Fixed points of monotone mappings and application to integral equations
- Mostafa Bachar^{1}Email author and
- Mohamed Amine Khamsi^{2, 3}
https://doi.org/10.1186/s13663-015-0362-x
© Bachar and Khamsi 2015
Received: 27 February 2015
Accepted: 23 June 2015
Published: 10 July 2015
Abstract
In this work, we discuss the existence of fixed points of monotone nonexpansive mappings defined on partially ordered Banach spaces. This work is a continuity of the previous works of Ran and Reurings, Nieto et al., and Jachimsky done for contraction mappings. As an application, we discuss the existence of solutions to an integral equations.
Keywords
MSC
1 Introduction
Banach’s contraction principle [1] is remarkable in its simplicity, yet it is perhaps the most widely applied fixed point theorem in all of analysis. This is because the contractive condition on the mapping is simple and easy to test, because it requires only a complete metric space for its setting, and because it finds almost canonical applications in the theory of differential and integral equations. Over the years, many mathematicians tried successfully to extend this fundamental theorem. Recently a version of this theorem has been given in partially ordered metric spaces [2, 3] (see also [4, 5]) and in metric spaces with a graph [6].
In this work, we discuss the case of nonexpansive mappings defined in partially ordered Banach spaces. Nonexpansive mappings are those which have Lipschitz constant equal to 1. The fixed point theory for such mappings is rich and varied. It finds many applications in nonlinear functional analysis [7]. It is worth mentioning that such investigation is new and has never been carried.
2 Monotone nonexpansive mappings
Let \((X, \|\cdot\|)\) be a Banach vector space. Assume that we have a partial order ⪯ defined on X such that order intervals are convex and τ-closed, where τ is a Hausdorff topology on X. Recall that an order interval is any of the subsets \([a,b] = \{x \in X; a \preceq x \preceq b\}\), \([a,{\rightarrow}) = \{x \in X; a \preceq x\}\), \(({\leftarrow},a] = \{x \in X; x \preceq a\}\) for any \(a,b \in X\).
Definition 2.1
- (1)
T is said to be monotone if \(T(x) \preceq T(y)\) whenever \(x \preceq y\) for any \(x, y \in C\).
- (2)T is said to be monotone nonexpansive if and only if T is monotone and$$\bigl\Vert T(x)-T(y)\bigr\Vert \leq\|x-y\|,\quad \mbox{whenever }x \preceq y. $$
Lemma 2.1
Proof
First note that if \(x \preceq y\) holds, then we have \(x \preceq\lambda x + (1-\lambda)y \preceq y\) for any \(x, y \in X\) since order intervals are convex. Therefore it is enough to only prove \(x_{n} \preceq T(x_{n})\) for any \(n \geq0\). By assumption, we have \(x_{0} \preceq T(x_{0})\). Assume that \(x_{n} \preceq T(x_{n})\) for \(n \geq1\). Then we have \(x_{n} \preceq\lambda x_{n} + (1-\lambda) T(x_{n}) \preceq T(x_{n})\), i.e., \(x_{n} \preceq x_{n+1} \preceq T(x_{n})\). Since T is monotone, we get \(T(x_{n}) \preceq T(x_{n+1})\). By induction, we conclude that the inequalities (KI) hold for any \(n \geq0\). Next let \(\{x_{\phi(n)}\}\) be a subsequence of \(\{x_{n}\}\) which τ-converges to z. Clearly, \(\{[x_{n}, {\rightarrow}); n \in\mathbb{N}\}\) is a decreasing family of sets. Consequently, if \(U_{z} \in\tau\) is a neighborhood of z, then \(U_{z} \cap[x_{n}, {\rightarrow}) \neq\emptyset\) for any \(n \in\mathbb{N}\). Therefore, z belongs to all sets \([x_{n},{\rightarrow})\) as they are closed. Let w be the τ-limit of another subsequence of \(\{x_{n}\}\). If \(U_{w} \in\tau\) is a neighborhood of w, then \(U_{w}\) contains many points from the sequence \(\{x_{n}\}\) since w is a τ-limit of one of its subsequences. Hence \(U_{w} \cap({\leftarrow}, z] \neq\emptyset\). Therefore, w belongs to \(({\leftarrow}, z]\) as it is closed, i.e., \(w \preceq z\). By reversing the roles of z and w, we get \(z \preceq w\). The properties of the partial order will force \(z = w\) as claimed. □
Remark 2.1
The following result is found in [10, 11].
Proposition 2.1
Proof
Before we state the main result of this work, let us recall the definition of Opial condition [12].
Definition 2.2
Now we are ready to state the main result of this section.
Theorem 2.1
Let X be a Banach space. Let τ be a topology on X such that X satisfies the τ-Opial condition. Let ⪯ be a partial order on X such that order intervals are convex and τ-closed. Let C be a bounded convex τ-compact nonempty subset of X. Let \(T: C \rightarrow C\) be a monotone nonexpansive mapping. Assume that there exists \(x_{0} \in C\) such that \(x_{0}\) and \(T(x_{0})\) are comparable. Then T has a fixed point.
Proof
The following results are direct consequences of Theorem 2.1.
Corollary 2.1
Let C be a bounded closed convex nonempty subset of \(l_{p}\), \(1 < p < +\infty\). Let τ be the weak topology. Consider the pointwise partial ordering in \(l_{p}\), i.e., \((\alpha_{n}) \preceq(\beta_{n})\) iff \(\alpha_{n} \leq\beta _{n}\) for all \(n \geq1\). Then any monotone nonexpansive mapping \(T: C \rightarrow C\) has a fixed point provided there exists a point \(x_{0} \in C\) such that \(x_{0}\) and \(T(x_{0})\) are comparable.
Remark 2.2
The case of \(p=1\) is not interesting for the weak topology since \(l_{1}\) is a Schur Banach space. But if we consider the weak* topology \(\sigma(l_{1}, c_{0})\) on \(l_{1}\) or the pointwise convergence topology, then \(l_{1}\) satisfies the Opial condition for these topologies. Note that these two topologies are Hausdorff. In this case we have a similar conclusion of Corollary 2.1 for \(l_{1}\).
- (i)
\(l_{p} = \{(\alpha_{n}) \in\mathbb{R}^{\mathbb{N}}, \sum_{n} |\alpha_{n}|^{p} < +\infty\}\) for \(1 \leq p < +\infty\);
- (ii)
\(c_{0} = \{(\alpha_{n}) \in\mathbb{R}^{\mathbb{N}}, \lim_{n \rightarrow+\infty} \alpha_{n} = 0\}\).
3 Application to integral equations
- (i)
g is in \(L^{2}([0,1],\mathbb{R})\),
- (ii)\(F:[0,1]\times[0,1]\times L^{2}([0,1],\mathbb{R}) \rightarrow\mathbb{R}\) is measurable and satisfies the conditionwhere \(t,s\in[0,1]\), and \(x, y \in L^{2}([0,1],\mathbb{R})\) such that \(y \leq x\).$$ 0 \leq F(t,s,x)-F(t,s,y)\leq x -y, $$(3.1)
Theorem 3.1
- (i)
the integral equation (IE) has a non-negative solution provided we assume that \(g(t)+\int^{1}_{0}F(t,s,0)\, ds \geq0\) for almost every \(t \in[0,1]\) (which implies \(J(0) \geq0\));
- (ii)
the integral equation (IE) has a non-positive solution provided we assume that \(g(t)+\int^{1}_{0}F(t,s,0)\, ds \leq0\) for almost every \(t \in[0,1]\) (which implies \(J(0) \leq0\)).
Declarations
Acknowledgements
The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for funding this Research group No. (RG-1435-079).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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