,z)=G(p\{x,y,z\})$, where p is a permutation of x, y, z (symmetry),

• (v) $G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$ (rectangle inequality).

A G-metric is said to be symmetric if $G(x,y,y)=G(y,x,x)$ for all $x,y\in X$.

Definition 2 ([3])

Suppose that $(X,G)$ is a G-metric space.

• (1) A sequence $\{x_n\}$ in X is said to be G-Cauchy sequence if, for each $\epsilon >0$, there exists a positive integer $n_0$ such that for all $n,m,l\ge n_0$, $G(x_n,x_m,x_l)<\epsilon$.

• (2) A sequence $\{x_n\}$ in X is said to be G-convergent to a point $x\in X$ if, for each $\epsilon >0$, there exists a positive integer $n_0$ such that for all $m,n\ge n_0$, $G(x_m,x_n,x)<\epsilon$.

Definition 3 ([4])

Let $(X,G)$ be a G-metric space. Then a function$\mathrm{\Omega }:X\times X\times X⟶[0,\mathrm{\infty })$ is called an Ω-distance on X if thefollowing conditions are satisfied:

• (a) $\mathrm{\Omega }(x,y,z)\le \mathrm{\Omega }(x,a,a)+\mathrm{\Omega }(a,y,z)$ for all $x,y,z,a\in X$,

• (b) $\mathrm{\Omega }(x,y,\cdot ),\mathrm{\Omega }(x,\cdot ,y):X\to [0,\mathrm{\infty })$ are lower semi-continuous for any $x,y\in X$,

• (c) for each $\epsilon >0$, there exists $\delta >0$ such that $\mathrm{\Omega }(x,a,a)\le \delta$ and $\mathrm{\Omega }(a,y,z)\le \delta$ imply $G(x,y,z)\le \epsilon$.

Example 4 ([4])

Suppose that $(X,d)$ is a metric space. Let $G:X^3⟶[0,\mathrm{\infty })$ be defined as follows:

$$G(x,y,z)=max\{d(x,y),d(y,z),d(x,z)\}$$

for all $x,y,z\in X$. Then one can easily show that$\mathrm{\Omega }=G$ is an Ω-distance on X.

Example 5 ([4])

Let $X=\mathbb{R}$ and $(X,G)$ be a G-metric, where

$$G(x,y,z)=\frac{1}{3}(|x-y|+|y-z|+|x-z|)$$

for all $x,y,z\in X$. If we define $\mathrm{\Omega }:{\mathbb{R}}^3⟶[0,\mathrm{\infty })$ as follows:

$$\mathrm{\Omega }(x,y,z)=\frac{1}{3}(|z-x|+|x-y|)$$

for all $x,y,z\in X$, then it is an Ω-distance on ℝ.

We refer, e.g., to [4, 11] for more details and examples on the topic.

Lemma 6[4]

Suppose that$(X,G)$is aG-metric space and Ω is an Ω-distanceonX. Let$\{x_n\}$, $\{y_n\}$be sequences inXand$\{{\alpha }_n\}$, $\{{\beta }_n\}$be sequences in$[0,\mathrm{\infty })$converging to zero and$x,y,z,a\in X$. Then

• (a) if$\mathrm{\Omega }(y,x_n,x_n)\le {\alpha }_n$and$\mathrm{\Omega }(x_n,y,z)\le {\beta }_n$for$n\in \mathbb{N}$, then$G(y,y,z)<\epsilon$, and hence$y=z$;

• (b) if$\mathrm{\Omega }(y_n,x_n,x_n)\le {\alpha }_n$and$\mathrm{\Omega }(x_n,y_m,z)\le {\beta }_n$for$m>n$, then$G(y_n,y_m,z)\to 0$, and hence$y_n\to z$;

• (c) if$\mathrm{\Omega }(x_n,x_m,x_l)\le {\alpha }_n$for any$l,m,n\in \mathbb{N}$with$n\le m\le l$, then$\{x_n\}$is aG-Cauchy sequence;

• (d) if$\mathrm{\Omega }(x_n,a,a)\le {\alpha }_n$for any$n\in \mathbb{N}$, then$\{x_n\}$is aG-Cauchy sequence.

Definition 7 ([4])

Suppose that $(X,G)$ is a G-metric space and Ω is anΩ-distance on X. $(X,G)$ is called Ω-bounded if there is a constant$C>0$ with $\mathrm{\Omega }(x,y,z)\le C$ for all $x,y,z\in X$.

Definition 8 Let $(X,\le )$ be a partially ordered set. A self-mapping$T:X\to X$ is said to be non-decreasing if, for$x,y\in X$,

$$x\le y⟹T(x)\le T(y).$$

The tripled $(X,G,\le )$ is called a partially ordered G-metric spaceif $(X,\le )$ is a partially ordered set endowed with aG-metric on X; see also [12, 13].

2 Fixed point theorems on partially ordered G-metric spaces

We start this section with the following classes of mappings:

$$\begin{array}{c}\mathrm{\Phi }=\{\varphi |\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })\text{ continuous, non-decreasing}\}\text{and }\hfill \\ \mathrm{\Psi }=\{\psi |\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })\text{ continuous, non-decreasing}\}\hfill \end{array}$$

with ${\varphi }^{-1}(\{0\})={\psi }^{-1}(\{0\})=\{0\}$.

Definition 9 Let $(X,\le )$ be a partially ordered space. Suppose that thereexists a G-metric on X such that $(X,G)$ is a complete G-metric space. A self-mapping$T:X\to X$ is said to be a generalized weak-contraction mappingif it satisfies the following condition:

$$\psi \left(\mathrm{\Omega }(Tx,T^{2}x,Ty)\right)\le \psi (\mathrm{\Omega }(x,Tx,y))-\varphi (\mathrm{\Omega }(x,Tx,y))\text{for all }x,y\in X,\text{ with }x\le y,$$

where $\psi \in \mathrm{\Psi }$ and $\varphi \in \mathrm{\Phi }$.

Theorem 10Let$(X,G,\le )$be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mapping$T:X\to X$is a generalized weak-contraction mapping, that is,

$$\psi \left(\mathrm{\Omega }(Tx,T^{2}x,Ty)\right)\le \psi (\mathrm{\Omega }(x,Tx,y))-\varphi (\mathrm{\Omega }(x,Tx,y))\text{for all}x,y\in X,\text{ with }x\le Tx,$$

with$\psi \in \mathrm{\Psi }$and$\varphi \in \mathrm{\Phi }$. Suppose also that$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}>0$for every$y\in X$with$y\ne Ty$. If there exists$x_0\in X$with$x_0\le T{x}_0$, thenThas a unique fixed point, say$u\in X$. Moreover, $\mathrm{\Omega }(u,u,u)=0$.

Proof If $x_0=T{x}_0$, then the proof is finished. Suppose that$x_0\ne T{x}_0$. Since $x_0\le T{x}_0$ and T is non-decreasing, we obtain

$$x_0\le T{x}_0\le T^2{x}_0\le \cdots \le T^{n+1}{x}_0\le \cdots .$$

Now, if for some $n\in \mathbb{N}$, $\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)=0$, then

$$\begin{array}{rcl}\psi \left(\mathrm{\Omega }(T^{n+1}{x}_0,T^{n+2}{x}_0,T^{n+2}{x}_0)\right)& \le & \psi \left(\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)\right)\\ -\varphi \left(\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)\right),\end{array}$$

then $\mathrm{\Omega }(T^{n+1}{x}_0,T^{n+2}{x}_0,T^{n+2}{x}_0)=0$. Due to [(a), Definition 3], we have$\mathrm{\Omega }(T^n{x}_0,T^{n+2}{x}_0,T^{n+2}{x}_0)=0$. On the other hand, by [(c), Definition 3], weeasily derive that $G(T^n{x}_0,T^{n+2}{x}_0,T^{n+2}{x}_0)=0$, which completes the proof.

Consequently, throughout the proof, we suppose that $\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)>0$ for all $n\in \mathbb{N}$. Hence, we have

$$\begin{array}{rcl}\psi \left(\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)\right)& \le & \psi \left(\mathrm{\Omega }(T^{n-1}{x}_0,T^n{x}_0,T^n{x}_0)\right)\\ -\varphi \left(\mathrm{\Omega }(T^{n-1}{x}_0,T^n{x}_0,T^n{x}_0)\right),\end{array}$$

(2.1)

which yields that

$$\psi \left(\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)\right)\le \psi \left(\mathrm{\Omega }(T^{n-1}{x}_0,T^n{x}_0,T^n{x}_0)\right).$$

As a result, we conclude that $\{\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)\}$ is non-increasing. Thus, there exists$r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)=r.$$

We shall show that $r=0$. Suppose, on the contrary, that$r>0$. Then we have $\varphi (r)>0$. Letting $n\to \mathrm{\infty }$ on (2.1), we obtain

$$\psi (r)\le \psi (r)-\varphi (r),$$

a contraction. Hence, we have

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)=0.$$

(2.2)

Recursively, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+t}{x}_0)=0$$

(2.3)

for every $t\in \mathbb{N}$.

Let $l\ge m\ge n$ with $m=n+k$ and $l=m+t$ ($k,t\in \mathbb{N}$). By the triangle inequality, we derive that

$$\begin{array}{rcl}\mathrm{\Omega }(T^n{x}_0,T^m{x}_0,T^l{x}_0)& \le & \mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)+\mathrm{\Omega }(T^{n+1}{x}_0,T^m{x}_0,T^l{x}_0)\\ \le & \mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,T^{n+1}{x}_0)+\mathrm{\Omega }(T^{n+1}{x}_0,T^{n+2}{x}_0,T^{n+2}{x}_0)\\ +\cdots +\mathrm{\Omega }(T^{m-1}{x}_0,T^m{x}_0,T^l{x}_0).\end{array}$$

Letting $n\to \mathrm{\infty }$ in the inequality above, by keeping the limits (2.2)and (2.3), we obtain

$$\underset{n,m,l\to \mathrm{\infty }}{lim}\mathrm{\Omega }(T^n{x}_0,T^m{x}_0,T^l{x}_0)=0.$$

Therefore, $\{T^n{x}_0\}$ is a G-Cauchy sequence. Since X isG-complete, $\{T^n{x}_0\}$ converges to a point $u\in X$. Now, for $\epsilon >0$ and by the lower semi-continuity of Ω,

$$\mathrm{\Omega }(T^n{x}_0,T^m{x}_0,u)\le \underset{p\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\mathrm{\Omega }(T^n{x}_0,T^m{x}_0,T^p{x}_0)\le \epsilon ,m\ge n,$$

and

$$\mathrm{\Omega }(T^n{x}_0,u,T^l{x}_0)\le \underset{p\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\mathrm{\Omega }(T^n{x}_0,T^p{x}_0,T^l{x}_0)\le \epsilon ,l\ge n.$$

Assume that $u\ne Tu$. Since $T^n{x}_0\le T^{n+1}{x}_0$,

$$0<inf\{\mathrm{\Omega }(T^n{x}_0,u,T^n{x}_0)+\mathrm{\Omega }(T^n{x}_0,u,T^{n+1}{x}_0)+\mathrm{\Omega }(T^n{x}_0,T^{n+1}{x}_0,u):n\in \mathbb{N}\}\le 3\epsilon ,$$

a contraction. Hence, we have $u=Tu$.

We shall show that u is the unique fixed point of T. Suppose, onthe contrary, that v is another fixed point of T. So, we have

$$\begin{array}{rcl}\psi (\mathrm{\Omega }(u,u,v))& =& \psi \left(\mathrm{\Omega }(Tu,T^{2}u,Tv)\right)\\ \le & \psi (\mathrm{\Omega }(u,Tu,v))-\varphi (\mathrm{\Omega }(u,Tu,v))\\ =& \psi (\mathrm{\Omega }(u,u,v))-\varphi (\mathrm{\Omega }(u,u,v))\\ <& \psi (\mathrm{\Omega }(u,u,v)),\end{array}$$

a contraction. Thus, the fixed point u is unique. Now, since$u=Tu$, we have

$$\begin{array}{rcl}\psi (\mathrm{\Omega }(u,u,u))& =& \psi \left(\mathrm{\Omega }(Tu,T^{2}u,Tu)\right)\\ \le & \psi (\mathrm{\Omega }(u,Tu,u))-\varphi (\mathrm{\Omega }(u,Tu,u))\\ =& \psi (\mathrm{\Omega }(u,u,u))-\varphi (\mathrm{\Omega }(u,u,u)).\end{array}$$

So, $\mathrm{\Omega }(u,u,u)=0$. □

Definition 11 Let $(X,\le )$ be a partially ordered space. Suppose that thereexists a G-metric on X such that $(X,G)$ is a complete G-metric space. A self-mapping$T:X\to X$ is said to be a weak-contraction mapping if itsatisfies the following condition:

$$\mathrm{\Omega }(Tx,T^{2}x,Ty)\le \mathrm{\Omega }(x,Tx,y)-\varphi (\mathrm{\Omega }(x,Tx,y))\text{for all }x,y\in X,\text{ with }x\le y,$$

where $\varphi \in \mathrm{\Phi }$.

Corollary 12Let$(X,G,\le )$be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mapping$T:X\to X$is a weak-contraction mapping, that is,

$$\mathrm{\Omega }(Tx,T^{2}x,Ty)\le \mathrm{\Omega }(x,Tx,y)-\varphi (\mathrm{\Omega }(x,Tx,y))\text{for all }x,y\in X,\text{ with }x\le Tx,$$

where$\varphi \in \mathrm{\Phi }$. Suppose also that$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}>0$for every$y\in X$with$y\ne Ty$. If there exists$x_0\in X$with$x_0\le T{x}_0$, thenThas a unique fixed point, say$u\in X$. Moreover, $\mathrm{\Omega }(u,u,u)=0$.

If we take $\varphi (t)=kt$, where $k\in [0,1)$, we derive Theorem 2.2 [4] as the following corollary.

Corollary 13Let$(X,G,\le )$be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that there exists$k\in [0,1)$such that

$$\mathrm{\Omega }(Tx,T^{2}x,Ty)\le k\mathrm{\Omega }(x,Tx,y)\text{for all }x,y\in X,\text{ with }x\le Tx.$$

Suppose also that$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}>0$for every$y\in X$with$y\ne Ty$. If there exists$x_0\in X$with$x_0\le T{x}_0$, thenThas a unique fixed point, say$u\in X$. Moreover, $\mathrm{\Omega }(u,u,u)=0$.

Definition 14 Let $(X,\le )$ be a partially ordered space. Suppose that thereexists a G-metric on X such that $(X,G)$ is a complete G-metric space. A self-mapping$T:X\to X$ is said to be a Ćirić-type contractionmapping if it satisfies that there exists $0\le k<1$ such that

$$\mathrm{\Omega }(Tx,T^{2}x,Ty)\le kM(x,x,y),$$

where

$$M(x,x,y)=max\{\mathrm{\Omega }(x,Tx,Tx),\mathrm{\Omega }(y,Ty,Ty),\frac{1}{2}\mathrm{\Omega }(x,Ty,Ty)\}$$

for all $x,y\in X$ with $x\le y$.

Theorem 15Let$(X,G,\le )$be a partially ordered completeG-metric space, and let Ω be anΩ-distance onX. Suppose that a non-decreasing self-mapping$T:X⟶X$is a Ćirić-type contraction mapping.

• (i) For every$x\in X$and$y\in X$with$y\ne T(y)$, $inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le T(x)\}>0$,

• (ii) There exists$x_0\in X$such that$x_0\le T(x_0)$,

thenThas a fixed pointuinXand$\mathrm{\Omega }(u,u,u)=0$.

Proof By assumption (ii), there exists $x_0\in X$ such that $x_0\le T(x_0)$. We fix $x_1\in X$ such that $x_1=T(x_0)$. Since T is a non-decreasing mapping,$T{x}_0\le T{x}_1$. There exists $x_2\in X$ such that $T{x}_1=x_2$. Recursively, we construct the sequence$\{x_n\}$ in the following way:

$$x_{n+1}=T{x}_n\le T{x}_{n+1}=x_{n+2}\text{for all }n\ge 0.$$

Since T is a Ćirić-type contraction mapping, by replacing$x=x_n$ and $y=x_{n+1}$, we get that

$$\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})=\mathrm{\Omega }(T{x}_n,T{x}_{n+1},T{x}_{n+1})\le kM(x_n,x_n,x_{n+1}),$$

(2.4)

where

$$\begin{array}{rcl}M(x_n,x_n,x_{n+1})& =& max\{\mathrm{\Omega }(x_n,T{x}_n,T{x}_n),\mathrm{\Omega }(x_{n+1},T{x}_{n+1},T{x}_{n+1}),\\ \frac{1}{2}\mathrm{\Omega }(x_n,T{x}_{n+1},T{x}_{n+1})\}\\ =& max\{\mathrm{\Omega }(x_n,x_{n+1},x_{n+1}),\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2}),\\ \frac{1}{2}\mathrm{\Omega }(x_n,x_{n+2},x_{n+2})\}\\ \le & max\{\mathrm{\Omega }(x_n,x_{n+1},x_{n+1}),\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2}),\\ \frac{1}{2}[\mathrm{\Omega }(x_n,x_{n+1},x_{n+1})+\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})]\}\\ =& max\{\mathrm{\Omega }(x_n,x_{n+1},x_{n+1}),\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})\}.\end{array}$$

Notice that if $M(x_n,x_n,x_{n+1})\le \mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})$, then (2.4) yields a contradiction since$k<1$.

Thus, $M(x_n,x_n,x_{n+1})\le \mathrm{\Omega }(x_n,x_{n+1},x_{n+1})$ and inequality (2.4) and $k<1$ turn into

$$\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})\le k\mathrm{\Omega }(x_n,x_{n+1},x_{n+1}).$$

(2.5)

Upon the discussion above, we conclude that the sequence $\{\mathrm{\Omega }(x_n,x_{n+1},x_{n+1})\}$ is non-increasing and bounded below. Therefore, thereexists $r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Omega }(x_n,x_{n+1},x_{n+1})=r.$$

We shall show that $r=0$. By a standard calculation, using inequality (2.5)and keeping $k<1$ in mind, we obtain ${lim}_{n\to \mathrm{\infty }}\mathrm{\Omega }(x_n,x_{n+1},x_{n+1})=0$. We claim that the sequence $\{x_n\}$ is G-Cauchy. Let $l\ge m\ge n$ with $m=n+k$ and $l=m+t$ ($k,t\in \mathbb{N}$). By the triangle inequality, we derive that

$$\begin{array}{rcl}\mathrm{\Omega }(x_n,x_m,x_l)& \le & \mathrm{\Omega }(x_n,x_{n+1},x_{n+1})+\mathrm{\Omega }(x_{n+1},x_m,x_l)\\ \le & \mathrm{\Omega }(x_n,x_{n+1},x_{n+1})+\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})+\cdots +\mathrm{\Omega }(x_{m-1},x_m,x_l).\end{array}$$

(2.6)

On the other hand, we have

$$\begin{array}{rcl}\mathrm{\Omega }(x_{m-1},x_m,x_{m+t})& \le & kM(x_{m-2},x_{m-2},x_{m+t-1})\\ =& kmax\{\mathrm{\Omega }(x_{m-2},x_{m-1},x_{m-1}),\mathrm{\Omega }(x_{m+t-1},x_{m+t},x_{m+t}),\\ \frac{1}{2}\mathrm{\Omega }(x_{m-2},x_{m+t},x_{m+t})\}\\ \le & kmax\{\mathrm{\Omega }(x_{m-2},x_{m-1},x_{m-1}),\mathrm{\Omega }(x_{m+t-1},x_{m+t},x_{m+t}),\\ \frac{1}{2}[\mathrm{\Omega }(x_{m-2},x_{m-1},x_{m-1})+\mathrm{\Omega }(x_{m-1},x_m,x_m)\\ +\cdots +\mathrm{\Omega }(x_{m+t-1},x_{m+t},x_{m+t})\left]\right\}.\end{array}$$

(2.7)

By combining expressions (2.6) and (2.7), we find that

$$\begin{array}{r}\mathrm{\Omega }(x_n,x_m,x_l)\\ \le \mathrm{\Omega }(x_n,x_{n+1},x_{n+1})+\mathrm{\Omega }(x_{n+1},x_{n+2},x_{n+2})+\cdots +\mathrm{\Omega }(x_{m-2},x_{m-1},x_{m-1})\\ +kmax\{\mathrm{\Omega }(x_{m-2},x_{m-1},x_{m-1}),\mathrm{\Omega }(x_{m+t-1},x_{m+t},x_{m+t}),\frac{1}{2}[\mathrm{\Omega }(x_{m-2},x_{m-1},x_{m-1})\\ +\mathrm{\Omega }(x_{m-1},x_m,x_m)+\cdots +\mathrm{\Omega }(x_{m+t-1},x_{m+t},x_{m+t})\left]\right\}.\end{array}$$

(2.8)

Taking $n\to \mathrm{\infty }$ in (2.8), we conclude that

$$\underset{n,m,l\to \mathrm{\infty }}{lim}\mathrm{\Omega }(x_n,x_m,x_l)=0,$$

and hence $\{x_n\}$ is a G-Cauchy sequence due to expression (c)of Lemma 6. Since X is G-complete, $\{x_n\}$ converges to a point $u\in X$. Thus, for $\epsilon >0$ and by the lower semi-continuity of Ω, we have

$$\mathrm{\Omega }(x_n,x_m,u)\le \underset{p\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\mathrm{\Omega }(x_n,x_m,x_p)\le \epsilon ,m\ge n,$$

and

$$\mathrm{\Omega }(x_n,u,x_l)\le \underset{p\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\mathrm{\Omega }(x_n,x_p,x_l)\le \epsilon ,l\ge n.$$

Assume that $u\ne Tu$. Since $x_{n+1}\le x_{n+2}$,

$$0<inf\{\mathrm{\Omega }(x_{n+1},u,x_{n+1})+\mathrm{\Omega }(x_{n+1},u,x_{n+2})+\mathrm{\Omega }(x_{n+1},x_{n+2},u):n\in \mathbb{N}\}\le 3\epsilon$$

for every $\epsilon >0$, that is a contraction. Therefore, we have$u=Tu$ and $\mathrm{\Omega }(u,u,u)=0$. □

Definition 16 Let $(X,\le )$ be a partially ordered space and$f,g:X\to X$. We say that g is an f-monotonemapping if

$$x,y\in X,f(x)\le f(y)⟹g(x)\le g(y).$$

Theorem 17Let$(X,G,\le )$be a partially ordered completeG-metric space, and let Ω be anΩ-distance onXsuch thatXis Ω-bounded. Let$f:X⟶X$and$g:f(X)⟶X$commute, fbe non-decreasing andgbe anf-monotone mapping such that:

• (a) $gf(X)\subseteq f^2(X)$;

• (b) $\mathrm{\Omega }(gfx,gy,g^{2}x)\le kM(x,x,y)$, where$M(x,x,y)=max\{\mathrm{\Omega }(f^{2}x,fy,fgx),\mathrm{\Omega }(fy,fy,gy),\mathrm{\Omega }(f^{2}x,f^{2}x,fgx)\}$for all$x,y\in X$with$f(x)\le f(y)$and$0\le k<1$;

• (c) for every$x\in X$and$z\in X$with$f^{2}z\ne gfz$,

  $$inf\{\mathrm{\Omega }(x,z,x)+\mathrm{\Omega }(x,x,z)+\mathrm{\Omega }(f^{2}x,gx,gfx):f^{2}x\le gfx\}>0;$$

• (d) there exists$x_0\in f(X)$such that$f(x_0)\le g(x_0)$;

thenfandghave a unique common fixed pointuinXand$\mathrm{\Omega }(u,u,u)=0$.

Proof Let $x_0\in f(X)$ such that $f(x_0)\le g(x_0)$. By part (a), we can choose $x_1\in f(X)$ such that $f(x_1)=g(x_0)$. Again from part (a), we can choose$x_2\in f(X)$ such that $f(x_2)=g(x_1)$. Continuing this process, we can construct sequences$\{x_n\}$ in $f(X)$ and $\{z_n\}$ in $f^2(X)$ such that

$$y_n=g{x}_n=f{x}_{n+1},$$

(2.9)

and

$$z_n=g{y}_{n-1}=gf{x}_n=fg{x}_n=f{y}_n.$$

(2.10)

Since $f(x_0)\le g(x_0)$ and $f(x_1)=g(x_0)$, we have $f(x_0)\le f(x_1)$. Then by Definition 16, $g(x_0)\le g(x_1)$. Continuing, we obtain

$$g{x}_n\le g{x}_{n+1},\mathrm{\forall }n\ge 0.$$

(2.11)

So, by (2.9) and (2.11), for all $t\ge 1$, $f{x}_n\le f{x}_{n+t}$. Now, for all $s\ge 0$,

$$\begin{array}{rcl}\mathrm{\Omega }(z_n,z_{n+s},z_{n+1})& =& \mathrm{\Omega }(gf{x}_n,g{x}_{n+s-1},g^2{x}_n)\\ \le & kmax\{\mathrm{\Omega }(f^2{x}_n,f{y}_{n+s-1},fg{x}_n),\mathrm{\Omega }(f{y}_{n+s-1},f{y}_{n+s-1},g{y}_{n+s-1}),\\ \mathrm{\Omega }(f^2{x}_n,f^2{x}_n,fg{x}_n)\}\\ =& kmax\{\mathrm{\Omega }(z_{n-1},z_{n+s-1},z_n),\mathrm{\Omega }(z_{n+s-1},z_{n+s-1},z_{n+s}),\\ \mathrm{\Omega }(z_{n-1},z_{n-1},z_n)\}.\end{array}$$

Then, for $s=0$,

$$\mathrm{\Omega }(z_n,z_n,z_{n+1})\le k\mathrm{\Omega }(z_{n-1},z_{n-1},z_n).$$

For $s=1$,

$$\mathrm{\Omega }(z_n,z_{n+1},z_{n+1})\le k^{1+1}max\{\mathrm{\Omega }(z_{n-1},z_n,z_n),\mathrm{\Omega }(z_{n-1},z_{n-1},z_n)\}.$$

For $s=2$,

$$\mathrm{\Omega }(z_n,z_{n+2},z_{n+1})\le k^{1+2}max\{\mathrm{\Omega }(z_{n-1},z_{n+1},z_n),\mathrm{\Omega }(z_{n-1},z_{n-1},z_n)\}$$

and

$$\begin{array}{rcl}\mathrm{\Omega }(z_{n-1},z_{n-1},z_n)& \le & kmax\{\mathrm{\Omega }(z_{n-2},z_{n-2},z_{n-1}),\mathrm{\Omega }(z_{n-2},z_{n-2},z_{n-1}),\mathrm{\Omega }(z_{n-2},z_{n-2},z_{n-1})\}\\ =& k\mathrm{\Omega }(z_{n-2},z_{n-2},z_{n-1})\\ ⋮\\ \le & k^{n-1}\mathrm{\Omega }(z_0,z_0,z_1).\end{array}$$

Therefore, for all $n\ge 1$ and $s\ge 0$,

$$\mathrm{\Omega }(z_n,z_{n+s},z_{n+1})\le k^{n+s}max\{\mathrm{\Omega }(z_{n-1},z_{n+s-1},z_n),\mathrm{\Omega }(z_0,z_0,z_1)\}.$$

(2.12)

Notice that if $\mathrm{\Omega }(z_n,z_{n+s},z_{n+1})\le k^{n+s}\mathrm{\Omega }(z_0,z_0,z_1)$, so for all $s\ge 0$, ${lim}_{n\to \mathrm{\infty }}\mathrm{\Omega }(z_n,z_{n+s},z_{n+1})=0$. If $\mathrm{\Omega }(z_n,z_{n+s},z_{n+1})\le k^{n+s}\mathrm{\Omega }(z_{n-1},z_{n+s-1},z_n)$, so $\{\mathrm{\Omega }(z_{n-1},z_{n+s-1},z_n)\}$ is non-increasing and bounded below. Therefore, thereexists $r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Omega }(z_{n-1},z_{n+s-1},z_n)=r.$$

We shall show that $r=0$. By a standard calculation, using inequality (2.12)and keeping $k<1$ in mind, we obtain ${lim}_{n\to \mathrm{\infty }}\mathrm{\Omega }(z_{n-1},z_{n+s-1},z_n)=0$. Now, for any $l\ge m\ge n$ with $m=n+k$ and $l=m+t$ ($k,t\in \mathbb{N}$), we have

$$\begin{array}{rcl}\mathrm{\Omega }(z_n,z_m,z_l)& \le & \mathrm{\Omega }(z_n,z_{n+1},z_{n+1})+\mathrm{\Omega }(z_{n+1},z_m,z_l)\\ \le & \mathrm{\Omega }(z_n,z_{n+1},z_{n+1})+\mathrm{\Omega }(z_{n+1},z_{n+2},z_{n+2})+\cdots +\mathrm{\Omega }(z_{m-1},z_m,z_l)\\ \le & \mathrm{\Omega }(z_n,z_{n+1},z_{n+1})+\mathrm{\Omega }(z_{n+1},z_{n+2},z_{n+2})+\cdots +\mathrm{\Omega }(z_{m-1},z_m,z_m)\\ +\mathrm{\Omega }(z_m,z_{m+1},z_{m+1})+\cdots +\mathrm{\Omega }(z_{m+t-1},z_m,z_{m+t}).\end{array}$$

So,

$$\underset{n,m,l\to \mathrm{\infty }}{lim}\mathrm{\Omega }(z_n,z_m,z_l)=0,$$

and consequently, by Part (3) of Lemma 6, $\{z_n\}$ is a G-Cauchy sequence. Since X isG-complete, $\{z_n\}$ converges to a point $z\in X$. Thus, for $\epsilon >0$ and by the lower semi-continuity of Ω, we have

$$\mathrm{\Omega }(z_n,z_m,z)\le \underset{p\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\mathrm{\Omega }(z_n,z_m,z_p)\le \epsilon ,m\ge n,$$

and

$$\mathrm{\Omega }(z_n,z,z_l)\le \underset{p\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\mathrm{\Omega }(z_n,z_p,z_l)\le \epsilon ,l\ge n.$$

Assume that $f^{2}z\ne gfz$. Since f is non-decreasing, we obtain

$$z_n=f^2{x}_{n+1}=f(f{x}_{n+1})\le f(f{x}_{n+2})=gf{x}_{n+1}=z_{n+1},$$

then $z_n\le z_{n+1}$. Also, for all $n\ge 1$,

$$\begin{array}{rcl}\mathrm{\Omega }(f^2{z}_n,g{z}_n,gf{z}_n)& =& \mathrm{\Omega }(gf{z}_{n-1},g{z}_n,g^2{z}_{n-1})\\ \le & kmax\{\mathrm{\Omega }(f^2{z}_{n-1},f{z}_n,fg{z}_{n-1}),\mathrm{\Omega }(f{z}_n,f{z}_n,g{z}_n),\\ \mathrm{\Omega }(f^2{z}_{n-1},f^2{z}_{n-1},fg{z}_{n-1})\}\\ =& kmax\{\mathrm{\Omega }(gf{z}_{n-2},g{z}_{n-1},g^2{z}_{n-2}),\mathrm{\Omega }(f{z}_n,f{z}_n,g{z}_n),\\ \mathrm{\Omega }(gf{z}_{n-2},gf{z}_{n-2},g^2{z}_{n-2})\}\\ \le & k^{3}max\{\mathrm{\Omega }(f^2{z}_{n-2},f{z}_{n-1},fg{z}_{n-2}),\mathrm{\Omega }(f{z}_{n-1},f{z}_{n-1},g{z}_{n-1}),\\ \mathrm{\Omega }(f^2{z}_{n-2},f^2{z}_{n-2},fg{z}_{n-2}),\mathrm{\Omega }(f{z}_n,f{z}_n,g{z}_n),\\ \mathrm{\Omega }(f^2{z}_{n-2},f^2{z}_{n-2},fg{z}_{n-2}),\mathrm{\Omega }(f^2{z}_{n-2},f^2{z}_{n-2},gf{z}_{n-2}),\\ \mathrm{\Omega }(f^2{z}_{n-2},f^2{z}_{n-2},fg{z}_{n-2})\}\\ =& k^{3}max\{\mathrm{\Omega }(f^2{z}_{n-2},f{z}_{n-1},fg{z}_{n-2}),\mathrm{\Omega }(f{z}_{n-1},f{z}_{n-1},g{z}_{n-1}),\\ \mathrm{\Omega }(f^2{z}_{n-2},f^2{z}_{n-2},fg{z}_{n-2}),\mathrm{\Omega }(f{z}_n,f{z}_n,g{z}_n)\}\\ ⋮\\ \le & k^{2n+1}max\{\mathrm{\Omega }(f^2{z}_1,g{z}_1,gf{z}_1),\mathrm{\Omega }(f^2{z}_1,f^2{z}_1,fg{z}_1),\\ \mathrm{\Omega }(f{z}_i,f{z}_i,g{z}_i),0\le i\le n\}\\ \le & k^{2n+1}C,\end{array}$$

where $C=max\{\mathrm{\Omega }(f^2{z}_1,g{z}_1,gf{z}_1),\mathrm{\Omega }(f^2{z}_1,f^2{z}_1,fg{z}_1),\mathrm{\Omega }(f{z}_i,f{z}_i,g{z}_i),0\le i\le n\}$, and consequently ${lim}_{n\to \mathrm{\infty }}\mathrm{\Omega }(f^2{z}_n,g{z}_n,gf{z}_n)=0$. Therefore,

$$0<inf\{\mathrm{\Omega }(z_n,z,z_n)+\mathrm{\Omega }(z_n,z_n,z)+\mathrm{\Omega }(f^2{z}_n,g{z}_n,gf{z}_n):n\in \mathbb{N}\}\le 3\epsilon$$

for every $\epsilon >0$, that is a contraction. So, we have$f^{2}z=gfz$. Then, by (b),

$$\begin{array}{rcl}\mathrm{\Omega }(g{f}^{2}z,g(gfz),g^{2}fz)& \le & kmax\{\mathrm{\Omega }(f^{2}fz,f(gfz),fg(fz)),\mathrm{\Omega }(f(gfz),f(gfz),g(gfz)),\\ \mathrm{\Omega }(f^2(fz),f^2(fz),fg(fz))\}.\end{array}$$

So, $\mathrm{\Omega }(g{f}^{2}z,g(gfz),g^{2}fz)=0$. Since X is Ω-bounded,$\mathrm{\Omega }(g{f}^{2}z,g(gfz),g^{2}fz)=0<M$. Similarly, $\mathrm{\Omega }(g{f}^{2}z,gfz,g^{2}fz)\le k\mathrm{\Omega }(f^{2}z,f^{2}z,f^{2}z)<M$. By part (c) of Definition 3,$G(g{f}^{2}z,gfz,g^{2}fz)=0$. Then $g^{2}fz=gfz$, which implies that $gfz$ is a fixed point for g. Now,

$$f(gfz)=g{f}^{2}z=g^{2}fz=gfz.$$

Then $u=gfz$ is a common fixed point of f andg.

Uniqueness. Assume that there exists $v\in X$ such that $fv=gv=v$. Hence, we have

$$\mathrm{\Omega }(v,v,v)\le k\mathrm{\Omega }(v,v,v),$$

and so $\mathrm{\Omega }(v,v,v)=\mathrm{\Omega }(u,u,u)=0$. Also, $\mathrm{\Omega }(v,u,v)=0$. Then, by Part (c) of Definition 3,$u=v$ and $\mathrm{\Omega }(u,u,u)=0$. □

The following corollary is a generalization of Theorem 2.1 [14].

Denote by Λ the set of all functions $\lambda :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ satisfying the following hypotheses:

• (i) λ is a Lebesgue-integrable mapping on each compact subset of $[0,+\mathrm{\infty })$,

• (ii) for every $\epsilon >0$, we have ${\int }_0^{\epsilon }\lambda (s)ds>0$,

• (iii) $\parallel \lambda \parallel <1$, where $\parallel \lambda \parallel$ denotes the norm of λ.

Now, we have the following corollary.

Corollary 18Let$(X,G,\le )$be a partially ordered completeG-metric space, let Ω be anΩ-distance onX, and let$T:X\to X$be a non-decreasing self-mapping. Suppose that$\psi \in \mathrm{\Psi }$and$\varphi \in \mathrm{\Phi }$such that

$${\int }_0^{\psi (\mathrm{\Omega }(Tx,T^{2}x,Ty))}\lambda (s)ds\le {\int }_0^{\psi (\mathrm{\Omega }(x,Tx,y))}\lambda (s)ds-{\int }_0^{\varphi (\mathrm{\Omega }(x,Tx,y))}\lambda (s)ds,$$

(2.13)

for all$x\le Tx$, $y\in X$, where$\lambda \in \mathrm{\Lambda }$. Also, for every$x\in X$,

$$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}>0$$

for every$y\in X$with$y\ne Ty$. If there exists$x_0\in X$with$x_0\le T{x}_0$, thenThas a unique fixed point.

Proof Define $\gamma :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ by $\gamma (t)={\int }_0^t\lambda (s)ds$, then from inequality (2.13), we have

$$\gamma \left(\psi \left(\mathrm{\Omega }(Tx,T^{2}x,Ty)\right)\right)\le \gamma \left(\psi (\mathrm{\Omega }(x,Tx,y))\right)-\gamma \left(\varphi (\mathrm{\Omega }(x,Tx,y))\right),$$

which can be written as

$${\psi }_1\left(\mathrm{\Omega }(Tx,T^{2}x,Ty)\right)\le {\psi }_1(\mathrm{\Omega }(x,Tx,y))-{\varphi }_1(\mathrm{\Omega }(x,Tx,y)),$$

where ${\psi }_1=\gamma \circ \psi$ and ${\varphi }_1=\gamma \circ \varphi$. Since the functions ${\psi }_1$ and ${\varphi }_1$ satisfy the properties of ψ andϕ, by Theorem 10, T has a unique fixedpoint. □

Corollary 19Let$(X,G,\le )$be a partially ordered completeG-metric space, let Ω be anΩ-distance onX, and let$T:X\to X$be a non-decreasing self-mapping. Suppose that thereexists$0\le k<1$such that

$${\int }_0^{\psi (\mathrm{\Omega }(Tx,T^{2}x,Ty))}k\lambda (s)ds\le {\int }_0^{M(x,x,y)}\lambda (s)ds$$

(2.14)

for all$x\le Tx$, $y\in X$, where

$$M(x,x,y)=max\{\mathrm{\Omega }(x,Tx,Tx),\mathrm{\Omega }(y,Ty,Ty),\frac{1}{2}\mathrm{\Omega }(x,Ty,Ty)\}$$

and$\lambda \in \mathrm{\Lambda }$. Also, for every$x\in X$,

$$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}>0$$

for every$y\in X$with$y\ne Ty$. If there exists$x_0\in X$with$x_0\le T{x}_0$, thenThas a unique fixed point.

3 Application

In this section, we give an existence theorem for a solution of the followingintegral equations:

$$x(t)={\int }_0^{1}K(t,s,x(s))ds+g(t),t\in [0,1].$$

(3.1)

Let $X=C([0,1])$ be the set of all continuous functions defined on$[0,1]$. Define $G:X\times X\times X\to \mathbb{R}$ by

$$G(x,y,z)=\parallel x-y\parallel +\parallel y-z\parallel +\parallel z-x\parallel ,$$

where $\parallel x\parallel =sup\{|x(t)|:t\in [0,1]\}$. Then $(X,G)$ is a complete G-metric space. Let$\mathrm{\Omega }=G$. Then Ω is an Ω-distance on X.Define an ordered relation ≤ on X by

$$x\le y\text{iff}x(t)\le y(t),\mathrm{\forall }t\in [0,1].$$

Then $(X,\le )$ is a partially ordered set. Now, we prove thefollowing result.

Theorem 20Suppose the following hypotheses hold:

• (1) $K:[0,1]\times [0,1]\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$and$g:[0,1]\to \mathbb{R}$are continuous mappings,

• (2) Kis non-decreasing in its first coordinate andgis non-decreasing,

• (3) There exists a continuous function $G:[0,1]\times [0,1]\to [0,+\mathrm{\infty })$ such that

  $$|K(t,s,u)-K(t,s,v)|\le G(t,s)|u-v|$$

for every comparable$u,v\in {\mathbb{R}}^{+}$and$s,t\in [0,1]$with${sup}_{t\in [0,1]}{\int }_0^{1}G(t,s)ds\le \frac{1}{2}$,

• (4) There exist continuous, non-decreasing functions$\varphi ,\psi :[0,\mathrm{\infty })\to (0,\mathrm{\infty })$with${\psi }^{-1}(\{0\})={\varphi }^{-1}(\{0\})=\{0\}$and$\psi (r)\le \psi (2r)-\varphi (2r)$for all$r\in [0,\mathrm{\infty })$.

Then the integral equation has a solution in$C([0,1])$.

Proof Define $Tx(t)={\int }_0^{1}K(t,s,x(s))ds+g(t)$. By hypothesis (2), we have that T isnon-decreasing.

Now, if

$$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}=0$$

for every $y\in X$ with $y\ne Ty$, then for each $n\in \mathbb{N}$, there exists $x_n\in C([0,1])$ with $x_n\le T{x}_n$ such that

$$\mathrm{\Omega }(x_n,y,x_n)+\mathrm{\Omega }(x_n,y,T{x}_n)+\mathrm{\Omega }(x_n,T{x}_n,y)\le \frac{1}{n}.$$

Then we have

$$\mathrm{\Omega }(x_n,y,T{x}_n)=\underset{t\in [0,1]}{sup}|x_n-y|+\underset{t\in [0,1]}{sup}|y-T{x}_n|+\underset{t\in [0,1]}{sup}|T{x}_n-x_n|\le \frac{1}{n}.$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n(t)=y(t),\underset{n\to \mathrm{\infty }}{lim}T{x}_n(t)=y(t).$$

By the continuity of K, we have

$$\begin{array}{rl}y(t)& =\underset{n\to \mathrm{\infty }}{lim}T{x}_n(t)={\int }_0^{1}K(t,s,\underset{n\to \mathrm{\infty }}{lim}{x}_n(s))ds+g(t)\\ ={\int }_0^{1}K(t,s,y(s))ds+g(t)=Ty(t),\end{array}$$

which is a contradiction. Therefore,

$$inf\{\mathrm{\Omega }(x,y,x)+\mathrm{\Omega }(x,y,Tx)+\mathrm{\Omega }(x,Tx,y):x\le Tx\}>0.$$

Now, for $x,y\in X$ with $x\le Tx$, we have

$$\begin{array}{rcl}\psi \left(\mathrm{\Omega }(Tx,T^{2}x,Ty)\right)& =& \psi (\underset{t\in [0,1]}{sup}|Tx(t)-T^{2}x(t)|+\underset{t\in [0,1]}{sup}|T^{2}x(t)-Ty(t)|\\ +\underset{t\in [0,1]}{sup}|Ty(t)-Tx(t)|)\\ \le & \psi (\underset{t\in [0,1]}{sup}{\int }_0^1|K(t,s,x(s))-K(t,s,Tx(s))|ds\\ +\underset{t\in [0,1]}{sup}{\int }_0^1|K(t,s,Tx(s))-K(t,s,y(s))|ds\\ +\underset{t\in [0,1]}{sup}{\int }_0^1|K(t,s,y(s))-K(t,s,x(s))|ds)\\ \le & \psi (\underset{t\in [0,1]}{sup}({\int }_0^{1}G(t,s)|x(s)-Tx(s)|ds)\\ +\underset{t\in [0,1]}{sup}({\int }_0^{1}G(t,s)|Tx(s)-y(s)|ds)\\ +\underset{t\in [0,1]}{sup}({\int }_0^{1}G(t,s)|y(s)-x(s)|ds))\\ \le & \psi (\underset{t\in [0,1]}{sup}\left(|x(t)-Tx(t)|\right)\underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)ds\\ +\underset{t\in [0,1]}{sup}\left(|Tx(t)-y(t)|\right)\underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)ds\\ +\underset{t\in [0,1]}{sup}\left(|y(t)-x(t)|\right)\underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)ds)\\ \le & \psi (\frac{1}{2}\underset{t\in [0,1]}{sup}\left(|x(t)-Tx(t)|\right)+\frac{1}{2}\underset{t\in [0,1]}{sup}\left(|Tx(t)-y(t)|\right)\\ +\frac{1}{2}\underset{t\in [0,1]}{sup}\left(|y(t)-x(t)|\right))\\ \le & \psi (\frac{1}{2}\mathrm{\Omega }(x,Tx,y))\le \psi (\mathrm{\Omega }(x,Tx,y))-\varphi (\mathrm{\Omega }(x,Tx,y)).\end{array}$$

Thus, by Theorem 10, there exists a solution $u\in C[0,1]$ of integral equation (3.1). □