Some multidimensional fixed point theorems on partially preordered $G^{\ast }$-metric spaces under $(\psi ,\phi )$-contractivity conditions

Abstract

In this paper we present some (unidimensional and) multidimensional fixed point results under $(\psi ,\phi )$-contractivity conditions in the framework of $G^{\ast }$-metric spaces, which are spaces that result from G-metric spaces (in the sense of Mustafa and Sims) omitting one of their axioms. We prove that these spaces let us consider easily the product of $G^{\ast }$-metrics. Our result clarifies and improves some recent results on this topic because, among other different reasons, we will not need a partial order on the underlying space. Furthermore, the way in which several contractivity conditions are proposed imply that our theorems cannot be reduced to metric spaces.

MSC: 46T99, 47H10, 47H09, 54H25.

1 Introduction

In the sixties, inspired by the mapping that associated the area of a triangle to its three vertices, Gähler [1, 2] introduced the concept of 2-metric spaces. Gähler believed that 2-metric spaces can be interpreted as a generalization of usual metric spaces. However, some authors demonstrated that there is no clear relationship between these notions. For instance, Ha et al. [3] showed that a 2-metric does not have to be a continuous function of its three variables. Later, inspired by the perimeter of a triangle rather than the area, Dhage [4] changed the axioms and presented the concept of D-metric. Different topological structures (see [5–7]) were considered in such spaces and, subsequently, several fixed point results were established. Unfortunately, most of their properties turned out to be false (see [8–10]). These considerations led to the concept of G-metric space introduced by Mustafa and Sims [11]. Since then, this theory has been expansively developed, paying a special attention to fixed point theorems (see, for instance, [12–28] and references therein).

The main aim of the present paper is to prove new unidimensional and multidimensional fixed point results in the framework of the G-metric spaces provided with a partial preorder (not necessarily a partial order). However, we need to overcome the well-known fact that the usual product of G-metrics is not necessarily a G-metric unless it comes from classical metrics (see [11], Section 4). Hence, we will omit one of the axioms that define a G-metric and we consider a new class of metrics, called $G^{\ast }$-metrics. As a consequence, our main results are valid in the context of G-metric spaces.

2 Preliminaries

Let n be a positive integer. Henceforth, X will denote a non-empty set and $X^n$ will denote the product space $X\times X\times \stackrel{n}{\cdots }\times X$. Throughout this manuscript, m and k will denote non-negative integers and $i,j,s\in \{1,2,\dots ,n\}$. Unless otherwise stated, ‘for all m’ will mean ‘for all $m\ge 0$’ and ‘for all i’ will mean ‘for all $i\in \{1,2,\dots ,n\}$’. Let ${\mathbb{R}}_0^{+}=[0,\mathrm{\infty })$.

Definition 1 We will say that ≼ is a partial preorder on X (or $(X,\preccurlyeq )$ is a preordered set or $(X,\preccurlyeq )$ is a partially preordered space) if the following properties hold.

• Reflexivity: $x\preccurlyeq x$ for all $x\in X$.

• Transitivity: If $x,y,z\in X$ verify $x\preccurlyeq y$ and $y\preccurlyeq z$, then $x\preccurlyeq z$.

Henceforth, let $\{\mathsf{A},\mathsf{B}\}$ be a partition of ${\mathrm{\Lambda }}_n=\{1,2,\dots ,n\}$, that is, $\mathsf{A}\cup \mathsf{B}={\mathrm{\Lambda }}_n$ and $\mathsf{A}\cap \mathsf{B}=\mathrm{\varnothing }$ such that A and B are non-empty sets. In the sequel, we will denote

$$\begin{array}{c}{\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}=\{\sigma :{\mathrm{\Lambda }}_n\to {\mathrm{\Lambda }}_n:\sigma (\mathsf{A})\subseteq \mathsf{A}\text{ and }\sigma (\mathsf{B})\subseteq \mathsf{B}\}\text{and}\hfill \\ {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}=\{\sigma :{\mathrm{\Lambda }}_n\to {\mathrm{\Lambda }}_n:\sigma (\mathsf{A})\subseteq \mathsf{B}\text{ and }\sigma (\mathsf{B})\subseteq \mathsf{A}\}.\hfill \end{array}$$

From now on, let $\mathrm{\Upsilon }=({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n)$ be an n-tuple of mappings from $\{1,2,\dots ,n\}$ into itself verifying ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$ if $i\in \mathsf{A}$ and ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$ if $i\in \mathsf{B}$.

If $(X,\preccurlyeq )$ is a partially preordered space, $x,y\in X$ and $i\in {\mathrm{\Lambda }}_n$, we will use the following notation:

$$x{\preccurlyeq }_{i}y\iff \{\begin{array}{ll}x\preccurlyeq y,& \text{if }i\in \mathsf{A},\\ x\succcurlyeq y,& \text{if }i\in \mathsf{B}.\end{array}$$

Consider on the product space $X^n$ the following partial preorder: for $\mathsf{X}=(x_1,x_2,\dots ,x_n),\mathsf{Y}=(y_1,y_2,\dots ,y_n)\in X^n$,

$$\mathsf{X}⊑\mathsf{Y}\iff x_i{\preccurlyeq }_i{y}_i\text{for all }i.$$

(1)

Notice that ⊑ depends on A and B. We say that two points X and Y are ⊑-comparable if $\mathsf{X}⊑\mathsf{Y}$ or $\mathsf{X}⊒\mathsf{Y}$.

Proposition 2 If $\mathsf{X}⊑\mathsf{Y}$ and $\sigma \in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}\cup {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$, then $(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (n)})$ and $(y_{\sigma (1)},y_{\sigma (2)},\dots ,y_{\sigma (n)})$ are ⊑-comparable. In particular,

$$\begin{array}{c}(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (n)})⊑(y_{\sigma (1)},y_{\sigma (2)},\dots ,y_{\sigma (n)})\mathit{\text{if}}\sigma \in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}},\hfill \\ (x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (n)})⊒(y_{\sigma (1)},y_{\sigma (2)},\dots ,y_{\sigma (n)})\mathit{\text{if}}\sigma \in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}.\hfill \end{array}$$

Proof Suppose that $x_i{\preccurlyeq }_i{y}_i$ for all i. Hence $x_{\sigma (i)}{\preccurlyeq }_{\sigma (i)}{y}_{\sigma (i)}$ for all i. Fix $\sigma \in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$. If $i\in \mathsf{A}$, then $\sigma (i)\in \mathsf{A}$, so $x_{\sigma (i)}{\preccurlyeq }_{\sigma (i)}{y}_{\sigma (i)}$ implies that $x_{\sigma (i)}\preccurlyeq y_{\sigma (i)}$, which means that $x_{\sigma (i)}{\preccurlyeq }_i{y}_{\sigma (i)}$. If $i\in \mathsf{B}$, then $\sigma (i)\in \mathsf{B}$, so $x_{\sigma (i)}{\preccurlyeq }_{\sigma (i)}{y}_{\sigma (i)}$ implies that $x_{\sigma (i)}\succcurlyeq y_{\sigma (i)}$, which means that $x_{\sigma (i)}{\preccurlyeq }_i{y}_{\sigma (i)}$. In any case, if $\sigma \in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$, then $x_{\sigma (i)}{\preccurlyeq }_i{y}_{\sigma (i)}$ for all i. It follows that $(x_{\sigma (1)},x_{\sigma (2)},\dots ,x_{\sigma (n)})⊑(y_{\sigma (1)},y_{\sigma (2)},\dots ,y_{\sigma (n)})$.

Now fix $\sigma \in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$. If $i\in \mathsf{A}$, then $\sigma (i)\in \mathsf{B}$, so $x_{\sigma (i)}{\preccurlyeq }_{\sigma (i)}{y}_{\sigma (i)}$ implies that $x_{\sigma (i)}\succcurlyeq y_{\sigma (i)}$, which means that $x_{\sigma (i)}{\succcurlyeq }_i{y}_{\sigma (i)}$. If $i\in \mathsf{B}$, then $\sigma (i)\in \mathsf{A}$, so $x_{\sigma (i)}{\preccurlyeq }_{\sigma (i)}{y}_{\sigma (i)}$ implies that $x_{\sigma (i)}\preccurlyeq y_{\sigma (i)}$, which means that $x_{\sigma (i)}{\succcurlyeq }_i{y}_{\sigma (i)}$. □

Let $F:X^n\to X$ be a mapping.

Definition 3 (Roldán et al. [20])

A point $(x_1,x_2,\dots ,x_n)\in X^n$ is called an ϒ-fixed point of the mapping F if

$$F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)})=x_i\text{for all }i.$$

(2)

Definition 4 (Roldán et al. [20])

Let $(X,\preccurlyeq )$ be a partially preordered space. We say that F has the mixed monotone property (w.r.t. $\{\mathsf{A},\mathsf{B}\}$) if F is monotone non-decreasing in the arguments of A and monotone non-increasing in the arguments of B, i.e., for all $x_1,x_2,\dots ,x_n,y,z\in X$ and all i,

$$y\preccurlyeq z\Rightarrow F(x_1,\dots ,x_{i-1},y,x_{i+1},\dots ,x_n){\preccurlyeq }_{i}F(x_1,\dots ,x_{i-1},z,x_{i+1},\dots ,x_n).$$

We will use the following results about real sequences in the proof of our main theorems.

Lemma 5 Let ${\{a_m^1\}}_{m\in \mathbb{N}},\dots ,{\{a_m^n\}}_{m\in \mathbb{N}}$ be n real lower bounded sequences such that ${\{max(a_m^1,\dots ,a_m^n)\}}_{m\in \mathbb{N}}\to \delta$. Then there exist $i_0\in \{1,2,\dots ,n\}$ and a subsequence ${\{a_{m(k)}^{i_0}\}}_{k\in \mathbb{N}}$ such that ${\{a_{m(k)}^{i_0}\}}_{k\in \mathbb{N}}\to \delta$.

Proof Let $b_m=max(a_m^1,a_m^2,\dots ,a_m^n)$ for all m. As $\{b_m\}$ is convergent, it is bounded. As $a_m^i\le b_m$ for all m and i, then every $\{a_m^i\}$ is bounded. As ${\{a_m^1\}}_{m\in \mathbb{N}}$ is a real bounded sequence, it has a convergent subsequence ${\{a_{{\sigma }_1(m)}^1\}}_{m\in \mathbb{N}}\to a_1$. Consider the subsequences ${\{a_{{\sigma }_1(m)}^2\}}_{m\in \mathbb{N}},{\{a_{{\sigma }_1(m)}^3\}}_{m\in \mathbb{N}},\dots ,{\{a_{{\sigma }_1(m)}^n\}}_{m\in \mathbb{N}}$, that are $n-1$ real bounded sequences, and the sequence ${\{b_{{\sigma }_1(m)}\}}_{m\in \mathbb{N}}$ that also converges to δ. As ${\{a_{{\sigma }_1(m)}^2\}}_{m\in \mathbb{N}}$ is a real bounded sequence, it has a convergent subsequence ${\{a_{{\sigma }_2{\sigma }_1(m)}^2\}}_{m\in \mathbb{N}}\to a_2$. Then the sequences ${\{a_{{\sigma }_2{\sigma }_1(m)}^3\}}_{m\in \mathbb{N}},{\{a_{{\sigma }_2{\sigma }_1(m)}^4\}}_{m\in \mathbb{N}},\dots ,{\{a_{{\sigma }_2{\sigma }_1(m)}^n\}}_{m\in \mathbb{N}}$ also are $n-2$ real bounded sequences and ${\{a_{{\sigma }_2{\sigma }_1(m)}^1\}}_{m\in \mathbb{N}}\to a_1$ and ${\{b_{{\sigma }_2{\sigma }_1(m)}\}}_{m\in \mathbb{N}}\to \delta$. Repeating this process n times, we can find n subsequences ${\{a_{\sigma (m)}^1\}}_{m\in \mathbb{N}},{\{a_{\sigma (m)}^2\}}_{m\in \mathbb{N}},\dots ,{\{a_{\sigma (m)}^n\}}_{m\in \mathbb{N}}$ (where $\sigma ={\sigma }_n\cdots {\sigma }_1$) such that ${\{a_{\sigma (m)}^i\}}_{m\in \mathbb{N}}\to a_i$ for all i. And ${\{b_{\sigma (m)}\}}_{m\in \mathbb{N}}\to \delta$. But

$${\{b_{\sigma (m)}\}}_{m\in \mathbb{N}}={\{max(a_{\sigma (m)}^n,\dots ,a_{\sigma (m)}^n)\}}_{m\in \mathbb{N}}\to max(a_1,\dots ,a_n),$$

so $\delta =max(a_1,\dots ,a_n)$ and there exists $i_0\in \{1,2,\dots ,n\}$ such that $a_{i_0}=\delta$. Therefore, there exist $i_0\in \{1,2,\dots ,n\}$ and a subsequence ${\{a_{\sigma (m)}^{i_0}\}}_{m\in \mathbb{N}}$ such that ${\{a_{\sigma (m)}^{i_0}\}}_{m\in \mathbb{N}}\to a_{i_0}=\delta$. □

Lemma 6 Let ${\{a_m\}}_{m\in \mathbb{N}}$ be a sequence of non-negative real numbers which has not any subsequence converging to zero. Then, for all $\epsilon >0$, there exist $\delta \in ]0,\epsilon [$ and $m_0\in \mathbb{N}$ such that $a_m\ge \delta$ for all $m\ge m_0$.

Proof Suppose that the conclusion is not true. Then there exists ${\epsilon }_0>0$ such that, for all $\delta \in ]0,{\epsilon }_0[$, there exists $m_0\in \mathbb{N}$ verifying $a_{m_0}<\delta$. Let $k_0\in \mathbb{N}$ be such that $1/k_0<{\epsilon }_0$. For all $k\in \mathbb{N}$, take ${\delta }_k=1/(k+k_0)\in ]0,{\epsilon }_0[$. Then there exists $m(k)\in \mathbb{N}$ verifying $0\le a_{m(k)}<{\delta }_k=1/(k+k_0)$. Taking limit when $k\to \mathrm{\infty }$, we deduce that ${lim}_{k\to \mathrm{\infty }}{a}_{m(k)}=0$. Then $\{a_m\}$ has a subsequence converging to zero (maybe, reordering $\{a_{m(k)}\}$), but this is a contradiction. □

Let

$$\mathrm{\Psi }=\{\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty }):\varphi \text{ is continuous, non-decreasing and }{\varphi }^{-1}(\{0\})=\{0\}\}.$$

Lemma 7 If $\psi \in \mathrm{\Psi }$ and $\{a_m\}\subset [0,\mathrm{\infty })$ verifies $\{\psi (a_m)\}\to 0$, then $\{a_m\}\to 0$.

Proof If the conclusion does not hold, there exists ${\epsilon }_0>0$ such that, for all $m_0\in \mathbb{N}$, there exists $m\ge m_0$ verifying $a_m\ge {\epsilon }_0$. This means that $\{a_m\}$ has a partial subsequence ${\{a_{m(k)}\}}_k$ such that $a_{m(k)}\ge {\epsilon }_0$. As ψ is non-decreasing, $\psi ({\epsilon }_0)\le \psi (a_{m(k)})$ for all $k\in \mathbb{N}$. Therefore, ${\{\psi (a_m)\}}_m$ has a subsequence ${\{\psi (a_{m(k)})\}}_k$ lower bounded by $\psi ({\epsilon }_0)>0$, but this is impossible since ${lim}_{m\to \mathrm{\infty }}\psi (a_m)=0$. □

Lemma 8 Let $\{a_m^1\},\{a_m^2\},\dots ,\{a_m^n\},\{b_m^1\},\{b_m^2\},\dots ,\{b_m^n\}\subset [0,\mathrm{\infty })$ be 2n sequences of non-negative real numbers and suppose that there exist $\psi ,\phi \in \mathrm{\Psi }$ such that

$$\begin{array}{c}\psi \left(a_{m+1}^i\right)\le (\psi -\phi )\left(b_m^i\right)\mathit{\text{for all}}i\mathit{\text{and all}}m,\mathit{\text{and}}\hfill \\ \psi \left(\underset{1\le i\le n}{max}{b}_m^i\right)\le \psi \left(\underset{1\le i\le n}{max}{a}_m^i\right)\mathit{\text{for all}}m.\hfill \end{array}$$

Then $\{a_m^i\}\to 0$ for all i.

Proof Let $c_m={max}_{1\le i\le n}{a}_m^i$ for all m. Then, for all m,

$$\begin{array}{rl}\psi (c_{m+1})& =\psi \left(\underset{1\le i\le n}{max}{a}_{m+1}^i\right)=\underset{1\le i\le n}{max}\psi \left(a_{m+1}^i\right)\le \underset{1\le i\le n}{max}[(\psi -\phi )\left(b_m^i\right)]\le \underset{1\le i\le n}{max}\psi \left(b_m^i\right)\\ =\psi \left(\underset{1\le i\le n}{max}{b}_m^i\right)\le \psi \left(\underset{1\le i\le n}{max}{a}_m^i\right)=\psi (c_m).\end{array}$$

Therefore, $\{\psi (c_m)\}$ is a non-increasing, bounded below sequence. Then it is convergent. Let $\mathrm{\Delta }\ge 0$ be such that $\{\psi (c_m)\}\to \mathrm{\Delta }$ and $\mathrm{\Delta }\le \psi (c_m)$. Let us show that $\mathrm{\Delta }=0$. Since

$$\left\{\underset{1\le i\le n}{max}\psi \left(a_m^i\right)\right\}=\left\{\psi \left(\underset{1\le i\le n}{max}{a}_m^i\right)\right\}=\{\psi (c_m)\}\to \mathrm{\Delta },$$

Lemma 5 guarantees that there exist $i_0\in \{1,2,\dots ,n\}$ and a partial subsequence ${\{a_{m(k)}^{i_0}\}}_{k\in \mathbb{N}}$ such that $\{\psi (a_{m(k)}^{i_0})\}\to \mathrm{\Delta }$. Moreover,

$$0\le \psi \left(a_{m(k)}^{i_0}\right)\le (\psi -\phi )\left(b_{m(k)-1}^{i_0}\right)\text{for all }k.$$

(3)

Consider the sequence ${\{b_{m(k)-1}^{i_0}\}}_{k\in \mathbb{N}}$. If this sequence has a partial subsequence converging to zero, then we can take limit in (3) when $k\to 0$ using that partial subsequence, and we deduce $\mathrm{\Delta }=0$. On the contrary, if ${\{b_{m(k)-1}^{i_0}\}}_{k\in \mathbb{N}}$ has not any partial subsequence converging to zero, Lemma 6 assures us that there exist $\delta \in ]0,1[$ and $k_0\in \mathbb{N}$ such that $b_{m(k)-1}^{i_0}\ge \delta$ for all $k\ge k_0$. Since φ is non-decreasing, $-\phi (b_{m(k)-1}^{i_0})\le -\phi (\delta )<0$. Then, by (3), for all $k\ge k_0$,

$$\begin{array}{rl}0& \le \psi \left(a_{m(k)}^{i_0}\right)\le (\psi -\phi )\left(b_{m(k)-1}^{i_0}\right)=\psi \left(b_{m(k)-1}^{i_0}\right)-\phi \left(b_{m(k)-1}^{i_0}\right)\le \psi \left(b_{m(k)-1}^{i_0}\right)-\phi (\delta )\\ \le \psi \left(\underset{1\le i\le n}{max}{b}_{m(k)-1}^i\right)-\phi (\delta )\le \psi \left(\underset{1\le i\le n}{max}{a}_{m(k)-1}^i\right)-\phi (\delta )=\psi (c_{m(k)-1})-\phi (\delta ).\end{array}$$

Taking limit as $k\to \mathrm{\infty }$, we deduce $\mathrm{\Delta }\le \mathrm{\Delta }-\phi (\delta )$, which is impossible. This proves that $\mathrm{\Delta }=0$. Since $\{\psi (c_m)\}\to \mathrm{\Delta }=0$, Lemma 7 implies that $\{c_m\}\to 0$, which is equivalent to $\{a_m^i\}\to 0$ for all i. □

Corollary 9 If $\psi ,\phi \in \mathrm{\Psi }$ and $\{a_m\},\{b_m\}\subset [0,\mathrm{\infty })$ verify $\psi (a_{m+1})\le (\psi -\phi )(b_m)$ and $\psi (b_m)\le \psi (a_m)$ for all m, then $\{a_m\}\to 0$.

Corollary 10 If $\psi ,\phi \in \mathrm{\Psi }$ and $\{a_m\}\subset [0,\mathrm{\infty })$ verifies $\psi (a_{m+1})\le \psi (a_m)-\phi (a_m)$ for all m, then $\{a_m\}\to 0$.

Definition 11 (Mustafa and Sims [11])

A generalized metric (or a G-metric) on X is a mapping $G:X^3\to {\mathbb{R}}_0^{+}$ verifying, for all $x,y,z\in X$:

• ($G_{\mathbf{1}}$) $G(x,x,x)=0$.

• ($G_{\mathbf{2}}$) $G(x,x,y)>0$ if $x\ne y$.

• ($G_{\mathbf{3}}$) $G(x,x,y)\le G(x,y,z)$ if $y\ne z$.

• ($G_{\mathbf{4}}$) $G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots$ (symmetry in all three variables).

• ($G_{\mathbf{5}}$) $G(x,y,z)\le G(x,a,a)+G(a,y,z)$ (rectangle inequality).

Let ${\{(X_i,G_i)\}}_{i=1}^n$ be a family of G-metric spaces, consider the product space $X=X_1\times X_2\times \cdots \times X_n$ and define $G^m$ and $G^s$ on $X^3$ by

$$G^m(\mathsf{X},\mathsf{Y},\mathsf{Z})=\underset{1\le i\le n}{max}{G}_i(x_i,y_i,z_i)\text{and}{G}^s(\mathsf{X},\mathsf{Y},\mathsf{Z})=\sum _{i=1}^n{G}_i(x_i,y_i,z_i)$$

for all $\mathsf{X}=(x_1,x_2,\dots ,x_n),\mathsf{Y}=(y_1,y_2,\dots ,y_n),\mathsf{Z}=(z_1,z_2,\dots ,z_n)\in X$.

A classical example of G-metric comes from a metric space $(X,d)$, where $G(x,y,z)=d_{xy}+d_{yz}+d_{zx}$ measures the perimeter of a triangle. In this case, property $(G_3)$ has an obvious geometric interpretation: the length of an edge of a triangle is less than or equal to its semiperimeter, that is, $2d_{xy}\le d_{xy}+d_{yz}+d_{zx}$. However, property $(G_3)$ implies that, in general, the major structures $G^m$ and $G^s$ are not necessarily G-metrics on $X_1\times X_2\times \cdots \times X_n$. Only when each $G_i$ is symmetric (that is, $G(x,x,y)=G(y,y,x)$ for all x, y), the product is also a G-metric (see [11]). But in this case, symmetric G-metrics can be reduced to usual metrics, which limits the interest in this kind of spaces.

In order to prove our main results, that are also valid in G-metric spaces, we will not need property $(G_3)$. Omitting this property, we consider a class of spaces for which $G^m$ and $G^s$ have the same initial metric structure. Then we present the following spaces.

3 $G^{\ast }$-metric spaces

Definition 12 A $G^{\ast }$-metric on X is a mapping $G:X^3\to {\mathbb{R}}_0^{+}$ verifying $(G_1)$, $(G_2)$, $(G_4)$ and $(G_5)$.

The open ball $B(x,r)$ of center $x\in X$ and radius $r>0$ in a $G^{\ast }$-metric space $(X,G)$ is

$$B(x,r)=\{y\in X:G(x,x,y)<r\}.$$

The following lemma is a characterization of the topology generated by a neighborhood system at each point.

Lemma 13 Let X be a set and, for all $x\in X$, let ${\beta }_x$ be a non-empty family of subsets of X verifying:

• 1. $x\in N$ for all $N\in {\beta }_x$.

• 2. For all $N_1,N_2\in {\beta }_x$, there exists $N_3\in {\beta }_x$ such that $N_3\subseteq N_1\cap N_2$.

• 3. For all $N\in {\beta }_x$, there exists $N^{\mathrm{\prime }}\in {\beta }_x$ such that for all $y\in N^{\mathrm{\prime }}$, there exists $N^{\mathrm{\prime }\mathrm{\prime }}\in {\beta }_y$ verifying $N^{\mathrm{\prime }\mathrm{\prime }}\subseteq N$.

Then there exists a unique topology τ on X such that ${\beta }_x$ is a neighborhood system at x.

Let $(X,G)$ be a $G^{\ast }$-metric space and consider the family ${\beta }_x=\{B(x,r):r>0\}$. It is clear that $x\in B(x,r)$ (by $(G_1)$, $G(x,x,x)=0$) and $N_3=B(x,min(r,s))\subseteq B(x,r)\cap B(x,s)$. Next, let $N=N^{\mathrm{\prime }}=B(x,r)\in {\beta }_x$ and let $y\in N^{\mathrm{\prime }}=B(x,r)$. We have to prove that there exists $s>0$ such that $N^{\mathrm{\prime }\mathrm{\prime }}=B(y,s)\subseteq B(x,r)=N$. Indeed, if $y=x$, then we can take $s=r>0$. On the contrary, if $y\ne x$, then $0<G(x,x,y)<r$ by $(G_2)$. Let $r^{\mathrm{\prime }}\in ]G(x,x,y),r[$ arbitrary and let $s=r-r^{\mathrm{\prime }}>0$ (that is, $r^{\mathrm{\prime }}+s=r$). Now we prove that $B(y,s)\subseteq B(x,r)$. Let $z\in B(y,s)$. Then, using $(G_4)$ and $(G_5)$,

$$G(x,x,z)=G(z,x,x)\stackrel{a=y}{\le }G(z,y,y)+G(y,x,x)=G(x,x,y)+G(y,y,z)<r^{\mathrm{\prime }}+s=r.$$

Then $z\in B(x,r)$ and, as a consequence, $B(y,s)\subseteq B(x,r)$. Lemma 13 guarantees that there exists a unique topology ${\tau }_G$ on X such that ${\beta }_x=\{B(x,r):r>0\}$ is a neighborhood system at each $x\in X$.

Next, let us show that ${\tau }_G$ is Hausdorff. Let $x,y\in X$ be two points such that $x\ne y$. By $(G_2)$, $r=G(x,x,y)>0$. We claim that $B(x,r/4)\cap B(y,r/4)=\mathrm{\varnothing }$. We reason by contradiction. Let $z\in B(x,r/4)\cap B(y,r/4)$, that is, $G(x,x,z)<r/4$ and $G(y,y,z)<r/4$. Using $(G_4)$ and $(G_5)$ twice

$$\begin{array}{rl}0& <r=G(x,x,y)=G(y,x,x)\le G(y,z,z)+G(z,x,x)=G(z,z,y)+G(x,x,z)\\ \le G(z,y,y)+G(y,z,y)+G(x,x,z)=G(y,y,z)+G(y,y,z)+G(x,x,z)\\ <\frac{r}{4}+\frac{r}{4}+\frac{r}{4}=\frac{3r}{4}<r,\end{array}$$

which is impossible. Then $B(x,r/4)\cap B(y,r/4)=\mathrm{\varnothing }$ and ${\tau }_G$ is Hausdorff.

A subset $A\subseteq X$ is G-open if for all $x\in A$ there exists $r>0$ such that $B(x,r)\subseteq A$. Following classic techniques, it is possible to prove that there exists a unique topology ${\tau }_G$ on X such that ${\beta }_x=\{B(x,r):r>0\}$ is a neighborhood system at each $x\in X$. Furthermore, ${\tau }_G$ is a Hausdorff topology. In this topology, we characterize the notions of convergent sequence and Cauchy sequence in the following way. Let $(X,G)$ be a $G^{\ast }$-metric space, let $\{x_m\}\subseteq X$ be a sequence and let $x\in X$.

• $\{x_m\}$ G-converges to x, and we will write $\{x_m\}\stackrel{G}{\to }x$ if ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m^{\mathrm{\prime }}},x)=0$, that is, for all $\epsilon >0$, there exists $m_0\in \mathbb{N}$ verifying that $G(x_m,x_{m^{\mathrm{\prime }}},x)<\epsilon$ for all $m,m^{\mathrm{\prime }}\in \mathbb{N}$ such that $m,m^{\mathrm{\prime }}\ge m_0$.

• $\{x_m\}$ is G-Cauchy if ${lim}_{m,m^{\mathrm{\prime }},m^{\mathrm{\prime }\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }\mathrm{\prime }}})=0$, that is, for all $\epsilon >0$, there exists $m_0\in \mathbb{N}$ verifying that $G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }\mathrm{\prime }}})<\epsilon$ for all $m,m^{\mathrm{\prime }},m^{\mathrm{\prime }\mathrm{\prime }}\in \mathbb{N}$ such that $m,m^{\mathrm{\prime }},m^{\mathrm{\prime }\mathrm{\prime }}\ge m_0$.

Lemma 14 Let $(X,G)$ be a $G^{\ast }$-metric space, let $\{x_m\}\subseteq X$ be a sequence and let $x\in X$. Then the following conditions are equivalent.

• (a) $\{x_m\}$ G-converges to x.

• (b) ${lim}_{m\to \mathrm{\infty }}G(x,x,x_m)=0$.

• (c) ${lim}_{m\to \mathrm{\infty }}G(x_m,x_m,x)=0$.

• (d) ${lim}_{m\to \mathrm{\infty }}G(x_m,x_m,x)=0$ and ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$.

• (e) ${lim}_{m\to \mathrm{\infty }}G(x,x,x_m)=0$ and ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$.

Notice that the condition ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$ is not strong enough to prove that $\{x_m\}$ G-converges to x.

Proposition 15 The limit of a G-convergent sequence in a $G^{\ast }$-metric space is unique.

Lemma 16 If $(X,G)$ is a $G^{\ast }$-metric space and $\{x_m\}\subseteq X$ is a sequence, then the following conditions are equivalent.

• (a) $\{x_m\}$ is G-Cauchy.

• (b) ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}})=0$.

• (c) ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m+1},x_{m^{\mathrm{\prime }}})=0$.

Remark 17 As a consequence, a sequence $\{x_m\}\subseteq X$ is not G-Cauchy if and only if there exist ${\epsilon }_0>0$ and two partial subsequences ${\{x_{n(k)}\}}_{k\in \mathbb{N}}$ and ${\{x_{m(k)}\}}_{k\in \mathbb{N}}$ such that $k<n(k)<m(k)<n(k+1)$, $G(x_{n(k)},x_{n(k)+1},x_{m(k)})\ge {\epsilon }_0$ and $G(x_{n(k)},x_{n(k)+1},x_{m(k)-1})<{\epsilon }_0$ for all k.

Definition 18 Let $(X,G)$ be a $G^{\ast }$-metric space and let ≼ be a preorder on X. We will say that $(X,G,\preccurlyeq )$ is regular non-decreasing (respectively, regular non-increasing) if for all ≼-monotone non-decreasing (respectively, non-increasing) sequence $\{x_m\}$ such that $\{x_m\}\stackrel{G}{\to }{z}_0$, we have that $x_m\preccurlyeq z_0$ (respectively, $x_m\succcurlyeq z_0$) for all m. We will say that $(X,G,\preccurlyeq )$ is regular if it is both regular non-decreasing and regular non-increasing.

Some authors said that $(X,G,\preccurlyeq )$ verifies the sequential monotone property if $(X,G,\preccurlyeq )$ is regular (see [20]). The notion of G-continuous mapping $F:X^n\to X$ follows considering on X the topology ${\tau }_G$ and in $X^n$ the product topology.

Definition 19 If $(X,G)$ is a $G^{\ast }$-metric space, we will say that a mapping $F:X^n\to X$ is G-continuous if for all n sequences $\{a_m^1\},\{a_m^2\},\dots ,\{a_m^n\}\subseteq X$ such that $\{a_m^i\}\stackrel{G}{\to }{a}_i\in X$ for all i, we have that $\{F(a_m^1,a_m^2,\dots ,a_m^n)\}\stackrel{G}{\to }F(a_1,a_2,\dots ,a_n)$.

In this topology, the notion of convergence is the following.

$$\begin{array}{rl}\{x_m\}\stackrel{G}{\to }x& \iff [\mathrm{\forall }B(x,r),\mathrm{\exists }{m}_0\in \mathbb{N}:(m\ge m_0\Rightarrow x_m\in B(x,r))]\\ \iff [\mathrm{\forall }\epsilon >0,\mathrm{\exists }{m}_0\in \mathbb{N}:(m\ge m_0\Rightarrow G(x,x,x_m)<\epsilon )]\\ \iff [\underset{m\to \mathrm{\infty }}{lim}G(x,x,x_m)=0].\end{array}$$

This property can be characterized as follows.

Lemma 20 Let $(X,G)$ be a $G^{\ast }$-metric space, let $\{x_m\}\subseteq X$ be a sequence and let $x\in X$. Then the following conditions are equivalent.

• (a) $\{x_m\}$ G-converges to x (that is, ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m^{\mathrm{\prime }}},x)=0$, which means that for all $\epsilon >0$, there exists $n_0\in \mathbb{N}$ such that $G(x_m,x_{m^{\mathrm{\prime }}},x)$ for all $m,m^{\mathrm{\prime }}\ge m_0$).

• (b) ${lim}_{m\to \mathrm{\infty }}G(x,x,x_m)=0$.

• (c) ${lim}_{m\to \mathrm{\infty }}G(x_m,x_m,x)=0$.

• (d) ${lim}_{m\to \mathrm{\infty }}G(x_m,x_m,x)=0$ and ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$.

• (e) ${lim}_{m\to \mathrm{\infty }}G(x,x,x_m)=0$ and ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$.

Proof [(a) ⇒ (c)] It is apparent using $m=m^{\mathrm{\prime }}$.

• [(c) ⇒ (b)] Using $(G_5)$, $G(x,x,x_m)\le G(x,x_m,x_m)+G(x_m,x,x_m)=2G(x_m,x_m,x)$.

• [(b) ⇒ (a)] Using $(G_4)$ and $(G_5)$,

  $$G(x_m,x_{m^{\mathrm{\prime }}},x)\le G(x_m,x,x)+G(x,x_{m^{\mathrm{\prime }}},x)\le 2max(G(x,x,x_m),G(x,x,x_{m^{\mathrm{\prime }}})).$$

• [(a) ⇒ (d),(e)] It is apparent using $m^{\mathrm{\prime }}=m$ and $m^{\mathrm{\prime }}=m+1$.

• [(d) ⇒ (c)] It is evident.

• [(e) ⇒ (b)] It is evident. □

Corollary 21 If $(X,G)$ is a G-metric space, then $\{x_m\}\stackrel{G}{\to }x$ if and only if ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$.

Proof We only need to prove that the condition is sufficient. Suppose that ${lim}_{m\to \mathrm{\infty }}G(x_m,x_{m+1},x)=0$. In a G-metric space, the following property holds (see [11]):

$$G(x,y,z)\le G(x,a,z)+G(a,y,z)\text{for all }x,y,z,a\in X.$$

Then, using $a=x_{m+1}$,

$$G(x,x,x_m)=G(x,x_{m+1},x_m)+G(x_{m+1},x,x_m)=2G(x_m,x_{m+1},x).$$

This proves (b) in the previous lemma. □

Proposition 22 The limit of a G-convergent sequence in a $G^{\ast }$-metric space is unique.

Proof Suppose that $\{x_m\}\stackrel{G}{\to }x$ and $\{x_m\}\stackrel{G}{\to }y$. Then

$$G(x,x,y)=G(y,x,x)\le G(y,x_m,x_m)+G(x_m,x,x).$$

By items (a) and (c) of Lemma 20, we deduce that $G(x,x,y)=0$, which means that $x=y$ by $(G_2)$. □

In the topology ${\tau }_G$, the notion of Cauchy sequence is the following.

$$\{x_m\}\text{ is }G\text{-Cauchy}\iff [\mathrm{\forall }\epsilon >0,\mathrm{\exists }{m}_0\in \mathbb{N}:(m,m^{\mathrm{\prime }},m^{\mathrm{\prime }\mathrm{\prime }}\ge m_0\Rightarrow G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }\mathrm{\prime }}})<\epsilon )].$$

This definition can be characterized as follows.

Lemma 23 If $(X,G)$ is a $G^{\ast }$-metric space and $\{x_m\}\subseteq X$ is a sequence, then the following conditions are equivalent.

• (a) $\{x_m\}$ is G-Cauchy.

• (b) ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}})=0$.

• (c) ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m+1},x_{m^{\mathrm{\prime }}})=0$.

Proof [(b) ⇒ (a)] Using $(G_5)$, $G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }\mathrm{\prime }}})\le G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}})+G(x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }\mathrm{\prime }}})$.

• [(a) ⇒ (c)] It is apparent using $m^{\mathrm{\prime }\mathrm{\prime }}=m+1$.

• [(c) ⇒ (b)] Let $\epsilon >0$ and let $m_0\in \mathbb{N}$ be such that $G(x_m,x_{m+1},x_{m^{\mathrm{\prime }}})<\epsilon /2$ for all $m,m^{\mathrm{\prime }}\ge m_0$. Then

  $$\begin{array}{c}{m}^{\mathrm{\prime }},m\ge m_0\Rightarrow G(x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}+1},x_m)<\epsilon /2,\hfill \\ m^{\mathrm{\prime }},m^{\mathrm{\prime }}+1\ge m_0\Rightarrow G(x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}+1},x_{m^{\mathrm{\prime }}+1})<\epsilon /2.\hfill \end{array}$$

Therefore, using $(G_4)$ and $(G_5)$,

$$\begin{array}{rcl}G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}})& =& G(x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}},x_m)\le G(x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}+1},x_{m^{\mathrm{\prime }}+1})+G(x_{m^{\mathrm{\prime }}+1},x_{m^{\mathrm{\prime }}},x_m)\\ <& \epsilon /2+\epsilon /2=\epsilon .\end{array}$$

Therefore, ${lim}_{m,m^{\mathrm{\prime }}\to \mathrm{\infty }}G(x_m,x_{m^{\mathrm{\prime }}},x_{m^{\mathrm{\prime }}})=0$. □

4 Product of $G^{\ast }$-metric spaces

Lemma 24 Let ${\{(X_i,G_i)\}}_{i=1}^n$ be a family of $G^{\ast }$-metric spaces, consider the product space $\mathbb{X}=X_1\times X_2\times \cdots \times X_n$ and define $G_n^{max}$ and $G_n^{sum}$ on ${\mathbb{X}}^3$ by

$$G_n^{max}(\mathsf{X},\mathsf{Y},\mathsf{Z})=\underset{1\le i\le n}{max}{G}_i(x_i,y_i,z_i)\mathit{\text{and}}{G}_n^{sum}(\mathsf{X},\mathsf{Y},\mathsf{Z})=\sum _{i=1}^n{G}_i(x_i,y_i,z_i)$$

for all $\mathsf{X}=(x_1,x_2,\dots ,x_n),\mathsf{Y}=(y_1,y_2,\dots ,y_n),\mathsf{Z}=(z_1,z_2,\dots ,z_n)\in \mathbb{X}$. Then the following statements hold.

• 1. $G_n^{max}$ and $G_n^{sum}$ are $G^{\ast }$-metrics on $\mathbb{X}$.

• 2. If $A_m=(a_m^1,a_m^2,\dots ,a_m^n)\in \mathbb{X}$ for all m and $A=(a_1,a_2,\dots ,a_n)\in \mathbb{X}$, then $\{A_m\}$ $G_n^{max}$-converges (respectively, $G_n^{sum}$-converges) to A if and only if each $\{a_m^i\}$ $G_i$-converges to $a_i$.

• 3. $\{A_m\}$ is $G_n^{max}$-Cauchy if and only if each $\{a_m^i\}$ is $G_i$-Cauchy.

• 4. $(\mathbb{X},G_n^{max})$ (respectively, $(\mathbb{X},G_n^{sum})$) is complete if and only if every $(X_i,G_i)$ is complete.

• 5. For all i, let ${⪯}_i$ be a preorder on $X_i$ and define $\mathsf{X}⪯\mathsf{Y}$ if and only if $x_i{⪯}_i{y}_i$ for all i. Then $(X,G_n^{max},⪯)$ is regular (respectively, regular non-decreasing, regular non-increasing) if and only if each factor $(X_i,G_i)$ is also regular (respectively, regular non-decreasing, regular non-increasing).

Proof Let us denote $G=G_n^{max}$. Taking into account that $G_n^{max}\le G_n^{sum}\le n{G}_n^{max}$, we will only develop the proof using G.

(1) It is a straightforward exercise to prove the following statements.

• $G(\mathsf{X},\mathsf{X},\mathsf{X})={max}_{1\le i\le n}{G}_i(x_i,x_i,x_i)={max}_{1\le i\le n}0=0$.

• If $\mathsf{Y}\ne \mathsf{Z}$, there exists $j\in \{1,2,\dots ,n\}$ such that $y_j\ne z_j$. Then $G(\mathsf{X},\mathsf{Y},\mathsf{Z})={max}_{1\le i\le n}{G}_i(x_i,y_i,z_i)\ge G_j(x_j,y_j,z_j)>0$.

• Symmetry in all three variables of G follows from symmetry in all three variables of each $G_i$.

• We have that

  $$\begin{array}{rl}G(\mathsf{X},\mathsf{Y},\mathsf{Z})& =\underset{1\le i\le n}{max}{G}_i(x_i,y_i,z_i)\le \underset{1\le i\le n}{max}[G_i(x_i,a_i,a_i)+G_i(a_i,y_i,z_i)]\\ \le \underset{1\le i\le n}{max}{G}_i(x_i,a_i,a_i)+\underset{1\le i\le n}{max}{G}_i(a_i,y_i,z_i)=G(\mathsf{X},\mathsf{A},\mathsf{A})+G(\mathsf{A},\mathsf{Y},\mathsf{Z}).\end{array}$$

(2) We use Lemma 20. Suppose that $\{A_m\}$ G-converges to A and let $\epsilon >0$. Then, for all $j\in \{1,2,\dots ,n\}$ and all m,

$$G_j(a_j,a_j,a_m^j)\le \underset{1\le i\le n}{max}{G}_i(a_i,a_i,a_m^i)=G(A,A,A_m).$$

Therefore, $\{a_m^j\}$ $G_j$-converges to $a_j$. Conversely, assume that each $\{a_m^i\}$ $G_i$-converges to $a_i$. Let $\epsilon >0$ and let $m_i\in \mathbb{N}$ be such that if $m\ge m_i$, then $G_i(a_i,a_i,a_m^i)<\epsilon$. If $m_0=max(m_1,m_2,\dots ,m_n)$ and $m,m^{\mathrm{\prime }}\ge m_0$, then $G(A,A,A_m)={max}_{1\le i\le n}{G}_i(a_i,a_i,a_m^i)<\epsilon$, so $\{A_m\}$ G-converges to A.

(3) We use Lemma 23. Suppose that $\{A_m\}$ is G-Cauchy and let $\epsilon >0$. Then, for all $j\in \{1,2,\dots ,n\}$ and all m, $m^{\mathrm{\prime }}$,

$$G_j(a_m^j,a_m^j,a_{m^{\mathrm{\prime }}}^j)\le \underset{1\le i\le n}{max}{G}_i(a_m^i,a_m^i,a_{m^{\mathrm{\prime }}}^i)=G(A_m,A_m,A_{m^{\mathrm{\prime }}}).$$

Therefore, $\{a_m^j\}$ is $G_j$-Cauchy. Conversely, assume that each $\{a_m^i\}$ is $G_i$-Cauchy. Let $\epsilon >0$ and let $m_i\in \mathbb{N}$ be such that if $m,m^{\mathrm{\prime }}\ge m_i$, then $G_i(a_m^j,a_m^j,a_{m^{\mathrm{\prime }}}^j)<\epsilon$. If $m_0=max(m_1,m_2,\dots ,m_n)$ and $m,m^{\mathrm{\prime }}\ge m_0$, then $G(A_m,A_m,A_{m^{\mathrm{\prime }}})={max}_{1\le i\le n}{G}_i(a_m^i,a_m^i,a_{m^{\mathrm{\prime }}}^i)<\epsilon$, so $\{A_m\}$ is G-Cauchy.

(4) It is an easy consequence of items 2 and 3 since

$$\begin{array}{rl}\{A_m\}G\text{-Cauchy}& \iff \text{each }\left\{a_m^i\right\}G\text{-Cauchy}\iff \text{each }\left\{a_m^i\right\}G\text{-convergent}\\ \iff \{A_m\}G\text{-convergent.}\end{array}$$

(5) A sequence $\{A_m\}$ on $\mathbb{X}$ is ⪯-monotone non-decreasing if and only if each sequence $\{a_m^i\}$ is ⪯-monotone non-decreasing. Moreover, $\{A_m\}$ G-converges to $A=(a_1,a_2,\dots ,a_n)\in \mathbb{X}$ if and only if each $\{a_m^i\}$ $G_i$-converges to $a_i$. Finally, $A_m⪯A$ if and only if $a_m^i{⪯}_i{a}_i$ for all i. Therefore, $(X,G_n^{max},⪯)$ is regular non-decreasing if and only if each factor $(X_i,G_i)$ is also regular non-decreasing. Other statements may be proved similarly. □

Taking $(X_i,G_i)=(X,G)$ for all i, we derive the following result.

Corollary 25 Let $(X,G)$ be a $G^{\ast }$-metric space and consider on the product space $X^n$ the mappings $G_n$ and $G_n^{\mathrm{\prime }}$ defined by

$$G_n(\mathsf{X},\mathsf{Y},\mathsf{Z})=\underset{1\le i\le n}{max}G(x_i,y_i,z_i)\mathit{\text{and}}{G}_n^{\mathrm{\prime }}(\mathsf{X},\mathsf{Y},\mathsf{Z})=\sum _{i=1}^{n}G(x_i,y_i,z_i)$$

for all $\mathsf{X}=(x_1,x_2,\dots ,x_n),\mathsf{Y}=(y_1,y_2,\dots ,y_n),\mathsf{Z}=(z_1,z_2,\dots ,z_n)\in X^n$.

• 1. $G_n$ and $G_n^{\mathrm{\prime }}$ are $G^{\ast }$-metrics on $X^n$.

• 2. If $A_m=(a_m^1,a_m^2,\dots ,a_m^n)\in X^n$ for all m and $A=(a_1,a_2,\dots ,a_n)\in X^n$, then $\{A_m\}$ $G_n$-converges (respectively, $G_n^{\mathrm{\prime }}$-converges) to A if and only if each $\{a_m^i\}$ G-converges to $a_i$.

• 3. $\{A_m\}$ is $G_n$-Cauchy (respectively, $G_n^{\mathrm{\prime }}$-Cauchy) if and only if each $\{a_m^i\}$ is G-Cauchy.

• 4. $(X,G_n)$ (respectively, $(X^n,G_n^{\mathrm{\prime }})$) is complete if and only if $(X,G)$ is complete.

• 5. If $(X,G)$ is ≼-regular, then $(X^n,G_n)$ is ⊑-regular.

5 Unidimensional fixed point result in partially preordered $G^{\ast }$-metric spaces

Theorem 26 Let $(X,\preccurlyeq )$ be a preordered set endowed with a $G^{\ast }$-metric G and let $T:X\to X$ be a given mapping. Suppose that the following conditions hold:

• (a) $(X,G)$ is complete.

• (b) T is non-decreasing (w.r.t. ≼).

• (c) Either T is G-continuous or $(X,G,\preccurlyeq )$ is regular non-decreasing.

• (d) There exists $x_0\in X$ such that $x_0\preccurlyeq T{x}_0$.

• (e) There exist two mappings $\psi ,\phi \in \mathrm{\Psi }$ such that, for all $x,y\in X$ with $x\preccurlyeq y$,

  $$\psi \left(G(Tx,Ty,T^{2}x)\right)\le \psi (G(x,y,Tx))-\phi (G(x,y,Tx)).$$

Then T has a fixed point. Furthermore, if for all $z_1,z_2\in X$ fixed points of T there exists $z\in X$ such that $z_1\preccurlyeq z$ and $z_2\preccurlyeq z$, we obtain uniqueness of the fixed point.

Proof Define $x_m=T^m{x}_0$ for all $m\ge 1$. Since T is non-decreasing (w.r.t. ≼), then $x_m\preccurlyeq x_{m+1}$ for all $m\ge 0$. Then

$$\begin{array}{rl}\psi (G(x_{m+1},x_{m+2},x_{m+2}))& =\psi \left(G(T{x}_m,T{x}_{m+1},T^2{x}_m)\right)\\ \le \psi (G(x_m,x_{m+1},T{x}_m))-\phi (G(x_m,x_{m+1},T{x}_m))\\ =\psi (G(x_m,x_{m+1},x_{m+1}))-\phi (G(x_m,x_{m+1},x_{m+1})).\end{array}$$

Applying Lemma 10, $\{G(x_m,x_{m+1},x_{m+1})\}\to 0$. Let us show that $\{x_m\}$ is G-Cauchy. Reasoning by contradiction, if $\{x_m\}$ is not G-Cauchy, by Remark 17, there exist ${\epsilon }_0>0$ and two partial subsequences $\{x_{n(k)}\}$ and $\{x_{m(k)}\}$ verifying $k<n(k)<m(k)<n(k+1)$,

$$G(x_{n(k)},x_{m(k)},x_{n(k)+1})>{\epsilon }_0\text{and}G(x_{n(k)},x_{m(k)-1},x_{n(k)+1})\le {\epsilon }_0\text{for all }k\ge 1.$$

(4)

Therefore

$$\begin{array}{rl}0& <\psi ({\epsilon }_0)\le \psi (G(x_{n(k)},x_{m(k)},x_{n(k)+1}))=\psi \left(G(T{x}_{n(k)-1},T{x}_{m(k)-1},T^2{x}_{n(k)-1})\right)\\ \le \psi (G(x_{n(k)-1},x_{m(k)-1},T{x}_{n(k)-1}))-\phi (G(x_{n(k)-1},x_{m(k)-1},T{x}_{n(k)-1}))\\ =\psi (G(x_{n(k)-1},x_{m(k)-1},x_{n(k)}))-\phi (G(x_{n(k)-1},x_{m(k)-1},x_{n(k)})).\end{array}$$

(5)

Consider the sequence of non-negative real numbers $\{G(x_{n(k)-1},x_{m(k)-1},x_{n(k)})\}$. If this sequence has a partial subsequence converging to zero, then we can take the limit in (5) using this partial subsequence and we would deduce $0<\psi ({\epsilon }_0)\le 0$, which is impossible. Then $\{G(x_{n(k)-1},x_{m(k)-1},x_{n(k)})\}$ cannot have a partial subsequence converging to zero. This means that there exist $\delta >0$ and $k_0\in \mathbb{N}$ such that

$$G(x_{n(k)-1},x_{m(k)-1},x_{n(k)})\ge \delta \text{for all }k\ge k_0.$$

Since φ is non-decreasing, $-\phi (G(x_{n(k)-1},x_{m(k)-1},x_{n(k)})\le -\phi (\delta )<0$. By (G₅) and (4),

$$\begin{array}{r}G(x_{n(k)-1},x_{m(k)-1},x_{n(k)})\\ =G(x_{n(k)-1},x_{n(k)},x_{m(k)-1})[x=x_{n(k)-1},y=x_{n(k)},z=x_{m(k)-1},a=x_{n(k)+1}]\\ \le G(x_{n(k)-1},x_{n(k)+1},x_{n(k)+1})+G(x_{n(k)+1},x_{n(k)},x_{m(k)-1})\\ =G(x_{n(k)-1},x_{n(k)+1},x_{n(k)+1})+G(x_{n(k)},x_{n(k)+1},x_{m(k)-1})\\ [x=x_{n(k)-1},y=z=x_{n(k)+1},a=x_{n(k)}]\\ \le G(x_{n(k)-1},x_{n(k)},x_{n(k)})+G(x_{n(k)},x_{n(k)+1},x_{n(k)+1})+G(x_{n(k)},x_{n(k)+1},x_{m(k)-1})\\ \le G(x_{n(k)-1},x_{n(k)},x_{n(k)})+G(x_{n(k)},x_{n(k)+1},x_{n(k)+1})+{\epsilon }_0.\end{array}$$

Since ψ is non-decreasing, it follows from (5) that

$$\begin{array}{rl}0& <\psi ({\epsilon }_0)\le \psi (G(x_{n(k)-1},x_{m(k)-1},x_{n(k)}))-\phi (G(x_{n(k)-1},x_{m(k)-1},x_{n(k)}))\\ \le \psi (G(x_{n(k)-1},x_{m(k)-1},x_{n(k)}))-\phi (\delta )\\ \le \psi (G(x_{n(k)-1},x_{n(k)},x_{n(k)})+G(x_{n(k)},x_{n(k)+1},x_{n(k)+1})+{\epsilon }_0)-\phi (\delta ).\end{array}$$

Taking limit when $k\to \mathrm{\infty }$, we deduce that $0<\psi ({\epsilon }_0)\le \psi ({\epsilon }_0)-\phi (\delta )<\psi ({\epsilon }_0)$, which is impossible. This contradiction finally proves that $\{x_m\}$ is G-Cauchy. Since $(X,G)$ is complete, there exists $z_0\in X$ such that $\{x_m\}\stackrel{G}{\to }{z}_0$.

Now suppose that T is G-continuous. Then $\{x_{m+1}\}=\{T{x}_m\}\stackrel{G}{\to }T{z}_0$. By the unicity of the limit, $T{z}_0=z_0$ and $z_0$ is a fixed point of T.

On the contrary, suppose that $(X,G,\preccurlyeq )$ is regular non-decreasing. Since $\{x_m\}\stackrel{G}{\to }{z}_0$ and $\{x_m\}$ is monotone non-decreasing (w.r.t. ≼), it follows that $x_m\preccurlyeq z_0$ for all m. Hence

$$\begin{array}{rl}\psi (G(x_{m+1},T{z}_0,x_{m+2}))& =\psi \left(G(T{x}_m,T{z}_0,T^2{x}_m)\right)\\ \le \psi (G(x_m,z_0,T{x}_m))-\phi (G(x_m,z_0,T{x}_m))\\ =\psi (G(x_m,x_{m+1},z_0))-\phi (G(x_m,x_{m+1},z_0)).\end{array}$$

Since $\{x_m\}\stackrel{G}{\to }{z}_0$, then $\{G(x_m,x_{m+1},z_0)\}\to 0$. Taking limit when $k\to \mathrm{\infty }$, we deduce that $\{\psi (G(x_{m+1},T{z}_0,x_{m+2}))\}\to 0$. By Lemma 7, $\{G(x_{m+1},x_{m+2},T{z}_0)\}\to 0$, so $\{x_m\}\stackrel{G}{\to }T{z}_0$ and we also conclude that $z_0$ is a fixed point of T.

To prove the uniqueness, let $z_1,z_2\in X$ be two fixed points of T. By hypothesis, there exists $z\in X$ such that $z_1\preccurlyeq z$ and $z_2\preccurlyeq z$. Let us show that $\{T^{m}z\}\stackrel{G}{\to }{z}_1$. Indeed,

$$\begin{array}{rl}\psi \left(G(z_1,z_1,T^{m+1}z)\right)& =\psi \left(G(T{z}_1,T{T}^{m}z,T^2{z}_1)\right)\\ \le \psi \left(G(z_1,T^{m}z,T{z}_1)\right)-\phi \left(G(z_1,T^{m}z,T{z}_1)\right)\\ =\psi \left(G(z_1,z_1,T^{m}z)\right)-\phi \left(G(z_1,z_1,T^{m}z)\right).\end{array}$$

By Lemma 10, we deduce $\{G(z_1,z_1,T^{m}z)\}\to 0$, that is, $\{T^{m}z\}\stackrel{G}{\to }{z}_1$. The same reasoning proves that $\{T^{m}z\}\stackrel{G}{\to }{z}_2$, so $z_1=z_2$. □

We particularize the previous theorem in two cases. If take $\psi (t)=t$ in Theorem 26, then we get the following results.

Corollary 27 Let $(X,\preccurlyeq )$ be a preordered set endowed with a $G^{\ast }$-metric G and let $T:X\to X$ be a given mapping. Suppose that the following conditions hold:

• (a) $(X,G)$ is complete.

• (b) T is non-decreasing (w.r.t. ≼).

• (c) Either T is G-continuous or $(X,G,\preccurlyeq )$ is regular non-decreasing.

• (d) There exists $x_0\in X$ such that $x_0\preccurlyeq T{x}_0$.

• (e) There exists a mapping $\phi \in \mathrm{\Psi }$ such that, for all $x,y\in X$ with $x\preccurlyeq y$,

  $$G(Tx,Ty,T^{2}x)\le G(x,y,Tx)-\phi (G(x,y,Tx)).$$

Then T has a fixed point. Furthermore, if for all $z_1,z_2\in X$ fixed points of T there exists $z\in X$ such that $z_1\preccurlyeq z$ and $z_2\preccurlyeq z$, we obtain uniqueness of the fixed point.

If take $\phi (t)=(1-k)t$ with $k\in [0,1)$ in Corollary 27, then we derive the following result.

Corollary 28 Let $(X,\preccurlyeq )$ be a preordered set endowed with a $G^{\ast }$-metric G and let $T:X\to X$ be a given mapping. Suppose that the following conditions hold:

• (a) $(X,G)$ is complete.

• (b) T is non-decreasing (w.r.t. ≼).

• (c) Either T is G-continuous or $(X,G,\preccurlyeq )$ is regular non-decreasing.

• (d) There exists $x_0\in X$ such that $x_0\preccurlyeq T{x}_0$.

• (e) There exists a constant $k\in [0,1)$ such that, for all $x,y\in X$ with $x\preccurlyeq y$,

  $$G(Tx,Ty,T^{2}x)\le kG(x,y,Tx).$$

Then T has a fixed point. Furthermore, if for all $z_1,z_2\in X$ fixed points of T there exists $z\in X$ such that $z_1\preccurlyeq z$ and $z_2\preccurlyeq z$, we obtain uniqueness of the fixed point.

6 Multidimensional ϒ-fixed point results in partially preordered $G^{\ast }$-metric spaces

In this section we extend Theorem 26 to an arbitrary number of variables. To do that, it is necessary to introduce the following notation. Given a mapping $F:X^n\to X$, we define ${\mathbb{F}}_{\mathrm{\Upsilon }}:X^n\to X^n$ by

$$\begin{array}{r}{\mathbb{F}}_{\mathrm{\Upsilon }}(x_1,x_2,\dots ,x_n)\\ =(F(x_{{\sigma }_1(1)},x_{{\sigma }_1(2)},\dots ,x_{{\sigma }_1(n)}),F(x_{{\sigma }_2(1)},x_{{\sigma }_2(2)},\dots ,x_{{\sigma }_2(n)}),\dots ,F(x_{{\sigma }_n(1)},x_{{\sigma }_n(2)},\dots ,x_{{\sigma }_n(n)})),\end{array}$$

and $F_{\mathrm{\Upsilon }}^2=F\circ {\mathbb{F}}_{\mathrm{\Upsilon }}:X^n\to X$ will be

$$\begin{array}{r}{F}_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n)\\ =F(F(x_{{\sigma }_1(1)},x_{{\sigma }_1(2)},\dots ,x_{{\sigma }_1(n)}),F(x_{{\sigma }_2(1)},x_{{\sigma }_2(2)},\dots ,x_{{\sigma }_2(n)}),\dots ,F(x_{{\sigma }_n(1)},x_{{\sigma }_n(2)},\dots ,x_{{\sigma }_n(n)}))\end{array}$$

for all $\mathsf{X}=(x_1,x_2,\dots ,x_n)\in X^n$.

Lemma 29

• 1. $\mathsf{Z}\in X^n$ is a ϒ-fixed point of F if and only if Z is a fixed point of ${\mathbb{F}}_{\mathrm{\Upsilon }}$ (that is, ${\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{Z}=\mathsf{Z}$).

• 2. If F has the mixed monotone property, then ${\mathbb{F}}_{\mathrm{\Upsilon }}$ is ⊑-monotone non-decreasing on $X^n$.

• 3. If $(X,G)$ is a $G^{\ast }$-metric space and F is G-continuous, then ${\mathbb{F}}_{\mathrm{\Upsilon }}:X^n\to X^n$ is $G_n$-continuous and $F_{\mathrm{\Upsilon }}^2=F\circ {\mathbb{F}}_{\mathrm{\Upsilon }}:X^n\to X$ is G-continuous.

6.1 A first multidimensional contractivity result

In this subsection we apply Theorem 26 considering $T={\mathbb{F}}_{\mathrm{\Upsilon }}$ defined on $(X^n,G_n,⊑)$. In order to do that, we notice that joining some of the previous results, we obtain the following consequences.

• If $(X,G)$ is complete, it follows from Corollary 25 that $(X^n,G_n)$ is also complete.

• By item 2 of Lemma 29, if F has the mixed monotone property, then ${\mathbb{F}}_{\mathrm{\Upsilon }}$ is ⊑-monotone non-decreasing on $X^n$.

• By item 3 of Lemma 29, if F is G-continuous, then ${\mathbb{F}}_{\mathrm{\Upsilon }}:X^n\to X^n$ is $G_n$-continuous and $F_{\mathrm{\Upsilon }}^2=F\circ {\mathbb{F}}_{\mathrm{\Upsilon }}:X^n\to X$ is G-continuous.

• If $(X,G,\preccurlyeq )$ is regular, it follows from Corollary 25 that $(X^n,G_n,⊑)$ is also regular.

• If $x_0^1,x_0^2,\dots ,x_0^n\in X$ are such that $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i, then ${\mathsf{X}}_0=(x_0^1,x_0^2,\dots ,x_0^n)\in X^n$ verifies ${\mathsf{X}}_0⊑{\mathbb{F}}_{\mathrm{\Upsilon }}({\mathsf{X}}_0)$.

We study how the contractivity condition

$$\psi \left(G_n({\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{X},{\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{Y},{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X})\right)\le (\psi -\phi )(G_n(\mathsf{X},\mathsf{Y},{\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{X}))\text{for all }\mathsf{X},\mathsf{Y}\in X^n\text{ such that }\mathsf{X}⊑\mathsf{Y}$$

may be equivalently established. Let $\mathsf{X}=(x_1,x_2,\dots ,x_n)\in X^n$ and let $z_i=F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)})\in X$ for all i. Then

$$\begin{array}{c}{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}={\mathbb{F}}_{\mathrm{\Upsilon }}(F(x_{{\sigma }_1(1)},x_{{\sigma }_1(2)},\dots ,x_{{\sigma }_1(n)}),F(x_{{\sigma }_2(1)},x_{{\sigma }_2(2)},\dots ,x_{{\sigma }_2(n)}),\dots ,\hfill \\ \phantom{{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}=}F(x_{{\sigma }_n(1)},x_{{\sigma }_n(2)},\dots ,x_{{\sigma }_n(n)}))\hfill \\ \phantom{{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}}={\mathbb{F}}_{\mathrm{\Upsilon }}(z_1,z_2,\dots ,z_n)\hfill \\ \phantom{{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}}=(F(z_{{\sigma }_1(1)},z_{{\sigma }_1(2)},\dots ,z_{{\sigma }_1(n)}),F(z_{{\sigma }_2(1)},z_{{\sigma }_2(2)},\dots ,z_{{\sigma }_2(n)}),\dots ,F(z_{{\sigma }_n(1)},z_{{\sigma }_n(2)},\dots ,z_{{\sigma }_n(n)}))\hfill \\ \phantom{{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}}=(F(F(x_{{\sigma }_{{\sigma }_1(1)}(1)},\dots ,x_{{\sigma }_{{\sigma }_1(1)}(n)}),F(x_{{\sigma }_{{\sigma }_1(2)}(1)},\dots ,x_{{\sigma }_{{\sigma }_1(2)}(n)}),\dots ,F(x_{{\sigma }_{{\sigma }_1(n)}(1)},\dots ,x_{{\sigma }_{{\sigma }_1(n)}(n)})),\hfill \\ F(F(x_{{\sigma }_{{\sigma }_2(1)}(1)},\dots ,x_{{\sigma }_{{\sigma }_2(1)}(n)}),F(x_{{\sigma }_{{\sigma }_2(2)}(1)},\dots ,x_{{\sigma }_{{\sigma }_2(2)}(n)}),\dots ,F(x_{{\sigma }_{{\sigma }_2(n)}(1)},\dots ,x_{{\sigma }_{{\sigma }_n(n)}(n)})),\dots ,\hfill \\ F(F(x_{{\sigma }_{{\sigma }_n(1)}(1)},\dots ,x_{{\sigma }_{{\sigma }_n(1)}(n)}),F(x_{{\sigma }_{{\sigma }_n(2)}(1)},\dots ,x_{{\sigma }_{{\sigma }_n(2)}(n)}),\dots ,F(x_{{\sigma }_{{\sigma }_n(n)}(1)},\dots ,x_{{\sigma }_{{\sigma }_n(n)}(n)})))\hfill \\ \phantom{{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}}=(F_{\mathrm{\Upsilon }}^2(x_{{\sigma }_1(1)},x_{{\sigma }_1(2)}\dots ,x_{{\sigma }_1(n)}),F_{\mathrm{\Upsilon }}^2(x_{{\sigma }_2(1)},x_{{\sigma }_2(2)}\dots ,x_{{\sigma }_2(n)}),\dots ,\hfill \\ \phantom{{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X}=}{F}_{\mathrm{\Upsilon }}^2(x_{{\sigma }_n(1)},x_{{\sigma }_n(2)}\dots ,x_{{\sigma }_n(n)})).\hfill \end{array}$$

It follows that

$$\begin{array}{c}{G}_n(\mathsf{X},\mathsf{Y},{\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{X})=\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))\text{and}\hfill \\ G_n({\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{X},{\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{Y},{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X})=\underset{1\le i\le n}{max}G(F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}),F(y_{{\sigma }_i(1)},y_{{\sigma }_i(2)},\dots ,y_{{\sigma }_i(n)}),\hfill \\ \phantom{G_n({\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{X},{\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{Y},{\mathbb{F}}_{\mathrm{\Upsilon }}^2\mathsf{X})=}{F}_{\mathrm{\Upsilon }}^2(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)}\dots ,x_{{\sigma }_i(n)})).\hfill \end{array}$$

Therefore, a possible version of Theorem 26 applied to $(X^n,G_n,⊑)$ taking $T={\mathbb{F}}_{\mathrm{\Upsilon }}$ is the following.

Theorem 30 Let $(X,G)$ be a complete $G^{\ast }$-metric space and let ≼ be a partial preorder on X. Let $\mathrm{\Upsilon }=({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n)$ be an n-tuple of mappings from $\{1,2,\dots ,n\}$ into itself verifying ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$ if $i\in \mathsf{A}$ and ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$ if $i\in \mathsf{B}$. Let $F:X^n\to X$ be a mapping verifying the mixed monotone property on X. Assume that there exist $\psi ,\phi \in \mathrm{\Psi }$ such that

$$\begin{array}{r}\underset{1\le i\le n}{max}\psi \left(G(F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}),F(y_{{\sigma }_i(1)},y_{{\sigma }_i(2)},\dots ,y_{{\sigma }_i(n)}),F_{\mathrm{\Upsilon }}^2(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)}\dots ,x_{{\sigma }_i(n)}))\right)\\ \le (\psi -\phi )\left(\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))\right)\end{array}$$

(6)

for which $x_i{\preccurlyeq }_i{y}_i$ for all i. Suppose either F is continuous or $(X,G,\preccurlyeq )$ is regular. If there exist $x_0^1,x_0^2,\dots ,x_0^n\in X$ verifying $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i, then F has, at least, one ϒ-fixed point.

6.2 A second multidimensional contractivity result

In this section we introduce a slightly different contractivity condition that cannot be directly deduced applying Theorem 26 to $(X,G_n,⊑)$ taking $T={\mathbb{F}}_{\mathrm{\Upsilon }}$, because the contractivity condition is weaker. Then we need to show a classical proof.

Theorem 31 Let $(X,G)$ be a complete $G^{\ast }$-metric space and let ≼ be a partial preorder on X. Let $\mathrm{\Upsilon }=({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n)$ be an n-tuple of mappings from $\{1,2,\dots ,n\}$ into itself verifying ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$ if $i\in \mathsf{A}$ and ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$ if $i\in \mathsf{B}$. Let $F:X^n\to X$ be a mapping verifying the mixed monotone property on X. Assume that there exist $\psi ,\phi \in \mathrm{\Psi }$ such that

$$\begin{array}{r}\psi \left(G(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n),F_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n))\right)\\ \le (\psi -\phi )\left(\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))\right)\end{array}$$

(7)

for which $(x_1,x_2,\dots ,x_n),(y_1,y_2,\dots ,y_n)\in X^n$ are ⊑-comparable. Suppose either F is continuous or $(X,G,\preccurlyeq )$ is regular. If there exist $x_0^1,x_0^2,\dots ,x_0^n\in X$ verifying $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i, then F has, at least, one ϒ-fixed point.

Notice that (6) and (7) are very different contractivity conditions. For instance, (6) would be simpler if the image of all ${\sigma }_i$ are sets with a few points.

Proof Define ${\mathsf{X}}_0=(x_0^1,x_0^2,\dots ,x_0^n)$ and let $x_1^i=F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i. If ${\mathsf{X}}_1=(x_1^1,x_1^2,\dots ,x_1^n)$, then $x_0^i{\preccurlyeq }_i{x}_1^i$ for all i is equivalent to ${\mathsf{X}}_0⊑{\mathsf{X}}_1={\mathbb{F}}_{\mathrm{\Upsilon }}({\mathsf{X}}_0)$. By recurrence, define $x_{m+1}^i=F(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)})$ for all i and all m, and we have that ${\mathsf{X}}_m⊑{\mathsf{X}}_{m+1}={\mathbb{F}}_{\mathrm{\Upsilon }}({\mathsf{X}}_m)$. This means that the sequence $\{{\mathsf{X}}_{m+1}={\mathbb{F}}_{\mathrm{\Upsilon }}({\mathsf{X}}_m)\}$ is ⊑-monotone non-decreasing. Since $(X^n,G_n,⊑)$ is complete, it is only necessary to prove that $\{{\mathsf{X}}_m\}$ is $G_n$-Cauchy in order to deduce that it is $G_n$-convergent. By item 3 of Lemma 24, it will be sufficient to prove that each sequence $\{x_m^i\}$ is G-Cauchy. Firstly, notice that ${\mathsf{X}}_{m+1}={\mathbb{F}}_{\mathrm{\Upsilon }}({\mathsf{X}}_m)$ means that

$$x_{m+1}^i=F(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)})\text{for all }i\text{ and all }m.$$

Hence

$$\begin{array}{c}{x}_{m+2}^i=F(x_{m+1}^{{\sigma }_i(1)},x_{m+1}^{{\sigma }_i(2)},\dots ,x_{m+1}^{{\sigma }_i(n)})\hfill \\ \phantom{x_{m+2}^i}=F(F(x_m^{{\sigma }_{{\sigma }_i(1)}(1)},x_m^{{\sigma }_{{\sigma }_i(1)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(1)}(n)}),F(x_m^{{\sigma }_{{\sigma }_i(2)}(1)},x_m^{{\sigma }_{{\sigma }_i(2)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(2)}(n)}),\dots ,\hfill \\ \phantom{x_{m+2}^i=}F(x_m^{{\sigma }_{{\sigma }_i(n)}(1)},x_m^{{\sigma }_{{\sigma }_i(n)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(n)}(n)}))=F_{\mathrm{\Upsilon }}^2(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}).\hfill \end{array}$$

Furthermore, for all m,

$$\begin{array}{rl}{F}_{\mathrm{\Upsilon }}^2({\mathsf{X}}_m)=& F_{\mathrm{\Upsilon }}^2(x_m^1,x_m^2,\dots ,x_m^n)\\ =& F(F(x_m^{{\sigma }_1(1)},x_m^{{\sigma }_1(2)},\dots ,x_m^{{\sigma }_1(n)}),F(x_m^{{\sigma }_2(1)},x_m^{{\sigma }_2(2)},\dots ,x_m^{{\sigma }_2(n)}),\dots ,\\ F(x_m^{{\sigma }_n(1)},x_m^{{\sigma }_n(2)},\dots ,x_m^{{\sigma }_n(n)}))\\ =& F(x_{m+1}^1,x_{m+1}^2,\dots ,x_{m+1}^n)=F({\mathsf{X}}_{m+1}).\end{array}$$

(8)

Therefore, for all i and all m,

$$\begin{array}{r}\psi \left(G(x_{m+1}^i,x_{m+2}^i,x_{m+2}^i)\right)\\ =\psi (G(F(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}),F(x_{m+1}^{{\sigma }_i(1)},x_{m+1}^{{\sigma }_i(2)},\dots ,x_{m+1}^{{\sigma }_i(n)}),F_{\mathrm{\Upsilon }}^2(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}))\\ \le (\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)},F(x_m^{{\sigma }_{{\sigma }_i(j)}(1)},x_m^{{\sigma }_{{\sigma }_i(j)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(j)}(n)}))\right)\\ =(\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)})\right).\end{array}$$

Since ψ is non-decreasing, for all i and all m,

$$\psi \left(\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)})\right)\le \psi \left(\underset{1\le j\le n}{max}G(x_m^j,x_{m+1}^j,x_{m+1}^j)\right).$$

Applying Lemma 8 using

$$a_m^i=G(x_m^i,x_{m+1}^i,x_{m+1}^i)\text{and}{b}_m^i=\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)})$$

for all i and all m, we deduce that

$$\left\{G(x_m^i,x_{m+1}^i,x_{m+1}^i)\right\}\to 0\text{for all }i,\text{that is},\{G_n({\mathsf{X}}_m,{\mathsf{X}}_{m+1},{\mathsf{X}}_{m+1})\}\to 0.$$

(9)

Next, we prove that every sequence $\{x_m^i\}$ is G-Cauchy reasoning by contradiction. Suppose that ${\{x_m^{i_1}\}}_{m\ge 0},\dots ,{\{x_m^{i_s}\}}_{m\ge 0}$ are not G-Cauchy ($s\ge 1$) and ${\{x_m^{i_{s+1}}\}}_{m\ge 0},\dots ,{\{x_m^{i_n}\}}_{m\ge 0}$ are G-Cauchy, being $\{i_1,\dots ,i_n\}=\{1,\dots ,n\}$. From Proposition 2, for all $r\in \{1,2,\dots ,s\}$, there exist ${\epsilon }_r>0$ and subsequences ${\{x_{n_r(k)}^{i_r}\}}_{k\in \mathbb{N}}$ and ${\{x_{m_r(k)}^{i_r}\}}_{k\in \mathbb{N}}$ such that, for all $k\in \mathbb{N}$,

$$\begin{array}{c}k<n_r(k)<m_r(k)<n_r(k+1),G(x_{n_r(k)}^{i_r},x_{n_r(k)+1}^{i_r},x_{m_r(k)}^{i_r})\ge {\epsilon }_r,\hfill \\ G(x_{n_r(k)}^{i_r},x_{n_r(k)+1}^{i_r},x_{m_r(k)-1}^{i_r})<{\epsilon }_r.\hfill \end{array}$$

Now, let ${\epsilon }_0=max({\epsilon }_1,\dots ,{\epsilon }_s)>0$ and ${\epsilon }_0^{\mathrm{\prime }}=min({\epsilon }_1,\dots ,{\epsilon }_s)>0$. Since ${\{x_m^{i_{s+1}}\}}_{m\ge 0},\dots ,{\{x_m^{i_n}\}}_{m\ge 0}$ are G-Cauchy, for all $j\in \{i_{s+1},\dots ,i_n\}$, there exists $n^j\in \mathbb{N}$ such that if $m,m^{\mathrm{\prime }}\ge n^j$, then $G(x_m^j,x_{m+1}^j,x_{m^{\mathrm{\prime }}}^j)<{\epsilon }_0^{\mathrm{\prime }}/8$. Define $n_0={max}_{j\in \{i_{s+1},\dots ,i_n\}}(n^j)$. Therefore, we have proved that there exists $n_0\in \mathbb{N}$ such that if $m,m^{\mathrm{\prime }}\ge n_0$ then

$$G(x_m^j,x_{m+1}^j,x_{m^{\mathrm{\prime }}}^j)<{\epsilon }_0^{\mathrm{\prime }}/4\text{for all }j\in \{i_{s+1},\dots ,i_n\}.$$

(10)

Next, let $q\in \{1,2,\dots ,s\}$ be such that ${\epsilon }_q={\epsilon }_0=max({\epsilon }_1,\dots ,{\epsilon }_s)$. Let $k_1\in \mathbb{N}$ be such that $n_0<n_q(k_1)$ and define $n(1)=n_q(k_1)$. Consider the numbers $n(1)+1,n(1)+2,\dots ,m_q(k_1)$ until finding the first positive integer $m(1)>n(1)$ verifying

$$\underset{1\le r\le s}{max}G(x_{n(1)}^{i_r},x_{n(1)+1}^{i_r},x_{m(1)}^{i_r})\ge {\epsilon }_0,G(x_{n(1)}^{i_j},x_{n(1)+1}^{i_j},x_{m(1)-1}^{i_j})<{\epsilon }_0\text{for all }j\in \{1,2,\dots ,s\}.$$

Now let $k_2\in \mathbb{N}$ be such that $m(1)<n_q(k_2)$ and define $n(2)=n_q(k_2)$. Consider the numbers $n(2)+1,n(2)+2,\dots ,m_q(k_2)$ until finding the first positive integer $m(2)>n(2)$ verifying

$$\begin{array}{c}\underset{1\le r\le s}{max}G(x_{n(2)}^{i_r},x_{n(2)+1}^{i_r},x_{m(2)}^{i_r})\ge {\epsilon }_0,\hfill \\ G(x_{n(2)}^{i_j},x_{n(2)+1}^{i_j},x_{m(2)-1}^{i_j})<{\epsilon }_0\text{for all }j\in \{1,2,\dots ,s\}.\hfill \end{array}$$

Repeating this process, we can find sequences such that, for all $k\ge 1$,

$$\begin{array}{c}{n}_0<n(k)<m(k)<n(k+1),\underset{1\le r\le s}{max}G(x_{n(k)}^{i_r},x_{n(k)+1}^{i_r},x_{m(k)}^{i_r})\ge {\epsilon }_0,\hfill \\ G(x_{n(k)}^{i_j},x_{n(k)+1}^{i_j},x_{m(k)-1}^{i_j})<{\epsilon }_0\text{for all }j\in \{1,2,\dots ,s\}.\hfill \end{array}$$

Note that by (10), $G(x_{n(k)}^{i_r},x_{n(k)+1}^{i_r},x_{m(k)}^{i_r}),G(x_{n(k)}^{i_r},x_{n(k)+1}^{i_r},x_{m(k)-1}^{i_r})<{\epsilon }_0^{\mathrm{\prime }}/4<{\epsilon }_0/2$ for all $r\in \{s+1,s+2,\dots ,n\}$, so

$$\begin{array}{r}\underset{1\le j\le n}{max}G(x_{n(k)}^j,x_{n(k)+1}^j,x_{m(k)}^j)=\underset{1\le r\le s}{max}G(x_{n(k)}^{i_r},x_{n(k)+1}^{i_r},x_{m(k)}^{i_r})\ge {\epsilon }_0\text{and}\\ G(x_{n(k)}^i,x_{n(k)+1}^i,g{x}_{m(k)-1}^i)<{\epsilon }_0\end{array}$$

(11)

for all $i\in \{1,2,\dots ,n\}$ and all $k\ge 1$. Next, for all k, let $i(k)\in \{1,2,\dots ,s\}$ be an index such that

$$G(x_{n(k)}^{i(k)},x_{n(k)+1}^{i(k)},x_{m(k)}^{i(k)})=\underset{1\le r\le s}{max}G(x_{n(k)}^{i_r},x_{n(k)+1}^{i_r},x_{m(k)}^{i_r})=\underset{1\le j\le n}{max}G(x_{n(k)}^j,x_{n(k)+1}^j,x_{m(k)}^j)\ge {\epsilon }_0.$$

Notice that, applying (G₅) twice and (11), for all k and all j,

$$\begin{array}{rl}G(x_{n(k)-1}^j,x_{n(k)}^j,x_{m(k)-1}^j)\le & G(x_{n(k)-1}^j,x_{n(k)}^j,x_{n(k)}^j)+G(x_{n(k)}^j,x_{n(k)}^j,x_{m(k)-1}^j)\\ \le & G(x_{n(k)-1}^j,x_{n(k)}^j,x_{n(k)}^j)+G(x_{n(k)}^j,x_{n(k)+1}^j,x_{n(k)+1}^j)\\ +G(x_{n(k)+1}^j,x_{n(k)}^j,x_{m(k)-1}^j)\\ \le & G(x_{n(k)-1}^j,x_{n(k)}^j,x_{n(k)}^j)+G(x_{n(k)}^j,x_{n(k)+1}^j,x_{n(k)+1}^j)+{\epsilon }_0.\end{array}$$

(12)

Applying Proposition 2 to guarantee that the following points are ⊑-comparable, the contractivity condition (7) assures us for all k

$$\begin{array}{rl}0<& \psi ({\epsilon }_0)\le \psi \left(G(x_{n(k)}^{i(k)},x_{n(k)+1}^{i(k)},x_{m(k)}^{i(k)})\right)=\psi \left(G(x_{n(k)}^{i(k)},x_{m(k)}^{i(k)},x_{n(k)+1}^{i(k)})\right)\\ =& \psi \left(G\right(F(x_{n(k)-1}^{{\sigma }_{i(k)}(1)},x_{n(k)-1}^{{\sigma }_{i(k)}(2)},\dots ,x_{n(k)-1}^{{\sigma }_{i(k)}(n)}),F(x_{m(k)-1}^{{\sigma }_{i(k)}(1)},x_{m(k)-1}^{{\sigma }_{i(k)}(2)},\dots ,x_{m(k)-1}^{{\sigma }_{i(k)}(n)}),\\ F_{\mathrm{\Upsilon }}^2(x_{n(k)-1}^{{\sigma }_{i(k)}(1)},x_{n(k)-1}^{{\sigma }_{i(k)}(2)},\dots ,x_{n(k)-1}^{{\sigma }_{i(k)}(n)})\left)\right)\\ \le & (\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)},F(x_{n(k)-1}^{{\sigma }_{i(k)}(1)},x_{n(k)-1}^{{\sigma }_{i(k)}(2)},\dots ,x_{n(k)-1}^{{\sigma }_{i(k)}(n)}))\right)\\ =& (\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)})\right)\\ =& (\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)})\right).\end{array}$$

(13)

Consider the sequence

$${\left\{\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)})\right\}}_{k\ge 1}.$$

(14)

If this sequence has a subsequence that converges to zero, then we can take limit when $k\to \mathrm{\infty }$ in (13) using this subsequence, so that we would have $0<\psi ({\epsilon }_0)\le \psi (0)-\phi (0)=0$, which is impossible since ${\epsilon }_0>0$. Therefore, the sequence (14) has no subsequence converging to zero. In this case, taking ${\epsilon }_0>0$ in Lemma 6, there exist $\delta \in ]0,{\epsilon }_0[$ and $k_0\in \mathbb{N}$ such that ${max}_{1\le j\le n}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)})\ge \delta$ for all $k\ge k_0$. It follows that, for all $k\ge k_0$, $-\phi ({max}_{1\le j\le n}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)}))\le -\phi (\delta )$. Thus, by (13) and (12),

$$\begin{array}{rl}0& <\psi ({\epsilon }_0)\le \psi \left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)})\right)-\phi \left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)})\right)\\ \le \psi \left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^{{\sigma }_{i(k)}(j)},x_{n(k)}^{{\sigma }_{i(k)}(j)},x_{m(k)-1}^{{\sigma }_{i(k)}(j)})\right)-\phi (\delta )\\ \le \psi \left(\underset{1\le j\le n}{max}G(x_{n(k)-1}^j,x_{n(k)}^j,x_{m(k)-1}^j)\right)-\phi (\delta )\\ \le \psi (\underset{1\le j\le n}{max}(G(x_{n(k)-1}^j,x_{n(k)}^j,x_{n(k)}^j)+G(x_{n(k)}^j,x_{n(k)+1}^j,x_{n(k)+1}^j))+{\epsilon }_0)-\phi (\delta ).\end{array}$$

(15)

Taking limit in (15) as $k\to \mathrm{\infty }$ and taking into account (9), we deduce that $0<\psi ({\epsilon }_0)\le \psi ({\epsilon }_0)-\phi (\delta )$, which is impossible. The previous reasoning proves that every sequence $\{x_m^i\}$ is G-Cauchy.

Corollary 25 guarantees that the sequence $\{{\mathbb{F}}_{\mathrm{\Upsilon }}^m({\mathsf{X}}_0)={\mathsf{X}}_m=(x_m^1,x_m^2,\dots ,x_m^n)\}$ is $G_n$-Cauchy. Since $(X^n,G_n)$ is complete (again by Corollary 25), there exists $\mathsf{Z}\in X^n$ such that $\{{\mathsf{X}}_m\}\stackrel{G_n}{\to }\mathsf{Z}$, that is, if $\mathsf{Z}=(z_1,z_2,\dots ,z_n)$ then

$$\left\{G(x_m^i,x_{m+1}^i,z_i)\right\}\to 0\text{for all }i.$$

(16)

Suppose that F is G-continuous. In this case, item 3 of Lemma 29 implies that ${\mathbb{F}}_{\mathrm{\Upsilon }}$ is $G_n$-continuous, so $\{{\mathsf{X}}_m\}\stackrel{G_n}{\to }\mathsf{Z}$ and $\{{\mathsf{X}}_{m+1}={\mathbb{F}}_{\mathrm{\Upsilon }}({\mathsf{X}}_m)\}\stackrel{G_n}{\to }{\mathbb{F}}_{\mathrm{\Upsilon }}(\mathsf{Z})$. By the unicity of the $G_n$-limit, ${\mathbb{F}}_{\mathrm{\Upsilon }}(\mathsf{Z})=\mathsf{Z}$, which means that Z is a ϒ-fixed point of F.

Suppose that $(X,G,\preccurlyeq )$ is regular. In this case, by Corollary 25, $(X^n,G_n,⊑)$ is also regular. Then, taking into account that $\{{\mathsf{X}}_m={\mathbb{F}}_{\mathrm{\Upsilon }}^m({\mathsf{X}}_0)\}$ is a ⊑-monotone non-decreasing sequence such that $\{{\mathsf{X}}_m\}\stackrel{G_n}{\to }\mathsf{Z}$, we deduce that ${\mathsf{X}}_m⊑\mathsf{Z}$ for all m. From Proposition 2, since $(x_m^1,x_m^2,\dots ,x_m^n)={\mathsf{X}}_m⊑\mathsf{Z}=(z_1,z_2,\dots ,z_n)$, then $(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)})$ and $(z_{{\sigma }_i(1)},z_{{\sigma }_i(2)},\dots ,z_{{\sigma }_i(n)})$ are ⊑-comparable for all i and all m. Notice that for all i and all m,

$$\begin{array}{rl}F(x_{m+1}^{{\sigma }_i(1)},x_{m+1}^{{\sigma }_i(2)},\dots ,x_{m+1}^{{\sigma }_i(n)})=& F(F(x_m^{{\sigma }_{{\sigma }_i(1)}(1)},x_m^{{\sigma }_{{\sigma }_i(1)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(1)}(n)}),\dots ,\\ F(x_m^{{\sigma }_{{\sigma }_i(n)}(1)},x_m^{{\sigma }_{{\sigma }_i(n)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(n)}(n)}))\\ =& F_{\mathrm{\Upsilon }}^2(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}).\end{array}$$

It follows from condition (7) and (8) that, for all i,

$$\begin{array}{r}\psi \left(G(F(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}),F(x_{m+1}^{{\sigma }_i(1)},x_{m+1}^{{\sigma }_i(2)},\dots ,x_{m+1}^{{\sigma }_i(n)}),F(z_{{\sigma }_i(1)},z_{{\sigma }_i(2)},\dots ,z_{{\sigma }_i(n)}))\right)\\ =\psi \left(G(F(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}),F(z_{{\sigma }_i(1)},z_{{\sigma }_i(2)},\dots ,z_{{\sigma }_i(n)}),F_{\mathrm{\Upsilon }}^2(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)}))\right)\\ \le (\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},z_{{\sigma }_i(j)},F(x_m^{{\sigma }_{{\sigma }_i(j)}(1)},x_m^{{\sigma }_{{\sigma }_i(j)}(2)},\dots ,x_m^{{\sigma }_{{\sigma }_i(j)}(n)}))\right)\\ =(\psi -\phi )\left(\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},z_{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)})\right)\le \psi \left(\underset{1\le j\le n}{max}G(x_m^{{\sigma }_i(j)},x_{m+1}^{{\sigma }_i(j)},z_{{\sigma }_i(j)})\right)\\ \le \psi \left(\underset{1\le j\le n}{max}G(x_m^j,x_{m+1}^j,z_j)\right).\end{array}$$

By (16) we deduce that

$$\left\{F(x_m^{{\sigma }_i(1)},x_m^{{\sigma }_i(2)},\dots ,x_m^{{\sigma }_i(n)})\right\}\to F(z_{{\sigma }_i(1)},z_{{\sigma }_i(2)},\dots ,z_{{\sigma }_i(n)})\text{for all }i,$$

which means that

$$\begin{array}{r}\{{\mathbb{F}}_{\mathrm{\Upsilon }}{\mathsf{X}}_m=(F(x_m^{{\sigma }_1(1)},x_m^{{\sigma }_1(2)},\dots ,x_m^{{\sigma }_1(n)}),\dots ,F(x_m^{{\sigma }_n(1)},x_m^{{\sigma }_n(2)},\dots ,x_m^{{\sigma }_n(n)}))\}\\ \stackrel{G_n}{\to }(F(z_{{\sigma }_1(1)},z_{{\sigma }_1(2)},\dots ,z_{{\sigma }_1(n)}),\dots ,F(z_{{\sigma }_n(1)},z_{{\sigma }_n(2)},\dots ,z_{{\sigma }_n(n)}))={\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{Z}.\end{array}$$

Since $\{{\mathbb{F}}_{\mathrm{\Upsilon }}{\mathsf{X}}_m={\mathsf{X}}_{m+1}\}\stackrel{G_n}{\to }\mathsf{Z}$, we conclude that ${\mathbb{F}}_{\mathrm{\Upsilon }}\mathsf{Z}=\mathsf{Z}$, that is, Z is a ϒ-fixed point of F. □

If we take $\psi (t)=t$ in Theorem 26, then we get the following results.

Corollary 32 Let $(X,G)$ be a complete $G^{\ast }$-metric space and let ≼ be a partial preorder on X. Let $\mathrm{\Upsilon }=({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n)$ be an n-tuple of mappings from $\{1,2,\dots ,n\}$ into itself verifying ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$ if $i\in \mathsf{A}$ and ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$ if $i\in \mathsf{B}$. Let $F:X^n\to X$ be a mapping verifying the mixed monotone property on X. Assume that there exists $\phi \in \mathrm{\Psi }$ such that

$$\begin{array}{c}G(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n),F_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n))\hfill \\ \le \underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))-\phi \left(\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))\right)\hfill \end{array}$$

for which $(x_1,x_2,\dots ,x_n),(y_1,y_2,\dots ,y_n)\in X^n$ are ⊑-comparable. Suppose either F is continuous or $(X,G,\preccurlyeq )$ is regular. If there exist $x_0^1,x_0^2,\dots ,x_0^n\in X$ verifying $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i, then F has, at least, one ϒ-fixed point.

If we take $\phi (t)=(1-k)t$ for all $t\ge 0$, with $k\in [0,1)$, in Corollary 32, then we derive the following result.

Corollary 33 Let $(X,G)$ be a complete $G^{\ast }$-metric space and let ≼ be a partial preorder on X. Let $\mathrm{\Upsilon }=({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n)$ be an n-tuple of mappings from $\{1,2,\dots ,n\}$ into itself verifying ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$ if $i\in \mathsf{A}$ and ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$ if $i\in \mathsf{B}$. Let $F:X^n\to X$ be a mapping verifying the mixed monotone property on X. Assume that there exists $k\in [0,1)$ such that

$$\begin{array}{r}G(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n),F_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n))\\ \le k\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))\end{array}$$

(17)

for which $(x_1,x_2,\dots ,x_n),(y_1,y_2,\dots ,y_n)\in X^n$ are ⊑-comparable. Suppose either F is continuous or $(X,G,\preccurlyeq )$ is regular. If there exist $x_0^1,x_0^2,\dots ,x_0^n\in X$ verifying $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i, then F has, at least, one ϒ-fixed point.

Example 34 Let $X=\{0,1,2,3,4\}$ and let G be the G-metric on X given, for all $x,y,z\in X$, by $G(x,y,z)=max(|x-y|,|x-z|,|y-z|)$. Then $(X,G)$ is complete and G generates the discrete topology on X. Consider on X the following partial order:

$$x,y\in X,x\preccurlyeq y\iff x=y\text{or}(x,y)=(0,2).$$

Define $F:X^n\to X$ by

$$F(x_1,x_2,\dots ,x_n)=\{\begin{array}{ll}0,& \text{if }{x}_1,x_2,\dots ,x_n\in \{0,1,2\},\\ 1,& \text{otherwise}.\end{array}$$

Then the following statements hold.

• 1. F is a G-continuous mapping.

• 2. If $y,z\in X$ verify $y\preccurlyeq z$, then either $y,z\in \{0,1,2\}$ or $y,z\in \{3,4\}$. In particular, $F(x_1,\dots ,x_{i-1},y,x_{i+1},\dots ,x_n)=F(x_1,\dots ,x_{i-1},y,x_{i+1},\dots ,x_n)$ and F has the mixed monotone property on X.

• 3. If $(x_1,x_2,\dots ,x_n),(y_1,y_2,\dots ,y_n)\in X^n$ are ⊑-comparable, then $F(x_1,x_2,\dots ,x_n)=F(y_1,y_2,\dots ,y_n)$. In particular, (17) holds for $k=1/2$.

  For simplicity, henceforth, suppose that n is even and let A (respectively, B) be the set of all odd (respectively, even) numbers in $\{1,2,\dots ,n\}$.

• 4. For a mapping $\sigma :{\mathrm{\Lambda }}_n\to {\mathrm{\Lambda }}_n$, we use the notation $\sigma \equiv (\sigma (1),\sigma (2),\dots ,\sigma (n))$ and consider

  $${\sigma }_i\equiv (i,i+1,\dots ,n-1,n,1,2,\dots ,i-1)\text{for all }i.$$

  Then ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}$ if i is odd and ${\sigma }_i\in {\mathrm{\Omega }}_{\mathsf{A},\mathsf{B}}^{\mathrm{\prime }}$ if i is even. Let $\mathrm{\Upsilon }=({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n)$.

• 5. Take $x_0^i=0$ if i is odd and $x_0^i=2$ if i is even. Then $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$ for all i.

Therefore, we can apply Corollary 33 to conclude that F has, at least, one ϒ-fixed point. To finish, we prove the previous statements.

If $\{x_m\}\stackrel{G}{\to }x$, then there exists $m_0\in \mathbb{N}$ such that $|x_m-x|=G(x,x,x_m)<1/2$ for all $m\ge m_0$. Since X is discrete, then $x_m=x$ for all $m\ge m_0$. This proves that ${\tau }_G$ is the discrete topology on X.

• 1. If $\{a_m^1\},\{a_m^2\},\dots ,\{a_m^n\}\subseteq X$ are n sequences such that $\{a_m^i\}\stackrel{G}{\to }{a}_i\in X$ for all i, then there exists $m_0\in \mathbb{N}$ such that $a_m^i=a_i$ for all $m\ge m_0$ and all i. Then $\{F(a_m^1,a_m^2,\dots ,a_m^n)\}\stackrel{G}{\to }F(a_1,a_2,\dots ,a_n)$ and F is G-continuous.

• 2. If $y,z\in X$ verify $y\preccurlyeq z$, the either $y=z$ (in this case, there is nothing to prove) or $(y,z)=(0,2)$. Then either $y,z\in \{0,1,2\}$ or $y,z\in \{3,4\}$. In particular,

  $$\begin{array}{rl}F(x_1,\dots ,x_{i-1},y,x_{i+1},\dots ,x_n)& =\left\{\begin{array}{ll}0& \text{if }{x}_1,\dots ,x_{i-1},y,x_{i+1},\dots ,x_n\in \{0,1,2\},\\ 1,& \text{otherwise}\end{array}\right\}\\ =F(x_1,\dots ,x_{i-1},z,x_{i+1},\dots ,x_n).\end{array}$$

  Hence F has the mixed monotone property on X.

• 3. Suppose that $(x_1,x_2,\dots ,x_n),(y_1,y_2,\dots ,y_n)\in X^n$ are ⊑-comparable, and we claim that $F(x_1,x_2,\dots ,x_n)=F(y_1,y_2,\dots ,y_n)$. Indeed, assume, for instance, that $x_i{\preccurlyeq }_i{y}_i$ for all i. By item 2, for all i, either $x_i,y_i\in \{0,1,2\}$ or $x_i,y_i\in \{3,4\}$. Then

  $$\begin{array}{rl}F(x_1,x_2,\dots ,x_n)& =\left\{\begin{array}{ll}0& \text{if }{x}_1,x_2,\dots ,x_n\in \{0,1,2\},\\ 1,& \text{otherwise}\end{array}\right\}\\ =\left\{\begin{array}{ll}0& \text{if }{y}_1,y_2,\dots ,y_n\in \{0,1,2\},\\ 1,& \text{otherwise}\end{array}\right\}\\ =F(y_1,y_2,\dots ,y_n).\end{array}$$

  If $x_i{\succcurlyeq }_i{y}_i$ for all i, the proof is similar. Next, we prove that (17) holds using $k=1/4$. If $(x_1,x_2,\dots ,x_n)\in X^n$, then $F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)})\in \{0,1\}\subset \{0,1,2\}$. Therefore

  $$\begin{array}{r}{F}_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n)\\ =F(F(x_{{\sigma }_1(1)},x_{{\sigma }_1(2)},\dots ,x_{{\sigma }_1(n)}),F(x_{{\sigma }_2(1)},x_{{\sigma }_2(2)},\dots ,x_{{\sigma }_2(n)}),\dots ,\\ F(x_{{\sigma }_n(1)},x_{{\sigma }_n(2)},\dots ,x_{{\sigma }_n(n)}))\\ =0.\end{array}$$

  Suppose that $(x_1,x_2,\dots ,x_n),(y_1,y_2,\dots ,y_n)\in X^n$ are ⊑-comparable. It follows that

  $$\begin{array}{r}G(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n),F_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n))\\ =max\left(\right|F(x_1,x_2,\dots ,x_n)-F(y_1,y_2,\dots ,y_n)|,\\ |F(x_1,x_2,\dots ,x_n)-0|,|F(y_1,y_2,\dots ,y_n)-0|)\\ =max(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n))\\ =\{\begin{array}{ll}0& \text{if }F(x_1,x_2,\dots ,x_n)=F(y_1,y_2,\dots ,y_n)=0,\\ 1,& \text{otherwise}.\end{array}\end{array}$$

  It is clear that (17) holds if the previous number is 0. On the contrary, suppose that

  $$G(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n),F_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n))=1.$$

  Then $F(x_1,x_2,\dots ,x_n)=1$ or $F(y_1,y_2,\dots ,y_n)=1$ (both cases are similar). Assume, for instance, that $F(x_1,x_2,\dots ,x_n)=1$. Then there exists $i_0\in \{1,2,\dots ,n\}$ such that $x_{i_0}\in \{3,4\}$. In particular

  $$|x_{i_0}-F(x_{{\sigma }_{i_0}(1)},x_{{\sigma }_{i_0}(2)},\dots ,x_{{\sigma }_{i_0}(n)})|\ge 3-1=2.$$

  Therefore

  $$\begin{array}{rl}\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)}))& \ge G(x_{i_0},y_{i_0},F(x_{{\sigma }_{i_0}(1)},x_{{\sigma }_{i_0}(2)},\dots ,x_{{\sigma }_{i_0}(n)}))\\ \ge |x_{i_0}-F(x_{{\sigma }_{i_0}(1)},x_{{\sigma }_{i_0}(2)},\dots ,x_{{\sigma }_{i_0}(n)})|\ge 2.\end{array}$$

  This means that

  $$\begin{array}{r}G(F(x_1,x_2,\dots ,x_n),F(y_1,y_2,\dots ,y_n),F_{\mathrm{\Upsilon }}^2(x_1,x_2,\dots ,x_n))\\ =1=\frac{1}{2}2\le \frac{1}{2}\underset{1\le i\le n}{max}G(x_i,y_i,F(x_{{\sigma }_i(1)},x_{{\sigma }_i(2)},\dots ,x_{{\sigma }_i(n)})).\end{array}$$

  Therefore, in this case, (17) also holds.

• 4. It is evident.

• 5. Since $x_0^i\in \{0,1,2\}$ for all i, then $F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})=0$ for all i. If i is odd, then $x_0^i=0{\preccurlyeq }_{i}0=F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$. If i is even, then $x_0^i=2\succcurlyeq 0=F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$, so $x_0^i{\preccurlyeq }_{i}F(x_0^{{\sigma }_i(1)},x_0^{{\sigma }_i(2)},\dots ,x_0^{{\sigma }_i(n)})$.