From now on, let us consider that \((X,d )\) is a complete metric space and \(k\in \mathbb{N}\).
Definition 2.1
(Extended \(M_{k}\) function)
For any \((u_{i})_{i=1}^{\infty },(v_{i})_{i=1}^{\infty }\in \prod_{i=1}^{ \infty } X\) and \(T:\prod_{i=1}^{\infty } X\to X\), we define the extended function \(M_{k}:\prod_{l=1}^{\infty }X \times \prod_{l=1}^{\infty }X\rightarrow X\) as follows:
$$\begin{aligned}& M_{k} \bigl((u_{i})_{i=1}^{\infty },(v_{i})_{i=1}^{\infty } \bigr)\\& \quad = \textstyle\begin{cases} {\max \lbrace \sup_{l\geqslant k} d (u_{l},v_{l} ),\sup_{l\geqslant k} d (u_{l},T(u_{i,\hat{l}})_{i=1}^{ \infty } ), \sup_{l\geqslant k} d (v_{l},T(v_{i, \hat{l}})_{i=1}^{\infty } ) \rbrace }, \\ \quad {\text{ if all supremum exists}} \\ {\max \lbrace d (u_{k},v_{k} ), d (u_{k},T(u_{i, \hat{k}})_{i=1}^{\infty } ), d (v_{k},T(v_{i,\hat{k}})_{i=1}^{ \infty } ) \rbrace }, \\ \quad {\text{ if one fails to exist}} \end{cases}\displaystyle \end{aligned}$$
Example 2.1
Consider \(X=[0,1]\). For \(k=1\), take the operator \(T:\prod_{i=1}^{\infty }X\to X\) defined by \(T((u_{i})_{i=1}^{\infty })=u_{k}=u_{1}\) and \(M_{k} ((u_{i})_{i=1}^{\infty },(v_{i})_{i=1}^{\infty } )= |u_{k}-v_{k}|=|u_{1}-v_{1}|\).
Definition 2.2
(\(H_{k}\) contraction)
An operator \(T:\prod_{l=1}^{\infty }X\rightarrow X\) is called an \(H_{k}\) contraction if and only if it satisfies the following inequality:
$$ d \bigl(T \bigl((u_{i,\widehat{k}})_{i=1}^{\infty } \bigr), T \bigl((v_{i, \widehat{k}})_{i=1}^{\infty } \bigr) \bigr) \leqslant \beta \bigl(M_{k} \bigl((u_{i,\widehat{k}})_{i=1}^{\infty },(v_{i,\widehat{k}})_{i=1}^{ \infty } \bigr) \bigr)M_{k} \bigl((u_{i,\widehat{k}})_{i=1}^{\infty },(v_{i, \widehat{k}})_{i=1}^{\infty } \bigr) $$
(1)
for all \(u_{1}, u_{2}, \ldots , u_{k}, v_{1}, v_{2}, \ldots , v_{k}\in X\), where \(\beta \in \mathbb{G}\).
Example 2.2
Consider \(X=[0,1]\). The operator \(T:\prod_{i=1}^{\infty }X\to X\) defined by \(T((x_{i})_{i=1}^{\infty }):=cx_{k}\) is an \(H_{k}\) contraction, where \(k\in \mathbb{N}\) is any fixed number and \(c\in (0,1)\).
Definition 2.3
(Kannan–Geraghty \(H_{k}\) contraction)
An operator \(T:\prod_{l=1}^{k} X\rightarrow X\) is called a Kannan–Geraghty contraction of dimension k if and only if it satisfies the following inequality:
$$ d \bigl(T \bigl((u_{i})_{i=1}^{k} \bigr), T \bigl((v_{i})_{i=1}^{k} \bigr) \bigr) \leqslant \frac{\beta (d(u_{k},v_{k}))}{2} \bigl(d \bigl(T \bigl((u_{i})_{i=1}^{k} \bigr), u_{k} \bigr)+ d \bigl(T \bigl((v_{i})_{i=1}^{k} \bigr), v_{k} \bigr) \bigr) $$
for all \(u_{1}, u_{2},\ldots ,u_{k}, v_{1}, v_{2}, \ldots , v_{k}\in X\), where \(\beta \in \mathbb{G}\).
Definition 2.4
(Extended Kannan–Geraghty \(H_{k}\) contraction)
An operator \(T:\prod_{l=1}^{\infty } X\rightarrow X\) is called an extended Kannan–Geraghty \(H_{k}\) contraction if and only if it satisfies the following:
$$ d \bigl(T \bigl((u_{i,\widehat{k}})_{i=1}^{\infty } \bigr),T \bigl((v_{i, \widehat{k}})_{i=1}^{\infty } \bigr) \bigr) \leqslant \frac{\beta (d(u_{k},v_{k}) )}{2} \bigl(d \bigl(T \bigl((u_{i, \widehat{k}})_{i=1}^{\infty } \bigr),u_{k} \bigr)+d \bigl(T \bigl((v_{i, \widehat{k}})_{i=1}^{\infty } \bigr),v_{k} \bigr) \bigr) $$
for all \(u_{1}, u_{2},\ldots , u_{k}, v_{1}, v_{2}, \ldots , v_{k}\in X\), where \(\beta \in \mathbb{G}\).
Definition 2.5
(Fisher–Geraghty \(H_{k}\) contraction)
An operator \(T:\prod_{l=1}^{k} X\rightarrow X\) is called a Fisher–Geraghty \(H_{k}\) contraction of dimension k if and only if it satisfies the following:
$$ d \bigl(T \bigl((u_{i})_{i=1}^{k} \bigr), T \bigl((v_{i})_{i=1}^{k} \bigr) \bigr) \leqslant \frac{\beta (d(u_{k},v_{k}))}{2} \bigl(d \bigl(T \bigl((u_{i})_{i=1}^{k} \bigr),v_{k} \bigr)+d \bigl(T \bigl((v_{i})_{i=1}^{k} \bigr),u_{k} \bigr) \bigr) $$
for all \(u_{1}, u_{2}, \ldots , u_{k}, v_{1}, v_{2}, \ldots , v_{k}\in X\), where \(\beta \in \mathbb{G}\).
Definition 2.6
(Extended Fisher–Geraghty \(H_{k}\) contraction)
An operator \(T:\prod_{l=1}^{\infty } X\rightarrow X\) is called an extended Kannan–Geraghty \(H_{k}\) contraction if and only if it satisfies the following:
$$ d \bigl(T \bigl((u_{i,\widehat{k}})_{i=1}^{\infty } \bigr),T \bigl((v_{i, \widehat{k}})_{i=1}^{\infty } \bigr) \bigr) \leqslant \frac{\beta (d(u_{k},v_{k}) )}{2} \bigl(d \bigl(T \bigl((u_{i, \widehat{k}})_{i=1}^{\infty } \bigr),v_{k} \bigr)+d \bigl(T \bigl((v_{i, \widehat{k}})_{i=1}^{\infty } \bigr),u_{k} \bigr) \bigr) $$
for all \(u_{1}, u_{2}, \ldots , u_{k}, v_{1}, v_{2}, \ldots , v_{k}\in X\), where \(\beta \in \mathbb{G}\).
Example 2.3
As an example for Definition 2.3 and Definition 2.5, we can consider, for \(X=[0,1]\), the operator \(T:\prod_{i=1}^{k}X\to X\) defined by \(T((x_{i})_{i=1}^{k})=cx_{k}\) for some \(c\in (0,\frac{1}{2})\).
Example 2.4
As an example for Definition 2.4 and Definition 2.6, for \(X=[0,1]\), take the operator \(T:\prod_{i=1}^{\infty } X\to X\) defined by \(T((x_{i})_{i=1}^{\infty })=cx_{k}\) for some \(c\in (0,\frac{1}{2})\).
The theory of Picard operators was developed by Rus and is used a lot in proving the existence and uniqueness of a solution of different types of integral or differential equations. For more details, see [39, 40].
Definition 2.7
(Picard operator)
Let \((X,d)\) be a metric space. An operator \(T : X \to X\) is a Picard operator if there exists \(x^{*}\in X\) such that \(FixT = \{x^{*}\}\) and the sequence \((T^{n}(x_{0}))_{n\in \mathbb{N}}\) converges to \(x^{*}\) for all \(x_{0}\in X\).
Inspired by the Picard operator definition, we extend this notion for the k-dimensional and infinite dimensional cases as follows.
Definition 2.8
(k-Picard sequence with respect to the operator T)
Let \(T:\prod_{i=1}^{k}X\to X\) be any operator, and let us choose \(x_{1}, x_{2}, \ldots , x_{k}\in X\). The k-Picard sequence with respect to the operator T based on the base point set \(\lbrace x_{1}, x_{2}, \ldots , x_{k}\rbrace \) is defined as \(x_{n+k}:=T((x_{n+i-1})_{i=1}^{k})\) for all \(n\geq k\).
Example 2.5
If we fix \(k=1\), then the base point set is singleton and the 1-Picard sequence with respect to T based on \(\lbrace x_{0}\rbrace \) is basically the Picard sequence of T based on the base point \(\lbrace x_{0}\rbrace \) defined by \(x_{n}:= T(x_{n-1})\) for all \(n\geqslant 1\) and some \(x_{0}\in X\).
Now, we define the following notions.
Definition 2.9
(Infinite k-Picard sequence with respect to the operator T)
Let \(T:\prod_{i=1}^{\infty }X\to X\) be any operator, and let us choose \(x_{1}, x_{2}, \ldots , x_{k},\ldots \in X\). The k-Picard sequence with respect to the operator T based on the base point set \(\lbrace x_{1}, x_{2}, \ldots , x_{k}, \ldots \rbrace \) is defined as \(x_{n+k}:=T((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty })\) for all \(n\geq k\).
Example 2.6
If we fix \(k=1\), then the base point set is singleton and the infinite 1- Picard sequence with respect to T based on the base point \(\lbrace x_{0}\rbrace \) is basically the sequence defined by \(x_{n}:= T((x_{n-1})_{n=1}^{\infty }) \ \forall n\geqslant 1\) for some \(x_{0}\in X\).
Let us give our first main fixed point result, which is a generalization of the Banach contraction principle with respect to the infinite-dimensional notion introduced in our paper.
Theorem 2.1
Let \((X,d )\) be a complete metric space. \(T:\prod_{l=1}^{\infty }X \rightarrow X\) is an \(H_{k}\) contraction for some \(k\in \mathbb{N}\). Then there exists \(u\in X\) such that \(T((u)_{i=1}^{\infty })=u\) and the infinite k-Picard sequence for T converges to u.
Proof
Let \(x_{1},x_{2}\ldots ,x_{k}\in X\). Then an infinite k-Picard sequence is defined as follows: \(\forall n\in \mathbb{N} \) we define
$$ x_{n+k}:=T \bigl((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty } \bigr). $$
(2)
We claim that \(\lbrace d (x_{n+k},x_{n+k+1} ) \rbrace _{n \in \mathbb{N}}\) is convergent. From Definition 2.2, one writes
$$\begin{aligned}& d (x_{n+k+1},x_{n+k+2} )=d \bigl(T \bigl((x_{n+i, \widehat{n+k}})_{i=1}^{\infty } \bigr), T \bigl((x_{n+i+1, \widehat{n+1+k}})_{i=1}^{\infty } \bigr) \bigr) \\& \quad \leqslant \beta \bigl(M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{ \infty }, (x_{n+1+i,\widehat{n+1+k}})_{i=1}^{\infty } \bigr) \bigr)M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty }, (x_{n+1+i, \widehat{n+1+k}})_{i=1}^{\infty } \bigr). \end{aligned}$$
Let
$$\begin{aligned}& M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty },(x_{n+i+1, \widehat{n+1+k}})_{i=1}^{\infty } \bigr) \\& \quad =\max \Bigl\lbrace \sup_{l\geqslant k} d (u_{l},v_{l} ),\sup_{l\geqslant k} d \bigl(u_{l},T(u_{i, \widehat{l}})_{i=1}^{\infty } \bigr),\sup_{l\geqslant k} d \bigl(v_{l},T(v_{i,\widehat{l}})_{i=1}^{\infty } \bigr) \Bigr\rbrace , \end{aligned}$$
where
$$\begin{aligned}& u_{l}:=x_{n+l},\quad \forall 1\leqslant l\leqslant k, \end{aligned}$$
(3)
$$\begin{aligned}& u_{l}:=x_{n+k},\quad \forall l> k, \end{aligned}$$
(4)
$$\begin{aligned}& v_{l}:=x_{n+l+1},\quad \forall 1\leqslant l\leqslant k, \end{aligned}$$
(5)
$$\begin{aligned}& u_{l}:=x_{n+k+1},\quad \forall l> k. \end{aligned}$$
(6)
But
$$\begin{aligned}& \sup_{l\geqslant k} d \bigl((u_{l},v_{l} )=d \bigl((x_{n+k},x_{n+k+1} ), \\& \sup_{l\geqslant k} d \bigl( \bigl(u_{l},T \bigl((u_{i,\widehat{l}})_{i=1}^{ \infty } \bigr) \bigr)=d ( \bigl(x_{n+k},T \bigl((x_{n+i, \widehat{n+k}})_{i=1}^{\infty } \bigr) \bigr), \\& \sup_{l\geqslant k} d \bigl( \bigl(v_{l},T \bigl((v_{i,\widehat{l}})_{i=1}^{ \infty } \bigr) \bigr)=d \bigl( \bigl(x_{n+k+1},T \bigl((x_{n+i+1, \widehat{n+k+1}})_{i=1}^{\infty } \bigr) \bigr). \end{aligned}$$
We have
$$\begin{aligned}& M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty },(x_{n+i+1, \widehat{n+k+1}})_{i=1}^{\infty } \bigr) \\& \quad =\max \bigl\lbrace d \bigl((x_{n+k},x_{n+k+1} ), d \bigl((x_{n+k},T \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty } \bigr) \bigr),d \bigl((x_{n+k+1},T \bigl((x_{n+1+i,\widehat{n+k+1}})_{i=1}^{\infty } \bigr) \bigr) \bigl\rbrace . \end{aligned}$$
That is,
$$\begin{aligned} &M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty },(x_{n+1+i, \widehat{n+k+1}})_{i=1}^{\infty } \bigr) \\ &\quad =\max \bigl\lbrace d ((x_{n+k},x_{n+k+1} ), d ((x_{n+k},x_{n+k+1} ),d ((x_{n+k+1},x_{n+k+2} ) \bigr\rbrace \\ &\quad =\max \bigl\lbrace d ((d_{n+k},d_{n+k+1} ),d ((x_{n+k+1},x_{n+k+2} ) \bigr\rbrace . \end{aligned}$$
Without loss of generality, assume that \(d ((x_{n+k+1},x_{n+k+2} )>0\) for each k.
If, for some k, we have \(M_{k} ((x_{n+i,\widehat{n+k}})_{i=1}^{\infty },(x_{n+i+1, \widehat{n+k+1}})_{i=1}^{\infty } )=d ((x_{n+k+1},x_{n+k+2} )\), then, from Definition 2.2, we get
$$\begin{aligned} 0&< d ((x_{n+k+1},x_{n+k+2} )=d ( \bigl(T \bigl((x_{n+i, \widehat{n+k}})_{i=1}^{\infty } \bigr),T \bigl((x_{n+i+1, \widehat{n+k+1}})_{i=1}^{\infty } \bigr) \bigr) \\ &\leqslant \beta \bigl(M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{ \infty },(x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty } \bigr) \bigr) M_{k} \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty },(x_{n+i+1, \widehat{n+k+1}})_{i=1}^{\infty } \bigr) \\ &=\beta (d \bigl((x_{n+k+1},x_{n+k+2} ) \bigr)d ((x_{n+k+1},x_{n+k+2} ) \\ &< d ((x_{n+k+1},x_{n+k+2} ). \end{aligned}$$
Then we get a contradiction. Hence, for each k, \(M_{k} ((x_{n+i,\widehat{n+k}})_{i=1}^{\infty },(x_{n+i+1, \widehat{n+k+1}})_{i=1}^{\infty } )=d ((x_{n+k},x_{n+k+1} )\). One writes
$$ d ((x_{n+k+1},x_{n+k+2} )\leqslant \beta (d \bigl((x_{n+k},x_{n+k+1} ) \bigr)d ((x_{n+k},x_{n+k+1} )< d ((x_{n+k},x_{n+k+1} ). $$
We deduce that \(\lbrace x_{n+k} \rbrace _{n\in \mathbb{N}}\) is a strictly decreasing sequence. So there is \((r\geqslant 0)\in \mathbb{R}\) so that
$$ \lim_{n\to \infty }d( x_{n+k},x_{n+k+1})=r. $$
We claim \(\lim_{n\to \infty }d( x_{n+k},x_{n+k+1})=0\). Suppose on the contrary that \(r>0\). We have
$$ d ((x_{n+k+1},x_{n+k+2} )\leqslant \beta (d \bigl((x_{n+k},x_{n+k+1} ) \bigr)d ((x_{n+k},x_{n+k+1} ). $$
That is,
$$ \frac{d ((x_{n+k+1},x_{n+k+2} )}{d ((x_{n+k},x_{n+k+1} )} \leqslant \beta (d \bigl((x_{n+k},x_{n+k+1} ) \bigr). $$
This implies that \(\lim_{n\to \infty }\beta (d ((x_{n+k},x_{n+k+1} ) )\geqslant 1\).
Since \(\beta \in \mathbb{G}\), necessarily \(\lim_{n\to \infty }\beta (d ((x_{n+k},x_{n+k+1} ) )=1\), and so
$$ \lim_{n\to \infty }d ((x_{n+k},x_{n+k+1} )=0. $$
(7)
We now claim that \(\lbrace x_{n+k} \rbrace _{n\in \mathbb{N}}\) is Cauchy, and we prove it by contradiction. Suppose on the contrary that there is \(\varepsilon >0\) such that we can find some subsequences \(\lbrace x_{m (q )+k} \rbrace _{p\in \mathbb{N}}\), \(\lbrace x_{n (q )+k} \rbrace _{p\in \mathbb{N}}\) with \(m (q )>n (q )>q\) such that, for every q, we have
$$ d ((x_{m (q )+k},x_{n (q )+k} ) \geqslant \varepsilon . $$
(8)
Moreover, corresponding to each \(n(q)\), we can choose least of such \(m (q )\) satisfying (8). Then
$$ d ((x_{m (q )+k-1},x_{n (q )+k} )< \varepsilon . $$
(9)
From (7),(9) and using the triangle inequality, we get
$$\begin{aligned} d ((x_{m (q )+k-1},x_{n (q )+k-1} ) &\leqslant d ((x_{m (q )+k-1},x_{n (q )+k} )+d ((x_{n (q )+k-1},x_{n (q )+k} ) \\ &< \varepsilon +d ((x_{n (q )+k-1},x_{n (q )+k} ) \end{aligned}$$
(10)
and
$$\begin{aligned} \varepsilon \leqslant& d ((x_{n (q )+k},x_{m (q )+k} ) \\ \leqslant& d ((x_{n (q )+k},x_{n (q )+k-1} )+d ((x_{n (q )+k-1},x_{m (q )+k-1} )+d ((x_{m (q )+k-1},x_{m (q )+k} ). \end{aligned}$$
(11)
Letting \(q\to \infty \) in (10) and using (11), we get
$$ \lim_{q\to \infty }d ((x_{m (q )+k-1},x_{n (q )+k-1} )=\varepsilon . $$
(12)
On the other hand, if
$$\begin{aligned}& M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \\& \quad =\max \Bigl\lbrace \sup_{l\geqslant k-1} d ((u_{l},v_{l} ),\sup_{l\geqslant k-1} d ( \bigl(u_{l},T \bigl((u_{i, \widehat{l}})_{i=1}^{\infty } \bigr) \bigr),\sup _{l \geqslant k-1} d ( \bigl(v_{l},T \bigl((v_{i,\widehat{l}})_{i=1}^{ \infty } \bigr) \bigr) \Bigr\rbrace , \end{aligned}$$
where
$$\begin{aligned}& u_{l}:=x_{n (q )+l-1},\quad \forall 1\leqslant l\leqslant k, \end{aligned}$$
(13)
$$\begin{aligned}& u_{l}:=x_{n (q )+k-1},\quad \forall l> k, \end{aligned}$$
(14)
$$\begin{aligned}& v_{l}:=x_{m (q )+l-1},\quad \forall 1\leqslant l\leqslant k, \end{aligned}$$
(15)
$$\begin{aligned}& u_{l}:=x_{m (q )+k-1},\quad \forall l> k \end{aligned}$$
(16)
then
$$\begin{aligned}& M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \\& \quad =\max \bigl\lbrace \xi ((x_{n(q)+k-1},x_{m(q)+k-1} ), d ( \bigl(x_{n(q)+k-1},T \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr), \\& d \bigl( \bigl(x_{m (q )+k-1},T \bigl((x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr) \bigr) \bigr\rbrace \\& \quad =\max \bigl\lbrace d ((x_{n(q)+k-1},x_{m(q)+k-1} ), d(x_{n(q)+k-1},x_{n(q)+k}), d(x_{m(q)+k-1},x_{m(q)+k}) \bigr\rbrace . \end{aligned}$$
Using (7) and (12) we get
$$ \lim_{q\to \infty }M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{ \infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr)= \varepsilon . $$
(17)
By (1) and (17) we get
$$\begin{aligned} &\varepsilon \leq d ((x_{n (q )+k},x_{m (q )+k} )=d \bigl( \bigl(T(x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{ \infty }, T(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr) \\ &\quad \leq \beta \bigl(M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{ \infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr) \\ &\qquad {}\times M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=0}^{ \infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \\ &\quad < M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr). \end{aligned}$$
(18)
If we suppose
$$\begin{aligned}& \sup_{l\geqslant k} d ((u_{l},v_{l} )=d ((x_{n(q)+k-1},x_{m(q)+k-1} ), \\& \sup_{l\geqslant k} d ( \bigl(u_{l},T(u_{i,\widehat{l}})_{i=1}^{ \infty } \bigr)=d ( \bigl(x_{n(q)+k-1},T \bigl((x_{n(q)+i-1, \widehat{n(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr), \\& \sup_{l\geqslant k} d ( \bigl(v_{l},T(v_{i,\widehat{l}})_{i=1}^{ \infty } \bigr)=d ( \bigl(x_{m(q)+k-1},T \bigl((x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr), \end{aligned}$$
then
$$\begin{aligned} & M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \\ &\quad =\max \bigl\lbrace d ((x_{n(q)+k},x_{m(q)+k} ),d ( \bigl(x_{n(q)+k-1}, T \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr), \\ &d \bigl( \bigl(x_{m (q )+k-1},T \bigl((x_{m(q)+i-1, \widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr) \bigr) \bigr\rbrace \\ &\quad =\max \bigl\lbrace d ((x_{n(q)+k}, x_{m(q)+k} ), d(x_{n(q)+k-1}, x_{n(q)+k}), d(x_{m(q)+k-1},x_{m(q)+k}) \bigr\rbrace . \end{aligned}$$
Letting \(q\to \infty \) and using (1), (7), (17), and (18), we get
$$ \lim_{q\to \infty }\beta \bigl(M_{k} \bigl((x_{n(q)+i-1, \widehat{n(q)+k-1}})_{i=1}^{\infty }, (x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{ \infty } \bigr) \bigr) $$
and
$$ \lim_{q\to \infty }M_{k} \bigl((x_{n(q)+i-1, \widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{ \infty } \bigr)\geqslant \varepsilon . $$
From (17)
$$ \beta \bigl(M_{k} \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{ \infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr)\geqslant 1. $$
Since \(\beta \in \mathbb{G}\), we get
$$ \lim_{q\to \infty }\beta \bigl(M_{k} \bigl((x_{n(q)+i-1, \widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}}) _{i=1}^{ \infty } \bigr) \bigr)= 1. $$
Then
$$ \lim_{q\to \infty }M_{k} \bigl((x_{n(q)+i-1, \widehat{n(q)+k-1}})_{i=1}^{\infty },(x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{ \infty } \bigr)=0. $$
We get a contradiction to (17). Hence, \(\{x_{n+k}\}_{n\in \mathbb{N}}\) is a Cauchy sequence.
Since \((X,d )\) is complete, there exists \(u\in X \) such that
$$ \lim_{n\to \infty }x_{n+k}=u. $$
(19)
We claim that \(d ((T ((u)_{i=1}^{\infty } ),u )=0\). If we suppose on the contrary that \(d ((u,T ((u)_{i=1}^{\infty } ) )>0\), then we have
$$\begin{aligned} &M \bigl((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty }, \bigl((u)_{i=1}^{ \infty } \bigr) \bigr)\\ &\quad =\max \bigl\lbrace d(x_{n+k-1},u),~d ( \bigl(x_{n+k-1},T(x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty } \bigr),d ( \bigl(u,T \bigl((u)_{i=1}^{ \infty } \bigr) \bigr) \bigr\rbrace \\ &\quad =\max \bigl\lbrace d ((x_{n+k-1},u ),d ((x_{n+k},x_{n+k-1} ),d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \bigr\rbrace . \end{aligned}$$
Letting \(n\to \infty \) and using (7), (19), we get
$$ \lim_{n\to \infty }M \bigl((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{ \infty },(u)_{i=1}^{\infty } \bigr)=d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr)\neq 0. $$
Then we have
$$\begin{aligned} &d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr)\\ &\quad \leqslant d ((x_{n+k},u )+d ( \bigl(x_{n+k},T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \\ &\quad =d ((x_{n+k},u )+d ( \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty } \bigr),T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \\ &\quad \leqslant d ((x_{n+k},u )+\beta \bigl(M \bigl( (x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty },(u)_{i=1}^{\infty } \bigr) \bigr)M \bigl( (x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty },(u)_{i=1}^{\infty } \bigr) \\ &\quad \leqslant d ((x_{n+k},u )+\beta (d \bigl( \bigl(u,T \bigl((u)_{i=1}^{ \infty } \bigr) \bigr) \bigr)d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr). \end{aligned}$$
Letting \(n\to \infty \) and using (19), we get \(\lim_{n\to \infty }\beta (d ((u,T((u)_{i=1}^{ \infty }) ) )\geqslant 1\).
Since \(\beta \in \mathbb{G}\), one writes \(\lim_{n\to \infty }\beta (d ((u,T((u)_{i=1}^{ \infty }) ) )=1\). Then we have \(\lim_{n\to \infty } d ((u, T((u)_{i=1}^{\infty }) )=0\), which implies \(d ((u,T((u)_{i=1}^{\infty }) )=0\). So we get a contradiction. Hence, \(d ((u,T((u)_{i=1}^{\infty }) )=0\), that is, \(T((u)_{i=1}^{\infty }))=u\). □
Remark 2.1
Theorem 2.1 is a proper generalization of Theorem 1.5 since in the case of the simplest operator on \(\prod_{i=1}^{\infty }X\to X\) the contraction condition of Theorem 1.5 is not applicable; on the other hand, the \(H_{k}\) contraction (see Definition 2.2) is easily applicable for the infinite case. Also, if we restrict the operator to any finite k dimension through an easy calculation, it is obvious that it is an equivalent statement of Theorem 1.5.
Theorem 2.2
. Let \((X,d )\) be a complete metric space and \(T:\prod_{l=1}^{\infty } X\to X\) be an extended Kannan–Geraghty \(H_{k}\) contraction for some \(k\in \mathbb{N}\). Then there is \(u\in X \) such that \(T((u)_{i=1}^{\infty })=u\), and for any \(x_{1}, \ldots , x_{k} \in X \), the infinite k-Picard sequence converges to u.
Proof
Let \(x_{1}, \ldots , x_{k}\in X \). For all \(n\in \mathbb{N}\), we define the infinite k-Picard sequence as follows:
$$ x_{n+k}:=T \bigl((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty } \bigr). $$
(20)
We claim that \(\lim_{n\to \infty }d(x_{n+k},x_{n+k+1})=0\). Then we have
$$\begin{aligned}& d ((x_{n+k+1},x_{n+k+2} )\\& \quad =d ( \bigl(T \bigl((x_{n+i, \widehat{n+k}})_{i=1}^{\infty } \bigr),T \bigl((x_{n+i+1, \widehat{n+1+k}})_{i=1}^{\infty } \bigr) \bigr) \\& \quad \leqslant \frac{\beta (d(x_{n+k},x_{n+k+1}) )}{2} \bigl(d \bigl(T \bigl((x_{n+i,\widehat{n+k}})_{i=1} ^{\infty } \bigr),x_{n+k} \bigr)+d \bigl(T \bigl((x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty } \bigr),x_{n+k+1} \bigr) \bigr) \\& \quad < \frac{1}{2} \bigl(d(x_{n+k},x_{n+k+1})+d(x_{n+k+1},x_{n+k+2}) \bigr). \end{aligned}$$
Thus, \(d(x_{n+k+1},x_{n+k+2})< d(x_{n+k},x_{n+k+1})\). This sequence is decreasing, so there is \(r\geq 0\) so that
$$ \lim_{n\to 0}d(x_{n+k},x_{n+k+1})=r. $$
(21)
We claim that \(r=0\). If we consider on the contrary and suppose \(r>0\), we have
$$\begin{aligned} \lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2}) \leqslant& \frac{\lim_{n\to \infty }\beta (d(x_{n+k},x_{n+k+1}) )}{2} \Bigl( \lim_{n\to \infty }d(x_{n+k},x_{n+k+1})\\ &{}+ \lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2}) \Bigr). \end{aligned}$$
Therefore,
$$ \frac{2\lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2})}{\lim_{n\to \infty } d(x_{n+k},x_{n+k+1})+\lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2})} \leqslant \lim_{n\to \infty }\beta \bigl(d(x_{n+k},x_{n+k+1}) \bigr). $$
From (21) we get \(\frac{2r}{2r}\leqslant \beta (d(x_{n+k},x_{n+k+1}) )\), and so \(\lim_{n\to \infty }\beta (d(x_{n+k},x_{n+k+1}) ) \geqslant 1\). Since \(\beta \in \mathbb{G}\), we have \(\lim_{n\to \infty }\beta (d(x_{n+k},x_{n+k+1}) ) \leqslant 1\). Using the well-known “sandwich theorem”, we obtain
$$ \lim_{n\to \infty }\beta \bigl(d(x_{n+k},x_{n+k+1}) \bigr)=1 \quad \implies\quad \lim_{n\to \infty }d(x_{n+k},x_{n+k+1})=0. $$
(22)
Further, we have
$$\begin{aligned} d(x_{n+k},x_{m+k}) =&d \bigl(T \bigl((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty } \bigr),T \bigl((x_{m+i-1, \widehat{m+k-1}})_{i=1}^{\infty } \bigr) \bigr) \\ \leqslant& \frac{\beta (d(x_{n+k-1},x_{m+k-1}) )}{2} \bigl(d \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty } \bigr),x_{n+k-1} \bigr) \bigr)\\ &{} +d \bigl(T \bigl((x_{m+i-1, \widehat{m+k-1}})_{i=1}^{\infty } \bigr),x_{m+k-1} \bigr)) \\ < &\frac{1}{2} \bigl(d(x_{n+k-1},x_{n+k})+d(x_{m+k-1},x_{m+k}) \bigr). \end{aligned}$$
For large enough \(n,m\in \mathbb{N}\), one has
$$ d(x_{n+k},x_{m+k})< \varepsilon $$
for fixed \(\varepsilon >0\). Then \(\{x_{n+k}\}_{n\in \mathbb{N}}\) is a Cauchy sequence. Since \((X,d)\) is a complete metric space, there exists \(u\in X\) such that
$$ \lim_{n\to \infty }x_{n+k}=u. $$
(23)
We claim that \(T((u)_{i=1}^{\infty })=u\). If we suppose on the contrary that \(d(u,T((u)_{i=1}^{\infty }))>0\). Then, by (21) and (23), for arbitrary \(\varepsilon >0\) and sufficiently large n, we get
$$\begin{aligned} d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \leqslant& d ((x_{n+k},u )+d \bigl(x_{n+k},T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \\ =&d ((x_{n+k},u )+d ( \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty } \bigr),T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \\ \leqslant& d ((x_{n+k},u )+ \frac{\beta (d(u,x_{n+k-1}) )}{2} \bigl(d \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty } \bigr),x_{n+k-1} \bigr)\\ &{}+d \bigl(T \bigl((u)_{i=1}^{\infty } \bigr),u \bigr) \bigr) \\ \leqslant& \frac{1}{2}~d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr)+ \frac{\varepsilon }{2}+\frac{\varepsilon }{2} \leqslant \frac{1}{2}~d(u,T \bigl((u)_{i=1}^{ \infty } \bigr)+ \varepsilon \\ \implies& \frac{1}{2}~d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \leqslant \varepsilon . ~\text{ This is a contradiction.} \end{aligned}$$
Hence, \(d(T((u)_{i=1}^{\infty }),u)=0\), that is, \(T((u)_{i=1}^{\infty })=u\). □
Next, we will provide a new result for a multivalued proper extension of Theorem 1.4, which is also a generalization of Kannan (Theorem 1.3) as a result of Theorem 2.2.
Corollary 2.1
Let \((X,d)\) be a complete metric space and T be a Kannan–Geraghty \(H_{k}\) contraction. Then T has a fixed point and every k-Picard sequence for T converges to u.
Proof
Let us choose \(x_{0}, \ldots , x_{k}\in X\). We define a k-Picard sequence by
$$ x_{n+k}:=T(x_{n},\ldots ,x_{n+k-1}). $$
(24)
If we follow the same steps as in the proof of Theorem 2.2, we get the required result. □
Remark 2.2
For \(k=1\), we get Theorem 1.4 in [3] which proves that Corollary 2.1 is a proper generalization of Theorem 1.4.
Theorem 2.3
Let \((X,d)\) be a complete metric space and \(T:\prod_{l=1}^{\infty } X\to X\) be an extended Fisher–Geraghty \(H_{k}\) contraction for some \(k\in \mathbb{N}\). Then there exists \(u\in X \) such that \(T((u)_{i=1}^{\infty })=u\), and for any \(x_{1}, x_{2}, \ldots x_{k}\), the infinite k-Picard sequence converges to u.
Proof
Let \(x_{1}, \ldots , x_{k}\in X\). For all \(n\in \mathbb{N} \), we define an infinite k- Picard sequence by
$$ x_{n+k}:=T \bigl((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty } \bigr). $$
(25)
We claim \(\lim_{n\to \infty }d(x_{n+k},d_{n+k+1})=0\).
Then we have
$$\begin{aligned} d ((x_{n+k+1},x_{n+k+2} ) =&d ( \bigl(T \bigl((x_{n+i, \widehat{n+k}})_{i=1}^{\infty } \bigr),T \bigl((x_{n+i+1, \widehat{n+1+k}})_{i=1}^{\infty } \bigr) \bigr) \\ \leqslant &\frac{\beta (d(x_{n+k},x_{n+k+1}) )}{2} \bigl(d \bigl(T \bigl((x_{n+i,\widehat{n+k}})_{i=1}^{\infty } \bigr),x_{n+k+1} \bigr)\\ &{}+d \bigl(T \bigl((x_{n+i,\widehat{n+k+1}})_{i=1}^{\infty } \bigr),x_{n+k} \bigr) \bigr) \\ < &\frac{1}{2} \bigl(d(x_{n+k+1},x_{n+k+1})+d(x_{n+k},x_{n+k+2}) \bigr) \\ < &\frac{1}{2} \bigl(d(x_{n+k},x_{n+k+1})+d(x_{n+k+1},x_{n+k+2}) \bigr). \end{aligned}$$
We have \(d(x_{n+k+1},x_{n+k+2})< d(x_{n+k},x_{n+k+1})\). This sequence is decreasing, so there is \(r\geq 0\) such that
$$ \lim_{n\to 0}d(x_{n+k},x_{n+k+1})=r. $$
(26)
We claim that \(r=0\). If we suppose on the contrary that \(r>0\). One has
$$\begin{aligned}& \lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2}) \\& \quad \leqslant \frac{\lim_{n\to \infty }\beta (d(x_{n+k},x_{n+k+1}) )}{2} \Bigl( \lim_{n\to \infty }d(x_{n+k},x_{n+k+1})+ \lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2}) \Bigr). \end{aligned}$$
Thus,
$$ \frac{2\lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2})}{\lim_{n\to \infty }d(x_{n+k}, x_{n+k+1})+\lim_{n\to \infty }d(x_{n+k+1},x_{n+k+2})} \leqslant \lim_{n\to \infty }\beta \bigl(d(x_{n+k},x_{n+k+1}) \bigr). $$
Using (26), we obtain \(\frac{2r}{2r}\leqslant \lim_{n\to \infty }\beta (d(x_{n+k},x_{n+k+1}) )\). Hence,
$$ \lim_{n\to \infty }\beta \bigl(d(x_{n+k},x_{n+k+1}) \bigr) \geqslant ~1. $$
Since \(\beta \in \mathbb{G}\), we have \(\lim_{n\to \infty }\beta (d(x_{n+k},x_{n+k+1}) ) \leqslant 1\).
At the limit, we have
$$ \lim_{n\to \infty }\beta \bigl(d(x_{n+k},x_{n+k+1}) \bigr)=1. $$
(27)
Then \(\lim_{n\to \infty }d(x_{n+k},x_{n+k+1})=0\). We claim that \(\lbrace x_{n+k} \rbrace _{n\in \mathbb{N}}\) is Cauchy, and we want to prove this by contradiction. Then, using the contrary, there exists \(\varepsilon >0\) such that we can find subsequences \(\lbrace x_{m (q )+k} \rbrace _{p\in \mathbb{N}}\), \(\lbrace x_{n (q )+k} \rbrace _{p \in \mathbb{N}}\) with \(m (q )>n (q )>q\) such that, for every q, we have
$$ d ((x_{m (q )+k},x_{n (q )+k} ) \geqslant \varepsilon . $$
(28)
Moreover, corresponding to each \(n(q)\) we can choose \(m (q )\) satisfying (28) so that
$$ d ((x_{m (q )+k-1},x_{n (q )+k} )< \varepsilon . $$
(29)
Using (28), (29) and the triangle inequality, we get
$$\begin{aligned} d ((x_{m (q )+k-1},x_{n (q )+k-1} ) \leqslant &d ((x_{m (q )+k-1},x_{n (q )+k} )+d ((x_{n (q )+k-1},x_{n (q )+k} ) \\ < &\varepsilon +\xi ((x_{n (q )+k-1},x_{n (q )+k} ) \end{aligned}$$
(30)
and
$$\begin{aligned} \varepsilon &\leqslant d ((x_{n (q )+k},x_{m (q )+k} ) \\ &\leqslant d ((x_{n (q )+k},x_{n (q )+k-1} )+d ( (x_{n (q )+k-1},x_{m (q )+k-1} )+d ((x_{m (q )+k-1},x_{m (q )+k} ). \end{aligned}$$
Letting \(q\to \infty \) in (30) and (27), we get
$$\begin{aligned}& \lim_{q\to \infty }d ((x_{m (q )+k-1},x_{n (q )+k-1} )=\varepsilon , \\& \begin{gathered} d(x_{n(q)+k},x_{m(q)+k})\\ \quad =d \bigl(T \bigl((x_{n(q)+i-1,\widehat{n(q)+k-1}})_{i=1}^{ \infty } \bigr),T \bigl((x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr) \bigr) \\ \quad \leqslant \frac{\beta (d(x_{n(q)+k-1},x_{m(q)+k-1}))}{2} \bigl(d \bigl(T \bigl((x_{n(q)+i-1, \widehat{n(q)+k-1}})_{i=1}^{\infty } \bigr),x_{m(q)+k-1} \bigr) \\ \qquad {}+d \bigl(T \bigl((x_{m(q)+i-1,\widehat{m(q)+k-1}})_{i=1}^{\infty } \bigr),x_{n(q)+k-1} \bigr) \bigr) \\ \quad < \frac{1}{2} \bigl(d(x_{n(q)+k},x_{m(q)+k-1})+d(x_{m(q)+k},x_{n(q)+k-1}) \bigr) \\ \quad \leqslant \frac{1}{2} \bigl(\varepsilon +d(x_{m(q)+k},x_{m(q)+k-1})+d(x_{m(q)+k-1},x_{n(q)+k})+d(x_{n(q)+k},x_{n(q)+k-1}) \bigr) \\ \quad \leqslant \frac{1}{2} \bigl(2\varepsilon +d(x_{m(q)+k},x_{m(q)+k-1})+d(x_{n(q)+k},x_{n(q)+k-1}) \bigr). \end{gathered} \end{aligned}$$
(31)
Then from (31) and (27) we get a contradiction. Then
$$ \{x_{n+k}\}_{n\in \mathbb{N}}\quad \text{is a Cauchy sequence.} $$
(32)
Since \((X,d)\) is complete, there exists \(u\in X\) such that
$$ \lim_{n\to \infty }x_{n+k}=u. $$
(33)
We claim that \(T((u)_{i=1}^{\infty })=u\). If we suppose on the contrary that \(d(u,T((u)_{i=1}^{\infty }))>0\), then, by (26) and (33), for arbitrary \(\varepsilon >0\) and sufficiently large n, we get
$$\begin{aligned} &d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr)\\ &\quad \leqslant d ((x_{n+k},u )+d \bigl(x_{n+k},T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \\ &\quad =d ((x_{n+k},u )+d ( \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1}^{\infty } \bigr),T \bigl((u)_{i=1}^{\infty } \bigr) \bigr) \\ &\quad \leqslant d \biggl((x_{n+k},u )+ \frac{\beta (\xi (u,x_{n+k-1}) )}{2} \bigl(d \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1} \bigr)^{\infty } \bigr),u \bigr)+d \bigl(T \bigl((u)_{i=1}^{\infty } \bigr),x_{n+k-1} \bigr) \biggr) \\ &\quad \leqslant d \biggl((x_{n+k},u )+ \frac{\beta (d(u,x_{n+k-1}) )}{2} \bigl(d \bigl(T \bigl((x_{n+i-1, \widehat{n+k-1}})_{i=1} \bigr)^{\infty } \bigr),u \bigr)\\ &\qquad {}+d \bigl(T \bigl((u)_{i=1}^{\infty } \bigr),u \bigr)+d(u,x_{n+k-1}) \biggr) \\ &\quad \leqslant \frac{1}{2}~d ( \bigl(u,T \bigl((u)_{i=1}^{\infty } \bigr) \bigr)+ \frac{\varepsilon }{3}+\frac{\varepsilon }{6} \\ &\quad \leqslant \frac{1}{2}~d(u,T \bigl((u)_{i=1}^{\infty } \bigr)+\frac{1}{2} \varepsilon . \end{aligned}$$
That is, \(d ((u,T((u)_{i=1}^{\infty }) )\leqslant \varepsilon \) for any arbitrary \(\varepsilon >0\). Hence, \(d(T((u)_{i=1}^{\infty }),u)=0\), which implies \(T((u)_{i=1}^{\infty })=u\). □
Corollary 2.2
Let \((X,d)\) be a complete metric space and T be a Fisher–Geraghty \(H_{k}\) contraction. Then T has a fixed point.
Proof
Choose any \(x_{0}, \ldots , x_{k}\in X\). Define
$$ x_{n+k}:=T(x_{n},\ldots ,x_{n+k-1}). $$
(34)
Using the same steps as in the proof of Theorem 2.3, we get the conclusion. □
Remark 2.3
For \(k=1\), we get a new type of extension of Theorem 1.3 proved in [2], and for any k, Corollary 2.2 also gives the multidimensional extension of the same theorem stated in [2].
Remark 2.4
The Banach fixed point theorem [4], Theorem 1.1, Theorem 1.2, Theorem 1.3, Theorem 1.4, and Theorem 1.5 are all applicable only in complete metric spaces as well as the theorems described in [1], but Theorem 2.1, Theorem 2.2, Theorem 2.3, which we have proved, are talking about the \(\mathrm{space}_{i=1}^{\infty }X\) (where X is a complete metric space) which is indeed metrizable, but might not be complete.