Abstract
In this paper, we use Suzuki-type contraction to prove three fixed point theorems for generalized contractions in an ordered space equipped with two metrics; we obtain some generalizations of the Kannan fixed point theorem. Our results on partially ordered metric spaces generalize and extend some results of Ran and Reurings as well as of Nieto and Rodríguez-López. To illustrate the effectiveness of our main result, we give an application to matrix equations which involves monotone mappings.
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1 Introduction
It is well known that the Banach fixed point theorem plays a very important part in the resolution of various problems in nonlinear analysis such as integral equations and various nonlinear problems. It also has applications in various scientific disciplines. This theorem knew intense generalizations by the introduction of various type of contractions. We cite for example the generalizations obtained by Suzuki [20], Kikkawa and Suzuki [10], Mot and Petrusel [15], Dorić and Lazović [6], Bose and Roychowdhury [2], Singh, and Swami, Mishra, Chugh and Kamal [19].
2 Methods
Let \((X, d)\) be a metric space and T be a self-mapping on X.
T is called a generalized Kannan mapping if there exists \(r\in [ 0,\frac{1}{2}\mathclose[\) such that
T is called a Chatterjea mapping if there exists \(r\in [ 0,\frac{1}{2}\mathclose[\) such that
Kannan [8] (resp. Chatterjea [5]) shows the following results: If \((X,d)\) is a complete metric space then every generalized Kannan (resp. Chatterjea) mapping) has a unique fixed point.
Theorem 2.1
(Kikkawa and Suzuki [10])
LetTbe a mapping on a complete metric space\((X,d)\)andϕbe a non-increasing function from\([0,1\mathclose[\)onto\(\mathopen]\frac{1}{2},1]\)defined by
Assume that there exists\(r\in [0,1\mathclose[\)such that, for all\(x,y\in X\),
ThenThas a unique fixed point\(x^{\ast }\)and\({\lim_{n\to +\infty } T^{n}x = x^{\ast }}\)holds for every\(x\in X\).
In 1981, Fisher presented some related fixed points theorems involving two mappings on metrics spaces under some conditions on contractions.
Theorem 2.2
(Fischer [7])
Let\((X,d)\)and\((Y,\delta )\)be two metric spaces such that\((X,d)\)is complete. Let\(T:X\rightarrow Y\)and\(S:Y\rightarrow X\)be two mappings such that, for all\((x,y)\in X\times Y\),
where\(c\in {}[ 0,1\mathclose[\). Then there exists a unique pair\((x^{\ast },y^{\ast })\in X\times Y\)such that\(Tx^{\ast }=y^{\ast }\)and\(Sy^{\ast }=x^{\ast }\). Consequently, \(STx^{\ast }=x^{\ast }\)and\(TSy^{\ast }=y^{\ast }\).
Some generalizations of this result have appeared in many different directions. For details, we refer to [1, 3, 4, 8] and [16].
Chaira and Marhrani proved the following result which generalized the theorem of Fischer.
Theorem 2.3
([4])
Let\((X,d)\)and\((Y,\delta )\)be two metric spaces such that\((X,d)\)is complete; let\(T:X\rightarrow Y\)and\(S:Y\rightarrow X\)be two mappings satisfying the following condition:
For all\(x,y\in X\), one of the conditions
- 1.
\(d(x,STx)\leq d(x,Sy)\),
- 2.
\(\delta (y,TSy)\leq \delta (y,Tx)\),
implies
where\(\alpha :[0,+\infty \mathclose[ \rightarrow {}[ 0,1\mathclose[\)satisfies\(\lim \sup_{t\rightarrow t_{0}^{+}}\alpha (t)<1\), for all\(t_{0}\in {}[ 0,+\infty \mathclose[ \). Then there exists a unique pair\((x^{\ast },y^{\ast })\in X\times Y\)such that\(Tx^{\ast }=y^{\ast }\)and\(Sy^{\ast }=x^{\ast }\). And consequently, \(STx^{\ast }=x^{\ast }\)and\(TSy^{\ast }=y^{\ast }\).
In this article, we give a generalization of these results in the framework of ordered space equipped with two metrics by using the function \(\psi _{r_{0}} : [0,1\mathclose[ \rightarrow \mathopen]0,1]\) defined by
for any \(r_{0} \in \mathopen]0,1\mathclose[\). Our results on partially ordered metric spaces generalize the Ran and Reurings ([17]), Nieto and Rodríguez-López ([13]) fixed point theorems.
We also give an application to matrix equations to illustrate our results.
Let us recall some basic notions which will be used in our main discussions.
Definition 2.4
Let X be a nonempty set. Then \((X,d,\preceq )\) is called an ordered metric space if:
- (i)
\((X,d )\) is a metric space,
- (ii)
\((X,\preceq )\) is a partially ordered set.
Definition 2.5
([18])
Let \((X,\preceq )\) be a partially ordered set. We say that:
- (i)
\(x, y \in X\) are comparable if \(x \preceq y\) or \(y \preceq x\) holds.
- (ii)
\(f:X\longrightarrow X\) is monotone nondecreasing if \(x,y\in X\), \(x \preceq y\Longrightarrow f(x)\preceq f(y)\).
Definition 2.6
It is said that a partially ordered metric space X verifies the property (P) if
- (i)
for any nondecreasing sequence \((x_{n})_{n} \subset X \) converging to \(x\in X\), we have \(x_{n} \preceq x\) for all n;
- (ii)
for any sequence \((x_{n})_{n}\subset X\) which converges to x and \(y \in X\) such that \(x_{n} \preceq y\) for all integer n, we have \(x \preceq y\).
3 Main results
Theorem 3.1
Let\((X,\preceq )\)be a partially ordered set endowed with two metricsdandδsuch that\((X,d)\)is complete satisfying the property (P) and\(r_{0}\in\mathopen]0,\frac{-1+\sqrt{5}}{2}]\). LetTbe a monotone nondecreasing mapping onX. If there exists\(r\in [0,1\mathclose[\)such that, for all\(x,y\in X\)comparable, one of the conditions
- (i)
\(\psi _{r_{0}}(r) d(x,Tx) \le d(x,y)\),
- (ii)
\(\psi _{r_{0}}(r) \delta (x,Tx) \le \delta (x,y)\),
implies
and if there exists\(x_{0}\in X\)such that\(x_{0}\preceq Tx_{0}\), then there exists\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and
Moreover, the fixed point is unique if the set of fixed points\({\mathcal{F}}(T)\)is totally ordered.
Proof
If \(T ( x_{0} ) =x_{0}\) then the proof is finished. Assume that \(T(x_{0})\ne x_{0}\).
Since \(x_{0}\prec T ( x_{0} ) \) and T is monotone nondecreasing, we obtain by induction that
Put \(x_{n}=T^{n}(x_{0})\) for all \(n\geq 0 \). Since \(x_{n} \preceq x_{n+1}\) for all \(n\in \mathbb{N}\), and for any integer n, we have \(\psi _{r_{0}}(r) d(x_{n},Tx_{n}) \le d (x_{n},x_{n+1})\), then
hence
which implies
Consequently, \((x_{n})_{n\ge 0}\) is a Cauchy sequence in \((X,d)\) and \((X,\delta )\). Then there exists an element \(x^{\ast }\) in X, such that \({ \lim_{n\to +\infty } d(x_{n},x^{\ast })= 0}\).
Since X satisfies the property (P) and \(x_{n}\rightarrow x^{\ast }\), then \(x_{n}\preceq x^{\ast }\) for all \(n\geq 0\). As T is a monotone nondecreasing, we obtain
and by our assumption \(x^{\ast }\preceq Tx^{\ast }\).
We assert that \(Tx^{\ast }=x^{\ast }\). Assume that this is not the case.
If there exists a nondecreasing function \(\phi :\mathbb{N}\longrightarrow \mathbb{N}\) such that
and since \(\lim_{n} d(x_{\phi (n)},Tx_{\phi (n)})=0\), we obtain \(\lim_{n} d(x_{\phi (n)},Tx^{\ast })=0\); this contradicts our hypothesis. Then
First case: \(r\in [0,r_{0}\mathclose[\). Since \(x_{n} \preceq Tx^{\ast }\) for all n, we have
and then
On the other hand \(x^{\ast }\preceq Tx^{\ast }\), then
Assume that
If \(\delta (x^{\ast },Tx^{\ast })\le d(x^{\ast },Tx^{\ast })\). We have
which is absurd. The same conclusion is obtained if \(d(x^{\ast },Tx^{\ast })\le \delta (x^{\ast },Tx^{\ast })\). These considerations imply that one of the inequalities
holds. For \(x=Tx^{\ast }\) and \(y=x^{\ast }\), we have
Using (2), we obtain
We deduce from these relations that
Since \(r^{2}+r< 1\), we conclude to a contradiction with our hypothesis \(Tx^{\ast }\ne x^{\ast }\).
Second case: \(r\in [r_{0},1\mathclose[\). Put \(A=\{ z\in X ; z\text{ and }x_{n}\text{ are comparable}, \forall n\in \mathbb{N}\}\). A is nonempty, since \(x_{0} \) and \(x^{\ast }\) belongs to A.
Let \(z\in A \setminus \{x^{\ast }\}\), we have
and
therefore
which leads to
The inequality \(d(z,Tz) \le d(z,x^{\ast })+d(x^{\ast },Tz) \le d(z,x^{\ast })+r d(z,Tz)\) implies
By (3), z and \(x^{\ast }\) are comparable. Then
If \((x_{n})_{n}\) is a stationary sequence, we have \(x^{\ast }=Tx^{\ast }\) and if not, there exists a subsequence \((x_{\phi (n)})_{n}\), such that \(x_{\phi (n)}\ne x^{\ast }\), for all n. And consequently, for \(z=x_{\phi (n)}\), we have
which leads to
These considerations permit us to conclude that \(x^{\ast }=Tx^{\ast }\). These conclusions contradict the fact \(Tx^{\ast }\ne x^{\ast }\).
And then we conclude that \(Tx^{\ast }=x^{\ast }\).
We have \(\lim_{n\to +\infty } \delta (x_{n},x^{\ast })=0\). If not, \((\delta (x_{n},x^{\ast }) )_{n}\) does not converge to 0, then
since \(x_{n} \preceq x^{\ast }\) for all n, consequently
Thus, \(\lim_{n\to +\infty } \delta (x_{n+1},x^{\ast })=0\), which is a contradiction. Hence, \(\lim_{n\to +\infty } \delta (x_{n},x^{\ast })=0\).
For uniqueness: Let \(z\in X\) such that \(Tz=z\).
We have \(d(z,Tz) = 0 \le d(z,x^{\ast })\), since \({\mathcal{F}}(T)\) is totally ordered,
□
Example 3.2
Let \(X=[0,1]^{2} \). We define a partial order “⪯” in X as
Define d and δ by
for all \(((x,x'), (y,y'))\in X^{2}\).
Thus \((X,d,\preceq )\) and \((X,\delta ,\preceq )\) are two ordered metrics spaces. And define \(T:X\longrightarrow X\) by
We put \(r_{0} = \frac{-1+\sqrt{5}}{2}\) and \(r= \frac{1}{4}\). The map T is nondecreasing, and the space X satisfies the property (P) for d and δ. For all comparable \(( x,x' ), ( y,y') \in X\) such that \((x,x') \preceq (y,y')\), we have
Hence
In the same way
Moreover, \((\frac{1}{2},\frac{1}{2}) \preceq T(\frac{1}{2},\frac{1}{2})\). Therefore, T satisfies the hypotheses of Theorem 3.1. Hence, T has a unique fixed point \(x^{\ast }=(0,0)\).
If we assume, in the above theorem, that \(d=\delta \) and α is a constant function, we obtain a generalization of a Kannan-type contraction [9].
Corollary 3.3
Let\((X,d,\preceq )\)be an ordered metric space such that\((X,d)\)is complete andTa nondecreasing mapping on\((X,\preceq )\). Assume that there exists\(r = 2\alpha \), where\(\alpha \in [0,\frac{1}{2}\mathclose[\)such that, for all comparable\(x,y\in X\),
where\(r_{0} \in\mathopen]0,\frac{-1+\sqrt{5}}{2}]\).
If there exists\(x_{0}\in X\)such that\(x_{0}\preceq Tx_{0}\)andXsatisfies the property (P) ford, then there exists an element\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and\({\lim_{n\to +\infty } T^{n}x_{0}=x^{\ast }}\). Moreover, the fixed point of T is unique if for all any pair\(\lbrace x, y \rbrace \subset X\)admits an upper bound or a lower bound.
Proof
We remark that, for all comparable \(x,y\in X\),
By Theorem 3.1, we conclude that T has a fixed point.
For the uniqueness, we suppose that there exist two fixed points \(x,y \in X\). From the hypothesis on x and y there exists z in X such that \(z\preceq x\) and \(z\preceq y\).
By the monotony of T,
for all \(n\geq 0\). Set \(z_{n}=T^{n}(z)\), and we have
for all integer n. Then, for all \(n\geq 0\),
By induction, we show that
Since \(0\leq \alpha < \frac{1}{2}\), we have \(0\leq \frac{\alpha }{1-\alpha } <1\). Hence,
In the same way we prove that
By the triangle inequality,
and by (5) and (6) we conclude that \(x=y\). □
Note that, if X is not necessarily provided with a partial order, then we find the same result as that of Theorem 3.1. with some modifications in the proof consisting basically of avoiding the use of the order.
Theorem 3.4
Letdandδbe two metrics onXsuch that\((X,d)\)is complete and\(r_{0}\in\mathopen]0,\frac{-1+\sqrt{5}}{2}]\). LetTbe a self-mapping onX. If there exists\(r\in [0,1\mathclose[\)such that, for all\(x,y\in X\), one of the conditions:
- (i)
\(\psi _{r_{0}}(r) d(x,Tx) \le d(x,y)\),
- (ii)
\(\psi _{r_{0}}(r) \delta (x,Tx) \le \delta (x,y)\),
implies
then there exists an unique element\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and
holds for every\(x\in X\).
Example 3.5
Let \(X =\{ (0,0), (4,0), (0,4),(5,0), (4,5), (5,4)\}\) and define T by
Define d and δ by
for all \(((x,x'), (y,y'))\in X^{2}\).
We put \(r_{0} = \frac{-1+\sqrt{5}}{2}\) and \(r= \frac{2}{\sqrt{5}}\). Let \(((x,x'), (y,y'))\in X^{2}\).
First case: \(((x,x'),(y,y')) \notin \{((4,5),(5,4)),((5,4),(4,5))\}\), we have
Second case: \((x,x')=(4,5)\) and \((y,y')=(5,4)\). We have
Note that
and
Since \(d(T(x,x'),T(y,y'))= 9 \) and \(\delta (T(x,x'),T(y,y'))= \frac{9\sqrt{5}}{2}\),
Similarly for \((x,x')=(5,4)\) and \((y,y')=(4,5)\).
Hence, T satisfies the hypotheses of Theorem 3.4 but we do not have
for all \((x,x'),(y,y') \in X\). Hence, T has a unique fixed point \(x^{\ast }=(0,0)\).
With the same arguments as in the proof of Theorem 3.1, we obtain the following.
Theorem 3.6
Let\((X,\preceq )\)be a partially ordered set endowed with two metricsdandδsuch that\((X,d)\)is complete and\(r_{0}\in\mathopen]0,2(\sqrt{2}-1)]\). LetTbe a nondecreasing mapping; and assume that there exists\(r\in [0,1\mathclose[\)such that, for all comparable\(x,y\in X\), one of the conditions:
- (i)
\(\psi _{r_{0}}(r) d(x,Tx) \le d(x,y)\),
- (ii)
\(\psi _{r_{0}}(r) \delta (x,Tx) \le \delta (x,y)\),
implies
If there exists\(x_{0}\in X\)such that\(x_{0}\preceq Tx_{0}\)andXsatisfies the property (P) ford, then there exists an element\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and\(\lim_{n\to +\infty } d(T^{n}x_{0},x^{\ast })=\lim_{n \to +\infty } \delta (T^{n}x_{0}, x^{\ast })= 0 \).
Proof
First step. If \(T ( x_{0} ) =x_{0}\), then the existence of a fixed point is proved. Suppose that \(T ( x_{0} ) \ne x_{0}\). Following the lines of the proof of Theorem 2.3 we find that \(\lbrace {x_{n}} \rbrace = \lbrace {T^{n}(x_{0})\rbrace }\) is a convergent sequence in X. Indeed, by our assumption \(x_{n}\) and \(x_{n+1}\) are comparable, for every \(n=0,1,2,\ldots\) , for \(x=x_{n}\) and \(y=x_{n+1}\), we have \(\psi _{r_{0}}(r)d(x_{n},Tx_{n}) \le d (x_{n},x_{n+1})\); then
and
We deduce that
It follows that \((x_{n})_{n\ge 0}\) is Cauchy sequence in \((X,d)\) and \((X,\delta )\); therefore there exists an element \(x^{\ast }\) of X such that \(\lim_{n} d(x_{n},x^{\ast })= 0 \).
Second step. Assume that \((\delta (x_{n},x^{\ast }) )_{n}\) does not converge to 0. Since \(x_{n} \preceq x^{\ast }\) for all n, and
we obtain
The latter inequality implies \(\lim_{n}\delta (x_{n},Tx^{\ast })=0\); and then \(x^{\ast }\ne Tx^{\ast }\).
It follows that \(d(x_{n},Tx_{n})\le d(x_{n}, Tx^{\ast })\), for large integers n; since \(x_{n} \preceq Tx^{\ast }\) for all n,
which gives \(T^{2}x^{\ast }=x^{\ast }\).
Otherwise for \(x=x^{\ast }\) and \(y=Tx^{\ast }\), we have \(x \preceq y\) and \(\delta (x^{\ast },Tx^{\ast }) \le \delta (x^{\ast },Tx^{\ast })\); which implies
Consequently, \(Tx^{\ast }=x^{\ast }\); this gives us a contradiction and permits us to conclude \(\lim_{n} \delta (x_{n}, x^{\ast })= 0\).
Third step. Assume that \(x^{\ast }\ne Tx^{\ast }\). Since \(x_{n} \preceq Tx^{\ast }\) and for large integers, we have
which implies
thus,
and then
For \(x=x^{\ast }\) and \(y=Tx^{\ast }\), we have
and
Assume that
If \(d(x^{\ast },Tx^{\ast })\le \delta (x^{\ast },Tx^{\ast })\), the inequality (9) implies
and if \(d(x^{\ast },Tx^{\ast })\ge \delta (x^{\ast },Tx^{\ast })\), the inequality (8) implies
These considerations permit us to conclude that one of the inequalities
is satisfied. And consequently
so
Therefore by (7)
It follows that
Similarly, we have
If \(\delta (x^{\ast },Tx^{\ast }) \le d(x^{\ast },Tx^{\ast })\), then (10) implies
Since \(\frac{r}{2-r}(1+ \frac{r}{2})<1\), for all \(r\in [0,r_{0}\mathclose[\), we obtain \(Tx^{\ast }=x^{\ast }\).
If \(d(x^{\ast },Tx^{\ast })\le \delta (x^{\ast },Tx^{\ast })\), (11) implies
which gives \(Tx^{\ast }=x^{\ast }\). These conclusions contradict the fact \(Tx^{\ast }\ne x^{\ast }\), if \(r\in [0,r_{0}\mathclose[\).
Now assume \(r\in [r_{0},1\mathclose[\) and let \(z\in A \setminus \{x^{\ast }\}\) where \(A=\lbrace z\in X / z \text{ and } x_{n} \text{ are comparable}, \forall n\in \mathbb{N}\rbrace \). For large integers, we have \(\psi _{r_{0}}(r) d(Tx_{n},x_{n}) \le d(x_{n},z)\) and \(x_{n} \) is comparable to z for all n.
Thus
In the limit case, we obtain
and consequently
If \(\delta (z,x^{\ast })\leq d(z,x^{\ast })\), we have
and if \(d(z,x^{\ast })\leq \delta (z,x^{\ast })\), we have
Otherwise, one of the inequalities
is satisfied.
If \((x_{n})_{n}\) is a stationary sequence, we have \(x^{\ast }=Tx^{\ast }\) and if not, there exists a subsequence \((x_{\phi (n)})_{n}\), such that \(x_{\phi (n)}\ne x^{\ast }\), for all n. And consequently, for \(z=x_{\phi (n)}\), we have
which leads to
These considerations permit us to conclude that \(x^{\ast }=Tx^{\ast }\). These conclusions contradict the fact \(Tx^{\ast }\neq x^{\ast }\).
And then we conclude that \(Tx^{\ast }=x^{\ast }\). □
Proposition 3.7
Under the same conditions of Theorem 3.6, if for any pair\(\lbrace x, y \rbrace \subset X\)admits an upper bound or a lower bound, thenThas a unique fixed point.
Proof
Suppose that there exist two fixed points \(x, y \in X\) of T, from the hypothesis there exists z in X such that \(x\preceq z\) and \(y\preceq z\).
By the monotony of T,
for all \(n\geq 0\). Set \(z_{n}=T^{n}(z)\), since
for all integer n, then
so
for all \(n\geq 0\). Therefore,
where
Hence,
In the same way we prove that
By the triangle inequality,
and by (12) and (13) we conclude that \(x=y\). □
Corollary 3.8
Let\((X,d,\preceq )\)be an ordered metric space such that\((X,d)\)is complete, andTa nondecreasing mapping onXand\(r_{0}\in [0,2(\sqrt{2}-1)\mathclose[\). Assume that there exists\(r\in [0,1\mathclose[\)such that, for all\((x,y)\in X^{2}\), \(\psi _{r_{0}}(r) d(x,Tx) \le d(x,y)\)implies
If there exists\(x_{0}\in X\)such that\(x_{0}\preceq Tx_{0}\)andXsatisfies the property (P) ford, then there exists an element\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and\({\lim_{n\to +\infty } T^{n}x_{0}=x^{\ast }}\). Moreover, the fixed point of T is unique if for all any pair\(\lbrace x, y \rbrace \subset X\)admits an upper bound or a lower bound.
Theorem 3.9
Let\((X,\preceq )\)be a partially ordered set endowed with two metricsdandδsuch that\((X,d)\)is complete. LetTbe a monotone nondecreasing mapping; and assume that there exists\(r\in [0,\frac{1}{2}\mathclose[\)such that, for all comparable\(x,y\in X\), one of the following conditions:
- (i)
\(d(x,Tx) \le d(x,y)\),
- (ii)
\(\delta (x,Tx) \le \delta (x,y)\),
implies
If there exists\(x_{0}\in X\)such that\(x_{0}\preceq Tx_{0}\)andXsatisfies the property (P) ford, then there exists\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and
Moreover, the fixed point of T is unique if for all any pair\(\lbrace x, y \rbrace \subset X\)admits an upper bound or a lower bound.
Proof
Let \(x_{0}\in X\), if \(T ( x_{0} ) =x_{0}\), then the proof is finished. Suppose that
First step. We define a sequence \((x_{n})_{n}\) by \(x_{n}= T^{n} x_{0}\), for \(n\ge 0\). Following the same lines of the proof of Theorem 3.1, the sequence \((x_{n})_{n}\) is monotone for the partial order ⪯.
We have \(d(x_{n},Tx_{n}) \le d (x_{n},x_{n+1})\), then
so
and then
We have \(\frac{r}{1-r} \in [0,1\mathclose[\), \((x_{n})_{n\ge 0}\) is a Cauchy sequence for d and δ Then there exists \(x^{\ast }\) in X such that \(\lim_{n}d(x_{n},x^{\ast })= 0 \).
Second step. Assume that \((\delta (x_{n},x^{\ast }) )_{n}\) does not converge to 0. Then
since \(x_{n} \preceq x^{\ast }\) for all n, which leads to
The second equation in (14) gives
we obtain \({\limsup_{n\to +\infty } \delta (x_{n},Tx^{\ast })= \liminf_{n\to +\infty } \delta (x_{n},Tx^{\ast })=0 }\); then \({\lim_{n\to +\infty } \delta (x_{n},Tx^{\ast })=0 }\) and \(Tx^{\ast }\neq x^{\ast }\); hence \(d(x_{n},Tx_{n})\le d(x_{n}, Tx^{\ast })\), for large integers. Since \(x_{n} \preceq Tx^{\ast }\) for all n, it follows that
and then \(T^{2}x^{\ast }=x^{\ast }\).
We put \(y=Tx^{*}\). From the property (P) \(x^{*} \preceq Tx^{*}\), and \(\delta (x^{*},Tx^{*}) \leqslant \delta (x^{*},y)\) we have
Consequently \(Tx^{\ast }= x^{\ast }\). This contradiction shows that \(\lim_{n}\delta (x_{n},x^{\ast })=0\).
Third step. Assume \(Tx^{\ast }\ne x^{\ast }\). As above, we have
and then
If n goes to infinity, we obtain
For \(y=Tx^{\ast }\), we have \(\delta (x^{\ast },Tx^{\ast }) \le \delta (x^{\ast },y)\), so
Assume that \(\delta (x^{\ast },Tx^{\ast }) \le d(x^{\ast },Tx^{\ast })\), we have
with
Then
Since \(r^{2}+r<1\), we obtain \(d(x^{\ast },Tx^{\ast })=0\).
If \(\delta (x^{\ast },Tx^{\ast }) \ge d(x^{\ast },Tx^{\ast })\); we permute d and δ, and we obtain \(\delta (x^{\ast },Tx^{\ast })=0\). These conclusions contradict the fact \(Tx^{\ast }\neq x^{\ast }\).
And then we conclude that \(Tx^{\ast }=x^{\ast }\).
For the uniqueness, we suppose that there exist two fixed points \(x, y \in X\), from the hypothesis, there exists z in X such that \(x\preceq z\) and \(y\preceq z\).
By the monotony of T,
for all \(n\geq 0\). Set \(z_{n}=T^{n}(z)\), since
for all integer n, we have
so
for all \(n\geq 0\). Therefore,
where
Hence,
In the same way we prove that
By the triangle inequality,
and (16), (17) we conclude that \(x=y\). □
Example 3.10
We consider the space \(X=[0,1]\) ordered by “⪯” which is the reverse order of the usual order between the reals (\(x\preceq y \Leftrightarrow x \geq y\)) and endowed with the usual metric d and the metric defined by
We define
Set \(r=\frac{4}{9}\). The map T is nondecreasing, X satisfies the property (P) and \(1\preceq T(1)\).
For all \(x, y\in X\) such that \(x\preceq y\), we have \(r>\frac{1}{6}\), then \(\frac{6r-1}{r-1} \geq 1\) and \(\sin (y^{3})\leq \frac{6r-1}{r-1}x\), hence
Since \((1-r)\sin (y^{3})\leq (1-r)y\leq (6r-1)x\),
Hence, T has a unique fixed point \(x^{*}=0\).
Corollary 3.11
Let\((X,d,\preceq )\)be an ordered metric space such that\((X,d)\)is complete andTa nondecreasing mapping on\((X,\preceq )\). If there exists\(r\in [0,\frac{1}{2}\mathclose[\)such that, for all comparable\(x,y\in X\),
If there exists\(x_{0}\in X\)such that\(x_{0}\preceq Tx_{0}\)andXsatisfies the property (P) ford, then there exists an element\(x^{\ast }\in X\)such that\(Tx^{\ast }= x^{\ast }\)and\({\lim_{n\to +\infty } T^{n}x_{0}=x^{\ast }}\). Moreover, the fixed point of T is unique if for all any pair\(\lbrace x, y \rbrace \subset X\)admits an upper bound or a lower bound.
4 Application
Motivated by [11], in this section we apply Theorem 3.9 to study the existence and uniqueness of solution for the following general nonlinear matrix equation:
where Q is \(n\times n\) positive define matrix, \(\mathcal{P} ( n ) \) is the set of all \(n\times n\) Hermitian positive-definite matrices, A is \(n\times n\) nonsingular matrix, \(A^{\ast }\) is the Hermitian transpose of the matrix A and \(F: \mathcal{E}(n)\rightarrow \mathcal{E}(n)\) is a self-adjoint operator such that \(\mathcal{E}(n)\) is a nonempty subset of \(\mathcal{P}(n)\). This type of matrix equation arises in control theory, ladder networks, dynamic programming, stochastic filtering and statistics, etc. For \(M,N\in \mathcal{P}(n)\), the notation \(M \prec N \) means that \(N-M\) is positive definite. We equipped \(\mathcal{P}(n)\) with the partial ordered “⪯” given by
We denote by \(\Vert \cdot \Vert \) the spectral norm, i.e., \(\Vert A \Vert =\sqrt{\rho (A^{\ast }A)}= \Vert A^{ \ast } \Vert \) where \(\rho (A^{\ast }A)\) is the largest eigenvalue of \(A^{\ast }A\).
Let \(r \in \mathbb{N}^{*}\), the function
is a metric on \(\mathcal{P}(n)\). In fact, it is easy to show that δ satisfies:
- (i)
symmetry, that is, \(\delta _{r} ( A,B ) =\delta _{r} ( A,B )\), \(\forall A,B\in \mathcal{P}(n)\).
- (ii)
the triangle inequality, that is, \(\delta _{r} ( A,B ) \leq \delta _{r} (A,Z ) + \delta _{r} ( Z,B )\), \(\forall A,B,Z\in \mathcal{P}(n)\).
Moreover, \(\delta _{r}\) satisfies the identity of indiscernibles, i.e., \(\delta _{r} ( A,B ) =0\Longleftrightarrow A=B\). Indeed, let \(A, B \in \mathcal{P}(n)\), we have
Let \(\lambda >0\) be an eigenvalue of A and X a vector in the eigenspace \(E_{\lambda } (A )\) associated with λ. We have
Since all the eigenvalues of B are strictly positive, \(B- \lambda e^{2i\frac{k\pi }{r}}I_{n}\) is an inversible matrix. Hence,
Thus, \(E_{\lambda }(A) \subset E_{\lambda }(B)\). In the same way we have \(E_{\lambda }(B) \subset E_{\lambda }(A)\). Therefore,
and since A and B are diagonalizable, \(A=B\).
In the sequel, we consider the space \(\mathcal{P}(n)\) equipped by the metric \(\delta _{q}\) and the Thompson metric \(d : \mathcal{P}(n) \times \mathcal{P}(n) \longrightarrow \mathbb{R}_{+}\) given by
where \(W (A/B )= \inf \lbrace \lambda >0 : A\leq \lambda B \rbrace = \lambda _{\max } (B^{-\frac{1}{2}} A B^{- \frac{1}{2}} )\). It is easy to verify that \((\mathcal{P}(n), d )\) is a complete metric space (see [14]).
In the following lemmas, we give some elegant properties of the Thompson metric which play an important role in the proof of our main result of this section.
Lemma 4.1
([12])
Let\(d : \mathcal{P}(n) \times \mathcal{P}(n) \longrightarrow \mathbb{R}_{+}\)be a Thompson metric on the open convex cone\(\mathcal{P}(n)\), then, for any\(A,B \in \mathcal{P}(n)\)and nonsingular matrixN, we have the following conditions:
- 1.
\(d (A,B )= d (A^{-1},B^{-1} )=d (N^{*}AN,N^{*}BN )\), where\(A^{-1}\), \(B^{-1}\)are the inversion of matricesAandB, respectively;
- 2.
\(d (A^{r},B^{r} )\leq rd (A,B )\), \(r \in [0,1]\);
- 3.
\(d (N^{*}A^{r}N,N^{*}B^{r}N )\leq \vert r \vert d (A,B )\), \(r \in [-1,1]\).
Lemma 4.2
([12])
For any\(A, B, C, D \in \mathcal{P}(n)\),
Especially, \(d (A+C,B+C )\leq d (A,B )\).
Lemma 4.3
([21])
For any\(A,B \in \mathcal{P}(n)\), if\(A \preceq B\), then\(A^{r} \preceq B^{r}\)for all\(r \in\mathopen]0,1]\), and\(B^{r} \preceq A^{r}\)for all\(r \in [-1,0\mathclose[\).
Theorem 4.4
Let\(X_{0} \in \mathcal{P}(n)\)and\(\mathcal{E}(n)= \{ X \in \mathcal{P}(n) : X_{0} \preceq X \} \). If the operatorFis nondecreasing and for all\(X, Y \in \mathcal{E}(n)\)comparable, one of the assertions
- (i)
\(\Vert \ln (X^{-\frac{1}{2}} ( Q+A^{*}F(X)A )^{ \frac{1}{q}} X^{-\frac{1}{2}} ) \Vert \leq \Vert \ln (X^{- \frac{1}{2}} Y X^{-\frac{1}{2}} ) \Vert \),
- (ii)
\(\Vert X^{q} - Q - A^{*}F(X)A \Vert \leq \Vert X^{q} -Y^{q} \Vert \),
implies the system
and\(X_{0}^{q} \preceq Q\), then the matrix equation (18) has a unique solution.
Proof
\(\mathcal{E}(n)\) is nonempty subset of \(\mathcal{P}(n)\). We show that \(\mathcal{E} (n)\) is closed for distance d. In fact, let \(( Y_{k} )_{k}\) be a sequence of \(\mathcal{E}(n)\) which converges to \(Y\in \mathcal{P}(n)\). If the set \(\{ k \in \mathbb{N}: Y_{k} = X_{0} \} \) is infinite, there exists a nondecreasing function \(\phi : \mathbb{N}\longrightarrow \mathbb{N}\) such that \(Y_{\phi (k)} = X_{0}\), \(\forall k \in \mathbb{N}\), then \(Y= X_{0} \in \mathcal{E}(n)\). If not, \(Y_{k} \neq X_{0} \), for large integers k, then
Since \(\mathcal{P}(n) \) is closed for distance d, because \((\mathcal{P}(n),d)\) is a complete metric space, \(\mathcal{P}(n) + X_{0} \) is closed. As \(Y_{k} \rightarrow Y\), then \(Y \in \mathcal{P}(n) +X_{0}\). That is to say, \(Y \in \mathcal{E} (n) \).
First step. Let \(T:\mathcal{E}(n) \longrightarrow \mathcal{E}(n)\) be a function given by
Since F is nondecreasing, T is well defined and conserves the partial ordering on \(\mathcal{E} ( n ) \). Indeed, let \(X \in \mathcal{E}(n)\), we have
As \(X^{q}_{0} \preceq Q\) and \(A^{*} F ( X_{0} )A \in \mathcal{P}(n)\), then \(X^{q}_{0} \preceq Q + A^{*} F ( X_{0} )A \). By Lemma 4.3 and (20), we have
Let \(B,C\in \mathcal{E}(n)\) such that \(B \preceq C\), we have \(( T ( C ) ) ^{q}- ( T ( B ) )^{q}=A^{*} ( F ( C ) - F ( B ) )A\), a positive semidefinite matrix, \(( T ( B ) ) ^{q} \preceq ( T ( C ) )^{q} \). Combining Lemma 4.3 and \(0\leq \frac{1}{q}<1\), then \(T ( B ) \preceq T ( C )\).
Second step. From the hypothesis, we have \(X_{0}^{q} \preceq Q\). Since \(A^{*}F ( X_{0} ) A\in \mathcal{P}(n) \),
Hence \(X_{0} \preceq T ( X_{0} ) \).
Next, we show that \(\mathcal{E}(n)\) satisfies the property (P) for d, given in Definition 2.6. In fact, for (i), let \(( X_{k} )_{k}\) be a nondecreasing sequence of \(\mathcal{E}(n)\) which converges to \(X\in \mathcal{E}(n)\).
Fix \(m \in \mathbb{N}\) arbitrary. For all \(k >m\),
Then \(\mathcal{P}(n) + X_{m} \) is closed, since \(\mathcal{P}(n) \) is closed for distance d. As \(X_{k}\rightarrow X\), then
Thus
For (ii), let \(( X_{k} ) _{k}\subset \mathcal{E}(n) \) and \(Y\in \mathcal{E}(n) \) such that \(X_{k}\rightarrow X\) and \(X_{k} \preceq Y\), for all integer k. Then \(X \preceq Y\). Indeed, if \(X_{k} = Y \) for an infinity of the integer k, then there exists a nondecreasing function \(\psi : \mathbb{N}\longrightarrow \mathbb{N}\) such that \(Y_{\psi (k)} = Y\), \(\forall k \in \mathbb{N}\), so \(X= Y\). If not, \(X_{k} \neq Y \), for large integers k, then
Since
for large integers k, then \(d(Y,A_{k} +X) \rightarrow 0\). Since \(A_{k} +X \in \mathcal{P}(n) +X\) and \(A_{k} +X \rightarrow Y \). Thus, \(Y \in \mathcal{P}(n) +X\), consequently \(X \preceq Y\).
Third step. Let X, Y be two comparable elements in \(\mathcal{E}(n)\). If we assume that
then one of the following assertions is verified:
- (i)
\(\Vert \ln (X^{-\frac{1}{2}} ( Q+A^{*}F(X)A )^{ \frac{1}{q}} X^{-\frac{1}{2}} ) \Vert \leq \Vert \ln (X^{- \frac{1}{2}} Y X^{-\frac{1}{2}} ) \Vert \),
- (ii)
\(\Vert X^{q} - Q - A^{*}F(X)A \Vert \leq \Vert X^{q} -Y^{q} \Vert \),
which implies
Using Lemmas 4.1 and 4.2, we have
And
On the other hand, we get
Also
Therefore, there exists \(r=\frac{1}{q}\in {}[ 0,\frac{1}{2}\mathclose[\) for all comparable elements \(X,Y\in \mathcal{E}(n)\), one of the conditions:
- (i)
\(d(X,T(X)) \le d(X,Y)\),
- (ii)
\(\delta _{q}(X,T(X)) \le \delta _{q}(X,Y)\),
implies
Thus, according to Theorem 3.9, we can easily draw the conclusion that there exists \(X^{*} \in \mathcal{E}(n)\) such that \(T(X^{*})=X^{*}\).
According to the properties of \(\mathcal{P} ( n )\), we find that every pair \(\lbrace X, Y \rbrace \subset \mathcal{E} ( n )\) admits an upper bound or a lower bound. In fact, let \(X,Y\in \mathcal{E}(n)\), then there exists \(Z=X+Y \in \mathcal{E} ( n )\) such that \(X\preceq Z\) and \(Y\preceq Z\). Hence T has a unique fixed point.
Therefore, \(X^{*}\) is the unique solution of the matrix equation (18). □
Example 4.5
We consider the space \(\mathcal{E}(n)=\{ X \in \mathcal{P}(n) : \frac{1}{2} I_{n} \preceq X \}\) equipped by the metric \(\delta _{3}\) and the Thompson metric d where \(I_{n}\) is the identity matrix. Consider the following general nonlinear matrix equation:
where \(Q=A=I_{n}\) and \(F:\mathcal{E}(n) \rightarrow \mathcal{E}(n)\) is a mapping defined by
- (1)
F is not a monotone nonexpansive mapping for the metric \(\delta _{3}\) and the Thompson metric d,
- (2)
F is monotone satisfies condition (19) for all \(X, Y \in \mathcal{E}(n) \) comparable.
Setting \(X=\frac{8}{15} I_{n}\), \(Y=\frac{1}{2} I_{n}\), we obtain
However, F is not a nonexpansive mapping for the Thompson metric d. Setting \(X=\frac{27}{50} I_{n}\), \(Y=\frac{1}{2} I_{n}\), we obtain
Then F is not a nonexpansive mapping for the metric \(\delta _{3}\). Now, we show that F is monotone satisfies condition (19). In fact, let \(X \in \mathcal{E}(n) \setminus \lbrace \frac{1}{2} I_{n} \rbrace \) and \(Y = \frac{1}{2} I_{n}\) such that \(Y \preceq X\), we have
and
Therefore
In the same way, we have
and
Then
Since F is monotone it satisfies (27) and (28) for all \(X, Y \in \mathcal{E}(n) \) comparable and there exists \(X_{0} = \frac{1}{2} I_{n} \) such that \(( X_{0} )^{3} \preceq Q \), then all the conditions of Theorem 4.4 are satisfied. Consequently, the matrix equation (26) has a unique solution \(X^{*}=\sqrt[3]{\frac{7}{4}}I_{n}\).
References
Aliouche, A., Fisher, B.: A related fixed point theorem for two pairs of mappings on two complete metric spacs. Hacet. J. Math. Stat. 34, 39–45 (2005) http://www.hjms.hacettepe.edu.tr/uploads/596a9157-b6cf-4ca5-afe9-e9ef9357d7d4.pdf
Bose, R.K., Roychowdhury, M.K.: Fixed point theorems for some generalized contractive multivalued mappings and fuzzy mappings. Mat. Vesn. 63(1), 7–26 (2011) http://www.emis.de/journals/MV/111/mv11102.pdf
Chaira, K., Marhrani, E.: Some related fixed points theorems for a pair of mapping on two metric spaces. Int. J. Pure Appl. Math. 93(2), 191–200 (2014) https://ijpam.eu/contents/2014-93-2/4/4.pdf
Chaira, K., Marhrani, E.: Some related fixed points theorems of metric spaces. Bull. Math. Anal. Appl. 7(2), 46–55 (2015) http://www.bmathaa.org/repository/docs/BMAA7-2-5.pdf
Chatterjea, S.K.: Fixed-point theorems. C. R. Acad. Bulg. Sci. 25, 727–790 (1972)
Doric, D., Lazovic, R.: Some Suzuki-type fixed point theorems for generalized multivalued mappings and applications. Fixed Point Theory Appl. 2011, Article ID 40 (2011). https://doi.org/10.1186/1687-1812-2011-40
Fisher, B.: Fixed point on two metric spaces. Glas. Mat. 16(36), 333–337 (1981)
Kannan, R.: Some results on fixed points. Bull. Calcutta Math. Soc. 60, 71–76 (1968)
Kannan, R.: Some results on fixed points II. Am. Math. Mon. 76, 405–408 (1969). https://doi.org/10.1080/00029890.1969.12000228
Kikkawa, M., Suzuki, T.: Three fixed point theorems for generalized contractions with constants in complete metric spaces. Nonlinear Anal., Theory Methods Appl. 69(9), 2942–2949 (2008). https://doi.org/10.1016/j.na.2007.08.064
Liao, A.P., Yao, G.Z., Duan, X.F.: Thompson metric method for solving a class of nonlinear matrix equation. Appl. Math. Comput. 216, 1831–1836 (2010). https://doi.org/10.1016/j.amc.2009.12.022
Lim, Y.: Solving the nonlinear matrix equation \(X = Q + \sum_{i}^{m} M_{i}^{*}X^{\delta _{i}}M_{i}\) via a contraction principle. Linear Algebra Appl. 430(4), 1380–1383 (2009). https://doi.org/10.1016/j.laa.2008.10.034
Nieto, J.J., Rodriguez-Lopez, R.: Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations. Order 22(3), 223–239 (2005). https://doi.org/10.1007/s11083-005-9018-5
Nussbaum, R.D.: Hilbert’s projective metric and iterated nonlinear maps. Mem. Am. Math. Soc. 391 (1988). https://doi.org/10.1090/memo/0391
Petrusel, A., Mot, G.: Fixed point theory for a new type of contractive multivalued operators. Nonlinear Anal., Theory Methods Appl. 70(9), 3371–3377 (2009). https://doi.org/10.1016/j.na.2008.05.005
Petrusel, A., Rus, I.A.: Fixed point theory in terms of a metric and of an order relation. Fixed Point Theory 20(2), 601–622 (2019). https://doi.org/10.24193/fpt-ro.2019.2.40
Ran, A.C.M., Reurings, M.C.B.: A fixed point theorem in partially ordered sets and some applications to matrix equations. Proc. Am. Math. Soc. 132(5), 1435–1443 (2004). https://doi.org/10.1090/S0002-9939-03-07220-4
Roman, S.: Lattices and Ordered Sets. Springer, Berlin (2008). ISBN 978-0-387-78901-9. https://doi.org/10.1007/978-0-387-78901-9
Singh, S.L., Mishra, S.N., Chugh, R., Kamal, R.: General common fixed point theorems and applications. J. Appl. Math. 2012, Article ID 902312 (2012). https://doi.org/10.1155/2012/902312
Suzuki, T.: A generalized Banach contraction principle that characterises metric completeness. Proc. Am. Math. Soc. 136(5), 1861–1869 (2008). https://doi.org/10.1090/S0002-9939-07-09055-7
Zhan, X.: Computing the extremal positive definite solutions of a matrix equation. SIAM J. Sci. Comput. 17(5), 1167–1174 (1996). https://doi.org/10.1137/S1064827594277041
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Rida, O., Karim, C. & El Miloudi, M. Related Suzuki-type fixed point theorems in ordered metric space. Fixed Point Theory Appl 2020, 7 (2020). https://doi.org/10.1186/s13663-020-00674-0
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DOI: https://doi.org/10.1186/s13663-020-00674-0