A Generalization of Fixed-Point Theorems for Mappings with a Contractive Iterate

Abstract

In this paper, a generalization of fixed-point mappings with an iterate at a point in complete metric spaces is shown, using the notions of T-closed sets and i-P-regularity/d-P-regularity, proving conditions for the existence and uniqueness of the fixed point. Examples are provided to illustrate the results and an application to coupled fixed points is shown.

1. Introduction

Banach’s contraction principle asserts that any contraction map $T:X\to X$, where $(X,\rho )$ is a complete metric space and T satisfies the contraction condition

$$\rho (Tx,Ty)\le \alpha \rho (x,y)$$

(1)

for all $x,y\in X$ and some $\alpha \in [0,1)$, has a unique fixed point, i.e., there exists a unique $x\in X$, so that $x=Tx$.

There are wide classes of generalizations on Banach’s contraction principles. They may be classified by changing the contractive-type condition (1), by changing the underlying space, by changing the fixed-point notion or, simultaneously, all of them.

Well known generalizations where the contractive-type condition (1) is altered are, for example, Kannan [1], Chaterjea [2], Zamfirescu [3], Hardy-Rogeres [4], Reich [5], Meir-Keeler [6] maps. Directions in the investigations where the underlying space X is changed are, for example, modular function spaces [7], b-metric spacers [8], partial metric spaces [9], fuzzy metric spaces [10], and super metric spacers [11]. A different approach has been proposed in [12,13] by introducing the concept of coupled fixed points in partially ordered metric spaces or in normed spaces ordered by a cone. A deep observation about the connection between fixed points and coupled fixed points, published simultaneously, was pointed out in [14,15,16]. A further generalization, by replacing the partial ordering with an abstract $\mathbb{P}\subset X\times X$ set, was proposed in [17,18].

A result widely used in the theory of Volterra equation states that if $(X,\rho )$ is a complete metric space and $T:X\to X$ is a map, so that for some $n\in \mathbb{N}$ the map $T^n$ is a contraction map, then there is a unique fixed point of T. A generalization of this well-known condition has been proposed in [19] by supposing that for any $x\in X$, there is $n\left(x\right)\in \mathbb{N}$, so that the inequality

$$\rho (T^{n\left(x\right)}x,T^{n\left(x\right)}y)\le \alpha \rho (x,y)$$

(2)

holds for all $y\in X$. The idea from [19] has been further investigated in [20,21] and the considered maps are now known as maps with a contractive iterate at a point.

We will try to generalize the ideas of coupled fixed points in partially ordered metric spaces [12], for maps $F:X\times X\to X$ that satisfy a generalization of type (2) by applying ideas from [17,18].

2. Preliminaries

We will denote with $\mathbb{N}$ and $\mathbb{R}$ the sets of natural and real numbers, respectively.

The set of all fixed points for a map $T:X\to X$ is denoted by $\mathrm{Fix}\left(T\right)$. Let $T:X\to X$ and $x_0\in X$ be arbitrarily chosen. The iterated sequence ${\left\{x_n\right\}}_{n=0}^{\infty }$ is defined inductively by $x_1=T{x}_0$ and if we have defined $x_{n-1}$, then $x_n=T{x}_{n-1}=T^n{x}_0$.

Let $(X,\rho )$ be a metric space. We say that two sequences ${\left\{x_n\right\}}_{n=1}^{\infty },{\left\{y_n\right\}}_{n=1}^{\infty }\subset X$ are Cauchy equivalent [18] if ${lim}_{n\to \infty }\rho (x_n,y_n)=0$. It is easy to observe that if ${\left\{x_n\right\}}_{n=1}^{\infty }\subset X$ is convergent to a point $x\in X$ and ${\left\{x_n\right\}}_{n=1}^{\infty },{\left\{y_n\right\}}_{n=1}^{\infty }\subset X$ are Cauchy equivalent, then ${\left\{y_n\right\}}_{n=1}^{\infty }\subset X$ is convergent to the same point $x\in X$.

There are two approaches that can be applied to the study of fixed points when the contraction-type conditions are required to be satisfied for only part of the elements. One is to provide the underlying space with a graph [22,23] that replaces the transitive property of the partial ordering; the other is to define the notion of a $\mathbb{P}\subset X\times X$ set [17,18]. The two approaches are identical, with the only difference being in their definition. We will adhere to the exposition proposed in [17,18]. We will stick to the exposition proposed in [17,18], which is clearer than [24]. The ideas proposed in [17,18,24] expand on the first articles dealing with relaxing the contractive condition in partially ordered metric spaces [25,26].

Definition 1 

([17,18]). Let X be a non-empty set, $\mathbb{P}\subset X\times X$ and $T:X\to X$ be a map. We say that $\mathbb{P}$ is T-closed whenever $(Tx,Ty)\in \mathbb{P}$, provided that there holds $(x,y)\in \mathbb{P}$.

Let us recall [12,13,25], where the ordered pair $(X,\preccurlyeq )$ is said to be a partially ordered set, provided that X is a set and ≼ is a partial order of the elements of X. We say that $x,y\in X$ are comparable if either $x\preccurlyeq y$ or $y\preccurlyeq x$, and we denote it by $x\asymp y$. We denote $x\succcurlyeq y$, if there holds $y\preccurlyeq x$. We denote $x\prec y$, when $x\preccurlyeq$ but $x\ne y$. Let $(X,\rho )$ be a metric space and ≼ be a partial order in X, then the ordered triple $(X,\rho ,\preccurlyeq )$ is called a partially ordered metric space.

Example 1 

([17]). Let $(X,\preccurlyeq )$ be a partially ordered set and $T:X\to X$ be an increasing map, i.e., for any $x,y\in X$, satisfying $x\preccurlyeq y$, there holds $Tx\preccurlyeq Ty$. Then, the set $\mathbb{P}=\left\{\right(x,y)\in X\times X:x\preccurlyeq y\}$ is a T-closed set.

Example 2 

([17]). Let $(X,\preccurlyeq )$ be a partially ordered set and $T:X\to X$ be a map, such that for any two comparable elements $x,y\in X$, their images are comparable, i.e., there holds $Tx\asymp Ty$. Then, the set $\mathbb{P}=\left\{\right(x,y)\in X\times X:x\asymp y\}$ is a T-closed set.

Example 3. 

Let us consider $\mathbb{R}$ and $f:\mathbb{R}\to \mathbb{R}$ be the function $f\left(x\right)=x^2$. Let

$$\mathbb{P}=\{(x,y)\in \mathbb{R}:there aren,m\in \mathbb{N},so thatx=2^{n}andy=2^m\}.$$

Then, $\mathbb{P}$ is an f-closed set because for any $x,y$, such that $x=2^n$ and $y=2^m$, there holds $f\left(x\right)=2^{2n}$ and $f\left(y\right)=2^{2m}$, i.e., $\left(f\right(x),f(y\left)\right)\in \mathbb{P}$.

Definition 2 

([17,18]). Let $(X,\rho )$ be a metric space and $\mathbb{P}\subset X\times X$. The triple $(X,\rho ,\mathbb{P})$ is said to be:

• i-$\mathbb{P}$-regular if for any sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$, convergent to $x^{*}$, such that $(x_n,x_{n+1})\in \mathbb{P}$ for all $n\in \mathbb{N}$, there holds $(x_n,x^{*})\in \mathbb{P}$ for all $n\in \mathbb{N}$
• d-$\mathbb{P}$-regular if for any sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$, convergent to $x^{*}$, such that $(x_{n+1},x_n)\in \mathbb{P}$ for all $n\in \mathbb{N}$, there holds $(x^{*},x_n)\in \mathbb{P}$, for all $n\in \mathbb{N}$.

Example 4. 

Let $(\mathbb{R},|·-·|,\le )$ and $\mathbb{P}=\left\{\right(x,y)\in \mathbb{R}\times \mathbb{R}:x\le y\}$. If ${lim}_{n\to \infty }{x}_n=x$ and $(x_n,x_{n+1})\in \mathbb{P}$, i.e., ${\left\{x_n\right\}}_{n=1}^{\infty }$ is an increasing sequence, then $x_n\le x$ for all $n\in \mathbb{N}$. Therefore, $(x_n,x)\in \mathbb{P}$ and $(\mathbb{R},|·-·|,\le )$ is i-$\mathbb{P}$-regular.

Let $(X,{\rho }_X,\preccurlyeq )$ and $(Y,{\rho }_Y,\preccurlyeq )$ be two partially ordered metric spaces. We can endow the Cartesian product $X\times Y$ with a partial order by saying that $(x,y)\preccurlyeq (u,v)$ if $x\preccurlyeq u$ and $y\succcurlyeq v$ [12] and with the metric $\rho ((x,y),(u,v))={\rho }_X(x,u)+{\rho }_Y(y,v)$ for $(x,y),(u,v)\in X\times Y$ [12,27].

Definition 3 

([28,29]). Let $X,Y$ be non-empty sets and $F:X\times Y\to X$ and $f:X\times Y\to Y$. The ordered pair of maps $(F,f)$ is called a semi-cyclic map.

Definition 4 

([28,29]). Let $X,Y$ be non-empty sets and the ordered pair of maps $(F,f)$ be a semi-cyclic map. The ordered pair $(x,y)\in X\times Y$ is called a coupled fixed point of $(F,f)$ if $x=F(x,y)$ and $y=f(x,y)$.

If $X\equiv Y$ and $f(x,y)=F(y,x)$, we obtain the definition for a coupled fixed point from [12,13].

The notion of semi-cyclic maps arises naturally from the investigation of market equilibrium in duopoly markets [27,30]. Let us have a duopoly market, i.e., there are two producers in the market that can change their productions, taking into account their results and the results of the other producer. Therefore, let the first player be able to produce quantities from a set X and the second one from the set Y. At any stage of the market, each one introduces a change to their production through their response function. Thus, we can speak about a response function of player one $F:X\times Y\to X$ and of player two $G:X\times Y\to Y$. A market equilibrium is attained if $x=F(x,y),y=G(x,y)$ [27,30], which generalizes the notion of a market equilibrium in a duopoly market, obtained through the maximization of the payoff functions of the two players, provided that they are rational ones.

It is worth mentioning that if we stick to the notion of fixed points from [12,13], we obtain only coupled fixed points $(x,y)$, satisfying $x=y$, and if we try to solve a system of equations $x=F(x,y),y=F(y,x)$ we can apply the known techniques only for systems of symmetric equations [31]. The system of symmetric equations arises in the market equilibrium in duopoly markets when $G(x,y)=F(y,x)$, i.e., both players have one and the same response functions. This can be commented as producers with one and the same technologies, one and the same policy to the market, and one and the same knowledge of the market’s demand functions will reach equilibrium productions at equal levels $x=y$ [27,30].

The idea to investigate semi-cyclic maps introduced in [31] allows for solving non-symmetric systems of equations.

Definition 5 

([29]). Let $(Z,\preccurlyeq )$ be a partially ordered set, $X,Y\subseteq Z$, $F:X\times Y\to X$ and $f:X\times Y\to Y$. The ordered pair $(F,f)$ is said to satisfy the mixed monotone property if

• for any $x_1,x_2\in X$, $y\in Y$, such that $x_1\preccurlyeq x_2$, there holds $F(x_1,y)\preccurlyeq F(x_2,y)$
• for any $y_1,y_2\in Y$, $x\in X$, such that $y_1\preccurlyeq y_2$, there holds $f(x,y_1)\succcurlyeq f(x,y_2)$.

If in Definition 5 $f(x,y)=F(y,x)$ and $Y=X$ we obtain the notion for a map F with the mixed monotone property [12], i.e., (2) in Definition 5 is replaced by $F(y_1,x)\succcurlyeq F(y_2,x)$.

The idea of coupled fixed points had been introduced in [13] for maps partially ordered by a cone Banach spaces and that had not been investigated for a long time, until the publication [12], where coupled fixed points were introduced in partially ordered metric spaces. There are many results and generalizations based on the ideas from [12,13] nowadays.

A different approach is suggested in [17,18,32], instead of considering a partial order, the authors considered a family of subsets $\mathbb{M}\subset X\times X\times X\times X$, satisfying a property that replaces the mixed monotone property.

Definition 6 

([17,18,32]). Let X be a set and $F:X\times X\to X$. A nonempty subset $\mathbb{M}\subset X\times X\times X\times X$ is said to be F-closed if for all $x,y,u,v\in X$ there holds $\left(F\right(x,y),F(y,x),F(u,v),F(v,u\left)\right)\in \mathbb{M}$, provided that $(x,y,u,v)\in M$.

Example 5 

([18,24]). Let $(X,\preccurlyeq )$ be a partially ordered set and $F:X\times X\to X$ be a map with the mixed monotone property. Let us consider $(X\times X,\preccurlyeq )$ with the partial order $(x,y)\preccurlyeq (u,v)$ if $x\preccurlyeq u$ and $y\succcurlyeq v$. Let us define $\mathbb{M}=\left\{\right(x,y,u,v)\in X\times X\times X\times X:(x,y)\preccurlyeq (u,v\left)\right\}$. Then, the inequalities $F(x,y)\preccurlyeq F(u,y)\preccurlyeq F(u,v)$ and $F(y,x)\succcurlyeq F(v,x)\succcurlyeq F(v,u)$ hold for any $(x,y)\preccurlyeq (u,v)$. Therefore, $\left(F\right(x,y),F(y,x\left)\right)\preccurlyeq \left(F\right(u,v),F(v,u\left)\right)$, i.e., $\left(F\right(x,y),F(y,x),F(u,v),F(v,u\left)\right)\in \mathbb{M}$ for any $(x,y)\preccurlyeq (u,v)$.

Definition 7 

([33]). Let X and Y be topological spaces. The graph of a map $T:X\to Y$ is the set $\mathrm{Gr}\left(T\right):=\left\{\right(x,Tx)\in X\times Y:x\in X\}$. It is said that T has a closed graph in $X\times Y$ if the graph of T is a closed subset of $X\times Y$, endowed with the product topology.

In a similar manner to [18], we define the transitive property on a set $\mathbb{P}$.

Definition 8. 

We say $\mathbb{P}$ has the transitive property if $(x,z)\in \mathbb{P}$, provided that $(x,y),(y,z)\in \mathbb{P}$.

3. Main Results

We will start with a theorem that generalizes the result from [19].

Theorem 1. 

Let $(X,\rho )$ be a complete metric space, $\mathbb{P}\subset X\times X$, $T:X\to X$ be a map and there hold

(i) 

$\mathbb{P}$ is T-closed and has the transitive property;

(ii) 

T either has a closed graph or the triple $(X,\rho ,\mathbb{P})$ is i-$\mathbb{P}$-regular;

(iii) 

there exists $x_0\in X$, such that $(x_0,T{x}_0)\in \mathbb{P}$;

(iv) 

there exists $\alpha \in [0,1)$, so that for any $x\in X$ there is $n\left(x\right)\in \mathbb{N}$, such that for all $(x,y)\in \mathbb{P}$ holds

$$\rho (T^{n\left(x\right)}\left(x\right),T^{n\left(x\right)}\left(y\right))\le \alpha \rho (x,y).$$

(3)

Then

(a) 

$\mathrm{Fix}\left(f\right)\ne \varnothing$ and for any arbitrary chosen $x_0\in X$, such that $(x_0,T{x}_0)\in \mathbb{P}$ the iterated sequence $x_n=T^n{x}_0$ converges to an element $x^{*}\in \mathrm{Fix}\left(T\right)$

(b) 

For any $x\in X$ and $x_0$, so that $(x_0,T{x}_0)\in \mathbb{P}$, satisfying $(x_0,x)\in \mathbb{P}$ or $(x,x_0)\in \mathbb{P}$, the sequences $x_n=T^n\left(x_0\right)$ and $u_n=T^n\left(x\right)$ are Cauchy equivalent and, hence, $u_n$ converges to $x^{*}\in Fix\left(f\right)$, where $x^{*}={lim}_{n\to \infty }{T}^n{x}_0$

(c) 

If $y^{*}\in \mathrm{Fix}\left(T\right)$ and either $(x_0,y^{*})\in \mathbb{P}$ or $(y^{*},x_0)\in \mathbb{P}$ or there is $z\in X$, so that either $(x_0,z),(y^{*},z)\in \mathbb{P}$ or $(z,x_0),(z,y^{*})\in \mathbb{P}$ then $y^{*}=x^{*}$

(d) 

If, additionally, we suppose that for every $x,y\in X$, such that neither $(x,y)\in \mathbb{P}$ nor $(y,x)\in \mathbb{P}$, there is $z\in X$, so that either $(x,z),(y,z)\in \mathbb{P}$ or $(z,x),(z,y)\in \mathbb{P}$ then $\mathrm{Fix}\left(T\right)=\left\{x^{*}\right\}$.

Proof. 

Let $z\in X$ be arbitrary. For the proof, we will need the inductively defined sequence ${\left\{(m_n,u_n)\right\}}_{n=0}^{\infty }$, where $m_n\in \mathbb{N}$ and ${\left\{u_n\right\}}_{n=0}^{\infty }$ is a subsequence of the iterated sequence ${\left\{z_n\right\}}_{n=0}^{\infty }$, by $u_0=z$, $m_0=n\left(z\right)$, $u_1=T^{m_0}\left(z\right)$, $m_1=n\left(u_1\right)$, $u_2=T^{m_1}\left(u_1\right)$, …, $m_n=n\left(u_n\right)$, $u_{n+1}=T^{m_n}\left(u_n\right)$. We will denote $m_n=m_n\left(z\right)$ and $s_i=s_i\left(z\right)={\sum }_{k=0}^i{m}_k$.

From the assumption of Theorem 1(iii), there is $x_0\in X$, so that $(x_0,T{x}_0)\in \mathbb{P}$.

Let us denote $r=sup\{\rho (T^n{x}_0,x_0):n\in \mathbb{N}\}$, where $x_n=T^n{x}_0$ is the iterated sequence.

We will prove first that $r<+\infty$.

From the assumption in Theorem 1(iv), there exists $n\left(x_0\right)$, so that (3) holds true. Let us denote

$$l=max\{\rho (T^k{x}_0,x_0):k=1,2,\cdots ,n\left(x_0\right)\}.$$

From the assumption in Theorem 1(iii), we have that $(x_0,T{x}_0)\in \mathbb{P}$. From the T-closeness of $\mathbb{P}$ it follows that $(T{x}_0,T^2{x}_0)\in \mathbb{P}$. Using the transitive property of $\mathbb{P}$, it follows that $(x_0,T^2{x}_0)\in \mathbb{P}$.

By induction we can show that $(x_0,T^n{x}_0)\in \mathbb{P}$ for all $n\in \mathbb{N}$.

For any $n\in \mathbb{N}$, there exists $s=s\left(n\right)\in \mathbb{N}\cup \left\{0\right\}$, so that $s·n\left(x_0\right)<n\le (s+1)n\left(x_0\right)$ and we can write the chain of inequalities

$$\begin{array}{ccc}\rho (T^n{x}_0,x_0)\hfill & \le \hfill & \rho (T^{n\left(x_0\right)}{T}^{n-n\left(x_0\right)}{x}_0,T^{n\left(x_0\right)}{x}_0)+\rho (T^{n\left(x_0\right)}{x}_0,x_0)\hfill \\ & \le \hfill & \alpha \rho (T^{n-n\left(x_0\right)}{x}_0,x_0)+l\hfill \\ & \le \hfill & \alpha \rho (T^{n\left(x_0\right)}{T}^{n-2n\left(x_0\right)}{x}_0,T^{n\left(x_0\right)}{x}_0)+\alpha d(T^{n\left(x_0\right)}{x}_0,x_0)+l\hfill \\ & \le \hfill & l+\alpha l+{\alpha }^{2}l+\cdots +{\alpha }^{s}l\hfill \\ & \le \hfill & l\sum _{k=0}^s{\alpha }^k\le {\displaystyle \frac{l}{1-\alpha }}<+\infty .\hfill \end{array}$$

Therefore $r<+\infty$.

Let us consider the sequence ${\left\{(m_n,z_n)\right\}}_{n=0}^{\infty }$, where $z_n=T\left(x_0\right)$ and $m_n=m_n\left(x_0\right)$.

As $(x_0,T^n{x}_0)\in \mathbb{P}$ for all $n\in \mathbb{N}$, we have that $(z_n,z_m)\in \mathbb{P}$ for all $n<m$, $n,m\in \mathbb{N}$.

We will show that the sequence ${\left\{z_n\right\}}_{n=0}^{\infty }$ is convergent.

Let us recall that $s_i={\sum }_{k=0}^i{m}_k$. We can write the chain of inequalities

$$\begin{array}{ccc}\rho (z_{n+1},z_n)\hfill & =\hfill & \rho (T^{s_n}{x}_0,T^{s_{n-1}}{x}_0)=\rho (T^{m_{n-1}}{T}^{s_{n-2}+m_n}{x}_0,T^{m_{n-1}}{T}^{s_{n-2}}{x}_0)\hfill \\ & \le \hfill & \alpha \rho (T^{s_{n-2}+m_n}{x}_0,T^{s_{n-2}}{x}_0)\le \cdots \le {\alpha }^n\rho (T^{m_n}{x}_0,x_0)\le {\alpha }^{n}r.\hfill \end{array}$$

From the inequality

$$\rho (z_m,z_n)\le \sum _{i=n}^{m-1}\rho (z_{i+1},z_i)\le \sum _{i=n}^{m-1}{\alpha }^{i}r\le {\displaystyle \frac{{\alpha }^n}{1-\alpha }}r$$

that holds for any $m>n$, it follows that the sequence ${\left\{z_n\right\}}_{n=1}^{\infty }$ is a Cauchy one. From the completeness of $(X,\rho )$, there exists $x^{*}$ so that ${lim}_{n\to \infty }\rho (z_n,x^{*})=0$.

For any $m\in \mathbb{N}$, there is $k=k\left(m\right)\in \mathbb{N}$, so that the inequality $s_k<m\le s_{k+1}$. Therefore, there is $j=j\left(m\right)\in \mathbb{N}$ so that $m=s_k+j>s_k$ and we obtain

$$\rho (T^m{x}_0,z_k)\le {\alpha }^k\rho (T^j{x}_0,x_0)\le {\alpha }^{k}r.$$

Thus, the sequences ${\left\{T^m{x}_0\right\}}_{m=1}^{\infty }$ and ${\left\{z_k\right\}}_{k=1}^{\infty }$ are Cauchy equivalent and, consequently, ${lim}_{m\to \infty }{T}^m{x}_0=x^{*}$.

Theorem 1(a): If T has a closed graph, from the assumption in Theorem 1(iii), we have that $(x_0,T{x}_0)\in \mathbb{P}$. From the T-closeness of $\mathbb{P}$, it follows that $(T{x}_0,T^2{x}_0)\in \mathbb{P}$ and by induction we have that $(T^n{x}_0,T^{n+1}{x}_0)=(x_n,x_{n+1})=(x_n,T{x}_n)\in \mathbb{P}$, $(x_n,T{x}_n)\in \mathrm{Gr}\left(T\right)$ for all $n\in \mathbb{N}$. Therefore, ${lim}_{n\to \infty }(x_n,x_{n+1})=(x^{*},x^{*})$ and using the assumption that T has a closed graph, it follows $(x^{*},x^{*})\in \mathrm{Gr}\left(T\right)$, i.e., $x^{*}=T{x}^{*}$.

If the triple $(X,\rho ,\mathbb{P})$ is i-$\mathbb{P}$-regular, then, since $(z_n,T{z}_n)\in \mathbb{P}$, we obtain that $(T{z}_n,x^{*})\in \mathbb{P}$.

From the inequality $\rho (z_k,T^{n\left(x^{*}\right)}{x}^{*})\le \alpha \rho (T^{s_{k-1}-n\left(x^{*}\right)}{x}_0,x^{*})$, having in mind that ${\left\{T^{s_{k-1}-n\left(x^{*}\right)}{x}_0\right\}}_{k=1}^{\infty }$ is a subsequence of ${\left\{T^m\left(x_0\right)\right\}}_{m=1}^{\infty }$, we obtain

$$\rho (x^{*},T^{n\left(x^{*}\right)}{x}^{*})=\underset{k\to \infty }{lim}\rho (z_k,T^{n\left(x^{*}\right)}{x}^{*})=0,$$

i.e., $x^{*}=T^{n\left(x^{*}\right)}{x}^{*}$. If we assume that $\rho (x^{*},T{x}^{*})>0$, we obtain

$$\rho (x^{*},T{x}^{*})=\rho (T^{n\left(x^{*}\right)}{x}^{*},T^{n\left(x^{*}\right)}T{x}^{*})\le \alpha \rho (x^{*},T{x}^{*})<\rho (x^{*},T{x}^{*}),$$

which is a contradiction. Therefore, $x^{*}\in \mathrm{Fix}\left(T\right)$.

Theorem 1(b): Without loss of generality, let $(x_0,x)\in \mathbb{P}$. For n large enough, there exist $q=q\left(n\right),r\in \mathbb{N}\cup \left\{0\right\}$, so that $n=qn\left(x_0\right)+r$, $0\le r<n\left(x_0\right)$ and ${lim}_{n\to \infty }q\left(n\right)=+\infty$. Let $x_n=T^n{x}_0$ and $u_n=T^{n}x$. We observe that

$$\underset{n\to \infty }{lim}\rho (x_n,u_n)=\underset{n\to \infty }{lim}\rho (T^n{x}_0,T^{n}x)\le \underset{n\to \infty }{lim}{\alpha }^{q\left(n\right)}\rho (T^r{x}_0,T^{r}x)\le \underset{n\to \infty }{lim}{\alpha }^{q\left(n\right)}L=0,$$

where $L=max\{d(T^k{x}_0,T^{k}x):k=1,2,\dots ,n\left(x_0\right)\}<+\infty$. Consequently, ${\left\{x_n\right\}}_{n=1}^{\infty }$${\left\{u_n\right\}}_{n=1}^{\infty }$ are Cauchy equivalent sequences. The proof of Theorem 1(b) is finished.

A more general claim that will be useful for the remaining proof of the theorem is

Proposition 1.

Given the conditions of Theorem 1, we have that any two sequences ${\left\{u_n\right\}}_{n=0}^{\infty }$ and ${\left\{v_n\right\}}_{n=0}^{\infty }$ are Cauchy equivalent, given that $(u_0,v_0)\in \mathbb{P}$ or $(v_0,u_0)\in \mathbb{P}$.

Proof. 

The proof is analogous to the one in Theorem 1(b). □

Theorem 1(c): If $(x_0,y^{*})\in \mathbb{P}$, then ${\left\{T^n\left(x_0\right)\right\}}_{n=0}^{\infty }$ and ${\left\{T^n\left(y^{*}\right)\right\}}_{n=0}^{\infty }$ are Cauchy equivalent. From ${lim}_{n\to \infty }{T}^n\left(x_0\right)=x^{*}$ and

$T^n\left(y^{*}\right)=y^{*}$ it follows that $x^{*}=y^{*}$.

If $(x_0,y^{*})\notin \mathbb{P}$ and $(y^{*},x_0)\notin \mathbb{P}$, let there hold either $(x_0,z),(y^{*},z)\in \mathbb{P}$ or $(z,x_0)$, $(z,y^{*})\in \mathbb{P}$ for some $z\in X$. Without loss of generality, let $(x_0,z),(y^{*},z)\in \mathbb{P}$. By Theorem 1(b), the iterated sequence ${\left\{z_n\right\}}_{n=0}^{\infty }$, where $z_n=T^{n}z$, is convergent to $x^{*}$ and the sequences ${\left\{T^n\left(y^{*}\right)\right\}}_{n=0}^{\infty }$ and ${\left\{T^n\left(z\right)\right\}}_{n=0}^{\infty }$ are Cauchy equivalent by Proposition 1. Using $T^n\left(y^{*}\right)=y^{*}$, it follows that $x^{*}=y^{*}$.

Theorem 1(d): Let there be two different fixed points $x^{*}$ and $y^{*}$ of T in X, i.e., $T{x}^{*}=x^{*}$, $T{y}^{*}=y^{*}$ and $x^{*}\ne y^{*}$. If $(x^{*},y^{*})\in \mathbb{P}$, then $T^n{x}^{*}$ and $T^n{y}^{*}$ are Cauchy equivalent by Proposition 1. As $T^n{x}^{*}=x^{*}$ and $T^n{y}^{*}=y^{*}$, we conclude that $x^{*}=y^{*}$.

Let there be two different fixed points, $x^{*}$ and $y^{*}$ of T in X, i.e., $T{x}^{*}=x^{*}$, $T{y}^{*}=y^{*}$ and $x^{*}\ne y^{*}$, but $(x^{*},y^{*})\notin \mathbb{P}$ and $(y^{*},x^{*})\notin \mathbb{P}$. If there is $z\in X$ so that either $(x^{*},z),(z,y^{*})\in \mathbb{P}$ or $(z,x^{*}),(y^{*},z)\in \mathbb{P}$, then by the transitive property of $\mathbb{P}$, it follows that $(x^{*},y^{*})\in \mathbb{P}$ or $(y^{*},x^{*})\in \mathbb{P}$, a contradiction with the assumption that $(x^{*},y^{*})\notin \mathbb{P}$ and $(y^{*},x^{*})\notin \mathbb{P}$.

Without loss of generality, let there hold $(x^{*},z),(y^{*},z)\in \mathbb{P}$. By Proposition 1, the sequences ${\left\{T^n{x}^{*}\right\}}_{n=1}^{\infty }$ and ${\left\{T^{n}z\right\}}_{n=1}^{\infty }$ are Cauchy equivalent, as are the sequences ${\left\{T^n{y}^{*}\right\}}_{n=1}^{\infty }$ and ${\left\{T^{n}z\right\}}_{n=1}^{\infty }$. As $T^n{x}^{*}=x^{*}$, it follows that $T^{n}z\to x^{*}$. The same holds true for $y^{*}$, i.e., $z_n\to y^{*}$. Therefore, $x^{*}=y^{*}$.

Remark 1. 

A similar result holds if instead of the triple $(X,\rho ,\mathbb{P})$ beingi-$\mathbb{P}$-regular and there existing an $x_0\in X$, such that $(x_0,T{x}_0)\in \mathbb{P}$, we assume that the triple $(X,\rho ,\mathbb{P})$ isd-$\mathbb{P}$-regular and that there exists $x_0\in X$, such that $(T{x}_0,x_0)\in \mathbb{P}$.

4. Illustrative Examples and Coupled Fixed Points

4.1. Examples

Example 6. 

Let us choose $k\in \mathbb{N}$ and define $T_k:\mathbb{C}\to \mathbb{C}$

$$T_{k}z=\left\{\begin{array}{c}\begin{array}{cc}z{e}^{i{\displaystyle \frac{\pi }{k}}}\hfill & \hfill z\notin \mathbb{R}\\ {\displaystyle \frac{z}{2}}{e}^{i{\displaystyle \frac{\pi }{k}}}\hfill & \hfill z\in \mathbb{R}.\end{array}\hfill \end{array}\right.$$

We will say that

$$z_1\preccurlyeq z_2\iff \left\{\begin{array}{cc}thereexistsm\in \mathbb{N},arg{z}_1-arg{z}_2={\displaystyle \frac{\pi m}{k}},\hfill & z_1,z_2\ne 0\hfill \\ forall{z}_2,\hfill & z_1=0.\hfill \end{array}\right.$$

Let

$$\mathbb{P}=\{(z,w)\in {\mathbb{C}}^2:z\preccurlyeq wandIm\left(z^k\right)=0\}.$$

If $(z,w)\in \mathbb{P}$, then $arg{T}_{k}z-arg{T}_{k}w=argz+{\displaystyle \frac{\pi }{k}}-argw-{\displaystyle \frac{\pi }{k}}=argz-argw={\displaystyle \frac{\pi m}{k}}$. Therefore, $\mathbb{P}$ is T-closed. It is easy to see that $\mathbb{P}$ is also transitive. Additionally, the triple $(\mathbb{C},|·-·|,\mathbb{P})$ is $i-\mathbb{P}-regular$. We will prove that after enough iterations, we can achieve $|T_k^{n\left(z_2\right)}{z}_1-T_k^{n\left(z_2\right)}{z}_2|\le {\displaystyle \frac{1}{2}}|z_1-z_2|$ for $(z_1,z_2)\in \mathbb{P}$.

If $z_1=z_2=0$, then the inequality is obvious.

Any $z_1,z_2\ne 0$ can be expressed as $z_j=r_j{e}^{i{\theta }_j},j=1,2$, because of $z_1\preccurlyeq z_2$, we have ${\theta }_1-{\theta }_2=\theta \in \mathbb{R}$. Then, we find that

$$|z_1-z_2|=|r_1{e}^{i{\theta }_1}-r_2{e}^{i{\theta }_2}|=|e^{i{\theta }_2}\left|\right|r_1{e}^{i\theta }-r_2|=|r_1{e}^{i\theta }-r_2|.$$

For all $z_1\ne 0$, $n\left(z_2\right)$ may be taken to be k. Then, we have two cases. Because $(z_1,z_2)\in \mathbb{P}$, we have that the difference of their arguments remains $\theta$ on every iteration of T and that both complex numbers will land on the real axis. We observe that

$$|T_k^{n\left(z_2\right)}{z}_1-T_k^{n\left(z_2\right)}{z}_2|=\left|{\displaystyle \frac{r_1}{2}}{e}^{i{\theta }_1+2\pi i}-{\displaystyle \frac{r_2}{2}}{e}^{i{\theta }_2+2\pi i}\right|={\displaystyle \frac{1}{2}}|r_1{e}^{i\theta }-r_2|={\displaystyle \frac{1}{2}}|z_1-z_2|$$

In the case where $z_1=0$, let us denote $arg{z}_2={\displaystyle \frac{\pi p}{k}},p\in \mathbb{N},0\le p<2k$ and let us put $q=sk-p$, where $s=\left\{\begin{array}{cc}1\hfill & p\le k\hfill \\ 2\hfill & p>k\hfill \end{array}\right.$. Then, $n\left(z_2\right)=q+1$ and

$$|T_k^{n\left(z_2\right)}{z}_2|=|T_{k}r{e}^{i{\displaystyle \frac{\pi p}{k}}}{e}^{i{\displaystyle \frac{\pi q}{n}}}|=|T_{k}r{e}^{i{\displaystyle \frac{\pi sk}{k}}}|={\displaystyle \frac{r}{2}}={\displaystyle \frac{1}{2}}\left|z_1\right|$$

In all cases, we see that $|T_k^{n\left(z_2\right)}{z}_1-T_k^{n\left(z_2\right)}{z}_2|\le {\displaystyle \frac{1}{2}}|z_1-z_2|$ and for all z, such that $(z,T_{k}z)\in \mathbb{P}$, the sequence $T_k^{n}z$ will converge to the fixed point $z=0$ (Figure 1).

Example 7. 

Define $T:[0,1]\to [0,1]$

$$Tx=\left\{\begin{array}{c}\begin{array}{cc}{\displaystyle \frac{n+2}{n+3}}\left(x-{\displaystyle \frac{1}{2^{n-1}}}\right)+{\displaystyle \frac{1}{2^n}},\hfill & \hfill x\in [{\displaystyle \frac{3n+5}{2^{n+1}(n+2)}},{\displaystyle \frac{1}{2^{n-1}}}]\cap \mathbb{Q},\\ {\displaystyle \frac{n+2}{n+3}}\left(x-{\displaystyle \frac{1}{2^{n-1}}}\right)+{\displaystyle \frac{1}{2^n}}+{\displaystyle \frac{1}{\sqrt{23}}},\hfill & \hfill x\in [{\displaystyle \frac{3n+5}{2^{n+1}(n+2)}},{\displaystyle \frac{1}{2^{n-1}}}]\cap {\mathbb{Q}}^c,\\ {\displaystyle \frac{1}{2^{n+1}}},\hfill & \hfill x\in [{\displaystyle \frac{1}{2^n}},{\displaystyle \frac{3n+5}{2^{n+1}(n+2)}}]\cap \mathbb{Q},\\ {\displaystyle \frac{1}{2^{n+1}}}+{\displaystyle \frac{1}{\sqrt{23}}},\hfill & \hfill x\in [{\displaystyle \frac{1}{2^n}},{\displaystyle \frac{3n+5}{2^{n+1}(n+2)}}]\cap {\mathbb{Q}}^c.\end{array}\hfill \end{array}\right.$$

Let $\mathbb{P}=[0,1]\cap \mathbb{Q}\times [0,1]\cap \mathbb{Q}$.

It is easy to see that $\mathbb{P}$ is $T-closed$ and transitive. Additionally, the triple $\left(\right[0,1],|·-·|,\mathbb{P})$ is both $i-\mathbb{P}-regular$ and $d-\mathbb{P}-regular$. It is shown in [19] that if $x\in [{\displaystyle \frac{1}{2^n}},{\displaystyle \frac{1}{2^{n-1}}}]$ and $y\in [{\displaystyle \frac{1}{2^m}},{\displaystyle \frac{1}{2^{m-1}}}]$, then T satisfies

$$|Tx-Ty|\le {\displaystyle \frac{n+3}{n+4}}|x-y|$$

and if we choose the contractive constant to be $\alpha ={\displaystyle \frac{1}{2}}$, then $n\left(x\right)$ may be taken as $n+3$, while $n\left(0\right)$ can be any integer greater than 1. One fixed point of the map is $x=0$ and because $(x_0,T{x}_0)\in \mathbb{P}$ for any $x_0\in [0,1]\cap \mathbb{Q}$, this is the fixed point of the sequence $T^n{x}_0$. However, as there does not exist $z\in [0,1]$, such that for $(x_0,x)\notin \mathbb{P}$, $(z,x),(z,x_0)$ or $(x,z),(z,x_0)\in \mathbb{P}$ we cannot guarantee the uniqueness of the fixed point. Indeed, there exists another fixed point $y={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{\sqrt{23}}}$. It is clear that y is irrational and it can be shown by simple calculations that ${\displaystyle \frac{1}{2^n}}\le y\le {\displaystyle \frac{3n+5}{2^{n+1}(n+2)}}$ for $n=2$ and $Ty={\displaystyle \frac{1}{2^{2+1}}}+{\displaystyle \frac{1}{\sqrt{23}}}=y$ (Figure 2).

4.2. Coupled Fixed Points

Following the ideas from [31], we will generalize the notion of a T-closed set in the context of a semi-cyclic ordered pair of maps $(F,f)$.

Definition 9. 

Let $X,Y$ be sets and $F:X\times Y\to X$, $f:X\times Y\to Y$. A nonempty subset $\mathbb{M}\subset X\times Y\times X\times Y$ is said to be $(F,f)$-closed if for all $(x,y),(u,v)\in X\times Y$ there holds $\left(F\right(x,y),f(x,y),F(u,v),f(u,v\left)\right)\in \mathbb{M}$, provided that $(x,y,u,v)\in M$.

If $X=Y$ and $f(x,y)=F(y,x)$ we obtain the notion of a F-closed set from [17,18,32].

In what follows, let $X,Y$ be two sets, $F:X\times Y\to X$ and $f:X\times Y\to Y$ be two maps and $\mathbb{M}\subset X\times Y\times X\times Y$ be an $(F,f)$-closed set. Let us denote $Z=X\times Y$ and

$$\mathbb{P}=\left\{\right(z,w)\in Z\times Z:z=(x,y),w=(u,v),(x,y,u,v)\in \mathbb{M}\}.$$

Let us consider the map $T_{(F,f)}(x,y)=(F(x,y),f(x,y))$ for all $(x,y)\in X\times X$. A deep observation in [17,18] shows that an ordered pair $(x,y)\in Z$ is a coupled fixed point for the ordered pair of semi-cyclic maps $(F,f)$ if and only if $(x,y)$ is a fixed point for the map $T_{(F,f)}$, i.e., $(x,y)=T_{(F,f)}(x,y)$ if and only if $x=F(x,y)$ and $y=f(x,y)$.

It is easy to see that the following implication holds $\mathbb{M}$ is $(F,f)$-closed if and only if $\mathbb{P}$ is $T_{(F,f)}$-closed.

In a similar manner to [18], we define the notion of a $\mathrm{i}$-$\mathbb{M}$-regular and $\mathrm{d}$-$\mathbb{M}$-regular triple.

Definition 10. 

Let $(X,{\rho }_X)$ and $(Y,{\rho }_Y)$ be two metric spaces, and $F:X\times Y\to X$ and $f:X\times Y\to Y$ be two maps. Let $T_{(F,f)}=(F,f):X\times Y\to X\times Y$ and $(X\times Y,\rho )$ be a metric space endowed with the metric $\rho ((x,y),(u,v))={\rho }_X(x,u)+\rho (y,v)$ and $\mathbb{M}\subset X\times Y\times X\times Y$. The triple $(X\times Y,\rho ,\mathbb{M})$ is said to be:

(a) 

i-$\mathbb{M}$-regular if for any convergent to $(x^{*},y^{*})\in X\times Y$ sequence ${\left\{(x_n,y_n)\right\}}_{n=1}^{\infty }$, with $(x_n,y_n,x_{n+1},y_{n+1})\in \mathbb{M}$ for all $n\in \mathbb{N}$, we have $(x_n,y_n,x^{*},y^{*})\in \mathbb{M}$, for all $n\in \mathbb{N}$.

(b) 

d-$\mathbb{M}$-regular if for any convergent to $(x^{*},y^{*})\in X\times Y$ sequence ${\left\{(x_n,y_n)\right\}}_{n=1}^{\infty }$, with $(x_{n+1},y_{n+1},x_n,y_n)\in \mathbb{M}$ for all $n\in \mathbb{N}$, we have $(x^{*},y^{*},x_n,y_n)\in \mathbb{M}$, for all $n\in \mathbb{N}$.

Let $(X,{\rho }_X)$ and $(Y,{\rho }_Y)$ be two metric spaces, and $F:X\times Y\to X$ and $f:X\times Y\to Y$ be two maps. Let $T_{(F,f)}=(F,f):X\times Y\to X\times Y$ and $(X\times Y,\rho )$ be a metric space endowed with the metric $\rho ((x,y),(u,v))={\rho }_X(x,u)+\rho (y,v)$. Let ${\mathbb{P}}_X\subset X\times X$, ${\mathbb{P}}_Y\subset Y\times Y$ and $\mathbb{M}\subset X\times Y\times X\times Y$. The triple $(X\times Y,\rho ,\mathbb{M})$ is i-$\mathbb{M}$-regular if and only if $(X,{\rho }_X,{\mathbb{P}}_X)$ is i-${\mathbb{P}}_X$-regular $(Y,{\rho }_Y,{\mathbb{P}}_Y)$ is i-${\mathbb{P}}_Y$-regular.

Let us introduce some notations. Let $X,Y$ be two sets, and $F:X\times Y\to X$ and $f:X\times Y\to Y$ be two maps. Let $T_{(F,f)}(x,y)=(F(x,y),f(x,y)):X\times Y\to X\times Y$. For any initially chosen point $z_0=(x_0,y_0)\in X\times Y$, we will construct inductively the iterated sequence ${\left\{z_n\right\}}_{n=1}^{\infty }$, where $z_n=(x_n,y_n)$ by $z_1=T_{(F,f)}(x_0,y_0)=(F(x_0,y_0),f(x_0,y_0))$, $x_1=F(x_0,y_0)$, $y_1=f(x_0,y_0)$. If we have constructed $x_{n-1}$ and $y_{n-1}$, then we put

$$\begin{array}{ccc}{z}_n\hfill & =\hfill & T_{(F,f)}(x_{n-1},y_{n-1})=(F(x_{n-1},y_{n-1}),f(x_{n-1},y_{n-1}))\hfill \\ & =\hfill & T_{(F,f)}^n(x_0,y_0)=(F^n(x_0,y_0),f^n(x_0,y_0)),\hfill \end{array}$$

where $x_n=F(x_{n-1},y_{n-1})=F^n(x_0,y_0)$, $y_n=f^n(x_0,y_0).$

The next result is a direct corollary of Theorem 1, where we use the notations $Z=X\times Y$ and $\rho ((x,y),(u,v))={\rho }_X(x,u)+{\rho }_Y(y,v)$.

Theorem 2. 

Let $(X,{\rho }_X)$ and $(Y,{\rho }_Y)$ be two complete metric space, $\mathbb{M}\subset X\times Y\times X\times Y$, $F:X\times Y\to X$ and $f:X\times Y\to Y$ be two maps. Suppose it holds

(I) 

$\mathbb{M}$ is $(F,f)$-closed and has the transitive property

(II) 

F and f have a closed graph or the triple $(Z\times Z,\rho ,\mathbb{M})$ is i-$\mathbb{M}$-regular;

(III) 

there exists $(x_0,y_0)\in X$, such that $(x_0,y_0,F(x_0,y_0),f(x_0,y_0))\in \mathbb{M}$;

(IV) 

there exists $\alpha \in [0,1)$, such that for all $(x,y)\in X$ there is a positive integer $n(x,y)$ such that for all $(x,y,u,v)\in \mathbb{M}$, we have

$$\rho ((F^{n(x,y)}(x,y),F^{n(x,y)}(u,v)),(f^{n(x,y)}(x,y),f^{n(x,y)}(u,v)))\le \alpha (\rho ((x,y),(u,v)).$$

Then, we have the following conclusions

(a) 

$(F,f)$ has at least one fixed point and the sequence

$$z_n=(x_n,y_n)=T_{(F,f)}(x_{n-1},y_{n-1})=(F^{n(x_{n-1},y_{n-1})}\left(z_{n-1}\right),f^{n(x_{n-1},y_{n-1})}\left(z_{n-1}\right))$$

$n\in \mathbb{N}$ converges to an element $(x^{*},y^{*})\in Fix\left((F,f)\right)$;

(b) 

For any $(x,y)\in X^2$ such that $(x_0,y_0,x,y)\in \mathbb{M}$ or $(x,y,x_0,y_0)\in \mathbb{M}$, the sequences $v_n=(F^n(x_0,y_0),f^n(x_0,y_0))$ and $u_n=(F^n(x_0,y_0),f^n(x_0,y_0))$ are Cauchy equivalent (i.e., $d(v_n^1,u_n^1)+d(v_n^2,u_n^2)\to 0\mathrm{as}n\to \infty$) and hence $(v_n,u_n)$ converges to the same point $(x^{*},y^{*})\in Fix\left((F,f)\right)$;

(c) 

If additionally we suppose that for every $(x,y)\in X^2$ for which neither $(x_0,y_0,x,y)\in \mathbb{M}$ nor $(x,y,x_0,y_0)\in \mathbb{M}$ there exists $(z,w)\in X^2$ such that $(x_0,y_0,z,w),(z,w,x,y)\in \mathbb{M}$, then $Fix\left((F,f)\right)=(x^{*},y^{*})$ and $(F^n(x,y),f^n(x,y))$ converges to $(x^{*},y^{*})$ as $n\to \infty$.

Proof. 

Let $Z=X\times Y$ endowed with the metric

$$\rho ((x,y),(u,v))={\rho }_X(x,u)+{\rho }_Y(y,v)\mathrm{for}\mathrm{all}(x,y),(u,v)\in X\times Y$$

defined via ${\rho }_X$ and ${\rho }_Y$. Let us consider the operator $T_{(F,f)}:Z\to Z$ given by

$$T_{(F,f)}(x,y)=(F(x,y),f(y,x))\mathrm{for}\mathrm{all}(x,y)\in X\times Y$$

and define $\mathbb{P}=\left\{\right(z,w)\in Z\times Z:z=(x,y),w=(u,v),(x,y,u,v)\in \mathbb{M}\}$. Because $\mathbb{M}$ is $(F,f)$-closed we obtain that $\mathbb{P}$ is $T_{(F,f)}$-closed. Furthermore, we observe that $T_{(F,f)}$ satisfies the following assumptions:

(a)

$T_{(F,f)}$ has a closed graph with respect to $\rho$

(b)

there exists $z_0=(x_0,y_0)\in Z$, such that $(z_0,T_{(F,f)}\left(z_0\right))\in \mathbb{P}$

(c)

there exists $\alpha \in (0,1)$, such that for all $(z,w)\in \mathbb{P}$, we have

$$\rho (T_{(F,f)}^n\left(z\right),T_{(F,f)}^n\left(w\right))\le \alpha \rho (z,w).$$

As the fixed points of $T_{(F,f)}$ are coupled fixed points for $(F,f)$, the conclusion follows by Theorem 1. □

Remark 2. 

A similar result holds if instead of the triple $(X,d,\mathbb{M})$ beingi-$\mathbb{M}$-regular and there existing $(x_0,y_0)\in X$, such that $(x_0,y_0,F(x_0,y_0),f(x_0,y_0))\in \mathbb{M}$, we assume that the triple $(X,d,\mathbb{M})$ isd-$\mathbb{P}$-regular and that there exists $(x_0,y_0)\in X$ such that $(F(x_0,y_0),f(x_0,y_0),x_0,y_0)\in \mathbb{M}$.