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Fixed points of monotone mappings and application to integral equations
Fixed Point Theory and Applicationsvolume 2015, Article number: 110 (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.
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.
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 τclosed, 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
Let C be a nonempty subset of X. Let $T: C \rightarrow C$ be a map.

(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
$$\bigl\Vert T(x)T(y)\bigr\Vert \leq\xy\,\quad \mbox{whenever }x \preceq y. $$
The point $x \in C$ is called a fixed point of T if $T(x) = x$. The set of fixed points of T will be denoted by $\operatorname{Fix}(T)$.
Throughout the paper we assume that C is convex and bounded not reduced to one point. Let $T: C \rightarrow C$ be a monotone nonexpansive mapping. Fix $\lambda\in(0,1)$ and $x_{0} \in C$. The Krasnoselskii [8, 9] iteration sequence $\{ x_{n}\} \subset C$ is defined by
The following lemma holds.
Lemma 2.1
Under the above assumptions, if we assume that $x_{0} \preceq T(x_{0})$, then we have
for any $n \geq0$. Moreover, if $\{x_{n}\}$ has two subsequences which τconverge to z and w respectively, then we must have $z =w$.
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 τconverges 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 τlimit 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 τlimit 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
Note that under the assumptions of Lemma 2.1, if we assume $T(x_{0}) \preceq x_{0}$, then we will have
for any $n \geq0$. The conclusion on the τconvergence limits of $\{x_{n}\}$ will also hold.
The following result is found in [10, 11].
Proposition 2.1
Under the above assumptions, we have
for any $i, n \in\mathbb{N}$. This inequality implies
Proof
The first part of this proposition is easy to prove via an induction argument on the index i. As for the second part, note that $\{\x_{n}  T(x_{n})\\}$ is decreasing. Indeed we have $x_{n+1}  x_{n} = (1\lambda )(T(x_{n})x_{n})$ for any $n \geq1$. Therefore $\{\x_{n}  T(x_{n})\\}$ is decreasing if and only if $\{\x_{n+1}  x_{n}\\}$ is decreasing, which holds since
for any $n \geq0$. Set $\lim_{n \rightarrow+\infty} \x_{n}  T(x_{n})\ = R $. Then we let $i \rightarrow+\infty$ in the inequality (GK) to obtain
for any $n \in\mathbb{N}$, where $\delta(C) = \sup\{\xy\, x, y \in C\} < +\infty$. Hence
which implies $R = 0$, i.e., $\lim_{n \rightarrow+\infty} \ x_{n}  T(x_{n})\ = 0 $. □
Before we state the main result of this work, let us recall the definition of Opial condition [12].
Definition 2.2
X is said to satisfy the τOpial condition if for any sequence $\{y_{n}\}$ in X which τconverges to y, we have
for any $z \in X$ such that $z \neq y$.
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 τOpial condition. Let ⪯ be a partial order on X such that order intervals are convex and τclosed. Let C be a bounded convex τcompact 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
Without loss of any generality, we assume that $x_{0} \preceq T(x_{0})$. Consider the (KIS) sequence $\{x_{n}\}$ which starts at $x_{0}$. Since C is τcompact, then $\{x_{n}\}$ will have a subsequence $\{x_{k_{n}}\}$ which τconverges to some point $w \in C$. Lemma 2.1 implies that $\{x_{n}\}$ τconverges to w and $x_{n} \preceq w$ for any $n \in\mathbb{N}$. Consider the type function
Then Proposition 2.1 implies $r(x) = \limsup_{n \rightarrow+\infty} \T(x_{n})  x\$ for any $x \in C$. Since T is monotone nonexpansive, we get
In fact we have $r(T(x)) \leq r(x)$ for any $x \in C$ such that $x_{n}$ and x are comparable for any $n \in\mathbb{N}$. Finally, if X satisfies the τOpial condition, then we must have $T(w) = w$, i.e., w is a fixed point of T. □
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 $l_{1}$.
Recall the definitions of $l_{p}$ and $c_{0}$ spaces:

(i)
$l_{p} = \{(\alpha_{n}) \in\mathbb{R}^{\mathbb{N}}, \sum_{n} \alpha_{n}^{p} < +\infty\}$ for $1 \leq p < +\infty$;

(ii)
$c_{0} = \{(\alpha_{n}) \in\mathbb{R}^{\mathbb{N}}, \lim_{n \rightarrow+\infty} \alpha_{n} = 0\}$.
Application to integral equations
Let us consider the following integral equation of the form
where

(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 condition
$$ 0 \leq F(t,s,x)F(t,s,y)\leq x y, $$(3.1)where $t,s\in[0,1]$, and $x, y \in L^{2}([0,1],\mathbb{R})$ such that $y \leq x$.
Recall that for any $u, v \in L^{2}([0,1],\mathbb{R})$, we have
Condition (3.1) represents the monotonicity of the flow of the integral equation. A comprehensive study of the monotonicity of the flow can be found in the book of Smith [13]. Assume that there exists a nonnegative function $h(\cdot,\cdot) \in L^{2}([0,1]\times[0,1])$ and $M<\frac{1}{2}$ such that
where $t,s\in[0,1]$ and $x \in L^{2}([0,1],\mathbb{R})$.
Let
where ρ is sufficiently large, i.e., B is the closed ball of $L^{2}([0,1],\mathbb{R})$ centered at 0 with radius ρ. Consider the operator defined by
and define the operator $J: L^{2}([0,1],\mathbb{R}) \rightarrow L^{2}([0,1],\mathbb{R})$ by
We have $J(B) \subset B$. Indeed let $x\in B$, then by using the CauchySchwarz inequality, condition (3.2) and the quadratic inequality $(a+b)^{2}\leq2a^{2}+2b^{2}$ for any $a,b \in \mathbb{R}$, we have
Since $M < 1/2$, choose ρ such that
we will get $J(x) \in B$ as claimed. Next we prove that J is monotone nonexpansive. First from condition (3.1), J is obviously monotone. Let $x, y \in L^{2}([0,1],\mathbb{R})$ such that $y \leq x$. Using the CauchySchwarz inequality, we have
which implies that J is a monotone nonexpansive operator as claimed. In order to use Theorem 2.1, we need to check its assumptions. First note that $X = L^{2}([0,1],\mathbb{R})$ is a Hilbert space. If we choose τ to be the weak topology, then X satisfies the weak Opial condition. It is easy to check that order intervals are convex. In order to show that order intervals are closed, we will show that if $\{u_{n}\}$ is a nonnegative sequence of elements in X which converges weakly to u, then u is positive. Let $a < 0$, then the set $A = \{t \in[0,1]; u(t) \leq a\}$ has measure 0. Indeed, we have
because of weak convergence. So
Hence $m(A) = 0$. Set
Then D has measure 0, which implies that $u(t) \geq0$ for almost every $t \in[0,1]$. Using Theorem 2.1, we get the following result.
Theorem 3.1
Under the above assumptions, we conclude that

(i)
the integral equation (IE) has a nonnegative 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 nonpositive 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 $J(0) \leq0$).
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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. (RG1435079).
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MSC
 46B20
 45D05
 47E10
 34A12
Keywords
 fixed point
 integral equation
 Krasnoselskii iteration
 Lebesgue measure
 monotone mapping
 nonexpansive mapping