A Coupled Fixed Point Problem Under a Finite Number of Equality Constraints

Abstract

Let $(E,\Vert \cdot \Vert )$ be a Banach space with a cone P. Let $F,\varphi _i: E\times E\rightarrow E$ ($i=1,2,\ldots ,r$) be a finite number of mappings. In this chapter, we provide sufficient conditions for the existence and uniqueness of solutions to the problem: Find $(x,y)\in E\times E$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} F(x,y)&{}=&{}x,\\ F(y,x)&{}=&{}y,\\ \varphi _i(x,y)&{}=&{}0_E,\,\, i=1,2,\ldots ,r, \end{array} \right. \end{aligned}$$

where $0_E$ is the zero vector of E. The main reference for this chapter is the paper [4].

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

Let $(E,\Vert \cdot \Vert )$ be a Banach space with a cone P. Let $F,\varphi _i: E\times E\rightarrow E$ ($i=1,2,\ldots ,r$) be a finite number of mappings. In this chapter, we provide sufficient conditions for the existence and uniqueness of solutions to the problem: Find $(x,y)\in E\times E$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} F(x,y)&{}=&{}x,\\ F(y,x)&{}=&{}y,\\ \varphi _i(x,y)&{}=&{}0_E,\,\, i=1,2,\ldots ,r, \end{array} \right. \end{aligned}$$

(8.1)

where $0_E$ is the zero vector of E. The main reference for this chapter is the paper [4].

8.1 Preliminaries

At first, let us recall some basic definitions and some preliminary results that will be used later. In this chapter, the considered Banach space $(E,\Vert \cdot \Vert )$ is supposed to be partially ordered by a cone P. Recall that a nonempty closed convex set $P\subset E$ is said to be a cone (see [2]) if it satisfies the following conditions:  

(P1):

$\lambda \ge 0,\, x\in P \Longrightarrow \lambda x\in P$;

(P2):

$-x,x\in P\Longrightarrow x=0_E$.

 

We define the partial order $\le _P$ in E induced by the cone P by

$$\begin{aligned} (x,y)\in E\times E,\quad x\le _P y \Longleftrightarrow y-x\in P. \end{aligned}$$

Definition 8.1

([1]) Let $\varphi : E\times E\rightarrow E$ be a given mapping. We say that $\varphi$ is level closed from the right if for every $e\in E$, the set

$$ {\text{ lev }\varphi }_{\le _P}(e):=\{(x,y)\in E\times E:\, \varphi (x,y)\le _P e\} $$

is closed.

Definition 8.2

Let $\varphi : E\times E\rightarrow E$ be a given mapping. We say that $\varphi$ is level closed from the left if for every $e\in E$, the set

$$ {\text{ lev }\varphi }_{\ge _P}(e):=\{(x,y)\in E\times E:\, e\le _P\varphi (x,y)\} $$

is closed.

We denote by $\varPsi$ the set of functions $\psi :[0,\infty )\rightarrow [0,\infty )$ satisfying the conditions:

($\varPsi _1$):

$\psi$ is nondecreasing;

($\varPsi _2$):

For all $t>0$, we have

$$ \sum _{k=0}^{\infty } \psi ^{k}(t)<\infty . $$

Here, $\psi ^k$ is the kth iterate of $\psi$.

The following properties are not difficult to prove.

Lemma 8.1

Let $\psi \in \varPsi$. Then

• (i) $\psi (t)<t$, $t>0$;

• (ii) $\psi (0)=0$;

• (iii) $\psi$ is continuous at $t=0$.

Example 8.1

As examples, the following functions belong to the set $\varPsi$:

• $\psi (t)=k\,t$, $k\in (0,1)$.

• $\psi (t)=\left\{ \begin{array}{lll} t/2 &{} \text{ if } &{} 0\le t\le 1,\\ 1/2 &{} \text{ if } &{} t>1. \end{array} \right.$

• $\psi (t)=\left\{ \begin{array}{lll} t/2 &{} \text{ if } &{} 0\le t<1,\\ t-1/3 &{} \text{ if } &{} t\ge 1. \end{array} \right.$

Now, we are ready to state and prove the main results of this chapter. This is the aim of the next section.

8.2 Main Results

Through this chapter, $(E,\Vert \cdot \Vert )$ is a Banach space partially ordered by a cone P and $0_E$ denotes the zero vector of E.

Let us start with the case of one equality constraint.

8.2.1 A Coupled Fixed Point Problem Under One Equality Constraint

We are interested with the existence and uniqueness of solutions to the problem: Find $(x,y)\in E\times E$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} F(x,y)&{}=&{}x,\\ F(y,x)&{}=&{}y,\\ \varphi (x,y)&{}=&{}0_E, \end{array} \right. \end{aligned}$$

(8.2)

where $F,\varphi : E\times E\rightarrow E$ are two given mappings.

The following theorem provides sufficient conditions for the existence and uniqueness of solutions to (8.2).

Theorem 8.1

Let $F,\varphi : E\times E\rightarrow E$ be two given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi$ is level closed from the right.

• (ii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi (x_0,y_0)\le _P 0_E$.

• (iii) For every $(x,y)\in E\times E$, we have

  $$ \varphi (x,y)\le _P 0_E \Longrightarrow \varphi (F(x,y),F(y,x))\ge _P 0_E. $$

• (iv) For every $(x,y)\in E\times E$, we have

  $$ \varphi (x,y)\ge _P 0_E \Longrightarrow \varphi (F(x,y),F(y,x))\le _P 0_E. $$

• (v) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v) \in E\times E$ with $\varphi (x,y)\le _P 0_E$, $\varphi (u,v)\ge _P 0_E$.

Then (8.2) has a unique solution.

Proof

Let $(x_0,y_0)\in E\times E$ be such that

$$ \varphi (x_0,y_0)\le _p 0_E. $$

Such a point exists from (ii). From (iii), we have

$$ \varphi (x_0,y_0)\le _P 0_E \Longrightarrow \varphi (F(x_0,y_0),F(y_0,x_0))\ge _P 0_E. $$

Define the sequences $\{x_n\}$ and $\{y_n\}$ in E by

$$ x_{n+1}=F(x_n,y_n),\,\, y_{n+1}=F(y_n,x_n),\quad n=0,1,2,\ldots $$

Then we have

$$ \varphi (x_1,y_1)\ge _P 0_E. $$

From (iv), we have

$$ \varphi (x_1,y_1)\ge _P 0_E \Longrightarrow \varphi (F(x_1,y_1),F(y_1,x_1))\le _P 0_E, $$

that is,

$$ \varphi (x_2,y_2)\le _P 0_E. $$

Again, using (iii), we get from the above inequality that

$$ \varphi (x_3,y_3)\ge _P 0_E. $$

Then, by induction, we obtain

$$\begin{aligned} \varphi (x_{2n},y_{2n})\le _P 0_E,\,\, \varphi (x_{2n+1},y_{2n+1})\ge _P 0_E,\quad n=0,1,2,\ldots \end{aligned}$$

(8.3)

Using (v) and (8.3), by symmetry, we obtain

$$\begin{aligned} \Vert x_{n+1}-x_n\Vert +\Vert y_{n+1}-y_n\Vert \le \psi \left( \Vert x_n-x_{n-1}\Vert +\Vert y_n-y_{n-1}\Vert \right) ,\quad n=1,2,3,\ldots \end{aligned}$$

(8.4)

From (8.4), since $\psi$ is a nondecreasing function, for every $n=1,2,3,\ldots$, we have

$$\begin{aligned} \Vert x_{n+1}-x_n\Vert +\Vert y_{n+1}-y_n\Vert\le & {} \psi \left( \Vert x_n-x_{n-1}\Vert +\Vert y_n-y_{n-1}\Vert \right) \nonumber \\\le & {} \psi ^2\left( \Vert x_{n-1}-x_{n-2}\Vert +\Vert y_{n-1}-y_{n-2}\Vert \right) \nonumber \\\le & {} \cdots \nonumber \\\le & {} \psi ^n \left( \Vert x_{1}-x_{0}\Vert +\Vert y_{1}-y_{0}\Vert \right) . \end{aligned}$$

(8.5)

Suppose that

$$ \Vert x_{1}-x_{0}\Vert +\Vert y_{1}-y_{0}\Vert =0. $$

In this case, we have

$$ x_0=x_1=F(x_0,y_0)\quad \text{ and }\quad y_0=y_1=F(y_0,x_0). $$

Moreover, from (iii), since $\varphi (x_0,y_0)\le _P 0_E$, we obtain $\varphi (x_1,y_1)=\varphi (x_0,y_0)\ge 0_E$. Since P is a cone, the two inequalities $\varphi (x_0,y_0)\le _P 0_E$ and $\varphi (x_0,y_0)\ge _P 0_E$ yield

$$ \varphi (x_0,y_0)=0_E. $$

Thus, we proved that in this case, $(x_0,y_0)\in E\times E$ is a solution to (8.2).

Now, we may suppose that $\Vert x_{1}-x_{0}\Vert +\Vert y_{1}-y_{0}\Vert \ne 0$. Set

$$ \delta =\Vert x_{1}-x_{0}\Vert +\Vert y_{1}-y_{0}\Vert >0. $$

From (8.5), we have

$$\begin{aligned} \Vert x_{n+1}-x_n\Vert \le \psi ^n (\delta ),\quad n=0,1,2,\ldots \end{aligned}$$

(8.6)

Using the triangular inequality and (8.6), for all $m=1,2,3,\ldots$, we have

$$\begin{aligned} \Vert x_n-x_{n+m}\Vert\le & {} \Vert x_{n}-x_{n+1}\Vert +\Vert x_{n+1}-x_{n+2}\Vert +\cdots +\Vert x_{n+m-1}-x_{n+m}\Vert \\\le & {} \psi ^{n}(\delta )+\psi ^{n+1}(\delta )+\cdots +\psi ^{n+m-1}(\delta )\\= & {} \sum _{i=n}^{n+m-1}\psi ^i(\delta )\\\le & {} \sum _{i=n}^\infty \psi ^i(\delta ). \end{aligned}$$

On the other hand, since $\sum _{k=0}^\infty \psi ^k(\delta )<\infty$, we have

$$ \sum _{i=n}^\infty \psi ^i(\delta )\rightarrow 0 \text{ as } n\rightarrow \infty , $$

which implies that $\{x_n\}$ is a Cauchy sequence in $(E,\Vert \cdot \Vert )$. The same argument gives us that $\{y_n\}$ is a Cauchy sequence in $(E,\Vert \cdot \Vert )$. As consequence, there exists a pair of points $(x^*,y^*)\in E\times E$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert x_n-x^*\Vert =\lim _{n\rightarrow \infty } \Vert y_n-y^*\Vert =0. \end{aligned}$$

(8.7)

From (8.3), we have

$$ \varphi (x_{2n},y_{2n})\le _P 0_E,\quad n=0,1,2,\ldots , $$

that is,

$$ (x_{2n},y_{2n})\in {\text{ lev }\varphi }_{\le _P}(0_E),\quad n=0,1,2,\ldots , $$

Since $\varphi$ is level closed from the right, passing to the limit as $n\rightarrow \infty$ and using (8.7), we obtain

$$ (x^*,y^*)\in {\text{ lev }\varphi }_{\le _P}(0_E), $$

that is,

$$\begin{aligned} \varphi (x^*,y^*)\le _P 0_E. \end{aligned}$$

(8.8)

Now, using (8.3), (8.8), and (v), we obtain

$$\begin{aligned}&\Vert F(x_{2n+1},y_{2n+1})-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-F(y_{2n+1},x_{2n+1})\Vert \\&\le \psi \left( \Vert x_{2n+1}-x^*\Vert +\Vert y_{2n+1}-y^*\Vert \right) , \end{aligned}$$

for all $n=0,1,2,\ldots$, which implies that

$$ \Vert x_{2n+2}\,{-}\,F(x^*,y^*)\Vert \,{+}\,\Vert F(y^*,x^*) \,{-}\, y_{2n+2}\Vert {\le } \psi \left( \!\Vert x_{2n+1}\,{-}\,x^*\Vert +\Vert y_{2n+1}-y^*\Vert \!\right) , $$

for all $n=0,1,2,\ldots$ Passing to the limit as $n\rightarrow \infty$, using (8.7), the continuity of $\psi$ at 0, and the fact that $\psi (0)=0$ (see Lemma 8.1), we get

$$ \Vert x^*-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-y^{*}\Vert =0, $$

that is,

$$ x^*=F(x^*,y^*)\quad \text{ and }\quad y^*=F(y^*,x^*). $$

This proves that $(x^*,y^*)\in E\times E$ is a coupled fixed point of F. Finally, using (8.8) and the fact that $(x^*,y^*)$ is a coupled fixed point of F, it follows from (iii) that

$$\begin{aligned} \varphi (x^*,y^*)\ge _P 0_E. \end{aligned}$$

(8.9)

Then (8.8) and (8.9) yield

$$ \varphi (x^*,y^*)=0_E. $$

Thus, we proved that $(x^*,y^*)\in E\times E$ is a solution to (8.2). Suppose now that $(u^*,v^*)\in E\times E$ is a solution to (8.2) with $(x^*,y^*)\ne (u^*,v^*)$. Using (v), we obtain

$$ \Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert \le \psi (\Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert ). $$

Since $\Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert >0$, from (i) of Lemma 8.1, we have

$$ \psi (\Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert )<\Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert . $$

Then

$$ \Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert <\Vert u^*-x^*\Vert +\Vert y^*-v^*\Vert , $$

which is a contradiction. As consequence, $(x^*,y^*)$ is the unique solution to (8.2).

Remark 8.1

Observe that the conclusion of Theorem 8.1 is still valid if we replace condition (i) by the following condition:

(i’) $\varphi$ is level closed from the left.

In fact, from (8.3), we have

$$ \varphi (x_{2n+1},y_{2n+1})\ge _P 0_E,\quad n=0,1,2,\ldots , $$

that is,

$$ (x_{2n+1},y_{2n+1})\in {\text{ lev }\varphi }_{\ge _P},\quad n=0,1,2,\ldots $$

Passing to the limit as $n\rightarrow \infty$ and using (8.7), we obtain

$$\begin{aligned} \varphi (x^*,y^*)\ge _P 0_E. \end{aligned}$$

(8.10)

Using (8.3), (8.10) and (v), we obtain

$$ \Vert F(x_{2n},y_{2n})-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-F(y_{2n},x_{2n})\Vert \le \psi \left( \Vert x_{2n}-x^*\Vert +\Vert y_{2n}-y^*\Vert \right) , $$

for all $n=0,1,2,\ldots$, which implies that

$$ \Vert x_{2n+1}-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-y_{2n+1}\Vert \le \psi \left( \Vert x_{2n}-x^*\Vert +\Vert y_{2n}-y^*\Vert \right) , $$

for all $n=0,1,2,\ldots$ Passing to the limit as $n\rightarrow \infty$, we get

$$ \Vert x^*-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-y^{*}\Vert =0, $$

which proves that $(x^*,y^*)\in E\times E$ is a coupled fixed point of F. Using (8.10) and the fact that $(x^*,y^*)$ is a coupled fixed point of F, it follows from (iv) that

$$\begin{aligned} \varphi (x^*,y^*)\le _P 0_E. \end{aligned}$$

(8.11)

Then (8.10) and (8.11) yield

$$ \varphi (x^*,y^*)=0_E. $$

Thus, $(x^*,y^*)\in E\times E$ is a solution to (8.2).

8.2.2 A Coupled Fixed Point Problem Under Two Equality Constraints

Here, we are interested with the existence and uniqueness of solutions to the following problem: Find $(x,y)\in E\times E$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} F(x,y)&{}=&{}x,\\ F(y,x)&{}=&{}y,\\ \varphi _1(x,y)&{}=&{}0_E,\\ \varphi _2(x,y)&{}=&{}0_E,\\ \end{array} \right. \end{aligned}$$

(8.12)

where $F,\varphi _1,\varphi _2: E\times E\rightarrow E$ are three given mappings.

We have the following result.

Theorem 8.2

Let $F,\varphi _1,\varphi _2: E\times E\rightarrow E$ be three given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi _i$ ($i=1,2$) is level closed from the right.

• (ii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi _i(x_0,y_0)\le _P 0_E$ ($i=1,2$).

• (iii) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\le _P 0_E,\, i=1,2 \Longrightarrow \varphi _i(F(x,y),F(y,x))\ge _P 0_E,\,i=1,2. $$

• (iv) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\ge _P 0_E,\,i=1,2 \Longrightarrow \varphi _i(F(x,y),F(y,x))\le _P 0_E,\,i=1,2. $$

• (v) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $\varphi _i(x,y)\le _P 0_E, \varphi _i(u,v)\ge _P 0_E$, $i=1,2$.

Then (8.12) has a unique solution.

Proof

Let $(x_0,y_0)\in E\times E$ be such that

$$ \varphi _i(x_0,y_0)\le _p 0_E,\quad i=1,2. $$

Then from (iii), we have

$$ \varphi _i(F(x_0,y_0),F(y_0,x_0))\ge _P 0_E,\quad i=1,2. $$

Define the sequences $\{x_n\}$ and $\{y_n\}$ in E by

$$ x_{n+1}=F(x_n,y_n),\,\, y_{n+1}=F(y_n,x_n),\quad n=0,1,2,\ldots $$

We have

$$ \varphi _i(x_1,y_1)\ge _P 0_E,\quad i=1,2. $$

Then from (iv), we obtain

$$ \varphi _i(x_2,y_2)\le _P 0_E,\quad i=1,2. $$

Again, using (iii), we get from the above inequality that

$$ \varphi _i(x_3,y_3)\ge _P 0_E,\quad i=1,2. $$

Then, by induction, we obtain

$$ \varphi _i(x_{2n},y_{2n})\le _P 0_E,\,\, \varphi _i(x_{2n+1},y_{2n+1})\ge _P 0_E,\quad i=1,2,\, n=0,1,2,\ldots $$

Then, using (v), we obtain

$$ \Vert x_{n+1}-x_n\Vert +\Vert y_{n+1}-y_n\Vert \le \psi \left( \Vert x_n-x_{n-1}\Vert +\Vert y_n-y_{n-1}\Vert \right) ,\quad n=1,2,3,\ldots $$

Now, we argue exactly as in the proof of Theorem 8.1 to show that $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences in $(E,\Vert \cdot \Vert )$. As consequence, there exists a pair of points $(x^*,y^*)\in E\times E$ such that

$$ \lim _{n\rightarrow \infty } \Vert x_n-x^*\Vert =\lim _{n\rightarrow \infty } \Vert y_n-y^*\Vert =0. $$

On the other hand, we have

$$ (x_{2n},y_{2n})\in {\text{ lev }\varphi _i}_{\le _P}(0_E),\quad i=1,2,\, n=0,1,2,\ldots , $$

Since $\varphi _i$ ($i=1,2$) is level closed from the right, passing to the limit as $n\rightarrow \infty$, we obtain

$$ (x^*,y^*)\in {\text{ lev }\varphi _i}_{\le _P}(0_E),\quad i=1,2, $$

that is,

$$ \varphi _i(x^*,y^*)\le _P 0_E,\quad i=1,2. $$

Then we have

$$\begin{aligned}&\Vert F(x_{2n+1},y_{2n+1})-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-F(y_{2n+1},x_{2n+1})\Vert \\&\le \psi \left( \Vert x_{2n+1}-x^*\Vert +\Vert y_{2n+1}-y^*\Vert \right) , \end{aligned}$$

for all $n=0,1,2,\ldots$, which implies that

$$ \Vert x_{2n+2}\,{-}\,F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-y_{2n+2}\Vert \le \psi \left( \Vert x_{2n+1}-x^*\Vert +\Vert y_{2n+1}-y^*\Vert \right) , $$

for all $n=0,1,2,\ldots$ Passing to the limit as $n\rightarrow \infty$, we get

$$ \Vert x^*-F(x^*,y^*)\Vert +\Vert F(y^*,x^*)-y^{*}\Vert =0, $$

that is,

$$ x^*=F(x^*,y^*)\quad \text{ and }\quad y^*=F(y^*,x^*). $$

This proves that $(x^*,y^*)\in E\times E$ is a coupled fixed point of F. Since $\varphi _i(x^*,y^*)\le _P 0_E$ for $i=1,2$, from (iii) we have

$$ \varphi _i(F(x^*,y^*),F(y^*,x^*))\ge _P 0_E,\quad i=1,2, $$

that is,

$$ \varphi _i(x^*,y^*)\ge _P 0_E,\quad i=1,2. $$

Finally, the two inequalities $\varphi _i(x^*,y^*)\le _P 0_E$ and $\varphi _i(x^*,y^*)\ge _P 0_E$, $i=1,2$ yield $\varphi _i(x^*,y^*)=0_E$, $i=1,2$. Then we proved that $(x^*,y^*)\in E\times E$ is a solution to (8.12). The uniqueness can be obtained using a similar argument as in the proof of Theorem 8.1.

Replace $\varphi _2$ in Theorem 8.2 by $-\varphi _2$, we obtain the following result.

Theorem 8.3

Let $F,\varphi _1,\varphi _2: E\times E\rightarrow E$ be three given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi _1$ is level closed from the right and $\varphi _2$ is level closed from the left.

• (ii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi _1(x_0,y_0)\le _P 0_E$ and $\varphi _2(x_0,y_0)\ge _p 0_E$.

• (iii) For every $(x,y)\in E\times E$ with $\varphi _1(x,y)\le _P 0_E$ and $\varphi _2(x,y)\ge _P 0_E$, we have

  $$ \varphi _1(F(x,y),F(y,x))\ge _P 0_E,\, \varphi _2(F(x,y),F(y,x))\le _P 0_E. $$

• (iv) For every $(x,y)\in E\times E$ with $\varphi _1(x,y)\ge _P 0_E$ and $\varphi _2(x,y)\le _P 0_E$, we have

  $$ \varphi _1(F(x,y),F(y,x))\le _P 0_E,\,\varphi _2(F(x,y),F(y,x))\ge _P 0_E. $$

• (v) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $\varphi _1(x,y)\le _P 0_E,\,\varphi _2(x,y)\ge _P 0_E,\, \varphi _1(u,v)\ge _P 0_E,\,\varphi _2(u,v)\le _P 0_E$.

Then (8.12) has a unique solution.

Replace $\varphi _1$ in Theorem 8.3 by $-\varphi _1$, we obtain the following result.

Theorem 8.4

Let $F,\varphi _1,\varphi _2: E\times E\rightarrow E$ be three given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi _i$ ($i=1,2$) is level closed from the left.

• (ii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi _i(x_0,y_0)\ge _P 0_E$ ($i=1,2$).

• (iii) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\le _P 0_E,\,i=1,2 \Longrightarrow \varphi _i(F(x,y),F(y,x))\ge _P 0_E,\,i=1,2. $$

• (iv) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\ge _P 0_E,\, i=1,2 \Longrightarrow \varphi _i(F(x,y),F(y,x))\le _P 0_E,\,i=1,2. $$

• (v) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $\varphi _i(x,y)\le _P 0_E,\, \varphi _i(u,v)\ge _P 0_E$, $i=1,2$.

Then (8.12) has a unique solution.

8.2.3 A Coupled Fixed Point Problem Under r Equality Constraints

Now, we argue exactly as in the proof of Theorem 8.2 to obtain the following existence result for (8.1).

Theorem 8.5

Let $F,\varphi _i: E\times E\rightarrow E$ ($i=1,2,\ldots ,r$) be $r+1$ given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi _i$ ($i=1,2,\ldots , r$) is level closed from the right.

• (ii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi _i(x_0,y_0)\le _P 0_E$ ($i=1,2,\ldots ,r$).

• (iii) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\le _P 0_E,\,i=1,2,\ldots ,r \Longrightarrow \varphi _i(F(x,y),F(y,x))\ge _P 0_E,\,i=1,2,\ldots ,r. $$

• (iv) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\ge _P 0_E,\,i=1,2,\ldots ,r \Longrightarrow \varphi _i(F(x,y),F(y,x))\le _P 0_E,\,i=1,2,\ldots r. $$

• (v) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $\varphi _i(x,y)\le _P 0_E,\, \varphi _i(u,v)\ge _P 0_E$, $i=1,2,\ldots ,r$.

Then (8.1) has a unique solution.

8.3 Some Consequences

In this section, we present some consequences following from Theorem 8.5.

8.3.1 A Fixed Point Problem Under Symmetric Equality Constraints

Let X be a nonempty set and let $F: X\times X\rightarrow X$ be a given mapping. Recall that that $x\in X$ is said to be a fixed point of F if $F(x,x)=x$.

Let $F,\varphi : E\times E\rightarrow E$ be given mappings. We consider the problem: Find $x\in E$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} F(x,x)&{}=&{}x,\\ \varphi (x,x)&{}=&{}0_E. \end{array} \right. \end{aligned}$$

(8.13)

We have the following result.

Corollary 8.1

Let $F,\varphi : E\times E\rightarrow E$ be two given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi$ is level closed from the right.

• (ii) $\varphi$ is symmetric, that is,

  $$ \varphi (x,y)=\varphi (y,x),\quad (x,y)\in E\times E. $$

• (iii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi (x_0,y_0)\le _P 0_E$.

• (iv) For every $(x,y)\in E\times E$, we have

  $$ \varphi (x,y)\le _P 0_E \Longrightarrow \varphi (F(x,y),F(y,x))\ge _P 0_E. $$

• (v) For every $(x,y)\in E\times E$, we have

  $$ \varphi (x,y)\ge _P 0_E \Longrightarrow \varphi (F(x,y),F(y,x))\le _P 0_E. $$

• (vi) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $\varphi (x,y)\le _P 0_E$ and $\varphi (u,v)\ge _P 0_E$.

Then (8.13) has a unique solution.

Proof

From Theorem 8.1, we know that (8.2) has a unique solution $(x^*,y^*)\in E\times E$. Since $\varphi$ is symmetric, $(y^*,x^*)$ is also a solution to (8.2). By uniqueness, we get $x^*=y^*$. Then $x^*\in E$ is the unique solution to (8.13).

Let $F,\varphi _i: E\times E\rightarrow E$ ($i=1,2,\ldots ,r$) be $r+1$ given mappings. We consider the problem: Find $x\in X$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} F(x,x)&{}=&{}x,\\ \varphi _i(x,x)&{}=&{}0_E,\,\, i=1,2,\ldots ,r. \end{array} \right. \end{aligned}$$

(8.14)

Similarly, from Theorem 8.5, we have the following result.

Corollary 8.2

Let $F,\varphi _i: E\times E\rightarrow E$ ($i=1,2,\ldots ,r$) be $r+1$ given mappings. Suppose that the following conditions are satisfied:

• (i) $\varphi _i$ ($i=1,2,\ldots , r$) is level closed from the right.

• (ii) $\varphi _i$ ($i=1,2,\ldots , r$) is symmetric.

• (iii) There exists $(x_0,y_0)\in E\times E$ such that $\varphi _i(x_0,y_0)\le _P 0_E$ ($i=1,2,\ldots ,r$).

• (iv) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\le _P 0_E,\,i=1,2,\ldots ,r \Longrightarrow \varphi _i(F(x,y),F(y,x))\ge _P 0_E,\,i=1,2,\ldots ,r. $$

• (v) For every $(x,y)\in E\times E$, we have

  $$ \varphi _i(x,y)\ge _P 0_E,\,i=1,2,\ldots ,r \Longrightarrow \varphi _i(F(x,y),F(y,x))\le _P 0_E,\,i=1,2,\ldots r. $$

• (vi) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $\varphi _i(x,y)\le _P 0_E,\, \varphi _i(u,v)\ge _P 0_E$, $i=1,2,\ldots ,r$.

Then (8.14) has a unique solution.

8.3.2 A Common Coupled Fixed Point Result

We need the following definition.

Definition 8.3

Let X be a nonempty set, $F: X\times X\rightarrow X$ and $g:X\rightarrow X$ be two given mappings. We say that the pair of elements $(x,y)\in X\times X$ is a common coupled fixed point of F and g if

$$ x=gx=F(x,y)\quad \text{ and }\quad y=gy=F(y,x). $$

We have the following common coupled fixed point result.

Corollary 8.3

Let $F: E\times E\rightarrow E$ and $g:E\rightarrow E$ be two given mappings. Suppose that the following conditions hold:

• (i) g is a continuous mapping.

• (ii) There exists $(x_0,y_0)\in E\times E$ such that

  $$ gx_0\le _p x_0\quad \text{ and }\quad gy_0\le _p y_0. $$

• (iii) For every $(x,y)\in E\times E$, we have

  $$ gx\le _P x,\,gy\le _p y \Longrightarrow gF(x,y)\ge _P F(x,y),\,gF(y,x)\ge _P F(y,x). $$

• (iv) For every $(x,y)\in E\times E$, we have

  $$ gx\ge _P x,\,gy\ge _P y \Longrightarrow gF(x,y)\le _P F(x,y),\,gF(y,x)\le _P F(y,x). $$

• (v) There exists some $\psi \in \varPsi$ such that

  $$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

  for all $(x,y),(u,v)\in E\times E$ with $gx\le _P x$, $gy\le _Py$ and $gu\ge _P u$, $gv\ge _Pv$.

Then F and g have a unique common coupled fixed point.

Proof

Let us consider the mappings $\varphi _1,\varphi _2: E\times E\rightarrow E$ defined by

$$ \varphi _1(x,y)=gx-x,\quad (x,y)\in E\times E $$

and

$$ \varphi _2(x,y)=gy-y,\quad (x,y)\in E\times E. $$

Observe that $(x,y)\in E\times E$ is a common coupled fixed point of F and g if and only if $(x,y)\in E\times E$ is a solution to (8.12). Note that since g is continuous, then $\varphi _i$ is level closed from the right (also from the left) for all $i=1,2$. Now, applying Theorem 8.2, we obtain the desired result.

8.3.3 A Fixed Point Result

We denote by $\widetilde{\varPsi }$ the set of functions $\psi :[0,\infty )\rightarrow [0,\infty )$ satisfying the following conditions:

(${\widetilde{\varPsi }}_1$):

$\psi \in \varPsi$.

(${\widetilde{\varPsi }}_2$):

For all $a,b\in [0,\infty )$, we have

$$ \psi (a)+\psi (b)\le \psi (a+b). $$

Example 8.2

As example, let us consider the function

$$ \psi (t)=\left\{ \begin{array}{lll} t/2 &{} \text{ if } &{} 0\le t<1,\\ t-1/3 &{} \text{ if } &{} t\ge 1. \end{array} \right. $$

It is not difficult to observe that $\psi \in \varPsi$. Now, let us consider an arbitrary pair $(a,b)\in [0,\infty )\times [0,\infty )$. We discuss three possible cases.

Case 1. If $(a,b)\in [0,1)\times [0,1)$.

In this case, we have $\psi (a)+\psi (b)=(a+b)/2$. On the other hand, we have $a+b\in [0,2)$. So, if $0\le a+b<1$, then $\psi (a)+\psi (b)=(a+b)/2=\psi (a+b)$. However, if $1\le a+b<2$, then $\psi (a+b)-\psi (a)-\psi (b)=(a+b)/2-1/3\ge 0$.

Case 2. If $(a,b)\in [0,1)\times [1,\infty )$.

In this case, we have $\psi (a)+\psi (b)=a/2+b-1/3\le a+b-1/3=\psi (a+b)$.

Case 3. If $(a,b)\in [1,\infty )\times [1,\infty )$.

In this case, we have $\psi (a)+\psi (b)=a+b-2/3\le a+b-1/3=\psi (a+b)$.

Therefore, we have $\psi \in \widetilde{\varPsi }$.

Note that the set $\varPsi$ is more large than the set $\widetilde{\varPsi }$. The following example illustrates this fact.

Example 8.3

Let us consider the function

$$ \psi (t)=\left\{ \begin{array}{lll} t/2 &{} \text{ if } &{} 0\le t\le 1,\\ 1/2 &{} \text{ if } &{} t> 1. \end{array} \right. $$

Clearly, we have $\psi \in \varPsi$. However,

$$ \psi (1+1)=1/2<1=\psi (1)+\psi (1), $$

which proves that $\psi \not \in \widetilde{\varPsi }$.

We have the following fixed point result.

Corollary 8.4

Let $T: E\rightarrow E$ be a given mapping. Suppose that there exists some $\psi \in \widetilde{\varPsi }$ such that

$$\begin{aligned} \Vert Tu-Tx\Vert \le \psi (\Vert u-x\Vert ),\quad (u,x)\in E\times E. \end{aligned}$$

(8.15)

Then T has a unique fixed point.

Proof

Let us define the mapping $F: E\times E\rightarrow E$ by

$$ F(x,y)=Tx,\quad (x,y)\in E\times E. $$

Let $g: E\rightarrow E$ be the identity mapping, that is,

$$ gx=x,\quad x\in E. $$

From (8.15), for all $(x,y),(u,v)\in E\times E$, we have

$$ \Vert Tu-Tx\Vert \le \psi (\Vert u-x\Vert ) $$

and

$$ \Vert Ty-Tv\Vert \le \psi (\Vert v-y\Vert ). $$

Then

$$ \Vert Tu-Tx\Vert +\Vert Ty-Tv\Vert \le \psi (\Vert u-x\Vert )+\psi (\Vert v-y\Vert ). $$

Using the property ($\widetilde{\varPsi }_2$), we obtain

$$ \Vert Tu-Tx\Vert +\Vert Ty-Tv\Vert \le \psi (\Vert u-x\Vert +\Vert v-y\Vert ),\quad (x,y), (u,v)\in E\times E. $$

From the definitions of F and g, we obtain

$$ \Vert F(u,v)-F(x,y)\Vert +\Vert F(y,x)-F(v,u)\Vert \le \psi \left( \Vert u-x\Vert +\Vert v-y\Vert \right) , $$

for all $(x,y),(u,v)\in E\times E$ with $gx\le _P x$, $gy\le _Py$ and $gu\ge _P u$, $gv\ge _Pv$. By Corollary 3.5, there exists a unique $(x^*,y^*)\in E\times E$ such that

$$ x^*=F(x^*,y^*)=Tx^*\quad \text{ and }\quad y^*=F(y^*,x^*)=Ty^*. $$

Suppose that $x^*\ne y^*$. By (8.15), we have

$$ \Vert x^*-y^*\Vert = \Vert Tx^*-Ty^*\Vert \le \psi (\Vert x^*-y^*))<\Vert x^*-y^*\Vert , $$

which is a contradiction. As consequence, $x^*\in E$ is the unique fixed point of T.

Remark 8.2

Taking

$$ \psi (t)=k t,\quad t\ge 0, $$

where $k\in (0,1)$ is a constant, we obtain from Corollary 8.4 the Banach contraction principle.

Finally, for other related results, we refer the reader to Jleli and Samet [3].