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# Some fixed point results in ordered ${G}_{p}$-metric spaces

Fixed Point Theory and Applications20132013:317

https://doi.org/10.1186/1687-1812-2013-317

• Accepted: 25 October 2013
• Published:

## Abstract

In this paper, first we present some coincidence point results for six mappings satisfying the generalized $\left(\psi ,\phi \right)$-weakly contractive condition in the framework of partially ordered ${G}_{p}$-metric spaces. Secondly, we consider α-admissible mappings in the setup of ${G}_{p}$-metric spaces. An example is also provided to support our results.

MSC:47H10, 54H25.

## Keywords

• coincidence point
• common fixed point
• generalized weak contraction
• generalized metric space
• partially weakly increasing mapping
• altering distance function

## 1 Introduction and mathematical preliminaries

Recently, Zand and Nezhad [1] have introduced a new generalized metric space, a ${G}_{p}$-metric space, as a generalization of both partial metric spaces [2] and G-metric spaces [3].

We will use the following definition of a ${G}_{p}$-metric space.

Definition 1.1 [4]

Let X be a nonempty set. Suppose that a mapping ${G}_{p}:X×X×X\to {\mathbb{R}}^{+}$ satisfies:

(${G}_{p}1$) $x=y=z$ if ${G}_{p}\left(x,y,z\right)={G}_{p}\left(z,z,z\right)={G}_{p}\left(y,y,y\right)={G}_{p}\left(x,x,x\right)$;

(${G}_{p}2$) ${G}_{p}\left(x,x,x\right)\le {G}_{p}\left(x,x,y\right)\le {G}_{p}\left(x,y,z\right)$ for all $x,y,z\in X$ with $z\ne y$;

(${G}_{p}3$) ${G}_{p}\left(x,y,z\right)={G}_{p}\left(p\left\{x,y,z\right\}\right)$, where p is any permutation of x, y, z (symmetry in all three variables);

(${G}_{p}4$) ${G}_{p}\left(x,y,z\right)\le {G}_{p}\left(x,a,a\right)+{G}_{p}\left(a,y,z\right)-{G}_{p}\left(a,a,a\right)$ for all $x,y,z,a\in X$ (rectangle inequality).

Then ${G}_{p}$ is called a ${G}_{p}$-metric and $\left(X,{G}_{p}\right)$ is called a ${G}_{p}$-metric space.

The ${G}_{p}$-metric ${G}_{p}$ is called symmetric if
${G}_{p}\left(x,x,y\right)={G}_{p}\left(x,y,y\right)$
(1)

holds for all $x,y\in X$. Otherwise, ${G}_{p}$ is an asymmetric ${G}_{p}$-metric.

Remark 1 In [1] (see also [5]), instead of (${G}_{p}2$), the following condition was used:

(${G}_{p}{2}^{\prime }$) ${G}_{p}\left(x,x,x\right)\le {G}_{p}\left(x,x,y\right)\le {G}_{p}\left(x,y,z\right)$ for all $x,y,z\in X$.

However, with this assumption, it is very easy to obtain that (1) holds for all $x,y\in X$, i.e., the respective space is symmetric. On the other hand, there are a lot of examples of non-symmetric G-metric spaces. Hence, the conclusion stated in [1, 5] that each G-metric space is a ${G}_{p}$-metric space (satisfying (${G}_{p}{2}^{\prime }$)) does not hold. With our assumption (${G}_{p}2$), this conclusion holds true.

The following are some easy examples of ${G}_{p}$-metric spaces.

Example 1.1 Let $X=\left[0,+\mathrm{\infty }\right)$, and let ${G}_{p}:{X}^{3}\to {\mathbb{R}}^{+}$ be given by ${G}_{p}\left(x,y,z\right)=max\left\{x,y,z\right\}$. Obviously, $\left(X,{G}_{p}\right)$ is a symmetric ${G}_{p}$-metric space which is not a G-metric space.

Example 1.2 Let $X=\left\{0,1,2,3,\dots \right\}$. Define ${G}_{p}:{X}^{3}\to X$ by
${G}_{p}\left(x,y,z\right)=\left\{\begin{array}{cc}x+y+z+1,\hfill & x\ne y\ne z,\hfill \\ x+z+1,\hfill & y=z\ne x,\hfill \\ y+z+1,\hfill & x=z\ne y,\hfill \\ x+z+1,\hfill & x=y\ne z,\hfill \\ 1,\hfill & x=y=z.\hfill \end{array}$

It is easy to see that $\left(X,{G}_{p}\right)$ is a symmetric ${G}_{p}$-metric space.

Example 1.3 [4]

Let $X=\left\{0,1,2,3\right\}$. Let
$\begin{array}{rl}A=& \left\{\left(1,0,0\right),\left(0,1,0\right),\left(0,0,1\right),\left(2,0,0\right),\left(0,2,0\right),\left(0,0,2\right),\left(3,0,0\right),\left(0,3,0\right),\left(0,0,3\right),\\ \left(1,2,2\right),\left(2,1,2\right),\left(2,2,1\right),\left(2,3,3\right),\left(3,2,3\right),\left(3,3,2\right)\right\},\\ B=& \left\{\left(0,1,1\right),\left(1,0,1\right),\left(1,1,0\right),\left(0,2,2\right),\left(2,0,2\right),\left(2,2,0\right),\left(0,3,3\right),\left(3,0,3\right),\left(3,3,0\right),\\ \left(2,1,1\right),\left(1,2,1\right),\left(1,1,2\right),\left(3,2,2\right),\left(2,3,2\right),\left(2,2,3\right)\right\}.\end{array}$
Define ${G}_{p}:{X}^{3}\to {\mathbb{R}}^{+}$ by

It is easy to see that $\left(X,{G}_{p}\right)$ is an asymmetric ${G}_{p}$-metric space.

Proposition 1.1 [1]

Every ${G}_{p}$-metric space $\left(X,{G}_{p}\right)$ defines a metric space $\left(X,{d}_{{G}_{p}}\right)$ where
${d}_{{G}_{p}}\left(x,y\right)={G}_{p}\left(x,y,y\right)+{G}_{p}\left(y,x,x\right)-{G}_{p}\left(x,x,x\right)-{G}_{p}\left(y,y,y\right)$

for all $x,y\in X$.

Proposition 1.2 [1]

Let X be a ${G}_{p}$-metric space. Then, for each $x,y,z,a\in X$, it follows that:
1. (1)

${G}_{p}\left(x,y,z\right)\le {G}_{p}\left(x,a,a\right)+{G}_{p}\left(y,a,a\right)+{G}_{p}\left(z,a,a\right)-2{G}_{p}\left(a,a,a\right)$;

2. (2)

${G}_{p}\left(x,y,z\right)\le {G}_{p}\left(x,x,y\right)+{G}_{p}\left(x,x,z\right)-{G}_{p}\left(x,x,x\right)$;

3. (3)

${G}_{p}\left(x,y,y\right)\le 2{G}_{p}\left(x,x,y\right)-{G}_{p}\left(x,x,x\right)$;

4. (4)

${G}_{p}\left(x,y,z\right)\le {G}_{p}\left(x,a,z\right)+{G}_{p}\left(a,y,z\right)-{G}_{p}\left(a,a,a\right)$, $a\ne z$.

Definition 1.2 [1]

Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-metric space. Let $\left\{{x}_{n}\right\}$ be a sequence of points of X.
1. 1.

A point $x\in X$ is said to be a limit of the sequence $\left\{{x}_{n}\right\}$, denoted by ${x}_{n}\to x$, if ${lim}_{n,m\to \mathrm{\infty }}{G}_{p}\left(x,{x}_{n},{x}_{m}\right)={G}_{p}\left(x,x,x\right)$.

2. 2.

$\left\{{x}_{n}\right\}$ is said to be a ${G}_{p}$-Cauchy sequence if ${lim}_{n,m\to \mathrm{\infty }}{G}_{p}\left({x}_{n},{x}_{m},{x}_{m}\right)$ exists (and is finite).

3. 3.

$\left(X,{G}_{p}\right)$ is said to be ${G}_{p}$-complete if every ${G}_{p}$-Cauchy sequence in X is ${G}_{p}$-convergent to $x\in X$.

Using the above definitions, one can easily prove the following proposition.

Proposition 1.3 [1]

Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-metric space. Then, for any sequence $\left\{{x}_{n}\right\}$ in X and a point $x\in X$, the following are equivalent:
1. (1)

$\left\{{x}_{n}\right\}$ is ${G}_{p}$-convergent to x.

2. (2)

${G}_{p}\left({x}_{n},{x}_{n},x\right)\to {G}_{p}\left(x,x,x\right)$ as $n\to \mathrm{\infty }$.

3. (3)

${G}_{p}\left({x}_{n},x,x\right)\to {G}_{p}\left(x,x,x\right)$ as $n\to \mathrm{\infty }$.

Lemma 1.1 [4]

If ${G}_{p}$ is a ${G}_{p}$-metric on X, then the mappings ${d}_{{G}_{p}},{d}_{{G}_{p}}^{\prime }:X×X\to {R}^{+}$, given by
${d}_{{G}_{p}}\left(x,y\right)={G}_{p}\left(x,y,y\right)+{G}_{p}\left(y,x,x\right)-{G}_{p}\left(x,x,x\right)-{G}_{p}\left(y,y,y\right)$
and
${d}_{{G}_{p}}^{\prime }\left(x,y\right)=max\left\{{G}_{p}\left(x,y,y\right)-{G}_{p}\left(x,x,x\right),{G}_{p}\left(y,x,x\right)-{G}_{p}\left(y,y,y\right)\right\},$

define equivalent metrics on X.

Proof $\frac{a+b}{2}\le max\left\{a,b\right\}\le a+b$ for all nonnegative real numbers a, b. □

Based on Lemma 2.2 of [6], Parvaneh et al. have proved the following essential lemma.

Lemma 1.2 [4]
1. (1)

A sequence $\left\{{x}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence in a ${G}_{p}$-metric space $\left(X,{G}_{p}\right)$ if and only if it is a Cauchy sequence in the metric space $\left(X,{d}_{{G}_{p}}\right)$.

2. (2)
A ${G}_{p}$-metric space $\left(X,{G}_{p}\right)$ is ${G}_{p}$-complete if and only if the metric space $\left(X,{d}_{{G}_{p}}\right)$ is complete. Moreover, ${lim}_{n\to \mathrm{\infty }}{d}_{{G}_{p}}\left(x,{x}_{n}\right)=0$ if and only if
$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(x,{x}_{n},{x}_{n}\right)& =\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},x,x\right)=\underset{n,m\to \mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},{x}_{n},{x}_{m}\right)\\ =\underset{n,m\to \mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},{x}_{m},{x}_{m}\right)={G}_{p}\left(x,x,x\right).\end{array}$

Lemma 1.3 [4]

Assume that ${x}_{n}\to x$ as $n\to \mathrm{\infty }$ in a ${G}_{p}$-metric space $\left(X,{G}_{p}\right)$ such that ${G}_{p}\left(x,x,x\right)=0$. Then, for every $y\in X$,
1. (i)

${lim}_{n\to \mathrm{\infty }}{G}_{p}\left({x}_{n},y,y\right)={G}_{p}\left(x,y,y\right)$,

2. (ii)

${lim}_{n\to \mathrm{\infty }}{G}_{p}\left({x}_{n},{x}_{n},y\right)={G}_{p}\left(x,x,y\right)$.

Lemma 1.4 [4]

Assume that $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are three sequences in a ${G}_{p}$-metric space X such that
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},x,x\right)=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},{x}_{n},{x}_{n}\right)={G}_{p}\left(x,x,x\right),\hfill \\ \underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({y}_{n},y,y\right)=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({y}_{n},{y}_{n},{y}_{n}\right)={G}_{p}\left(y,y,y\right)\hfill \end{array}$
and
$\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{n},z,z\right)=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{n},{z}_{n},{z}_{n}\right)={G}_{p}\left(z,z,z\right).$
Then
1. (i)

${lim}_{n\to \mathrm{\infty }}{G}_{p}\left({x}_{n},{y}_{n},{z}_{n}\right)={G}_{p}\left(x,y,z\right)$ and

2. (ii)

${lim}_{n\to \mathrm{\infty }}{G}_{p}\left({x}_{n},{x}_{n},y\right)={G}_{p}\left(x,x,y\right)$

for every $y,z\in X$.

Lemma 1.5 [5]

Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-metric space. Then
1. (A)

If ${G}_{p}\left(x,y,z\right)=0$, then $x=y=z$.

2. (B)

If $x\ne y$, then ${G}_{p}\left(x,y,y\right)>0$.

Definition 1.3 [1]

Let $\left({X}_{1},{G}_{1}\right)$ and $\left({X}_{2},{G}_{2}\right)$ be two ${G}_{p}$-metric spaces, and let $f:\left({X}_{1},{G}_{1}\right)\to \left({X}_{2},{G}_{2}\right)$ be a mapping. Then f is said to be ${G}_{p}$-continuous at a point $a\in {X}_{1}$ if for a given $\epsilon >0$, there exists $\delta >0$ such that $x,y\in {X}_{1}$ and ${G}_{1}\left(a,x,y\right)<\delta +{G}_{1}\left(a,a,a\right)$ imply that ${G}_{2}\left(f\left(a\right),f\left(x\right),f\left(y\right)\right)<\epsilon +{G}_{2}\left(f\left(a\right),f\left(a\right),f\left(a\right)\right)$. The mapping f is ${G}_{p}$-continuous on ${X}_{1}$ if it is ${G}_{p}$-continuous at all $a\in {X}_{1}$.

Proposition 1.4 [1]

Let $\left({X}_{1},{G}_{1}\right)$ and $\left({X}_{2},{G}_{2}\right)$ be two ${G}_{p}$-metric spaces. Then a mapping $f:{X}_{1}\to {X}_{2}$ is ${G}_{p}$-continuous at a point $x\in {X}_{1}$ if and only if it is ${G}_{p}$-sequentially continuous at x; that is, whenever $\left\{{x}_{n}\right\}$ is ${G}_{p}$-convergent to x, $\left\{f\left({x}_{n}\right)\right\}$ is ${G}_{p}$-convergent to $f\left(x\right)$.

The concept of an altering distance function was introduced by Khan et al. [7] as follows.

Definition 1.4 The function $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is called an altering distance function if the following properties are satisfied:
1. 1.

ψ is continuous and nondecreasing.

2. 2.

$\psi \left(t\right)=0$ if and only if $t=0$.

A self-mapping f on X is called a weak contraction if the following contractive condition is satisfied:
$d\left(fx,fy\right)\le d\left(x,y\right)-\phi \left(d\left(x,y\right)\right),$

for all $x,y\in X$, where φ is an altering distance function.

The concept of a weakly contractive mapping was introduced by Alber and Guerre-Delabrere [8] in the setup of Hilbert spaces. Rhoades [9] considered this class of mappings in the setup of metric spaces and proved that a weakly contractive mapping defined on a complete metric space has a unique fixed point.

Zhang and Song [10] introduced the concept of a generalized φ-weakly contractive mapping as follows.

Definition 1.5 Self-mappings f and g on a metric space X are called generalized φ-weak contractions if there exists a lower semicontinuous function $\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ with $\phi \left(0\right)=0$ and $\phi \left(t\right)>0$ for all $t>0$ such that for all $x,y\in X$,
$d\left(fx,gy\right)\le N\left(x,y\right)-\phi \left(N\left(x,y\right)\right),$
where
$N\left(x,y\right)=max\left\{d\left(x,y\right),d\left(x,fx\right),d\left(y,gy\right),\frac{1}{2}\left[d\left(x,gy\right)+d\left(y,fx\right)\right]\right\}.$

Based on the above definition, they proved the following common fixed point result.

Theorem 1.1 [10]

Let $\left(X,d\right)$ be a complete metric space. If $f,g:X\to X$ are generalized φ-weakly contractive mappings, then there exists a unique point $u\in X$ such that $u=fu=gu$.

So far, many authors extended Theorem 1.1 (see [1113] and [14]). Moreover, Ðorić [12] generalized it by the definition of generalized $\left(\psi ,\phi \right)$-weak contractions.

Definition 1.6 Two mappings $f,g:X\to X$ are called generalized $\left(\psi ,\phi \right)$-weakly contractive if there exist two maps $\psi ,\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ such that
$\psi \left(d\left(fx,gy\right)\right)\le \psi \left(N\left(x,y\right)\right)-\phi \left(N\left(x,y\right)\right),$

for all $x,y\in X$, where N and φ are as in Definition 1.5 and $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is an altering distance function.

Theorem 1.2 [12]

Let $\left(X,d\right)$ be a complete metric space, and let $f,g:X\to X$ be generalized $\left(\psi ,\phi \right)$-weakly contractive self-mappings. Then there exists a unique point $u\in X$ such that $u=fu=gu$.

Recently, many researchers have focused on different contractive conditions in various metric spaces endowed with a partial order and studied fixed point theory in the so-called bi-structured spaces. For more details on fixed point results, their applications, comparison of different contractive conditions and related results in ordered various metric spaces, we refer the reader to [1529] and the references mentioned therein.

Let X be a nonempty set and $f:X\to X$ be a given mapping. For every $x\in X$, let ${f}^{-1}\left(x\right)=\left\{u\in X:fu=x\right\}$.

Definition 1.7 [24]

Let $\left(X,⪯\right)$ be a partially ordered set, and let $f,g,h:X\to X$ be given mappings such that $fX\subseteq hX$ and $gX\subseteq hX$. We say that f and g are weakly increasing with respect to h if for all $x\in X$, we have
and

If $f=g$, we say that f is weakly increasing with respect to h.

If $h=I$ (the identity mapping on X), then the above definition reduces to that of a weakly increasing mapping [30] (see also [24, 31]).

Definition 1.8 A partially ordered ${G}_{p}$-metric space $\left(X,⪯,{G}_{p}\right)$ is said to have the sequential limit comparison property if for every nondecreasing sequence (nonincreasing sequence) $\left\{{x}_{n}\right\}$ in X, ${x}_{n}\to x$ implies that ${x}_{n}⪯x$ ($x⪯{x}_{n}$).

The aim of this paper is to prove some coincidence and common fixed point theorems for weakly $\left(\psi ,\phi \right)$-contractive mappings in partially ordered ${G}_{p}$-metric spaces.

## 2 Main results

Let $\left(X,⪯,{G}_{p}\right)$ be an ordered ${G}_{p}$-metric space and $f,g,h,R,S,T:X\to X$ be six self-mappings. Throughout this paper, unless otherwise stated, for all $x,y,z\in X$, let
$\begin{array}{rl}M\left(x,y,z\right)=& max\left\{{G}_{p}\left(Tx,Ry,Sz\right),\\ {G}_{p}\left(Tx,fx,fx\right),{G}_{p}\left(Ry,gy,gy\right),{G}_{p}\left(Sz,hz,hz\right),\\ \frac{{G}_{p}\left(Tx,Tx,fx\right)+{G}_{p}\left(Ry,Ry,gy\right)+{G}_{p}\left(Sz,Sz,hz\right)}{3}\right\}.\end{array}$
Theorem 2.1 Let $\left(X,⪯,{G}_{p}\right)$ be a partially ordered ${G}_{p}$-metric space with the sequential limit comparison property. Let $f,g,h,R,S,T:X\to X$ be six mappings such that $f\left(X\right)\subseteq R\left(X\right)$, $g\left(X\right)\subseteq S\left(X\right)$ and $h\left(X\right)\subseteq T\left(X\right)$, and RX, SX and TX are ${G}_{p}$-complete subsets of X. Suppose that for comparable elements $Tx,Ry,Sz\in X$, we have
$\psi \left(2{G}_{p}\left(fx,gy,hz\right)\right)\le \psi \left(M\left(x,y,z\right)\right)-\phi \left(M\left(x,y,z\right)\right),$
(2)

where $\psi ,\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ are altering distance functions. Then the pairs $\left(f,T\right)$, $\left(g,R\right)$ and $\left(h,S\right)$ have a coincidence point ${z}^{\ast }$ in X provided that the pairs $\left(f,T\right)$, $\left(g,R\right)$ and $\left(h,S\right)$ are weakly compatible and the pairs $\left(f,g\right)$, $\left(g,h\right)$ and $\left(h,f\right)$ are partially weakly increasing with respect to R, S and T, respectively. Moreover, if $R{z}^{\ast }$, $S{z}^{\ast }$ and $T{z}^{\ast }$ are comparable, then ${z}^{\ast }\in X$ is a coincidence point of f, g, h, R, S and T.

Proof Let ${x}_{0}$ be an arbitrary point of X. Choose ${x}_{1}\in X$ such that $f{x}_{0}=R{x}_{1}$, ${x}_{2}\in X$ such that $g{x}_{1}=S{x}_{2}$ and ${x}_{3}\in X$ such that $h{x}_{2}=T{x}_{3}$. This can be done as $f\left(X\right)\subseteq R\left(X\right)$, $g\left(X\right)\subseteq S\left(X\right)$ and $h\left(X\right)\subseteq T\left(X\right)$.

Continuing this way, construct a sequence $\left\{{z}_{n}\right\}$ defined by ${z}_{3n+1}=R{x}_{3n+1}=f{x}_{3n}$, ${z}_{3n+2}=S{x}_{3n+2}=g{x}_{3n+1}$ and ${z}_{3n+3}=T{x}_{3n+3}=h{x}_{3n+2}$ for all $n\ge 0$. The sequence $\left\{{z}_{n}\right\}$ in X is said to be a Jungck-type iterative sequence with initial guess ${x}_{0}$.

As ${x}_{1}\in {R}^{-1}\left(f{x}_{0}\right)$, ${x}_{2}\in {S}^{-1}\left(g{x}_{1}\right)$ and ${x}_{3}\in {T}^{-1}\left(h{x}_{2}\right)$ and the pairs $\left(f,g\right)$, $\left(g,h\right)$ and $\left(h,f\right)$ are partially weakly increasing with respect to R, S and T, respectively, we have
$R{x}_{1}=f{x}_{0}⪯g{x}_{1}=S{x}_{2}⪯h{x}_{2}=T{x}_{3}⪯f{x}_{3}=R{x}_{4}.$

Continuing this process, we obtain $R{x}_{3n+1}⪯S{x}_{3n+2}⪯T{x}_{3n+3}$ for all $n\ge 0$.

We will complete the proof in three steps.

Step I. We will prove that $\left\{{z}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence. First, we show that ${lim}_{k\to \mathrm{\infty }}{G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right)=0$.

Define ${G}_{{p}_{k}}={G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right)$. Suppose ${G}_{{p}_{{k}_{0}}}=0$ for some ${k}_{0}$. Then ${z}_{{k}_{0}}={z}_{{k}_{0}+1}={z}_{{k}_{0}+2}$. In the case that ${k}_{0}=3n$, then ${z}_{3n}={z}_{3n+1}={z}_{3n+2}$ gives ${z}_{3n+1}={z}_{3n+2}={z}_{3n+3}$. Indeed,
$\begin{array}{rl}\psi \left(2{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)& =\psi \left(2{G}_{p}\left(f{x}_{3n},g{x}_{3n+1},h{x}_{3n+2}\right)\right)\\ \le \psi \left(M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right)-\phi \left(M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right),\end{array}$
where
$\begin{array}{r}M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left(T{x}_{3n},R{x}_{3n+1},S{x}_{3n+2}\right),{G}_{p}\left(T{x}_{3n},f{x}_{3n},f{x}_{3n}\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left(R{x}_{3n+1},g{x}_{3n+1},g{x}_{3n+1}\right),{G}_{p}\left(S{x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{G}_{p}\left(T{x}_{3n},T{x}_{3n},f{x}_{3n}\right)+{G}_{p}\left(R{x}_{3n+1},R{x}_{3n+1},g{x}_{3n+1}\right)+{G}_{p}\left(S{x}_{3n+2},S{x}_{3n+2},h{x}_{3n+2}\right)}{3}\right\}\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right),{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+1}\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+2}\right),{G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+3}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{G}_{p}\left({z}_{3n},{z}_{3n},{z}_{3n+1}\right)+{G}_{p}\left({z}_{3n+1},{z}_{3n+1},{z}_{3n+2}\right)+{G}_{p}\left({z}_{3n+2},{z}_{3n+2},{z}_{3n+3}\right)}{3}\right\}\\ \phantom{\rule{1em}{0ex}}=max\left\{0,0,0,{G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+3}\right),\frac{0+0+{G}_{p}\left({z}_{3n+2},{z}_{3n+2},{z}_{3n+3}\right)}{3}\right\}\\ \phantom{\rule{1em}{0ex}}={G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+3}\right)\\ \phantom{\rule{1em}{0ex}}\le 2{G}_{p}\left({z}_{3n+2},{z}_{3n+2},{z}_{3n+3}\right)\\ \phantom{\rule{1em}{0ex}}=2{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right).\end{array}$
Thus
$\psi \left(2{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)\le \psi \left(2{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)-\phi \left({G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+3}\right)\right)$

implies that $\phi \left({G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+3}\right)\right)=0$, that is, ${z}_{3n+1}={z}_{3n+2}={z}_{3n+3}$. Similarly, if ${k}_{0}=3n+1$, then ${z}_{3n+1}={z}_{3n+2}={z}_{3n+3}$ gives ${z}_{3n+2}={z}_{3n+3}={z}_{3n+4}$. Also, if ${k}_{0}=3n+2$, then ${z}_{3n+2}={z}_{3n+3}={z}_{3n+4}$ implies that ${z}_{3n+3}={z}_{3n+4}={z}_{3n+5}$. Consequently, the sequence $\left\{{z}_{k}\right\}$ becomes constant for $k\ge {k}_{0}$, hence $\left\{{z}_{k}\right\}$ is ${G}_{p}$-Cauchy.

Suppose that
${z}_{k}\ne {z}_{k+1}\ne {z}_{k+2}$
(3)
for each k. We now claim that the following inequality holds:
${G}_{p}\left({z}_{k+1},{z}_{k+2},{z}_{k+3}\right)\le {G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right)=M\left({x}_{k},{x}_{k+1},{x}_{k+2}\right)$
(4)

for each $k=1,2,3,\dots$ .

Let $k=3n$ and for $n\ge 0$, ${G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)>{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right)>0$. Then, as $T{x}_{3n}⪯R{x}_{3n+1}⪯S{x}_{3n+2}$, using (2) we obtain that
$\begin{array}{rcl}\psi \left({G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)& \le & \psi \left(2{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)\\ =& \psi \left(2{G}_{p}\left(f{x}_{3n},g{x}_{3n+1},h{x}_{3n+2}\right)\right)\\ \le & \psi \left(M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right)-\phi \left(M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right),\end{array}$
(5)
where
$\begin{array}{r}M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left(T{x}_{3n},R{x}_{3n+1},S{x}_{3n+2}\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left(T{x}_{3n},f{x}_{3n},f{x}_{3n}\right),{G}_{p}\left(R{x}_{3n+1},g{x}_{3n+1},g{x}_{3n+1}\right),{G}_{p}\left(S{x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{G}_{p}\left(T{x}_{3n},T{x}_{3n},f{x}_{3n}\right)+{G}_{p}\left(R{x}_{3n+1},R{x}_{3n+1},g{x}_{3n+1}\right)+{G}_{p}\left(S{x}_{3n+2},S{x}_{3n+2},h{x}_{3n+2}\right)}{3}\right\}\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+1}\right),{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+2}\right),{G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+3}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{G}_{p}\left({z}_{3n},{z}_{3n},{z}_{3n+1}\right)+{G}_{p}\left({z}_{3n+1},{z}_{3n+1},{z}_{3n+2}\right)+{G}_{p}\left({z}_{3n+2},{z}_{3n+2},{z}_{3n+3}\right)}{3}\right\}\\ \phantom{\rule{1em}{0ex}}\le max\left\{{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right),{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right),\\ \phantom{\rule{2em}{0ex}}\frac{2{G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right)+{G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)}{3}\right\}\\ \phantom{\rule{1em}{0ex}}={G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right).\end{array}$
Hence (5) implies that
$\psi \left({G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)\le \psi \left({G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\right)-\phi \left(M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right),$
which is possible only if $M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)=0$, that is, ${G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right)=0$. A contradiction to (3). Hence, ${G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)\le {G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right)$ and
$M\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)={G}_{p}\left({z}_{3n},{z}_{3n+1},{z}_{3n+2}\right).$

Therefore, (4) is proved for $k=3n$.

Similarly, it can be shown that
${G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+4}\right)\le {G}_{p}\left({z}_{3n+1},{z}_{3n+2},{z}_{3n+3}\right)=M\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)$
(6)
and
${G}_{p}\left({z}_{3n+3},{z}_{3n+4},{z}_{3n+5}\right)\le {G}_{p}\left({z}_{3n+2},{z}_{3n+3},{z}_{3n+4}\right)=M\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right).$
(7)
Hence, $\left\{{G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right)\right\}$ is a nonincreasing sequence of nonnegative real numbers. Therefore, there is $r\ge 0$ such that
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right)=r.$
(8)
Since
${G}_{p}\left({z}_{k+1},{z}_{k+2},{z}_{k+3}\right)\le M\left({x}_{k},{x}_{k+1},{x}_{k+2}\right)\le {G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right),$
(9)
taking the limit as $k\to \mathrm{\infty }$ in (9), we obtain
$\underset{k\to \mathrm{\infty }}{lim}M\left({x}_{k},{x}_{k+1},{x}_{k+2}\right)=r.$
(10)
Taking the limit as $n\to \mathrm{\infty }$ in (5), using (8), (10) and the continuity of ψ and φ, we have $\psi \left(r\right)\le \psi \left(r\right)-\phi \left(r\right)$. Therefore, $\phi \left(r\right)=0$. Hence
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+2}\right)=0$
(11)
from our assumptions about φ. Also, from Definition 1.1, part (${G}_{p}2$), we have
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{k},{z}_{k+1},{z}_{k+1}\right)=0,$
(12)
and since ${G}_{p}\left(x,y,y\right)\le 2{G}_{p}\left(x,x,y\right)$ for all $x,y\in X$, we have
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{k},{z}_{k},{z}_{k+1}\right)=0.$
(13)
Step II. We now show that $\left\{{z}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence in X. Therefore, we will show that
$\underset{m,n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{m},{z}_{n},{z}_{n}\right)=0.$
Because of (11), (12) and (13), it is sufficient to show that
$\underset{m,n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m},{z}_{3n},{z}_{3n}\right)=0,$

i.e., we prove that $\left\{{z}_{3n}\right\}$ is ${G}_{p}$-Cauchy.

Suppose the opposite. Then there exists $\epsilon >0$ for which we can find subsequences $\left\{{z}_{3m\left(k\right)}\right\}$ and $\left\{{z}_{3n\left(k\right)}\right\}$ of $\left\{{z}_{3n}\right\}$ such that $n\left(k\right)>m\left(k\right)\ge k$ and
${G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)\ge \epsilon ,$
(14)
and $n\left(k\right)$ is the smallest number such that the above statement holds; i.e.,
${G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)-3},{z}_{3n\left(k\right)-3}\right)<\epsilon .$
(15)
From the rectangle inequality and (15), we have
$\begin{array}{r}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)-3},{z}_{3n\left(k\right)-3}\right)+{G}_{p}\left({z}_{3n\left(k\right)-3},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)\\ \phantom{\rule{1em}{0ex}}<\epsilon +{G}_{p}\left({z}_{3n\left(k\right)-3},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)\\ \phantom{\rule{1em}{0ex}}<\epsilon +{G}_{p}\left({z}_{3n\left(k\right)-3},{z}_{3n\left(k\right)-2},{z}_{3n\left(k\right)-2}\right)+{G}_{p}\left({z}_{3n\left(k\right)-2},{z}_{3n\left(k\right)-1},{z}_{3n\left(k\right)-1}\right)\\ \phantom{\rule{2em}{0ex}}+{G}_{p}\left({z}_{3n\left(k\right)-1},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right).\end{array}$
(16)
Taking limit as $k\to \mathrm{\infty }$ in (16), from (12) and (14) we obtain that
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)=\epsilon .$
(17)
Using the rectangle inequality and (${G}_{p}2$), we have
$\begin{array}{r}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)+{G}_{p}\left({z}_{3n\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+1}\right)+{G}_{p}\left({z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)\\ \phantom{\rule{2em}{0ex}}+{G}_{p}\left({z}_{3n\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)+{G}_{p}\left({z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{2em}{0ex}}+{G}_{p}\left({z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)+{G}_{p}\left({z}_{3n\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right).\end{array}$
(18)
Taking limit as $k\to \mathrm{\infty }$, we have
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\le \epsilon \le \underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right).$
Finally,
$\begin{array}{r}{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3m\left(k\right)},{z}_{3m\left(k\right)}\right)+{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3m\left(k\right)},{z}_{3m\left(k\right)}\right)+{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)},{z}_{3n\left(k\right)}\right)\\ \phantom{\rule{2em}{0ex}}+{G}_{p}\left({z}_{3n\left(k\right)},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right).\end{array}$
(19)
Taking limit as $k\to \mathrm{\infty }$ and using (17), we have
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)\le \epsilon .$
Consider,
$\begin{array}{r}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3m\left(k\right)+1},{z}_{3m\left(k\right)+1}\right)+{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3m\left(k\right)+1},{z}_{3m\left(k\right)+1}\right)+{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+3},{z}_{3n\left(k\right)+3}\right)\\ \phantom{\rule{2em}{0ex}}+{G}_{p}\left({z}_{3n\left(k\right)+3},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}\le {G}_{p}\left({z}_{3m\left(k\right)},{z}_{3m\left(k\right)+1},{z}_{3m\left(k\right)+1}\right)+{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)\\ \phantom{\rule{2em}{0ex}}+{G}_{p}\left({z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right).\end{array}$
(20)
Taking limit as $k\to \mathrm{\infty }$ and using (11), (12) and (13), we have
$\epsilon \le \underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right).$
Therefore,
$\underset{k\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)=\epsilon .$
(21)
As $T{x}_{m\left(k\right)}⪯R{x}_{n\left(k\right)+1}⪯S{x}_{n\left(k\right)+2}$, so from (2) we have
$\begin{array}{r}\psi \left({G}_{p}\left({z}_{3m\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)\right)\\ \phantom{\rule{1em}{0ex}}=\psi \left({G}_{p}\left(f{x}_{3m\left(k\right)},g{x}_{3n\left(k\right)+1},h{x}_{3n\left(k\right)+2}\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \psi \left(M\left({x}_{3m\left(k\right)},{x}_{3n\left(k\right)+1},{x}_{3n\left(k\right)+2}\right)\right)-\phi \left(M\left({x}_{3m\left(k\right)},{x}_{3n\left(k\right)+1},{x}_{3n\left(k\right)+2}\right)\right),\end{array}$
(22)
where
$\begin{array}{r}M\left({x}_{3m\left(k\right)},{x}_{3n\left(k\right)+1},{x}_{3n\left(k\right)+2}\right)\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left(T{x}_{3m\left(k\right)},R{x}_{3n\left(k\right)+1},S{x}_{3n\left(k\right)+2}\right),{G}_{p}\left(T{x}_{3m\left(k\right)},f{x}_{3m\left(k\right)},f{x}_{3m\left(k\right)}\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left(R{x}_{3n\left(k\right)+1},g{x}_{3n\left(k\right)+1},g{x}_{3n\left(k\right)+1}\right),{G}_{p}\left(S{x}_{3n\left(k\right)+2},h{x}_{3n\left(k\right)+2},h{x}_{3n\left(k\right)+2}\right),\\ \phantom{\rule{2em}{0ex}}\frac{\begin{array}{c}{G}_{p}\left(T{x}_{3m\left(k\right)},T{x}_{3m\left(k\right)},f{x}_{3m\left(k\right)}\right)+{G}_{p}\left(R{x}_{3n\left(k\right)+1},R{x}_{3n\left(k\right)+1},g{x}_{3n\left(k\right)+1}\right)\\ +{G}_{p}\left(S{x}_{3n\left(k\right)+2},S{x}_{3n\left(k\right)+2},h{x}_{3n\left(k\right)+2}\right)\end{array}}{3}\right\}\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right),{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3m\left(k\right)+1},{z}_{3m\left(k\right)+1}\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left({z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+2}\right),{G}_{p}\left({z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3},{z}_{3n\left(k\right)+3}\right),\\ \phantom{\rule{2em}{0ex}}\frac{\begin{array}{c}{G}_{p}\left({z}_{3m\left(k\right)},{z}_{3m\left(k\right)},{z}_{3m\left(k\right)+1}\right)+{G}_{p}\left({z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+1},{z}_{3n\left(k\right)+2}\right)\\ +{G}_{p}\left({z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+2},{z}_{3n\left(k\right)+3}\right)\end{array}}{3}\right\}.\end{array}$
Taking limit as $k\to \mathrm{\infty }$ and using (12), (13), (17), (21) in (22), we have
$\psi \left(\epsilon \right)\le \psi \left(\epsilon \right)-\phi \left(\epsilon \right)<\psi \left(\epsilon \right),$

a contradiction. Hence, $\left\{{z}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence.

Step III. We will show that f, g, h, R, S and T have a coincidence point.

Since $\left\{{z}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence in the complete ${G}_{p}$-metric space X, from Lemma 1.2, $\left\{{z}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(X,{d}_{{G}_{p}}\right)$. Completeness of $\left(X,{G}_{p}\right)$ yields that $\left(X,{d}_{{G}_{p}}\right)$ is also complete. Then there exists ${z}^{\ast }\in X$ such that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{{G}_{p}}\left({z}_{n},{z}^{\ast }\right)=0.$
(23)

Now, since ${lim}_{m,n\to \mathrm{\infty }}{G}_{p}\left({z}_{m},{z}_{n},{z}_{n}\right)=0$, (23) and part (2) of Lemma 1.2 yield that ${G}_{p}\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0$.

Since $R\left(X\right)$ is ${G}_{p}$-complete and $\left\{{z}_{3n+1}\right\}\subseteq R\left(X\right)$, there exists $u\in X$ such that ${z}^{\ast }=Ru$ and
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3n+1},{z}_{3n+1},Ru\right)\\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(R{x}_{3n+1},R{x}_{3n+1},Ru\right)=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(f{x}_{3n},f{x}_{3n},Ru\right)=G\left(Ru,Ru,Ru\right)=0.\end{array}$
(24)
By similar arguments, there exist $v,w\in X$ such that ${z}^{\ast }=Sv=Tw$ and
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3n+2},{z}_{3n+2},{z}^{\ast }\right)\\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(S{x}_{3n+2},S{x}_{3n+2},{z}^{\ast }\right)=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(g{x}_{3n+1},g{x}_{3n+1},{z}^{\ast }\right)=G\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0\end{array}$
(25)
and
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({z}_{3n+3},{z}_{3n+3},{z}^{\ast }\right)\\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(T{x}_{3n+3},T{x}_{3n+3},{z}^{\ast }\right)=\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(h{x}_{3n+2},h{x}_{3n+2},{z}^{\ast }\right)=G\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0.\end{array}$
(26)

Now, we prove that w is a coincidence point of f and T.

Since $S{x}_{3n+2}\to {z}^{\ast }=Tw=Ru$ as $n\to \mathrm{\infty }$, so $S{x}_{3n+2}⪯Tw=Ru$. Therefore, from (2), we have
$\begin{array}{r}\psi \left({G}_{p}\left(fw,gu,h{x}_{3n+2}\right)\right)\le \psi \left(M\left(w,u,{x}_{3n+2}\right)\right)-\phi \left(M\left(w,u,{x}_{3n+2}\right)\right),\end{array}$
(27)
where
$\begin{array}{r}M\left(w,u,{x}_{3n+2}\right)\\ \phantom{\rule{1em}{0ex}}=max\left\{{G}_{p}\left(Tw,Ru,S{x}_{3n+2}\right),G\left(Tw,fw,fw\right),\\ \phantom{\rule{2em}{0ex}}{G}_{p}\left(Ru,gu,gu\right),G\left(S{x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right),\\ \phantom{\rule{2em}{0ex}}\frac{{G}_{p}\left(Tw,Tw,fw\right)+G\left(Ru,Ru,gu\right)+{G}_{p}\left(S{x}_{3n+2},S{x}_{3n+2},h{x}_{3n+2}\right)}{3}\right\}.\end{array}$
Taking limit as $n\to \mathrm{\infty }$ in (27), as $G\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0$, from Lemma 1.3, we obtain that
$\begin{array}{r}\psi \left({G}_{p}\left(fw,gu,{z}^{\ast }\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \psi \left({G}_{p}\left(fw,gu,{z}^{\ast }\right)\right)\\ \phantom{\rule{2em}{0ex}}-\phi \left(max\left\{{G}_{p}\left({z}^{\ast },fw,fw\right),{G}_{p}\left({z}^{\ast },gu,gu\right),\frac{{G}_{p}\left({z}^{\ast },{z}^{\ast },fw\right)+{G}_{p}\left({z}^{\ast },{z}^{\ast },gu\right)}{3}\right\}\right),\end{array}$

which implies that $gu=fw={z}^{\ast }=Tw=Ru$.

As f and T are weakly compatible, we have $f{z}^{\ast }=fTw=Tfw=T{z}^{\ast }$. Thus ${z}^{\ast }$ is a coincidence point of f and T.

Similarly it can be shown that ${z}^{\ast }$ is a coincidence point of the pairs $\left(g,R\right)$ and $\left(h,S\right)$.

Now, let $R{z}^{\ast }$, $S{z}^{\ast }$ and $T{z}^{\ast }$ be comparable. By (2) we have
$\psi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)\le \psi \left(M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\right)-\phi \left(M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\right),$
(28)
where
$\begin{array}{rl}M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=& max\left\{{G}_{p}\left(T{z}^{\ast },R{z}^{\ast },S{z}^{\ast }\right),\\ {G}_{p}\left(T{z}^{\ast },f{z}^{\ast },f{z}^{\ast }\right),{G}_{p}\left(R{z}^{\ast },g{z}^{\ast },g{z}^{\ast }\right),{G}_{p}\left(S{z}^{\ast },h{z}^{\ast },h{z}^{\ast }\right),\\ \frac{{G}_{p}\left(T{z}^{\ast },T{z}^{\ast },f{z}^{\ast }\right)+{G}_{p}\left(R{z}^{\ast },R{z}^{\ast },g{z}^{\ast }\right)+{G}_{p}\left(S{z}^{\ast },S{z}^{\ast },h{z}^{\ast }\right)}{3}\right\}\\ =& {G}_{p}\left(T{z}^{\ast },R{z}^{\ast },S{z}^{\ast }\right)={G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right).\end{array}$
Hence (28) gives
$\psi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)\le \psi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)-\phi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)=0.$

Therefore $f{z}^{\ast }=g{z}^{\ast }=h{z}^{\ast }=T{z}^{\ast }=R{z}^{\ast }=S{z}^{\ast }$. □

Theorem 2.2 Let $\left(X,⪯,{G}_{p}\right)$ be a partially ordered complete ${G}_{p}$-metric space. Let $f,g,h:X\to X$ be three mappings. Suppose that for every three comparable elements $x,y,z\in X$, we have
$\psi \left(2{G}_{p}\left(fx,gy,hz\right)\right)\le \psi \left(M\left(x,y,z\right)\right)-\phi \left(M\left(x,y,z\right)\right),$
(29)
where
$\begin{array}{rl}M\left(x,y,z\right)=& max\left\{{G}_{p}\left(x,y,z\right),\\ {G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,gy,gy\right),{G}_{p}\left(z,hz,hz\right),\\ \frac{{G}_{p}\left(x,x,fx\right)+{G}_{p}\left(y,y,gy\right)+{G}_{p}\left(z,z,hz\right)}{3}\right\}\end{array}$

and $\psi ,\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ are altering distance functions. Let f, g, h be continuous and the pairs $\left(f,g\right)$, $\left(g,h\right)$ and $\left(h,f\right)$ be partially weakly increasing. Then f, g and h have a common fixed point ${z}^{\ast }$ in X.

Proof Let ${x}_{0}$ be an arbitrary point and ${x}_{3n+1}=f{x}_{3n}$, ${x}_{3n+2}=g{x}_{3n+1}$ and ${x}_{3n+3}=h{x}_{3n+2}$ for all $n\ge 0$.

Following the proof of the previous theorem, we can show that there exists ${z}^{\ast }\in X$ such that
${G}_{p}\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0$
(30)
and
$\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left({x}_{3n},{x}_{3n},{z}^{\ast }\right)=0.$
(31)
Continuity of f yields that
$\underset{n\to \mathrm{\infty }}{lim}{G}_{p}\left(f{x}_{3n},f{x}_{3n},f{z}^{\ast }\right)={G}_{p}\left(f{z}^{\ast },f{z}^{\ast },f{z}^{\ast }\right).$
(32)
By the rectangle inequality, we have
${G}_{p}\left(f{z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\le {G}_{p}\left(f{z}^{\ast },f{x}_{3n},f{x}_{3n}\right)+{G}_{p}\left({x}_{3n+1},{z}^{\ast },{z}^{\ast }\right)$
(33)
and
${G}_{p}\left(f{z}^{\ast },f{z}^{\ast },{z}^{\ast }\right)\le {G}_{p}\left({z}^{\ast },f{x}_{3n},f{x}_{3n}\right)+{G}_{p}\left(f{x}_{3n},f{z}^{\ast },f{z}^{\ast }\right).$
(34)
Taking limit as $n\to \mathrm{\infty }$ in (33) and (34), from (30) we obtain
${G}_{p}\left(f{z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\le {G}_{p}\left(f{z}^{\ast },f{z}^{\ast },f{z}^{\ast }\right)$
and
${G}_{p}\left(f{z}^{\ast },f{z}^{\ast },{z}^{\ast }\right)\le {G}_{p}\left(f{z}^{\ast },f{z}^{\ast },f{z}^{\ast }\right).$

Similar inequalities are obtained for g and h.

On the other hand, as ${z}^{\ast }⪯{z}^{\ast }⪯{z}^{\ast }$, using (29) we obtain that
$\begin{array}{rl}\psi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)& \le \psi \left(2{G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)\\ \le \psi \left(M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\right)-\phi \left(M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\right),\end{array}$
(35)
where
$\begin{array}{rl}M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=& max\left\{{G}_{p}\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right),\\ {G}_{p}\left({z}^{\ast },f{z}^{\ast },f{z}^{\ast }\right),{G}_{p}\left({z}^{\ast },g{z}^{\ast },g{z}^{\ast }\right),{G}_{p}\left({z}^{\ast },h{z}^{\ast },h{z}^{\ast }\right),\\ \frac{{G}_{p}\left({z}^{\ast },{z}^{\ast },f{z}^{\ast }\right)+{G}_{p}\left({z}^{\ast },{z}^{\ast },g{z}^{\ast }\right)+{G}_{p}\left({z}^{\ast },{z}^{\ast },h{z}^{\ast }\right)}{3}\right\}\\ \le & max\left\{{G}_{p}\left(f{z}^{\ast },f{z}^{\ast },f{z}^{\ast }\right),{G}_{p}\left(g{z}^{\ast },g{z}^{\ast },g{z}^{\ast }\right),{G}_{p}\left(h{z}^{\ast },h{z}^{\ast },h{z}^{\ast }\right)\right\}.\end{array}$
(36)
We consider three cases as follows:
1. 1.

$f{z}^{\ast }=g{z}^{\ast }=h{z}^{\ast }$.

2. 2.

$f{z}^{\ast }\ne g{z}^{\ast }\ne h{z}^{\ast }$.

3. 3.

a. $f{z}^{\ast }=g{z}^{\ast }\ne h{z}^{\ast }$, or b. $f{z}^{\ast }\ne g{z}^{\ast }=h{z}^{\ast }$.

For case 1, by (36), $M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\le {G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)$.

For case 2, by (${G}_{p}2$), $M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\le {G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)$.

Now, from (35),
$\psi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)\le \psi \left({G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)-\phi \left(M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\right),$
(37)

hence $M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0$. Therefore, ${z}^{\ast }=f{z}^{\ast }=g{z}^{\ast }=h{z}^{\ast }$.

On the other hand, for case 3, part a, by (${G}_{p}2$), $M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\le 2{G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)$ and hence from (35), we have
$\psi \left(2{G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)\le \psi \left(2{G}_{p}\left(f{z}^{\ast },g{z}^{\ast },h{z}^{\ast }\right)\right)-\phi \left(M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)\right),$
(38)

hence $M\left({z}^{\ast },{z}^{\ast },{z}^{\ast }\right)=0$. Therefore, ${z}^{\ast }=f{z}^{\ast }=g{z}^{\ast }=h{z}^{\ast }$.

Now, let ${x}^{\ast }$ and ${y}^{\ast }$ as two fixed points of f, g and h be comparable. So, from (29) we have
$\begin{array}{rl}\psi \left(2{G}_{p}\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right)& =\psi \left(2{G}_{p}\left(f{x}^{\ast },g{x}^{\ast },h{y}^{\ast }\right)\right)\\ \le \psi \left(M\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right)-\phi \left(M\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right),\end{array}$
(39)
where
$\begin{array}{rl}M\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)=& max\left\{{G}_{p}\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right),\\ {G}_{p}\left({x}^{\ast },f{x}^{\ast },f{x}^{\ast }\right),{G}_{p}\left({x}^{\ast },g{x}^{\ast },g{x}^{\ast }\right),{G}_{p}\left({y}^{\ast },h{y}^{\ast },h{y}^{\ast }\right),\\ \frac{{G}_{p}\left({x}^{\ast },{x}^{\ast },f{x}^{\ast }\right)+{G}_{p}\left({x}^{\ast },{x}^{\ast },g{x}^{\ast }\right)+{G}_{p}\left({y}^{\ast },{y}^{\ast },h{y}^{\ast }\right)}{3}\right\}\\ \le & 2{G}_{p}\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right).\end{array}$
Hence (39) gives
$\psi \left(2{G}_{p}\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right)\le \psi \left(2{G}_{p}\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right)-\phi \left(M\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right).$

Therefore, $\phi \left(M\left({x}^{\ast },{x}^{\ast },{y}^{\ast }\right)\right)=0$ and hence ${x}^{\ast }={y}^{\ast }$. □

The following corollaries are special cases of the above results.

Corollary 2.1 Let $\left(X,⪯,{G}_{p}\right)$ be a partially ordered complete ${G}_{p}$-metric space. Let $f:X\to X$ be a mapping such that for every three comparable elements $x,y,z\in X$, we have
$\psi \left(2{G}_{p}\left(fx,fy,fz\right)\right)\le \psi \left(M\left(x,y,z\right)\right)-\phi \left(M\left(x,y,z\right)\right),$
(40)
where
$\begin{array}{rl}M\left(x,y,z\right)=& max\left\{{G}_{p}\left(x,y,z\right),\\ {G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right),{G}_{p}\left(z,fz,fz\right),\\ \frac{{G}_{p}\left(x,x,fx\right)+{G}_{p}\left(y,y,fy\right)+{G}_{p}\left(z,z,fz\right)}{3}\right\}\end{array}$
and $\psi ,\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ are altering distance functions. Then f has a fixed point in X provided that $fx⪯f\left(fx\right)$ for all $x\in X$ and either
1. a.

f is continuous, or

2. b.

X has the sequential limit comparison property.

Moreover, f has a unique fixed point provided that the fixed points of f are comparable.

Taking $y=z$ in Corollary 2.1, we obtain the following common fixed point result.

Corollary 2.2 Let $\left(X,⪯,{G}_{p}\right)$ be a partially ordered complete ${G}_{p}$-metric space, and let f be a self-mapping on X such that for every comparable elements $x,y\in X$,
$\psi \left(2{G}_{p}\left(fx,fy,fy\right)\right)\le \psi \left(M\left(x,y,y\right)\right)-\phi \left(M\left(x,y,y\right)\right),$
(41)
where
$M\left(x,y,y\right)=max\left\{{G}_{p}\left(x,y,y\right),G\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right),\frac{{G}_{p}\left(x,x,fx\right)+2{G}_{p}\left(y,y,fy\right)}{3}\right\},$
and $\psi ,\phi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ are altering distance functions. Then f has a fixed point in X provided that $fx⪯f\left(fx\right)$ for all $x\in X$ and either
1. a.

f is continuous, or

2. b.

X has the sequential limit comparison property.

## 3 Fixed point results via an α-admissible mapping with respect to η in ${G}_{p}$-metric spaces

Samet et al. [32] defined the notion of α-admissible mappings and proved the following result.

Definition 3.1 Let T be a self-mapping on X and $\alpha :X×X\to \left[0,+\mathrm{\infty }\right)$ be a function. We say that T is an α-admissible mapping if
$x,y\in X,\phantom{\rule{1em}{0ex}}\alpha \left(x,y\right)\ge 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\alpha \left(Tx,Ty\right)\ge 1.$

Denote with Ψ the family of all nondecreasing functions $\psi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\psi }^{n}\left(t\right)<+\mathrm{\infty }$ for all $t>0$, where ${\psi }^{n}$ is the n th iterate of ψ.

Theorem 3.1 Let $\left(X,d\right)$ be a complete metric space and T be an α-admissible mapping. Assume that
$\alpha \left(x,y\right)d\left(Tx,Ty\right)\le \psi \left(d\left(x,y\right)\right),$
(42)
where $\psi \in \mathrm{\Psi }$. Also suppose that the following assertions hold:
1. (i)

there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},T{x}_{0}\right)\ge 1$;

2. (ii)

either T is continuous or for any sequence $\left\{{x}_{n}\right\}$ in X with $\alpha \left({x}_{n},{x}_{n+1}\right)\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$, we have $\alpha \left({x}_{n},x\right)\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Then T has a fixed point.

For more details on α-admissible mappings, we refer the reader to [3337].

Very recently, Salimi et al. [38] modified and generalized the notions of α-ψ-contractive mappings and α-admissible mappings as follows.

Definition 3.2 [38]

Let T be a self-mapping on X and $\alpha ,\eta :X×X\to \left[0,+\mathrm{\infty }\right)$ be two functions. We say that T is an α-admissible mapping with respect to η if
$x,y\in X,\phantom{\rule{1em}{0ex}}\alpha \left(x,y\right)\ge \eta \left(x,y\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\alpha \left(Tx,Ty\right)\ge \eta \left(Tx,Ty\right).$

Note that if we take $\eta \left(x,y\right)=1$, then this definition reduces to Definition 3.1. Also, if we take $\alpha \left(x,y\right)=1$, then we say that T is an η-subadmissible mapping.

The following result properly contains Theorem 3.1 and Theorems 2.3 and 2.4 of [37].

Theorem 3.2 [38]

Let $\left(X,d\right)$ be a complete metric space and T be an α-admissible mapping with respect to η. Assume that
$x,y\in X,\phantom{\rule{1em}{0ex}}\alpha \left(x,y\right)\ge \eta \left(x,y\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}d\left(Tx,Ty\right)\le \psi \left(M\left(x,y\right)\right),$
(43)
where $\psi \in \mathrm{\Psi }$ and
$M\left(x,y\right)=max\left\{d\left(x,y\right),\frac{d\left(x,Tx\right)+d\left(y,Ty\right)}{2},\frac{d\left(x,Ty\right)+d\left(y,Tx\right)}{2}\right\}.$
Also, suppose that the following assertions hold:
1. (i)

there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},T{x}_{0}\right)\ge \eta \left({x}_{0},T{x}_{0}\right)$;

2. (ii)

either T is continuous or for any sequence $\left\{{x}_{n}\right\}$ in X with $\alpha \left({x}_{n},{x}_{n+1}\right)\ge \eta \left({x}_{n},{x}_{n+1}\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$, we have $\alpha \left({x}_{n},x\right)\ge \eta \left({x}_{n},x\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Then T has a fixed point.

In fact, the Banach contraction principle and Theorem 3.2 hold for the following example, but Theorem 3.1 does not hold.

Example 3.1 [38]

Let $X=\left[0,\mathrm{\infty }\right)$ be endowed with the usual metric $d\left(x,y\right)=|x-y|$ for all $x,y\in X$, and let $T:X\to X$ be defined by $Tx=\frac{1}{4}x$. Also, define $\alpha :{X}^{2}\to \left[0,\mathrm{\infty }\right)$ by $\alpha \left(x,y\right)=3$ and $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ by $\psi \left(t\right)=\frac{1}{2}t$.

Theorem 3.3 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space, f be a continuous α-admissible mapping with respect to η on X, there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},f{x}_{0}\right)\ge \eta \left({x}_{0},f{x}_{0}\right)$ and if any sequence $\left\{{x}_{n}\right\}$ in X converges to a point x, then we have $\alpha \left(x,x\right)\ge \eta \left(x,x\right)$. Assume that
$\begin{array}{r}\alpha \left(x,y\right)\ge \eta \left(x,y\right)\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}\end{array}$
(44)

for all $x,y\in X$, where $0\le r<1$. Then f has a fixed point.

Proof Let ${x}_{0}\in X$ and define a sequence $\left\{{x}_{n}\right\}$ by ${x}_{n}={f}^{n}{x}_{0}$ for all $n\in \mathbb{N}$. Since f is an α-admissible mapping with respect to η and $\alpha \left({x}_{0},{x}_{1}\right)=\alpha \left({x}_{0},f{x}_{0}\right)\ge \eta \left({x}_{0},f{x}_{0}\right)=\eta \left({x}_{0},{x}_{1}\right)$, we deduce that $\alpha \left({x}_{1},{x}_{2}\right)=\alpha \left(f{x}_{0},f{x}_{1}\right)\ge \eta \left(f{x}_{0},f{x}_{1}\right)=\eta \left({x}_{1},{x}_{2}\right)$. Continuing this process, we get $\alpha \left({x}_{n},{x}_{n+1}\right)\ge \eta \left({x}_{n},{x}_{n+1}\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Now, from (44) we have
$\begin{array}{r}{G}_{p}\left(f{f}^{n}{x}_{0},{f}^{2}{f}^{n}{x}_{0},{f}^{2}{f}^{n}{x}_{0}\right)\\ \phantom{\rule{1em}{0ex}}\le rmax\left\{{G}_{p}\left({f}^{n}{x}_{0},f{f}^{n}{x}_{0},f{f}^{n}{x}_{0}\right),{G}_{p}\left(f{f}^{n}{x}_{0},{f}^{2}{f}^{n}{x}_{0},{f}^{2}{f}^{n}{x}_{0}\right)\right\},\end{array}$
which implies
${G}_{p}\left({f}^{n+1}{x}_{0},{f}^{n+2}{x}_{0},{f}^{n+2}{x}_{0}\right)\le r{G}_{p}\left({f}^{n}{x}_{0},{f}^{n+1}{x}_{0},{f}^{n+1}{x}_{0}\right).$
(45)
Continuing the above process, we can obtain
${G}_{p}\left({f}^{n}{x}_{0},{f}^{n+1}{x}_{0},{f}^{n+1}{x}_{0}\right)\le r{G}_{p}\left({f}^{n-1}{x}_{0},{f}^{n}{x}_{0},{f}^{n}{x}_{0}\right)\le \cdots \le {r}^{n}{G}_{p}\left({x}_{0},f{x}_{0},f{x}_{0}\right).$
(46)
Then, for any $m>n$, by (46) we get
$\begin{array}{rl}{G}_{p}\left({f}^{n}{x}_{0},{f}^{m}{x}_{0},{f}^{m}{x}_{0}\right)\le & {G}_{p}\left({f}^{n}{x}_{0},{f}^{n+1}{x}_{0},{f}^{n+1}{x}_{0}\right)+{G}_{p}\left({f}^{n+1}{x}_{0},{f}^{m}{x}_{0},{f}^{m}{x}_{0}\right)\\ \le & {G}_{p}\left({f}^{n}{x}_{0},{f}^{n+1}{x}_{0},{f}^{n+1}{x}_{0}\right)+{G}_{p}\left({f}^{n+1}{x}_{0},{f}^{n+2}{x}_{0},{f}^{n+2}{x}_{0}\right)\\ +{G}_{p}\left({f}^{n+2}{x}_{0},{f}^{m}{x}_{0},{f}^{m}{x}_{0}\right)\\ \le & G\left({f}^{n}{x}_{0},{f}^{n+1}{x}_{0},{f}^{n+1}{x}_{0}\right)+{G}_{p}\left({f}^{n+1}{x}_{0},{f}^{n+2}{x}_{0},{f}^{n+2}{x}_{0}\right)\\ +{G}_{p}\left({f}^{n+2}{x}_{0},{f}^{n+3}{x}_{0},{f}^{n+3}{x}_{0}\right)+\cdots +{G}_{p}\left({f}^{m-1}{x}_{0},{f}^{m}{x}_{0},{f}^{m}{x}_{0}\right)\\ \le & \frac{{r}^{n}}{1-r}{G}_{p}\left({x}_{0},f{x}_{0},f{x}_{0}\right).\end{array}$

This implies that ${lim}_{m,n\to +\mathrm{\infty }}{G}_{p}\left({f}^{n}{x}_{0},{f}^{m}{x}_{0},{f}^{m}{x}_{0}\right)=0$, that is, $\left\{{x}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence.

Since $\left\{{x}_{n}\right\}$ is a ${G}_{p}$-Cauchy sequence in the complete ${G}_{p}$-metric space X, from Lemma 1.2, $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the metric space $\left(X,{d}_{{G}_{p}}\right)$. Completeness of $\left(X,{G}_{p}\right)$ yields that $\left(X,{d}_{{G}_{p}}\right)$ is also complete. Then there exists $z\in X$ such that
$\underset{n\to \mathrm{\infty }}{lim}{d}_{{G}_{p}}\left({x}_{n},z\right)=0.$
(47)
Since ${lim}_{m,n\to +\mathrm{\infty }}{G}_{p}\left({x}_{n},{x}_{m},{x}_{m}\right)=0$, from Lemma 1.2 we get
$\underset{n\to +\mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},z,z\right)=\underset{n\to +\mathrm{\infty }}{lim}{G}_{p}\left({x}_{n},{x}_{n},z\right)={G}_{p}\left(z,z,z\right)=0.$
(48)
From the continuity of f, we have
$\underset{n\to +\mathrm{\infty }}{lim}{G}_{p}\left({x}_{n+1},fz,fz\right)={G}_{p}\left(fz,fz,fz\right),$
and hence we get
${G}_{p}\left(z,fz,fz\right)\le \underset{n\to +\mathrm{\infty }}{lim}G\left(z,{x}_{n+1},{x}_{n+1}\right)+\underset{n\to +\mathrm{\infty }}{lim}G\left({x}_{n+1},fz,fz\right)={G}_{p}\left(fz,fz,fz\right).$
So, we get that ${G}_{p}\left(z,fz,fz\right)\le {G}_{p}\left(fz,fz,fz\right)$. Since the opposite inequality always holds, we get that
${G}_{p}\left(z,fz,fz\right)={G}_{p}\left(fz,fz,fz\right).$
As $\alpha \left(z,z\right)\ge \eta \left(z,z\right)$ we have
${G}_{p}\left(z,fz,fz\right)={G}_{p}\left(fz,fz,fz\right)\le rmax\left\{{G}_{p}\left(z,z,z\right),{G}_{p}\left(z,fz,fz\right),{G}_{p}\left(z,fz,fz\right)\right\},$
(49)

where $0\le r<1$. Hence, ${G}_{p}\left(z,fz,fz\right)\le r{G}_{p}\left(z,fz,fz\right)$. Thus, ${G}_{p}\left(z,fz,fz\right)=0$, that is, $z=fz$. □

If in Theorem 3.3 we take $\eta \left(x,y\right)=1$, then we deduce the following corollary.

Corollary 3.1 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space, f be a continuous α-admissible mapping on X, and there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},f{x}_{0}\right)\ge 1$. Assume that
$\alpha \left(x,y\right)\ge 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}$

for all $x,y\in X$, where $0\le r<1$, and if any sequence $\left\{{x}_{n}\right\}$ in X converges to a point x, then we have $\alpha \left(x,x\right)\ge 1$. Then f has a fixed point.

If in Theorem 3.3 we take $\alpha \left(x,y\right)=1$, then we deduce the following corollary.

Corollary 3.2 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space, f be a continuous η-subadmissible mapping on X, and there exists ${x}_{0}\in X$ such that $\eta \left({x}_{0},f{x}_{0}\right)\le 1$. Assume that
$\eta \left(x,y\right)\le 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}$
(50)

for all $x,y\in X$, where $0\le r<1$, and if any sequence $\left\{{x}_{n}\right\}$ in X converges to a point x, then we have $1\ge \eta \left(x,x\right)$. Then f has a fixed point.

In the following theorem, we omit the continuity of the mapping f.

Theorem 3.4 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space and f be an α-admissible mapping with respect to η on X such that
$\begin{array}{r}\alpha \left(x,y\right)\ge \eta \left(x,y\right)\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}\end{array}$
(51)
for all $x,y\in X$, where $0\le r<1$. Assume that the following conditions hold:
1. (i)

there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},f{x}_{0}\right)\ge \eta \left({x}_{0},f{x}_{0}\right)$;

2. (ii)

if $\left\{{x}_{n}\right\}$ is a sequence in X such that $\alpha \left({x}_{n},{x}_{n+1}\right)\ge \eta \left({x}_{n},{x}_{n+1}\right)$ for all n and ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$, then $\alpha \left({x}_{n},x\right)\ge \eta \left({x}_{n},x\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Then f has a fixed point.

Proof Let ${x}_{0}\in X$ be such that $\alpha \left({x}_{0},f{x}_{0}\right)\ge \eta \left({x}_{0},f{x}_{0}\right)$ and define a sequence $\left\{{x}_{n}\right\}$ in X by ${x}_{n}={f}^{n}{x}_{0}=f{x}_{n-1}$ for all $n\in \mathbb{N}$. Following the proof of Theorem 3.1, we have $\alpha \left({x}_{n},{x}_{n+1}\right)\ge \eta \left({x}_{n},{x}_{n+1}\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and there exists $x\in X$ such that ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$. Hence, from (ii) we deduce that $\alpha \left({x}_{n},x\right)\ge \eta \left({x}_{n},x\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Hence, by (51), it follows that for all n,
${G}_{p}\left({x}_{n+1},fx,fx\right)\le rmax\left\{{G}_{p}\left({x}_{n},x,x\right),{G}_{p}\left({x}_{n},{x}_{n+1},{x}_{n+1}\right),{G}_{p}\left(x,fx,fx\right)\right\}.$

Taking the limit as $n\to +\mathrm{\infty }$ in the above inequality, from Lemma 1.3 we obtain $\left(1-r\right)G\left(x,fx,fx\right)\le 0$, which implies that $x=fx$. □

Corollary 3.3 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space and f be an α-admissible mapping on X such that
$\alpha \left(x,y\right)\ge 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}$
(52)
for all $x,y\in X$, where $0\le r<1$. Assume that the following conditions hold:
1. (i)

there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},f{x}_{0}\right)\ge 1$;

2. (ii)

if $\left\{{x}_{n}\right\}$ is a sequence in X such that $\alpha \left({x}_{n},{x}_{n+1}\right)\ge 1$ for all n and ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$, then $\alpha \left({x}_{n},x\right)\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Then f has a fixed point.

Example 3.2 Let $X=\left[0,+\mathrm{\infty }\right)$ and ${G}_{p}\left(x,y,z\right)=max\left\{x,y,z\right\}$ be a ${G}_{p}$-metric on X. Define $f:X\to X$ by
and $\alpha :X×X\to \left[0,+\mathrm{\infty }\right)$ by

Now, we prove that all the hypotheses of Corollary 3.3 are satisfied and hence f has a fixed point.

Let $x,y\in X$, if $\alpha \left(x,y\right)\ge 1$, then $x,y\in \left[0,1\right]$. On the other hand, for all $x\in \left[0,1\right]$, we have $fx\le 1$ and hence $\alpha \left(fx,fy\right)\ge 1$. This implies that f is an α-admissible mapping on X. Obviously, $\alpha \left(0,f0\right)\ge 1$.

Now, if $\left\{{x}_{n}\right\}$ is a sequence in X such that $\alpha \left({x}_{n},{x}_{n+1}\right)\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$, then $\left\{{x}_{n}\right\}\subseteq \left[0,1\right]$ and hence $x\in \left[0,1\right]$. This implies that $\alpha \left({x}_{n},x\right)\ge 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

If $\alpha \left(x,y\right)\ge 1$, then $x,y\in \left[0,1\right]$. Hence,
$\begin{array}{rl}{G}_{p}\left(fx,fy,fy\right)& =max\left\{fx,fy\right\}=max\left\{\frac{x}{24},\frac{y}{24}\right\}\\ \le \frac{1}{12}max\left\{x,y\right\}\\ \le \frac{1}{12}max\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}.\end{array}$

Thus, all the conditions of Corollary 3.3 are satisfied and therefore f has a fixed point ($x=0$).

Corollary 3.4 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space and f be an η-subadmissible mapping on X such that
$\eta \left(x,y\right)\le 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}$
for all $x,y\in X$, where $0\le r<1$. Assume that the following conditions hold:
1. (i)

there exists ${x}_{0}\in X$ such that $\eta \left({x}_{0},f{x}_{0}\right)\le 1$;

2. (ii)

if $\left\{{x}_{n}\right\}$ is a sequence in X such that $\eta \left({x}_{n},{x}_{n+1}\right)\le 1$ for all n and ${x}_{n}\to x$ as $n\to +\mathrm{\infty }$, then $\eta \left({x}_{n},x\right)\le 1$ for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Then f has a fixed point.

## 4 Consequences

Theorem 4.1 Let $\left(X,{G}_{p}\right)$ be a ${G}_{p}$-complete ${G}_{p}$-metric space, f be a continuous α-admissible mapping on X, and there exists ${x}_{0}\in X$ such that $\alpha \left({x}_{0},f{x}_{0}\right)\ge 1$. Assume that
$\alpha \left(x,y\right){G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}$
(53)

for all $x,y\in X$, where $0\le r<1$ and if any sequence $\left\{{x}_{n}\right\}$ in X converges to a point x, then we have $\alpha \left(x,x\right)\ge \eta \left(x,x\right)$. Then f has a fixed point.

Proof Assume that $\alpha \left(x,y\right)\ge 1$, then from (53) we get
${G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}.$
That is,
$\alpha \left(x,y\right)\ge 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{G}_{p}\left(fx,fy,fy\right)\le rmax\left\{{G}_{p}\left(x,y,y\right),{G}_{p}\left(x,fx,fx\right),{G}_{p}\left(y,fy,fy\right)\right\}.$

Hence all the conditions of Corollary 3.1 hold and f has a fixed point. □

Similarly, we can deduce the following results.

Theorem 4.2