Abstract
In this paper, the concept of ordered contractive mappings is introduced on cone metric spaces over Banach algebras, and some existence theorems of fixed points for such mappings are obtained. As an application of a result, an example is given at the end of the paper.
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1 Introduction
The study of fixed points for nonlinear mappings is an important subject of nonlinear function analysis, and the theory of fixed points is frequently applied to nonlinear integral equations and differential equations (see [1–5] and the references given there). Besides, the work of Caristi [6], in which a partial ordering was introduced in metric spaces by a function and a fixed point theorem was proved, is worthy of attention. In 2007, in the work of Huang and Zhang [7], the concept of cone metric spaces as a generalization of general metric spaces was introduced, in which the distance \(d(x, y)\) of x and y is defined to be a vector in an ordered Banach space E. It was also proved that the Banach contraction principle remains true in the setting of cone metric spaces. After that, based on the work of Huang and Zhang [7], the fixed point results of some mappings with certain contractive property on cone metric spaces appeared like mushrooms after rain (see [8–14] and the references therein and [15–23]). Among those works, the results of [13, 14] attract much attention since they give an answer to the natural problem that whether cone metric spaces are equivalent to metric spaces in terms of the existence of fixed points of the involved mappings. Concretely, the authors showed that any cone metric space \((X, d)\) is equivalent to a usual metric space \((X, d^{*})\) if the real-valued metric function \(d^{*}\) is defined by a nonlinear scalarization function \(\xi_{e}\) (see [13]) or by a Minkowski functional \(q_{e}\) (see [14]).
In 2013, in order to generalize the Banach contraction principle to a more general form, Liu and Xu [24, 25] introduced the concept of cone metric spaces over Banach algebras by replacing Banach spaces with Banach algebras and proved some fixed point theorems of generalized Lipschitz mappings with weaker and natural conditions on the generalized Lipschitz constant k by means of spectral radius and pointed out that it is significant to introduce the concept of cone metric spaces over Banach algebras because it can be proved that cone metric spaces over Banach algebras are not equivalent to metric spaces in terms of the existence of fixed points of the generalized Lipschitz mappings (see [25]). Liu and Xu [25] showed that their results could not be reduced to a consequence of corresponding results in metric spaces by means of the methods in the literature. In 2014, Xu and Radenović [26] keenly discovered that the proofs of the main results of [25] strongly depend on the condition that the underlying solid cone is normal. Naturally, they considered that whether the conclusions of [25] remain valid if ‘normal’ is deleted from the hypotheses. By means of some properties of spectral radius, they proved that the main results of [25] still hold without the assumption of normality of the cone involved. Hence Xu and Radenović [26] improved the results of [25].
In this paper, on the basis of [25, 26], we introduce a partial ordering given by a continuous function in cone metric spaces over Banach algebras as well as the concept of ordered contractive mappings that differ from the known contractive mappings, and present several fixed point results of such mappings under some natural conditions. Finally, as an application of one of our results, we give a concrete example.
2 Preliminaries
Consistent with Huang and Zhang [7] and Liu and Xu [25], the following definitions and results are needed in the sequel.
Let \(\mathcal{A}\) always be a real Banach algebra, that is, \(\mathcal {A}\) is a real Banach space in which an operation of multiplication is defined, subject to the following properties (for all \(x, y, z \in\mathcal{A}\), \(\alpha\in\mathbb{R}\)):
-
(1)
\((xy)z = x(yz)\);
-
(2)
\(x(y + z)=xy + xz\) and \((x + y)z = xz + yz\);
-
(3)
\(\alpha(xy) = (\alpha x)y = x(\alpha y)\);
-
(4)
\(\|xy\|\leq\|x\|\|y\|\).
Here and subsequently, we assume that a Banach algebra has a unit (i.e., a multiplicative identity) e such that \(ex = xe = x\) for all \(x\in \mathcal{A}\). \(x \in \mathcal{A}\) is said to be invertible if there is an inverse element \(y\in \mathcal{A}\) such that \(xy = yx = e\). The inverse of x is denoted by \(x^{-1}\). We refer the reader to [25] for more details.
A non-empty closed convex subset P of a Banach algebra \(\mathcal{A}\) is called a cone if
-
(i)
\(\{\theta, e\}\subset P\);
-
(ii)
\(\alpha P+\beta P \subset P\);
-
(iii)
\(P^{2}=P P \subset P\);
-
(iv)
\(P \cap(-P)=\{\theta\}\),
where θ denotes the null of the Banach algebra \(\mathcal{A}\).
Fix a cone \(P\subset\mathcal{A}\), a partial ordering ‘⪯’ with respect to P can be defined by \(x\preceq y\) if and only if \(y-x\in P\). \(x\prec y\) stands for \(x\preceq y\) and \(x\neq y\). \(x\ll y\) stands for \(y-x\in\operatorname{int}(P)\), here \(\operatorname{int}(P)\) denotes the interior of P. P is called a solid cone if \(\operatorname{int}(P) \neq \emptyset\). P is called normal if there exists a positive constant N such that for all \(x, y\in\mathcal{A}\), \(\theta\preceq x \preceq y \Rightarrow\|x\|\leq N\|y\|\).
Definition 2.1
([25])
Let X be a non-empty set and \(\mathcal{A}\) be a real Banach algebra. Suppose that the mapping \(d: X\times X\rightarrow\mathcal{A}\) satisfies:
-
(1)
\(\theta\preceq d(x,y)\) for all \(x, y\in X\) and \(d(x,y)=\theta\) if and only if \(x=y\);
-
(2)
\(d(x,y)=d(y,x)\) for all \(x,y\in X\);
-
(3)
\(d(x,y)\preceq d(x,z)+d(z,y)\) for all \(x,y,z\in X\).
Then d is called a cone metric on X and \((X, d)\) is called a cone metric space over the Banach algebra \(\mathcal{A}\).
See [25] for some examples of cone metric spaces over Banach algebras.
Definition 2.2
([25])
Let \((X, d)\) be a cone metric space over the Banach algebra \(\mathcal {A}\), \(x\in X\) and let \(\{x_{n}\}\) be a sequence in X. Then:
-
(1)
\(\{x_{n}\}\) converges to x if for each \(c\in\mathcal{A}\) with \(\theta\ll c\), there is a natural number N such that \(d(x_{n},x)\ll c\) for all \(n>N\). We denote this by \(\lim_{n\rightarrow\infty}x_{n}=x\) or \(x_{n}\rightarrow x\).
-
(2)
\(\{x_{n}\}\) is a Cauchy sequence if for each \(c\in\mathcal{A}\) with \(\theta\ll c\), there is a natural number N such that \(d(x_{n},x_{m})\ll c\) for all \(n, m>N\).
-
(3)
\((X, d)\) is a complete cone metric space if every Cauchy sequence is convergent.
Definition 2.3
Let \((X,d)\) be a cone metric space over the Banach algebra \(\mathcal {A}\) and \(\varphi: X \rightarrow\mathcal{A}\) be a mapping. A relation ‘⩽’ (for the sake of differing from the partial ordering ‘⪯’ in \(\mathcal{A}\), we denote it by ‘⩽’) in X is defined as follows:
Clearly ‘⩽’ is a partial ordering in X and \(x\leqslant y\) implies \(\varphi(x)\succeq\varphi(y)\). Here we call it the partial ordering induced by φ. Meanwhile, \((X, d)\) is called a partial ordering cone metric space over the Banach algebra \(\mathcal{A}\).
Definition 2.4
Let \((X,d)\) be a partial ordering cone metric space over the Banach algebra \(\mathcal{A}\). We say that \(x, y \in X \) are comparable if \(x\leqslant y\) or \(y \leqslant x\) holds.
If \(x \leqslant y\) and \(x\neq y\), we write \(x< y\). We write \(x=x\vee y\) if \(y\leqslant x\) and write \(y=x\vee y\) if \(x\leqslant y\). The same notions as those in Definition 2.4 can be defined in \(\mathcal{A}\) as follows: For any \(x, y \in\mathcal{A}\), if \(x\preceq y\) or \(y \preceq x\) holds, we say that x and y are comparable. Let \(u, v \in\mathcal {A}\). Similarly, we write \(v=u\vee v\) if \(u\preceq v\) and write \(u=v\vee u\) if \(v\preceq u\). From Definition 2.3, it is evident that \(\varphi(x)\) and \(\varphi(y)\) are comparable in \(\mathcal{A}\) if \(x, y \in X \) are comparable.
Remark 2.1
Suppose that \(\mathcal{A}\) is a real Banach algebra, \(u, v, w \in\mathcal{A}\). The following results are clear.
-
(i)
If u and v are comparable, then \(u-v\) and \(v-u\) are comparable and \(\theta\preceq(u-v)\vee(v-u)\).
-
(ii)
If u and v, u and w together with v and w are comparable, then
$$(u-v)\vee(v-u)\preceq\bigl((u-w)\vee(w-u)\bigr)+\bigl((w-v)\vee(v-w)\bigr). $$
Definition 2.5
Let \((X,d)\) be a cone metric space over the Banach algebra \(\mathcal {A}\) and \(A: X \rightarrow X\) be a mapping. We say that A is continuous if for any \(\{x_{n}\}\subset X\), \(x_{n}\rightarrow x\) implies \(Ax_{n}\rightarrow Ax\) (\(n\rightarrow\infty\)).
In the rest of this section, we always assume that \(\mathcal{A}\) is a real Banach algebra and \((X,d)\) is a complete partial ordering cone metric space over \(\mathcal{A}\) with the partial ordering ‘⩽’ induced by φ, where \(\varphi: X\rightarrow\mathcal{A}\) is continuous, P is a solid cone of \(\mathcal{A}\) which gives the partial ordering ‘⪯’ in \(\mathcal{A}\).
Definition 2.6
A mapping \(A: X \rightarrow X\) is said to be φ-ordered contractive if there exists \(k\in P\) with \(0\leq r(k)<1\) such that for any \(x, y \in X\), if x and y are comparable, then Ax and Ay are comparable and
Remark 2.2
See Example 3.1 for a support example of Definition 2.3, Definition 2.4, Definition 2.5 and Definition 2.6.
Lemma 2.1
([7])
Let \(\{x_{n}\}\) and \(\{y_{n}\}\) be two sequences in X and \(x_{n}\rightarrow x_{0}\), \(y_{n}\rightarrow y_{0}\) as \(n\rightarrow\infty\). Then \(d(x_{n}, y_{n})\rightarrow d(x_{0},y_{0})\) (\(n\rightarrow\infty\)).
Lemma 2.2
Let \(u, v_{n} \in X\) (\(n=1,2,\ldots \)). If for any natural number n, u and \(v_{n}\) are comparable and \(v_{n}\rightarrow v_{0}\) (\(n\rightarrow\infty\)), then u and \(v_{0}\) are comparable.
Proof
Since for any natural number n, u and \(v_{n}\) are comparable, there exists a subsequence \(\{v_{n_{k}}\}\) of \(\{v_{n}\}\) such that for any k, \(u \leqslant v_{n_{k}}\) or \(v_{n_{k}} \leqslant u\). Without loss of generality, we assume that for any k, \(u\leqslant v_{n_{k}}\). As \(v_{n}\rightarrow v_{0}\) when \(n\rightarrow\infty\), we have \(v_{n_{k}}\rightarrow v_{0}\) as \(k\rightarrow\infty\). By Lemma 2.1 and the continuity of φ and noting \(u\leqslant v_{n_{k}}\) for all \(k\geq1\), we have
So \(u \leqslant v_{0}\). That is u and \(v_{0}\) are comparable. □
Lemma 2.3
Let \(u_{n}, v_{n} \in X\) (\(n=0,1,2,\ldots \)). If for any natural number n, \(u_{n}\) and \(v_{n}\) are comparable and \(u_{n}\rightarrow u_{0}\), \(v_{n}\rightarrow v_{0}\) (\(n\rightarrow\infty\)), then \(u_{0}\) and \(v_{0}\) are comparable. In particular, if for any natural number n, \(u_{n}\leqslant v_{n}\) and \(u_{n}\rightarrow u_{0}\), \(v_{n}\rightarrow v_{0}\) (\(n\rightarrow\infty\)), then \(u_{0}\leqslant v_{0}\).
Proof
Because the proof is similar to that of Lemma 2.2, we omit it. □
Lemma 2.4
([26])
Let x, y be vectors in \(\mathcal {A}\). If x and y commute, then the spectral radius r satisfies the following properties:
-
(i)
\(r(xy)\leq r(x)r(y)\);
-
(ii)
\(r(x+y)\leq r(x)+r(y)\);
-
(iii)
\(|r(x)-r(y)|\leq r(x-y)\).
Lemma 2.5
([26])
Let \(k\in\mathcal{A}\). If \(0\leq r(k)<1\), then \(e-k\) is invertible and \(r((e-k)^{-1})\leq(1-r(k))^{-1}\).
3 Main results
In this section, we will present the main results and their proofs. For simplicity, we always assume that \(\mathcal{A}\) is a real Banach algebra and \((X,d)\) is a complete partial ordering cone metric space over \(\mathcal{A}\) with the partial ordering ‘⩽’ induced by φ, where \(\varphi: X\rightarrow\mathcal{A}\) is continuous. Let P be a normal solid cone of \(\mathcal{A}\) which gives the partial ordering ‘⪯’ in \(\mathcal{A}\).
Theorem 3.1
If a continuous mapping \(A: X\rightarrow X\) is φ-ordered contractive and there exists \(x_{0}\in X\) such that \(x_{0}\) and \(Ax_{0}\) are comparable, then A has a fixed point \(x^{*}\) in X, that is, \(Ax^{*}=x^{*}\). Furthermore, the iterative sequence \(x_{n}=Ax_{n-1}\) (\(n=1,2,\ldots\)) converges to \(x^{*}\) and
where k is the element in P satisfying (2.1).
Proof
Put
Since \(x_{0}\) and \(Ax_{0}\) are comparable and A is φ-ordered contractive, \(x_{1}=Ax_{0}\) and \(x_{2}=Ax_{1}\) are comparable. By induction, it is not difficult to prove that for any natural number n, \(x_{n}\) and \(x_{n+1}\) are comparable. As A is φ-ordered contractive, there exists \(k\in P\) with \(0\leq r(k)<1\) such that for any natural number n,
By the definition of ‘⪯’ and the ordered contractive property of A, we get that
For any pair of natural numbers \(n, m\) with \(m>n\), we have
Then by the same argument as that used in Theorem 3.1 of [26], we can prove that \(\{x_{n}\}\) is a Cauchy sequence. Since X is complete, there exists \(x^{*}\in X\) such that \(x_{n}\rightarrow x^{*}\) (\(n\rightarrow\infty\)). The continuity of A implies that \(x_{n+1}=Ax_{n}\rightarrow Ax^{*}\) (\(n\rightarrow\infty\)). So \(Ax^{*}=x^{*}\), that is, \(x^{*}\) is a fixed point of A. Furthermore, by Lemma 2.1, we have
□
Theorem 3.2
If \(A: X\rightarrow X\) is φ-ordered contractive and there exists \(x_{0}\in X\) such that for any natural number n, \(x_{0}\) and \(A^{n}x_{0}\) are comparable, then the same conclusions as those of Theorem 3.1 hold.
Proof
Define \(\{x_{n}\}\) by \(x_{n}=Ax_{n-1}\) (\(n=1, 2, \ldots \)). By a proof similar to that of Theorem 3.1, one can prove easily that \(\{x_{n}\}\) is a Cauchy sequence. The completeness of X implies that there is \(x^{*}\in X\) such that \(x_{n}\rightarrow x^{*}\) (\(n\rightarrow\infty\)). Next, we show that \(x^{*}\) is a fixed point of A.
For any pair of natural numbers \(n, m\) with \(n< m\), \(x_{0}\) and \(x_{m-n}=A^{m-n}x_{0}\) are comparable according to the given conditions. As A is φ-ordered contractive, \(Ax_{0}\) and \(Ax_{m-n}\) are comparable. Continue this progress, we get that \(x_{n}=A^{n}x_{0}\) and \(x_{m}=A^{n}x_{m-n}=A^{m}x_{0}\) are comparable. Let \(m\rightarrow\infty\), Lemma 2.2 shows that for any natural number n, \(x_{n}\) and \(x^{*}\) are comparable. Because A is φ-ordered contractive, \(Ax_{n}\) and \(Ax^{*}\) are comparable and there exists \(k\in P\) with \(0\leq r(k)<1\) such that
Noting the ordered contractive property of A, we have
Let \(u_{n}=d(x_{n+1}, Ax^{*})\), \(v_{n}=k[(\varphi(x_{n})-\varphi(x^{*}))\vee(\varphi (x^{*})-\varphi(x_{n}))]\), \(n=0,1,2,\ldots \) . Then, for any natural number n, \(u_{n}\leqslant v_{n}\). Notice that \(u_{n}=d(x_{n+1}, Ax^{*})\rightarrow d(x^{*},Ax^{*})\) and the continuity of φ, by Lemma 2.3, we have \(d(x^{*},Ax^{*})\preceq k[\varphi(x^{*})-\varphi(x^{*})]=\theta\), so \(Ax^{*}=x^{*}\).
Moreover, we get that
Hence Lemma 2.1 gives that
□
Theorem 3.3
Let \(A: X \rightarrow X\) be a continuous mapping satisfying the following conditions:
-
(i)
There exist \(\lambda_{1}, \lambda_{2}\in P\) with \(0\leq r(\lambda _{1})+r(\lambda_{2})<1\) such that for any comparable pair \(x, y\in X\), Ax and Ay are comparable. Moreover, if x and Ax, y and Ay are comparable, then
$$\begin{aligned}& \bigl(\varphi(Ax)-\varphi(Ay)\bigr)\vee\bigl(\varphi(Ay)-\varphi(Ax)\bigr) \\& \quad \preceq \lambda_{1}\bigl[\bigl(\varphi(Ax)-\varphi(x)\bigr)\vee\bigl( \varphi(x)-\varphi (Ax)\bigr)\bigr] \\& \qquad {}+\lambda_{2}\bigl[\bigl(\varphi(Ay)-\varphi(y)\bigr)\vee\bigl( \varphi(y)-\varphi(Ay)\bigr)\bigr]. \end{aligned}$$ -
(ii)
There exists \(x_{0} \in X\) such that \(x_{0}\) and \(Ax_{0}\) are comparable.
Then A has a fixed point \(x^{*}\) in X. Furthermore, the iterative sequence \(x_{n}=Ax_{n-1}\) (\(n=1,2,\ldots\)) converges to \(x^{*}\) and
Proof
Let \(x_{n}=Ax_{n-1}\) (\(n=1, 2, \ldots\)). Since \(x_{0}\) and \(Ax_{0}\) are comparable, according to the given condition (i) and by induction, it is easy to verify that for any natural number n, \(x_{n}\) and \(x_{n+1}=Ax_{n}\) are comparable. Therefore
which yields that
Lemma 2.4 together with Lemma 2.5 shows that \(r((e-\lambda _{2})^{-1}\lambda_{1})\leq r((e-\lambda_{2})^{-1})r(\lambda_{1})\leq (1-r(\lambda_{2}))^{-1}r(\lambda_{1})\). As \(r(\lambda_{1})+r(\lambda_{2})<1\), \(r((e-\lambda_{2})^{-1}\lambda_{1})\leq (1-r(\lambda_{2}))^{-1}r(\lambda_{1})< 1\). Thus by the same proof as used in Theorem 3.1, we obtain that \(\{x_{n}\}\) is a Cauchy sequence. As X is complete, there exists \(x^{*}\in X\) such that \(x_{n}\rightarrow x^{*}\) (\(n\rightarrow\infty\)). The continuity of A implies \(x_{n+1}=Ax_{n}\rightarrow Ax^{*}\) (\(n\rightarrow\infty\)). So \(Ax^{*}=x^{*}\). Furthermore, by Lemma 2.1, we have
□
Theorem 3.4
Suppose that \(A: X \rightarrow X\) is a mapping satisfying the following conditions:
-
(i)
There exist \(\lambda_{1}, \lambda_{2} \in P\) with \(0\leq r(\lambda _{1})+r(\lambda_{2})<1\) such that for any comparable pair \(x, y\in X\), Ax and Ay are comparable. Moreover, if x and Ax, y and Ay are comparable, then
$$\begin{aligned}& \bigl(\varphi(Ax)-\varphi(Ay)\bigr)\vee\bigl(\varphi(Ay)-\varphi(Ax)\bigr) \\& \quad \preceq\lambda_{1}\bigl[\bigl(\varphi(Ax)-\varphi(x)\bigr)\vee\bigl( \varphi(x)-\varphi(Ax)\bigr)\bigr] \\& \qquad {}+\lambda_{2}\bigl[\bigl(\varphi(Ay)-\varphi(y)\bigr)\vee\bigl( \varphi(y)-\varphi(Ay)\bigr)\bigr]. \end{aligned}$$ -
(ii)
There exists \(x_{0} \in X\) such that for any natural number n, \(x_{0}\) and \(A^{n}x_{0}\) are comparable.
Then the same conclusions as those in Theorem 3.3 hold.
Proof
Set \(x_{n}=Ax_{n-1}\) (\(n=1, 2, \ldots\)). It follows by the same method as in Theorem 3.3 that \(\{x_{n}\}\) is a Cauchy sequence and there exists \(x^{*}\) in X such that \(x_{n}\rightarrow x^{*}\) (\(n\rightarrow\infty\)). Next we show that \(x^{*}\) is a fixed point of A.
Also, by a proof similar to that of Theorem 3.2 and by Lemma 2.2, we get that for any natural number n, \(x_{n}\) and \(x_{n+1}=Ax_{n}\) are comparable, \(x_{n}\) and \(x^{*}\) are also comparable. According to the given condition (i), for any natural number n, \(Ax_{n}\) and \(Ax^{*}\) are comparable. Let \(n\rightarrow\infty\), by Lemma 2.2, \(x^{*}\) and \(Ax^{*}\) are comparable. So
Let \(n\rightarrow\infty\), the continuity of φ together with Lemma 2.3 implies that
Note that \(0\leq r(\lambda_{2})<1\), by Proposition 3.5(ii) of [26], we have
The partial ordering in X gives that
It is evident from (3.4) and (3.5) that \(Ax^{*}=x^{*}\), so \(x^{*}\) is a fixed point of A.
By the same argument as in Theorem 3.3 and the partial ordering in X, we have
From Lemma 2.1, we have
□
Theorem 3.5
Assume that \(A: X \rightarrow X\) is continuous and satisfies the following:
-
(i)
There exist \(\lambda_{1}, \lambda_{1}\in P\) with \(0\leq r(\lambda _{2})<\frac{1}{2}\) such that for any comparable pair \(x, y\in X\), Ax and Ay are comparable. Moreover, if x and Ax, y and Ay are comparable, then
$$\begin{aligned}& \bigl(\varphi(Ax)-\varphi(Ay)\bigr)\vee\bigl(\varphi(Ay)-\varphi(Ax)\bigr) \\& \quad \preceq \lambda_{1}\bigl[\bigl(\varphi(Ax)-\varphi(y)\bigr)\vee\bigl( \varphi(y)-\varphi (Ax)\bigr)\bigr] \\& \qquad {}+\lambda_{2}\bigl[\bigl(\varphi(Ay)-\varphi(x)\bigr)\vee\bigl( \varphi(x)-\varphi(Ay)\bigr)\bigr]. \end{aligned}$$ -
(ii)
There exists \(x_{0} \in X\) such that \(x_{0}\) and \(Ax_{0}\), \(x_{0}\) and \(A^{2}x_{0}\) are comparable.
Then A has a fixed point \(x^{*}\) in X. Furthermore, the iterative sequence \(x_{n}=Ax_{n-1}\) (\(n=1,2,\ldots\)) converges to \(x^{*}\) and
Proof
Pick \(x_{n}=Ax_{n-1}\) (\(n=1, 2, \ldots\)). Since \(x_{0}\) and \(Ax_{0}\), \(x_{0}\) and \(A^{2}x_{0}\) are comparable, for any natural number n, \(x_{n}\) and \(x_{n+1}\) as well as \(x_{n}\) and \(x_{n+2}\) are comparable. Therefore \(\varphi(x_{n})\) and \(\varphi(x_{n+1})\), \(\varphi(x_{n})\) and \(\varphi (x_{n+2})\) are comparable. It follows from Remark 2.1 that
Therefore
The definition of the partial ordering in X gives that
Since \(0 \leq r(\lambda_{2})<\frac{1}{2}\), \(\{x_{n}\}\) is a Cauchy sequence. Hence there exists \(x^{*} \in X\) such that \(x_{n}\rightarrow x^{*}\) (\(n\rightarrow\infty\)). The continuity of A shows that
that is, \(x^{*}\) is a fixed point of A. From Lemma 2.1, we have
□
Theorem 3.6
Let \(A: X \rightarrow X\) be a mapping and satisfy the following conditions:
-
(i)
There exist \(\lambda_{1}, \lambda_{2}\in P\) with \(0\leq r(\lambda_{2})<1\) such that for any comparable pair \(x, y\in X\), Ax and Ay are comparable. Moreover, if x and Ax, y and Ay are also comparable, then
$$\begin{aligned}& \bigl(\varphi(Ax)-\varphi(Ay)\bigr)\vee\bigl(\varphi(Ay)-\varphi(Ax)\bigr) \\& \quad \preceq \lambda_{1}\bigl[\bigl(\varphi(Ax)-\varphi(y)\bigr)\vee\bigl( \varphi(y)-\varphi (Ax)\bigr)\bigr] \\& \qquad {}+\lambda_{2}\bigl[\bigl(\varphi(Ay)-\varphi(x)\bigr)\vee\bigl( \varphi(x)-\varphi(Ay)\bigr)\bigr]. \end{aligned}$$ -
(ii)
There exists \(x_{0} \in X\) such that for any natural number n, \(x_{0}\) and \(A^{n}x_{0}\) are comparable.
Then the same conclusions as those in Theorem 3.5 hold.
Proof
Define \(\{x_{n}\}\) by \(x_{n}=Ax_{n-1}\) (\(n=1, 2, \ldots\)). As in the proof of Theorem 3.4, there exists \(x^{*}\in X\) such that \(x_{n}\rightarrow x^{*}\) (\(n\rightarrow \infty\)) and
Let \(n\rightarrow\infty\), the continuity of φ implies that
Note that \(0\leq r(\lambda_{2})<1\), by Proposition 3.5(ii) of [26], we have
Since \(x^{*}\) and \(Ax^{*}\) are comparable,
From (3.7) and (3.8), we have \(Ax^{*}=x^{*}\), so \(x^{*}\) is a fixed point of A.
By the same method as in Theorem 3.3, we have
Lemma 2.1 shows that
□
Example 3.1
Let \(\mathcal{A}=\mathbb{R}^{2}\). For each \(x=(x_{1}, x_{2})\in\mathcal{A}\), let \(\|x\|=|x_{1}|+|x_{2}|\). The multiplication is defined by
Then \(\mathcal{A}\) is a Banach algebra with unit \(e=(1, 1)\). Let \(P =\{(x_{1}, x_{2})\in\mathbb{R}^{2}| x_{1}\geq0, x_{2}\geq0 \}\) and \(X = \mathbb{R}^{2}\). A metric d on X is defined by
Then \((X, d)\) is a complete cone metric space over the Banach algebra \(\mathcal{A}\).
Now define mapping \(T : X\rightarrow X\) by
Then T has a fixed point in X.
In fact, let \(k=(\frac{2}{e^{2}}, \frac{2}{\pi})\in P\) and define \(\varphi : X\rightarrow\mathcal{A}\) by \(\varphi(x, y)=(-e^{x}-x, -\arctan (y+1)-y)\). Then \(k\in P\) with \(0< r(k)<1\) and it is not difficult to verify that φ is continuous on X and the partial ordering in X can be induced by φ. Clearly, T is continuous and for any comparable pair \(x=(x_{1}, x_{2}), y=(y_{1}, y_{2})\in X\), Tx and Ty are comparable. Moreover, by simple calculations, we can obtain that
Thus T is φ-ordered contractive. Take \(x_{0}=(-1, 0)\), then \(x_{0}=(-1,0)\leq Tx_{0}= (\ln(e^{-3}+1), \tan (\frac{1}{2} ) )\), that is, \(x_{0}\) and \(Tx_{0}\) are comparable. Hence by Theorem 3.1, T has a fixed point in X.
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Yin, J., Wang, T. & Yan, Q. Fixed point theorems of ordered contractive mappings on cone metric spaces over Banach algebras. Fixed Point Theory Appl 2015, 48 (2015). https://doi.org/10.1186/s13663-015-0301-x
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DOI: https://doi.org/10.1186/s13663-015-0301-x