We will start with some concepts and results which are useful in our approach.
Definition 2.1
A multiply metric function \(\triangle(a_{1},a_{2},\ldots,a_{N})\) is a continuous N variable nonnegative real function with the domain
$$\bigl\{ (a_{1},a_{2},\ldots,a_{N})\in R^{N}: a_{i}\geq0, i\in\{1,2,3, \ldots,N\} \bigr\} $$
which satisfies the following conditions:

(1)
\(\triangle(a_{1},a_{2},\ldots,a_{N})\) is nondecreasing for each variable \(a_{i}\), \(i\in\{1,2,3, \ldots,N \}\);

(2)
\(\triangle(a_{1}+b_{1},a_{2}+b_{2},\ldots,a_{N}+b_{N})\leq \triangle(a_{1},a_{2},\ldots,a_{N})+\triangle(b_{1},b_{2},\ldots,b_{N})\);

(3)
\(\triangle(a,a,\ldots,a)=a\);

(4)
\(\triangle(a_{1},a_{2},\ldots,a_{N})\rightarrow0 \Leftrightarrow a_{i}\rightarrow0\), \(i\in\{1,2,3,\ldots, N \}\), for all \(a_{i},b_{i}, a \in\mathbb{R}\), \(i\in\{1,2,3, \ldots,N \}\), where \(\mathbb{R}\) denotes the set of all real numbers.
The following are some basic examples of multiply metric functions.
Example 2.2
(1) \(\triangle_{1}(a_{1},a_{2},\ldots,a_{N})=\frac{1}{N} \sum_{i=1}^{N}a_{i}\). (2) \(\triangle_{2}(a_{1},a_{2},\ldots,a_{N})=\frac{1}{h} \sum_{i=1}^{N}q_{i} a_{i}\), where \(q_{i}\in[0,1)\), \(i\in\{1,\ldots, N \}\), and \(0< h:= \sum_{i=1}^{N}q_{i}<1\).
Example 2.3
\(\triangle_{3}(a_{1},a_{2},\ldots,a_{N})=\sqrt{\frac{1}{N} \sum_{i=1}^{N}a_{i}^{2}}\).
Example 2.4
\(\triangle_{4}(a_{1},a_{2},\ldots,a_{N})=\max \{a_{1},a_{2},\ldots,a_{N}\}\).
An important concept is now presented.
Definition 2.5
Let \((X,d)\) be a metric space, \(T: X^{N}\rightarrow X\) be a N variable mapping, an element \(p\in X\) is called a multivariate fixed point (or a fixed point of order N; see [20]) of T if
In the following, we prove the following theorem, which generalizes the Banach contraction principle.
Theorem 2.6
Let
\((X,d)\)
be a complete metric space, \(T: X^{N}\rightarrow X\)
be an
N
variable mapping that satisfies the following condition:
$$d(Tx,Ty)\leq h \triangle\bigl(d(x_{1},y_{1}), d(x_{2},y_{2}), \ldots, d(x_{N},y_{N}) \bigr), \quad\forall x,y \in X^{N}, $$
where △ is a multiply metric function,
$$x=(x_{1},x_{2}, \ldots, x_{N}) \in X^{N},\qquad y=(y_{1},y_{2}, \ldots, y_{N}) \in X^{N}, $$
and
\(h \in(0,1)\)
is a constant.
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned} &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &p_{3}=(Tp_{2},Tp_{2},\ldots,Tp_{2}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned}$$
converges, in the multiply metric △, to
\((p,p,\ldots ,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).
Proof
We define a two variable function D on \(X^{N}\) by the following relation:
$$D\bigl((x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2},\ldots,y_{N})\bigr)=\triangle \bigl(d(x_{1},y_{1}),d(x_{2},y_{2}), \ldots,d(x_{N},y_{N}) \bigr) $$
for all \((x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2},\ldots,y_{N})\in X^{N}\). Next we show that D is a metric on \(X^{N}\). The following two conditions are obvious:

(i)
\(D((x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2},\ldots,y_{N}))=0 \Leftrightarrow (x_{1},x_{2},\ldots,x_{N})= (y_{1},y_{2},\ldots,y_{N})\);

(ii)
\(D( (y_{1},y_{2},\ldots,y_{N})), (x_{1},x_{2},\ldots ,x_{N})=D((x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2},\ldots,y_{N}))\), for all \((x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2},\ldots,y_{N})\in X^{N}\).
Next we prove the triangular inequality. For all
$$(x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2}, \ldots,y_{N}), (z_{1},z_{2},\ldots,z_{N}) \in X^{N}, $$
from the definition of △, we have
$$\begin{aligned} &D\bigl((x_{1},x_{2},\ldots,x_{N}), (y_{1},y_{2},\ldots,y_{N})\bigr) \\ &\quad=\triangle\bigl(d(x_{1},y_{1}),d(x_{2},y_{2}), \ldots,d(x_{N},y_{N}) \bigr) \\ &\quad\leq\triangle\bigl(d(x_{1},z_{1})+d(z_{1},y_{1}),d(x_{2},z_{2})+d(z_{2},y_{2}), \ldots,d(x_{N},z_{N})+d(z_{N},y_{N}) \bigr) \\ &\quad\leq\triangle\bigl(d(x_{1},z_{1}),d(x_{2},z_{2}), \ldots,d(x_{N},z_{N}) \bigr) +\triangle\bigl(d(z_{1},y_{1}),d(z_{2},y_{2}), \ldots,d(z_{N},y_{N}) \bigr) \\ &\quad=D\bigl((x_{1},x_{2},\ldots,x_{N}), (z_{1},z_{2},\ldots,z_{N})\bigr)+D \bigl((z_{1},z_{2},\ldots,z_{N}), (y_{1},y_{2},\ldots,y_{N})\bigr). \end{aligned}$$
Next we prove that \((X^{N},D)\) is a complete metric space. Let \(\{p_{n}\}\subset X^{N}\) be a Cauchy sequence, then we have
$$\lim_{n,m\rightarrow\infty} D(p_{n},p_{m})=\lim _{n,m\rightarrow\infty}\triangle \bigl(d(x_{1,n},x_{1,m}),d(x_{2,n},x_{2,m}), \ldots,d(x_{N,n},x_{N,m})\bigr)=0, $$
where
$$p_{n}=(x_{1,n},x_{2,n},x_{3,n}, \ldots,x_{N,n}),\qquad p_{m}=(x_{1,m},x_{2,m},x_{3,m}, \ldots,x_{N,m}). $$
From the definition of △, we have
$$\lim_{n,m\rightarrow\infty}d(x_{i,n},x_{i,m})=0, $$
for all \(i\in\{1,2,3,\ldots,N \}\). Hence each \(\{x_{i,n}\} \) (\(i\in\{1,2,3, \ldots,N \}\)) is a Cauchy sequence. Since \((X,d)\) is a complete metric space, there exist \(x_{1},x_{2},x_{3},\ldots,x_{N} \in X\) such that \(\lim_{n\rightarrow\infty}d(x_{i,n}, x_{i})=0\) for all \(i\in\{1,2,3,\ldots,N \}\). Therefore
$$\lim_{n\rightarrow\infty}D(p_{n},x)=0, $$
where
$$x=(x_{1},x_{2},x_{3},\ldots,x_{N})\in X^{N}, $$
which implies that \((X^{N},D)\) is a complete metric space.
We define a mapping \(T^{*}:X^{N}\rightarrow X^{N}\) by the following relation:
$$T^{*}(x_{1},x_{2},\ldots,x_{N})= \bigl(T(x_{1},x_{2},\ldots,x_{N}),T(x_{1},x_{2}, \ldots,x_{N}),\ldots,T(x_{1},x_{2}, \ldots,x_{N})\bigr), $$
for all \((x_{1},x_{2},\ldots,x_{N})\in X^{N}\). Next we prove that \(T^{*}\) is a contraction mapping from \((X^{N},D)\) into itself. Observe that, for any
$$x=(x_{1},x_{2},\ldots,x_{N}), y=(y_{1},y_{2}, \ldots,y_{N})\in X^{N}, $$
we have
$$\begin{aligned} D\bigl(T^{*}x,T^{*}y\bigr)&=\triangle\bigl(d(Tx,Ty),d(Tx,Ty),\ldots,d(Tx,Ty)\bigr) \\ &=d(Tx,Ty) \\ &\leq h \triangle\bigl(d(x_{1},y_{1}),d(x_{2},y_{2}), \ldots,d(x_{N},y_{N})\bigr) \\ &=h D(x,y). \end{aligned}$$
By the Banach contraction mapping principle, there exists a unique element \(u \in X^{N}\) such that \(u=T^{*}u=(Tu,Tu,\ldots,Tu)\) and, for any \(u_{0}=(x_{1},x_{2},\ldots,x_{N})\in X^{N}\), the iterative sequence \(u_{n+1}=T^{*}u_{n}\) converges to u. That is,
$$\begin{aligned} &u_{1}=(Tu_{0},Tu_{0},\ldots,Tu_{0}), \\ &u_{2}=(Tu_{1},Tu_{1},\ldots,Tu_{1}), \\ &u_{3}=(Tu_{2},Tu_{2},\ldots,Tu_{2}), \\ &\cdots \\ & u_{n+1}=(Tu_{n},Tu_{n},\ldots,Tu_{n}), \\ &\cdots \end{aligned}$$
converges to \(u\in X^{N}\). By the structure of \(\{u_{n}\}\), we know that there exists a unique element \(p\in X\) such that \(u=(p,p,\ldots, p)\) and hence the iterative sequence \(\{Tu_{n}\}\) converges to \(p \in X\). By
$$T^{*}u=u=(p,p,\ldots,p),\qquad Tu=T(p,p,\ldots,p), \qquad T^{*}u=(Tu,Tu,\ldots,Tu), $$
we obtain \(p=T(p,p,\ldots,p)\), that is, p is the unique multivariate fixed point of T. This completes the proof. □
Notice that taking \(N=1\), \(\triangle(a)=a\) in Theorem 2.6, we obtain Banach’s contraction principle.
Some other consequences of the above general result are the following corollaries.
Corollary 2.7
Let
\((X,d)\)
be a complete metric space, \(T: X^{N}\rightarrow X\)
be a
N
variables mapping satisfying the following condition:
$$d(Tx,Ty)\leq\frac{h}{N}\sum_{i=1}^{N}d(x_{i},y_{i}), \quad 0< h< 1, $$
where
$$x=(x_{1},x_{2}, \ldots, x_{N}) \in X^{N}, \qquad y=(y_{1},y_{2}, \ldots, y_{N}) \in X^{N}. $$
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned} &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &p_{3}=(Tp_{2},Tp_{2},\ldots,Tp_{2}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned}$$
converges, in the multiply metric, to
\((p,p,\ldots,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).
Notice that the above corollary is related to the wellknown Prešić’s fixed point theorem (see [21]).
Prešić’s theorem
Let
\((X,d)\)
be a complete metric space, N
be a given natural number, and
\(T:X^{N}\to X\)
be an operator, such that, for all
\(x_{1}, \ldots, x_{N}, x_{N+1}\in X\), we have
$$d\bigl(T(x_{1},x_{2}, \ldots x_{N}),T(x_{2}, \ldots, x_{N}, x_{N+1})\bigr)\le q_{1} d(x_{1},x_{2})+ \cdots+q_{N} d(x_{N},x_{N+1}), $$
where
\(q_{1}, \ldots, q_{N}\in\mathbb{R}_{+}\)
with
\(q_{1}+ \cdots+ q_{N}<1\).
Then there exists a unique multivariate fixed point
\(p\in X\)
and
p
is the limit of the sequence
\((x_{n})\)
given by
$$x_{n+k}:=T(x_{n}, \ldots, x_{n+k1}), \quad\textit{for } n\ge1, $$
independently of the initial
N
values.
Choosing \(\Delta:=\Delta_{2}\), \(h:= \sum_{i=1}^{N}q_{i}\), and \(x=(x_{1},x_{2},\ldots,x_{N}), y=(x_{2},x_{3},\ldots ,x_{N+1})\in X^{N}\), the contraction condition given in Theorem 2.6 leads to Prešić’s contraction type condition.
Corollary 2.8
Let
\((X,d)\)
be a complete metric space and
\(T: X^{N}\rightarrow X\)
be a
N
variable mapping which satisfies the following condition:
$$d(Tx,Ty)\leq h \sqrt{\frac{1}{N}\sum _{i=1}^{N}d(x_{i},y_{i})^{2}}, \quad 0< h< 1, $$
where
$$x=(x_{1},x_{2}, \ldots, x_{N}) \in X^{N}, \qquad y=(y_{1},y_{2}, \ldots, y_{N}) \in X^{N}. $$
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned} &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &p_{3}=(Tp_{2},Tp_{2},\ldots,Tp_{2}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned}$$
converges, in the multiply metric, to
\((p,p,\ldots,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).
Corollary 2.9
Let
\((X,d)\)
be a complete metric space and
\(T: X^{N}\rightarrow X\)
be a
N
variable mapping which satisfies the following condition:
$$d(Tx,Ty)\leq h \max \bigl\{ d(x_{1},y_{1}),d(x_{2},y_{2}), \ldots,d(x_{N},y_{N})\bigr\} ,\quad 0< h< 1, $$
where
$$x=(x_{1},x_{2}, \ldots, x_{N}) \in X^{N}, \qquad y=(y_{1},y_{2}, \ldots, y_{N}) \in X^{N}. $$
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned} &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned}$$
converges, in the multiply metric, to
\((p,p,\ldots,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).
Notice also here that the above corollary is related to a multivariate fixed point theorem of Ćirić and Prešić (see [22]), which reads as follows.
ĆirićPrešić’s theorem
Let
\((X,d)\)
be a complete metric space, N
be a given natural number, and
\(T:X^{N}\to X\)
be an operator, such that, for all
\(x_{1}, \ldots, x_{N}, x_{N+1}\in X\), we have
$$d\bigl(T(x_{1},x_{2}, \ldots x_{N}),T(x_{2}, \ldots, x_{N}, x_{N+1})\bigr)\le h \max\bigl\{ d(x_{1},x_{2}), d(x_{2},x_{3}), \ldots, d(x_{N},x_{N+1}) \bigr\} , $$
where
\(0< h<1\).
Then there exists a multivariate fixed point
\(p\in X\)
and
p
is the limit of the sequence
\((x_{n})\)
given by
$$x_{n+k}:=T(x_{n}, \ldots, x_{n+k1}), \quad\textit{for } n\ge1, $$
independently of the initial
N
values.
If in addition, we suppose that on the diagonal
\(\operatorname{Diag}\subset X^{N}\)
we have
$$d\bigl(T(u,\ldots, u), T(v,\ldots, v)\bigr)< d(u,v), \quad\textit{for all } u,v\in X \textit{ with } u\ne v, $$
then the multivariate fixed point is unique.
Choosing \(\Delta:=\Delta_{4}\), \(h\in(0,1)\), and \(x=(x_{1},x_{2},\ldots,x_{N}), y=(x_{2},x_{3},\ldots ,x_{N+1})\in X^{N}\), the contraction condition given in Theorem 2.6 leads to the above ĆirićPrešić’s contraction type condition.
It is worth to mention that the above results are in connection with a very interesting multivariate fixed point principle proved by Tasković in [23]. More precisely, Tasković’s result is as follows.
Tasković’s theorem
Let
\((X,d)\)
be a complete metric space, N
be a given natural number, \(f:\mathbb{R}^{N}\to\mathbb{R}\)
be a continuous, increasing, and semihomogeneous function (in the sense that
\(f(\lambda a_{1}, \ldots, \lambda a_{N})\le\lambda f(a_{1}, \ldots, a_{N})\), for any
\(\lambda, a_{1}, \ldots a_{N}\in\mathbb{R}\)) and let
\(T:X^{k}\to X\)
be an operator, such that, for all
\(x=(x_{1}, \ldots , x_{N}), y=(y_{1}, \ldots, y_{N})\in X^{k}\), we have
$$d\bigl(T(x),T(y)\bigr)\le\biglf\bigl(a_{1} d(x_{1},y_{1}), \ldots, a_{N} d(x_{N},y_{N})\bigr)\bigr, $$
where
\(a_{1}, \ldots, a_{N}\in\mathbb{R}_{+}\)
with
\(f(a_{1}, \ldots, a_{N})<1\).
Then there exists a unique multivariate fixed point
\(p\in X\)
and
p
is the limit of the sequence
\((x_{n})\)
given by
$$x_{n+k}:=T(x_{n}, \ldots, x_{n+k1}), \quad\textit{for } n\ge1, $$
independently of the initial
N
values.
Notice here that \(\triangle(a_{1},\ldots, a_{n}):=f(a_{1},\ldots, a_{n})\) satisfies part of the axioms of the multiply metric. More connections with the above mentioned results will be given in a forthcoming paper.
The following result is another multivariate fixed point theorem for a class of generalized contraction mappings related to the SY theorem. The proof of it can be obtained by Theorem SY, in the same way as was used in the proof of Theorem 2.6.
Theorem 2.10
Let
\((X,d)\)
be a complete metric space and
\(T: X^{N}\rightarrow X\)
be a
N
variable mapping which satisfies the following condition:
$$\psi\bigl( d(Tx,Ty)\bigr)\leq\phi\bigl( \triangle\bigl(d(x_{1},y_{1}), d(x_{2},y_{2}), \ldots, d(x_{N},y_{N}) \bigr)\bigr), $$
where △ is a multiply metric function,
$$x=(x_{1},x_{2}, \ldots, x_{N}) \in X^{N},\qquad y=(y_{1},y_{2}, \ldots, y_{N}) \in X^{N}, $$
and
\(\psi, \phi: [0, +\infty) \rightarrow[0, +\infty)\)
are two functions satisfying the conditions:
$$\begin{aligned}& (1)\quad \psi(a)\leq\phi(b) \quad\Rightarrow\quad a \leq b; \\& (2)\quad \textstyle\begin{cases} \psi(a_{n})\leq\phi(b_{n}) \\ a_{n}\rightarrow\varepsilon,\qquad b_{n}\rightarrow\varepsilon \end{cases}\displaystyle \quad\Rightarrow\quad\varepsilon=0. \end{aligned}$$
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned} &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned}$$
converges, in the multiply metric, to
\((p,p,\ldots,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).
In [7], Su and Yao also gave some examples of functions \(\psi(t)\), \(\phi(t)\). Here we recall some of them.
Example 2.11
([7])
The following functions satisfy the conditions (1) and (2) of Theorem 2.10.
$$(\mathrm{a}) \quad \textstyle\begin{cases} \psi_{1}(t)=t, \\ \phi_{1}(t)=\alpha t, \end{cases} $$
where \(0<\alpha<1\) is a constant.
$$\begin{aligned}& (\mathrm{b})\quad \textstyle\begin{cases} \psi_{2}(t)=t^{2}, \\ \phi_{2}(t)=\ln(t^{2}+1), \end{cases}\displaystyle \\& (\mathrm{c})\quad \textstyle\begin{cases} \psi_{3}(t)=t, \\ \phi_{3}(t)= \textstyle\begin{cases} t^{2}, & 0\leq t\leq\frac{1}{2},\\ t\frac{3}{8},& \frac{1}{2}< t< +\infty, \end{cases}\displaystyle \end{cases}\displaystyle \\& (\mathrm{d}) \quad \textstyle\begin{cases} \psi_{4}(t)= \textstyle\begin{cases} t, & 0\leq t\leq1,\\ t\frac{1}{2}, & 1< t< +\infty, \end{cases}\displaystyle \\ \phi_{4}(t)= \textstyle\begin{cases} \frac{t}{2}, & 0\leq t\leq1,\\ t\frac{4}{5}, & 1< t< +\infty, \end{cases}\displaystyle \end{cases}\displaystyle \\& (\mathrm{e}) \quad \textstyle\begin{cases} \psi_{5}(t)= \textstyle\begin{cases} t, & 0\leq t\leq1,\\ \alpha t^{2}, & 1\leq t< +\infty, \end{cases}\displaystyle \\ \phi_{5}(t)= \textstyle\begin{cases} t^{2}, & 0\leq t< 1,\\ \beta t, & 1< t< +\infty, \end{cases}\displaystyle \end{cases}\displaystyle \end{aligned}$$
where \(0<\beta< \alpha\) are constants.
For example, if we choose \(\psi_{5}(t)\), \(\phi_{5}(t)\) in Theorem 2.10, then we can get the following result.
Theorem 2.12
Let
\((X,d)\)
be a complete metric space. Let
\(T: X^{N}\rightarrow X\)
be a
N
variables mapping such that
$$\begin{aligned}& 0\leq d(Tx,Ty)< 1 \quad\Rightarrow\quad d(Tx,Ty)\leq\bigl(\triangle \bigl(d(x_{1},y_{1}), d(x_{2},y_{2}), \ldots, d(x_{N},y_{N})\bigr)\bigr)^{2}, \\& d(Tx,Ty)\geq1 \quad\Rightarrow\quad\alpha\bigl(d(Tx,Ty)\bigr)^{2}\leq \beta\triangle \bigl(d(x_{1},y_{1}), d(x_{2},y_{2}), \ldots, d(x_{N},y_{N})\bigr), \end{aligned}$$
for any
\(x=(x_{1},x_{2},x_{3}, \ldots,x_{N}), y=(y_{1},y_{2},y_{3}, \ldots,y_{N}) \in X^{N}\).
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned} &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned}$$
converges, in the multiply metric, to
\((p,p,\ldots,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).
Using the following notions it is easy to prove another consequence of our main results.
Remark 2.13
Let \(\psi, \phi: [0, +\infty) \rightarrow[0, +\infty)\) be two functions satisfying the conditions:

(i)
\(\psi(0)=\phi(0)\);

(ii)
\(\psi(t)>\phi(t)\), \(\forall t>0\);

(iii)
ψ is lower semicontinuous and ϕ is upper semicontinuous.
Then \(\psi(t)\), \(\phi(t)\) satisfy the above mentioned conditions (1) and (2).
Corollary 2.14
Let
\((X,d)\)
be a complete metric space. Let
\(T: X^{N}\rightarrow X\)
be a
N
variable mapping such that, for any
\(x=(x_{1},x_{2},x_{3}, \ldots,x_{N}), y=(y_{1},y_{2},y_{3}, \ldots,y_{N}) \in X^{N}\), we have
$$\psi\bigl(d(Tx,Ty)\bigr)\leq\phi\bigl(\triangle\bigl(d(x_{1},y_{1}), d(x_{2},y_{2}), \ldots, d(x_{N},y_{N}) \bigr)\bigr), $$
where
\(\psi, \phi: [0, +\infty) \rightarrow[0, +\infty)\)
are two functions with the conditions (i), (ii), and (iii).
Then
T
has a unique multivariate fixed point
\(p\in X\)
and, for any
\(p_{0} \in X^{N}\), the iterative sequence
\(\{p_{n}\}\subset X^{N}\)
defined by
$$\begin{aligned}[b] &p_{1}=(Tp_{0},Tp_{0},\ldots,Tp_{0}), \\ &p_{2}=(Tp_{1},Tp_{1},\ldots,Tp_{1}), \\ &\cdots \\ & p_{n+1}=(Tp_{n},Tp_{n},\ldots,Tp_{n}), \\ &\cdots \end{aligned} $$
converges, in the multiply metric, to
\((p,p,\ldots,p)\in X^{N}\)
and the iterative sequence
\(\{Tp_{n}\}\subset X\)
converges, with respect to
d, to
\(p \in X\).