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A fixed point problem under two constraint inequalities
Fixed Point Theory and Applications volume 2016, Article number: 18 (2016)
Abstract
Let \((X,d)\) be a metric space. Suppose that the set X is equipped with two partial orders ⪯_{1} and ⪯_{2}. Let \(T,A,B,C,D: X\to X\) be given operators. We provide sufficient conditions for the existence of a fixed point of T satisfying the two constraint inequalities: \(Ax\preceq_{1} Bx\) and \(Cx\preceq_{2} Dx\).
Introduction and basic definitions
Recently there have been many developments concerning the existence of fixed points for operators defined in a metric space equipped with a partial order. This trend was initiated by Turinici [1]. Next, Ran and Reurings [2] extended the Banach contraction principle to continuous monotone operators defined in a partially ordered metric space. They also presented some applications to the existence of positive solutions to certain classes of nonlinear matrix equations. The result obtained in [2] was extended and generalized by many authors in different directions (see [3–11] and the references therein).
Let \((X,d)\) be a metric space. Suppose that the set X is endowed with two partial orders ⪯_{1} and ⪯_{2}. Let us consider five selfoperators \(T,A,B,C,D: X\to X\). In this paper, we are concerned with the following problem: Find \(x\in X\) such that
We obtain sufficient conditions for the existence of at least one solution to (1.1).
The following definitions will be used throughout the paper.
Definition 1.1
Let X be a nonempty set. Let ⪯ be a binary relation on X. We say that ⪯ is a partial order on X if the following conditions are satisfied:

(i)
For every \(x\in X\), we have \(x\preceq x\).

(ii)
For every \(x,y,z\in X\), we have
$$x\preceq y, \quad y\preceq z\quad \Longrightarrow\quad x\preceq z. $$ 
(iii)
For every \(x,y\in X\), we have
$$x\preceq y,\quad y\preceq x\quad \Longrightarrow \quad x=y. $$
Definition 1.2
Let \((X,d)\) be a metric space and ⪯ be a partial order on X. We say that the partial order ⪯ is dregular if the following condition is satisfied: For every sequences \(\{a_{n}\},\{b_{n}\}\subset X\), we have
where \((a,b)\in X\times X\).
Example 1.3
Let \((X,\\cdot\)\) be a Banach space and P be a cone on X. Let us consider the partial order on X defined by
Consider the metric d on X defined by
Then \(\preceq_{P}\) is dregular. In fact, suppose that \(\{a_{n}\}\) and \(\{b_{n}\}\) are two sequences in X such that
and
for some \((a,b)\in X\times X\). Since \(b_{n}a_{n}\in P\) for all n and the sequence \(\{b_{n}a_{n}\}\) converges to \(ba\), the closure of the cone P yields \(ba\in P\), that is, \(a\preceq_{P} b\).
Definition 1.4
Let X be a nonempty set endowed with two partial orders ⪯_{1} and ⪯_{2}. Let \(T,A,B,C,D: X\to X\) be given operators. We say that the operator T is \((A,B,C,D,\preceq_{1},\preceq_{2})\)stable, if the following condition is satisfied:
Example 1.5
Let \(X=\mathbb{R}\) and consider the standard order ≤ on X. Let \(A,B,C,D: X\to X\) be the operators defined by
Then the operator T is \((A,B,C,D,\leq,\leq)\)stable. In fact, let \(x\in X\) be such that
Then
which yields
that is,
Example 1.6
Let \(X=\mathbb{R}^{2}\). We consider the two partial orders ⪯_{1} and ⪯_{2} on X defined by
and
Let us consider the operators \(T,A,B,C,D: X\to X\) defined by
for all \((a,b)\in X\). We claim that T is \((A,B,C,D,\preceq_{1},\preceq _{2})\)stable. In order to prove this claim, we take \(x=(a,b)\in X\) such that
which is equivalent (from the definition of ⪯_{1}) to
On the other hand, under the above inequality, we have
and
Clearly (from the definition of ⪯_{2}), we have
Then our claim is proved.
Now, we are ready to state and prove our main result.
Main result
Let us denote by Φ the set of functions \(\varphi: [0,\infty)\to [0,\infty)\) satisfying the following conditions:
 (\(\Phi_{1}\)):

φ is a lower semicontinuous function.
 (\(\Phi_{2}\)):

\(\varphi^{1}(\{0\})=\{0\}\).
Our main result in this paper is giving by the following theorem.
Theorem 2.1
Let \((X,d)\) be a complete metric space endowed with two partial orders ⪯_{1} and ⪯_{2}. Let \(T,A,B,C,D: X\to X\) be given operators. Suppose that the following conditions are satisfied:

(i)
\(\preceq_{i}\) is dregular, \(i=1,2\).

(ii)
A, B, C, and D are continuous.

(iii)
There exists \(x_{0}\in X\) such that
$$Ax_{0}\preceq_{1} Bx_{0}. $$ 
(iv)
T is \((A,B,C,D,\preceq_{1},\preceq_{2})\)stable.

(v)
T is \((C,D,A,B,\preceq_{2},\preceq_{1})\)stable.

(vi)
There exists \(\varphi\in\Phi\) such that
$$Ax\preceq_{1} Bx,\quad Cy\preceq_{2} Dy\quad \Longrightarrow \quad d(Tx,Ty)\leq d(x,y)\varphi\bigl(d(x,y)\bigr). $$
Then:

(I)
The sequence \(\{T^{n}x_{0}\}\) converges to some \(x^{*}\in X\) satisfying
$$Ax^{*}\preceq_{1} Bx^{*} \quad \textit{and} \quad Cx^{*} \preceq_{2} Dx^{*}. $$ 
(II)
The point \(x^{*}\in X\) is a solution to (1.1).
Proof
Let us prove (I). Let \(x_{0}\in X\) be such that
Such a point exists from (iii). Let us consider the sequence \(\{x_{n}\} \subset X\) defined by
Since T is \((A,B,C,D,\preceq_{1},\preceq_{2})\)stable (see (iv)), we have
that is,
Then we have
Since T is \((C,D,A,B,\preceq_{2},\preceq_{1})\)stable (see (v)),
that is,
Again, T is \((A,B,C,D,\preceq_{1},\preceq_{2})\)stable; this yields
that is,
Thus we have
By induction, we get
Using (2.1) and (vi), by symmetry, we have
which yields
Then \(\{d(x_{n+1},x_{n})\}\) is a decreasing sequence of positive numbers. Therefore, there exists some \(r\geq0\) such that
From (2.2), we have
which yields
Using (2.3) and the lower semicontinuity of φ, we obtain
which implies that
Since \(\varphi^{1}(\{0\})=\{0\}\), we get \(r=0\), i.e.,
Now, we show that \(\{x_{n}\}\) is a Cauchy sequence in \((X,d)\). Suppose that \(\{x_{n}\}\) is not a Cauchy sequence. Then there exists some \(\varepsilon>0\) for which we find two sequences of positive integers \(\{ m(k)\}\) and \(\{n(k)\}\) such that, for all positive integers k,
From (2.5), we have
Then, for all k, we have
Passing to the limit as \(k\to\infty\) and using (2.4), we get
On the other hand, we have
Then from (2.6), we have
We have also
Then from (2.6), we have
Similarly,
Then from (2.7), we have
Observe that, for all k, there exists a positive integer \(0\leq i(k)\leq1\) such that
From (2.1), for all \(k>1\), we have
or
Then from (vi), by symmetry, we have
that is,
Set
We consider two cases.
• Case 1. \(\Lambda=\infty\).
From (2.10), we get
which gives us
Using (2.9), (2.6), and the lower semicontinuity of φ, we obtain
which yields
Since \(\varphi^{1}(\{0\})=\{0\}\), we get \(\varepsilon=0\), which is a contradiction with \(\varepsilon>0\).
• Case 2. \(\Lambda<\infty\).
In this case, we have \(\Delta=\infty\). Moreover, from (2.10), we have
which gives us
Using (2.7), (2.8), and the lower semicontinuity of φ, we obtain
which yields \(\varepsilon\in\varphi^{1}(\{0\})=\{0\}\), that is, a contradiction with \(\varepsilon>0\).
Therefore, we deduce that \(\{x_{n}\}\) is a Cauchy sequence in \((X,d)\). Since \((X,d)\) is complete, there exists some \(x^{*}\in X\) such that
On the other hand, from (2.1), we have
Using the continuity of A and B, it follows from (2.11) that
Since ⪯_{1} is dregular, we get
Similarly, from (2.1), we have
Using the continuity of C and D, it follows from (2.11) that
Since ⪯_{2} is dregular, we get
Thus we proved (I).
Now, let us prove (II). Using (2.12), (2.13), (2.1), and (vi), we have
that is,
Then
Using the lower semicontinuity of φ, the fact that \(\varphi (0)=0\), and (2.11), we obtain
which yields
Therefore, from (2.12), (2.13), and (2.14), we deduce that \(x^{*}\in X\) is a solution to (1.1). This ends the proof. □
In the next section, we present some consequences following from Theorem 2.1.
Some consequences
A fixed point problem under one constraint equality
Here, we are concerned with the following problem: Find \(x\in X\) such that
where \(T,A,B: X\to X\) are given operators and \((X,d)\) is a metric space endowed with a certain partial order ⪯. Observe that (3.1) is equivalent to (1.1) with
Then from Theorem 2.1, we obtain the following result.
Corollary 3.1
Let \((X,d)\) be a complete metric space endowed with a certain partial order ⪯. Let \(T,A,B: X\to X\) be three given operators. Suppose that the following conditions are satisfied:

(i)
⪯ is dregular.

(ii)
A and B are continuous.

(iii)
There exists \(x_{0}\in X\) such that
$$Ax_{0}\preceq Bx_{0}. $$ 
(iv)
For all \(x\in X\), we have
$$Ax\preceq Bx\quad \Longrightarrow \quad BTx\preceq ATx. $$ 
(v)
For all \(x\in X\), we have
$$Bx\preceq Ax\quad \Longrightarrow \quad ATx\preceq BTx. $$ 
(vi)
There exists \(\varphi\in\Phi\) such that
$$Ax\preceq Bx, \quad By\preceq Ay \quad \Longrightarrow \quad d(Tx,Ty)\leq d(x,y)\varphi \bigl(d(x,y)\bigr). $$
Then:

(I)
The sequence \(\{T^{n}x_{0}\}\) converges to some \(x^{*}\in X\) satisfying \(Ax^{*}=Bx^{*}\).

(II)
The point \(x^{*}\in X\) is a solution to (3.1).
A common fixed point problem
Let us consider the following problem: Find \(x\in X\) such that
where \(T,B: X\to X\) are given operators and \((X,d)\) is a metric space endowed with a certain partial order ⪯. Observe that (3.2) is equivalent to (3.1) with \(A=I_{X}\), the identity mapping on X. So, take \(A=I_{X}\) in Corollary 3.1, We obtain the following result.
Corollary 3.2
Let \((X,d)\) be a complete metric space endowed with a certain partial order ⪯. Let \(T,B: X\to X\) be two given operators. Suppose that the following conditions are satisfied:

(i)
⪯ is dregular.

(ii)
B is continuous.

(iii)
There exists \(x_{0}\in X\) such that
$$x_{0}\preceq Bx_{0}. $$ 
(iv)
For all \(x\in X\), we have
$$x\preceq Bx \quad \Longrightarrow \quad BTx\preceq Tx. $$ 
(v)
For all \(x\in X\), we have
$$Bx\preceq x\quad \Longrightarrow \quad Tx\preceq BTx. $$ 
(vi)
There exists \(\varphi\in\Phi\) such that
$$x\preceq Bx, \quad By\preceq y \quad \Longrightarrow \quad d(Tx,Ty)\leq d(x,y) \varphi\bigl(d(x,y)\bigr). $$
Then:

(I)
The sequence \(\{T^{n}x_{0}\}\) converges to some \(x^{*}\in X\) satisfying \(x^{*}=Bx^{*}\).

(II)
The point \(x^{*}\in X\) is a solution to (3.2).
Next, we present an example that illustrates the above result.
Example 3.3
Let \(X\subset\mathbb{R}^{2}\) be the set defined by
Let ⪯ be the partial order on X defined by
We endow X with the metric d defined by
Let \(T,B: X\to X\) be the mappings defined by
Observe that
However,
Let \(u\in X\) be such that \(u\preceq Bu\).
For \(u=(4,0)\), we have
For \(u=(0,0)\), we have
For \(u=(3,0)\), we have
Then we have
Let \(v\in X\) be such that \(Bv\preceq v\).
For \(v=(4,0)\), we have
For \(v=(6,0)\), we have
Then we have
Now, let \((u,v)\in X\times X\) be such that
We have
For \((u,v)=((4,0),(4,0))\), we have
For \((u,v)=((4,0),(6,0))\), we have
For \((u,v)=((0,0),(4,0))\), we have
For \((u,v)=((0,0),(6,0))\), we have
For \((u,v)=((3,0),(4,0))\), we have
For \((u,v)=((3,0),(6,0))\), we have
Then we have
where \(\varphi(t)=\frac{t}{2}\), \(t\geq0\). So, all the required conditions of Corollary 3.2 are satisfied. Observe that \(x^{*}=(4,0)\) is a common fixed point of T and B. Observe also that
which shows that T is not a contraction on X.
Taking \(B=T\) in Corollary 3.2, we obtain the following fixed point result.
Corollary 3.4
Let \((X,d)\) be a complete metric space endowed with a certain partial order ⪯. Let \(T: X\to X\) be a given operator. Suppose that the following conditions are satisfied:

(i)
⪯ is dregular.

(ii)
T is continuous.

(iii)
There exists \(x_{0}\in X\) such that
$$x_{0}\preceq Tx_{0}. $$ 
(iv)
For all \(x\in X\), we have
$$x\preceq Tx \quad \Longrightarrow \quad T^{2}x\preceq Tx. $$ 
(v)
For all \(x\in X\), we have
$$Tx\preceq x\quad \Longrightarrow \quad Tx\preceq T^{2}x. $$ 
(vi)
There exists \(\varphi\in\Phi\) such that
$$x\preceq Tx,\quad Ty\preceq y \quad \Longrightarrow \quad d(Tx,Ty)\leq d(x,y) \varphi\bigl(d(x,y)\bigr). $$
Then the sequence \(\{T^{n}x_{0}\}\) converges to a fixed point of T.
Conclusion
In this paper, we obtained sufficient conditions for the existence of a fixed point of a certain operator under two constraint inequalities with respect to two partial orders. The used technique can also be adapted for any finite number of constraint inequalities and other contractive conditions. An interesting question is the existence of a best proximity point of a certain operator under constraint inequalities. Such a question will be studied in a future work.
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Acknowledgements
The authors extend their appreciation to Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia).
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Jleli, M., Samet, B. A fixed point problem under two constraint inequalities. Fixed Point Theory Appl 2016, 18 (2016). https://doi.org/10.1186/s1366301605049
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DOI: https://doi.org/10.1186/s1366301605049
Keywords
 fixed point
 constraint inequalities
 partial order