Existence and uniqueness of fixed points for monotone operators in partially ordered Banach spaces and applications

Abstract

In this paper, we employ partial order method, cone theory, and the techniques of measure of weak noncompactness to prove several new theorems on the existence and the uniqueness of fixed points or coupled fixed points for operators satisfying some monotonicity assumptions. Our conclusions generalize and improve several well-known results. As an application, we investigate the existence of a unique solution for a class nonlinear second-order ordinary differential equations. We also discuss the existence of a unique solution for a system of integral equations.

1 Introduction

Fixed point theorems are powerful tools in the analysis of many applied problems including differential equations, dynamical systems, mathematical economics (game theory, equilibrium problems, and optimization problems), and mathematical modeling. The study of fixed point theorems based on the order-theoretical approach is one of most attractive and interesting branches of fixed point theory. Actually, many beautiful and important fixed point theorems were obtained by the use of order technique, see [16, 18, 21] and the references therein. This study has taken two different directions. One direction is to impose compactness type conditions that guarantee only existence of minimal and maximal solutions. The other direction is to utilize convexity or order contractiveness type conditions that ensure existence and uniqueness of a fixed point. In both directions, the majority of the fixed point theorems require a space that is ordered by a normal cone.

In this work, we combine the advantages of the first approach with those of the second one to draw new and meaningful conclusions about fixed points for monotone mapping defined on a Banach space partially ordered by a cone (not necessarily normal). We shall particularly indicate how techniques from the two approaches may be combined with the techniques of measures of weak noncompactness to improve and extend several well-known fixed point and coupled fixed point results for mappings satisfying certain monotonicity conditions and to handle existence and uniqueness problems for a certain class of nonlinear systems with lack of compactness.

In the second half of our work, we provide several examples to illustrate the effectiveness and the applicability of our new approach.

The organization of the paper is as follows. In Sect. 2, definitions and preliminary results are given. In Sect. 3, we establish some new fixed point theorems for monotone operators under weak topology features. Section 4 is concerned with coupled fixed point theorems for mixed monotone operators. In Sect. 5, we investigate the existence of a unique solution for a class nonlinear second-order ordinary differential equations. The last section is devoted to discuss the existence and uniqueness of solution for a system of integral equations.

2 Preliminaries

The aim of this preparatory section is to collect, mainly for the purpose of later reference and to fix some notation, several basic facts on ordered Banach spaces.

Definition 2.1

Let E be a Banach space with norm $\left\| . \right\|$ and P be a subset of E. Then, P is called a cone if:

• (i) P is closed, non-empty and $P\ne \{0\}$;

• (ii) if $a, b \ge 0$ and $x,y\,\in P$, then $ax+by\in P$;

• (iii) if $x\in P$ and $-x\in P$, then $x=0$.

By an ordered Banach space (OBS for short), we mean a pair (E, P) where E is a real Banach space and P is a cone on E. As is well known, the cone P induces a partial order $\le$ on E which is given by $x\le y$ if and only if $y - x \in P.$ An alternative notation for $x\le y$ is $y\ge x.$ Moreover, we use the notation $x<y$ to indicate that $x\le y$, but $x \ne y.$ For $x, y \in E$, we call the set $[x, y] =\{z\in E :x \le z \le y\}$ the order interval between x and y. The cone P in the ordered Banach space E is called total or spatial if $P-P$ is dense in E and it is called generating or reproducing if $E=P-P.$ The cone P is called normal if there is a number $\delta > 0$, such that for all $x,y\in E$, we have $0\le x\le y$ implies $\left\| x \right\| \le \delta \left\| y \right\| .$ We note that if P is normal, every order interval is norm bounded.

Additionally, we say that a set $M\subset E$ is upper bounded with respect to P if there is a $y \in E$, such that $x \le y$ for all $x \in M;$ the element y is called an upper bound for M. In the same way, we can define a lower bound. If the set of upper bounds of M has a minimal element z,  then z is called the least upper bound of M;  it is denoted by $\sup M.$ In the same way, we can define the greatest lower bound, $\inf M.$ A cone P is called minihedral, if each two-element set $M = \{x, y\}$ has a least upper bound, $\sup \{x, y\}.$

The following result provides a useful characterization of generating cones.

Lemma 2.1

[13, Lemma 2.2] Let E be an OBS with a cone P. The following assertions are equivalent :

• (i) P is generating.

• (ii) Every pair $x, y \in E$ has a lower bound.

• (iii) Every pair $x, y \in E$ has an upper bound.

• (iv) For all $x\in E,$ there exists $u\ge 0$, such that $x\le u.$

• (v) For all $x\in E,$ there exists $u\le 0$, such that $x\ge u.$

Remark 2.1

Lemma 2.1 implies that a minihedral cone is always generating. The converse is not true; for example, let $E = C^1[0, 1],$ $P = \{x \in E:x(t) \ge 0, \forall t \in [0, 1]\},$ then P is generating but not minihedral.

Let $E^\prime$ be the dual space of an OBS E. We set:

$$\begin{aligned} P^\prime =\{\varphi \in E^\prime :\left\langle \varphi ,x\right\rangle \ge 0 \,\hbox { for all}\, x\in P\}. \end{aligned}$$

The set $P^\prime$ is a closed convex subset of $E^\prime$ which is invariant under multiplication by positive scalars. It should be highlighted that if P is normal, then $E^\prime =P^\prime -P^\prime$ (see [2, Theorem 2.26]). Furthermore, it is a common knowledge that $P^\prime \cap (-P^\prime )=\{0\}$ if and only if the cone P is total. In this case, $P^\prime$ renders the dual space $E^\prime$ also an OBS, and $P^\prime$ is called the dual cone of P.

For separation of two subsets in a normed space, we use the following version of the Hahn–Banach theorem.

Theorem 2.1

[23, p. 133] Let M and N be non-empty convex subsets in a real normed space E. Assume that M is open and $M \cap N = \emptyset .$ Then, there exists a functional $\varphi \in E^\prime$ and a real number $\gamma$, such that $\left\langle \varphi ,x\right\rangle <\gamma \le \left\langle \varphi ,y\right\rangle$ for all $x\in M$ and all $y\in N.$

We always have the following characterization of positive elements in E,  no matter whether P is total or not. This result has been stated and proved in [14] and [16, Theorem 1.4.1]. We outline the proof for the reader’s convenience.

Theorem 2.2

Let E be an OBS and let $x_0 \in E.$ Then, $x_0\in P$ if and only if $\left\langle \varphi ,x_0\right\rangle \ge 0$ for all $\varphi \in P^\prime .$

Proof

The direct implication being obvious, let us prove the converse. To do so, take any $x_0\in E$ with $x_0\notin P.$ Then, there exists an open convex neighborhood V of $x_0$, such that $V\cap P=\emptyset ,$ and by applying Theorem 2.1, we find a functional $\varphi \in E^\prime$ and a real number $\alpha$, such that:

$$\begin{aligned} \left\langle \varphi , x_0 \right\rangle <\alpha \le \left\langle \varphi ,x\right\rangle \,\,\hbox { for all}\,\, x \in P. \end{aligned}$$

Since $0 \in P,$ we conclude that $\alpha \le 0,$ so $\left\langle \varphi , x_0 \right\rangle <0.$ It only remains to show that $\varphi \in P^\prime .$ Let $x\in P.$ For every $n\in {\mathbb {N}}$, we have $nx\in P,$ so $n\left\langle \varphi , x \right\rangle \ge \alpha .$ Dividing by n and letting $n\rightarrow +\infty ,$ we can see that $\left\langle \varphi , x \right\rangle \ge 0,$ so $\varphi \in P^\prime$. $\square$

Corollary 2.1

[14, Corollary 2.3] Let E be an OBS and let x be a non-zero vector in E. Then, there exists $\varphi \in P^\prime$, such that $\left\langle \varphi , x \right\rangle \ne 0.$

Proof

If $\left\langle \varphi , x \right\rangle = 0$ for all $\varphi \in P^\prime ,$ then it follows from Theorem 2.2 that $x \ge 0$ and $-x \ge 0,$ so $x = 0$. $\square$

Definition 2.2

Let E be an OBS and let $T :D \subset E \rightarrow E$ be a map. We say that T is increasing (or isotone) if $Tu \le Tv$ for $u, v \in D$ with $u\le v.$ The map T is said to be decreasing (or anti-isotone) if $Tv \le Tu$ for $u, v \in D$ with $u\le v.$

Definition 2.3

Let E be a Banach space. An operator $T :D\subset E\rightarrow E$ is said to be:

• (i) (monotone) continuous if for every (monotone) sequence $(x_n)_n$ of D which converges strongly to some $x\in D,$ the sequence $(Tx_n)$ converges strongly to Tx.

• (ii) (monotone) subcontinuous if for every (monotone) sequence $(x_n)_n$ of D which converges strongly to some $x\in D,$ the sequence $(Tx_n)$ converges weakly to Tx.

We remark in passing that monotone non-expansive mappings, which have recently been of considerable research interest (see, for example [25]), provide natural examples of monotone continuous mappings.

Now, we present some basic facts regarding measures of weak noncompactness in Banach spaces, which we will need in the sequel. We emphasize that the interplay between fixed point theory and measures of weak noncompactness is very powerful and fruitful [10]. Let ${\mathcal {B}}(E)$ denote the family of all non-empty bounded subsets of E and let ${\mathcal {K}}(E)$ (resp. ${\mathcal {W}}(E)$ ) denote the collection of all relatively compact (resp. weakly relatively compact) subsets of E.

Following [5], we will adopt the following definition.

Definition 2.4

A mapping $\mu :{\mathcal {B}}(E) \longrightarrow {\mathbb {R}}^+$ is said to be a measure of weak noncompactness (MWNC for short) if it satisfies the following conditions :

($1^{\circ }$):

Consistency : The family $\ker \mu = \{M\in {\mathcal {B}}(E):\mu (M)=0\}$ is non-empty and $\ker \mu \subset {\mathcal {W}}(E).$

($2^{\circ }$):

Monotonicity : $M_1 \subset M_2 \Longrightarrow \mu (M_1) \le \mu (M_2)$.

($3^{\circ }$):

Invariance under passage to the closed convex hull : $\mu (\overline{co} (M))= \mu (M)$.

($4^{\circ }$):

Convexity : $\mu (\lambda \, M_1+(1-\lambda )M_2) \le \lambda \mu (M_1)+ (1- \lambda ) \,\mu (M_2)$ for $\lambda \in [0,1]$.

($5^{\circ }$):

Generalized Cantor intersection property : If $(M_n)_{n \ge 1}$ is a sequence of non-empty, weakly closed subsets of E with $M_1$ bounded and $M_1 \supseteq M_2 \supseteq \cdots \supseteq M_n \supseteq \cdots$ and such that $\lim _{n \rightarrow +\infty }\,\mu (M_n)= 0$, then the set $M_{\infty }:=\bigcap _{n=1}^{\infty }\,M_n$ is non-empty.

From these axioms, it follows readily that $\mu (\overline{M^w})=\mu (M),$ where $\overline{M^w}$ stands for the weak closure of M. In addition, the set $M_{\infty }$ belongs to $\ker \,\mu$, since $\mu (M_{\infty }) \le \mu (M_n)$ for every n and $\lim _{n \rightarrow +\infty }\,\mu (M_n)= 0$.

In applications, there are measures of weak noncompactness satisfying some additional handy conditions.

($6^{\circ }$):

Fullness : $\mu (M)= 0$ if and only if $M\in {\mathcal {W}}(E).$

($7^{\circ }$):

Subadditivity : $\mu (M_1+M_2) \le \mu (M_1)+\mu (M_2),\,\,\forall M_1,M_2\in {\mathcal {B}}(E).$

($8^{\circ }$):

Homogeneity : $\mu (\lambda \,M)=|\lambda |\mu (M),\,\forall \lambda \in {\mathbb {R}},\,\forall M\in {\mathcal {B}}(E).$

($9^{\circ }$):

Maximum property : $\mu (M_1\cup M_2)=\max (\mu (M_1),\mu (M_2)), \forall M_1,M_2\in {\mathcal {B}}(E).$

($10^{\circ }$):

Non-singularity : $\mu (M\cup \{x\})=\mu (M), \forall M\in {\mathcal {B}}(E),\forall x\in E.$

($11^{\circ }$):

Sublinearity : $\mu$ is homogenous and subadditive.

($12^{\circ }$):

Regularity : $\mu$ is full and sublinear, and has the maximum property.

We refer the reader to [10] for a review on measures of weak noncompactness (see also [4]). We would like to stress that one of the most frequently exploited measures of weak noncompactness was defined by De Blasi [11] as follows:

$$\begin{aligned} \beta (M)=\inf \{r>0:\,\,\hbox {there exists}\,\, W \,\,\hbox {weakly compact such that}\,\, M\subseteq W+B_r\}, \end{aligned}$$

for each bounded subset M of E;  Here, $B_r$ stands for the closed ball of E centered at origin with radius r. We point out that the measure $\beta$ is regular.

For convenience purposes, we list some necessary results below for completeness. The following lemma exhibits the relation between MWNCs on E and ${\mathcal {C}}(I, E),$ where I is a compact interval of ${\mathbb {R}}$ and E is a Banach space.

Lemma 2.2

[24] Let H be a bounded, equicontinuous subset of ${\mathcal {C}}(I, E)$. Then, $t \in I\mapsto \beta (H(t))$ is continuous and:

$$\begin{aligned} \beta _{{\mathcal {C}}} (H)=\sup _{t\in I}\beta (H(t)) =\beta (H(I)), \end{aligned}$$

where $\beta$ stands for the De Blasi measure of weak noncompactness on E and $\beta _{{\mathcal {C}}}$ refers to the De Blasi measure of weak noncompactness on ${\mathcal {C}}(I,E).$

The following lemma allows us to interchange integrals and measures of weak noncompactness under appropriate conditions.

Lemma 2.3

[10] Let H be a bounded, equicontinuous subset of ${\mathcal {C}}(I, E)$. Then, for any $t\in I$, one has the inequality:

$$\begin{aligned} \beta \left( \int _{0}^{t}H(s)\mathrm{d}s \right) \le \int _{0}^{t} \beta \left( H(s) \right) \mathrm{d}s, \end{aligned}$$

where:

$$\begin{aligned} \int _{0}^{t}H(s)\mathrm{d}s :=\left\{ \int _{0}^{t}h(s)\mathrm{d}s\,,\;h\in H \right\} . \end{aligned}$$

For our purpose, we introduce the following definition.

Definition 2.5

A mapping $\mu :{\mathcal {B}}(E)\rightarrow {\mathbb {R}}^+$ is said to be a generalized measure of weak noncompactness in E (GMWNC for short) (resp. generalized measure of noncompactness (GMNC for short)), if it satisfies the requirements:

• (i) The family $\ker \mu :=\{M\in {\mathcal {B}}(E):\mu (M)=0\}$ is non-empty and $\ker \mu \subset {\mathcal {W}}(E)$ (resp. $\ker \mu \subset {\mathcal {K}}(E)$).

• (ii) For every $x\in E$ and $M\in {\mathcal {B}}(E)$, we have:

  $$\begin{aligned} \mu (\{x\}\cup M)=\mu ( M). \end{aligned}$$

Remark 2.2

It is worth noticing that every non-singular MWNC is a GMWNC. The converse is not true. For example, in every Banach space E,  the map:

$$\begin{aligned} \mu (M)=\left\{ \begin{array}{ll} 0&{}\quad \hbox {if}\quad M\,\hbox {is a finite set,}\\ \\ 1&{}\quad \hbox {otherwise,} \end{array}\right. \end{aligned}$$

is a GMWNC. However, $\mu$ is not a MWNC. Indeed, for any two-point set $M=\{x,y\}$ ($x\ne y$), we have $co(M)=\{\lambda x+(1-\lambda )y,\,\,\lambda \in [0,1]\}.$ Thus, $\mu (M)=0$ and $\mu (co(M))=1\ne \mu (M).$

We also note the following fact for later reference. We refer the reader to [18] or [19] for a proof.

Lemma 2.4

Let E be an OBS with a normal cone. Suppose that $(u_n)_{n\in {\mathbb {N}}}$ is a monotone sequence which have a subsequence converging weakly to some $u\in E$. Then, the full sequence $(u_n)_{n\in {\mathbb {N}}}$ converges strongly to u.

We close this section by stating the following monotone iterative principle for increasing operators in ordered Banach spaces. This result was proved in [1].

Theorem 2.3

Let E be an OBS with a normal cone P. Let $u_0,v_0\in E$ with $u_0<v_0$ and $A:[u_0,v_0]\rightarrow E$ be a monotone subcontinuous increasing operator satisfying the following:

$$\begin{aligned} u_0\le Au_0,\,\,Av_0\le v_0. \end{aligned}$$

(2.1)

Suppose that A verifies

$({\mathcal {P}}(n_0)) :$ There exists an integer $n_0\ge 1$, such that, for any monotone order bounded sequence $V= \{x_n\}$ of E,  and for any finite subset F of E of cardinal $n_0,$ we have:

$$\begin{aligned} V= F\cup A^{n_0}(V)\,\,\hbox {implies}\,\, V \,\,\hbox {is weakly relatively compact}. \end{aligned}$$

Then, A has a minimal fixed point $u_*$ and a maximal fixed point $u^*$ in $[u_0,v_0]$ and:

$$\begin{aligned} u_*=\lim _{n\rightarrow \infty } u_n\quad \hbox {and}\quad u^*=\lim _{n\rightarrow \infty } v_n, \end{aligned}$$

(2.2)

where $u_n=Au_{n-1}$ and $v_n=Av_{n-1},\,\,n=1,2,\ldots$:

$$\begin{aligned} u_0\le u_1\le \cdots u_*\le u^*\le \cdots \le v_n\le \cdots \le v_1\le v_0. \end{aligned}$$

(2.3)

3 Fixed point theorems for monotone operators

In this section, we consider the following condition:

$$\begin{aligned} \mathbf{({\mathcal {H}}2)}\quad \left\{ \begin{array}{l} \hbox {for any}\, \varphi \in P^\prime \,\hbox { there is} \,\, c=c(\varphi )\in [0,1)\,\hbox { such that for all }\, u,v\in E \\ \\ \displaystyle \,\hbox {with}\, v\le u \,\hbox {we have} \,\,\, \left\langle \varphi ,\, Au-Av\right\rangle \le c\left\langle \varphi , u-v\right\rangle . \end{array}\right. \end{aligned}$$

Our main purpose in the immediate sequel is to prove some new fixed point theorems for monotone mappings satisfying the condition $({\mathcal {H}}2).$ Before proceeding further with the first main theorem, we obtain some auxiliary results.

Lemma 3.1

Let E be an OBS with a cone P and let $\{u_n\}$ be a monotone sequence of elements of E. If the set $\{u_n:n\ge 0\}$ is weakly relatively compact (resp. relatively compact), then the full sequence $\{u_n\}$ is weakly (resp. strongly) convergent.

Proof

Without loss of generality, we may assume that the sequence $\{u_n\}$ is increasing. Suppose that $\{u_n:n\ge 0\}$ is weakly relatively compact. Then, there exists a subsequence $\{u_{\tau (n)}\}$ of $\{u_n\}$ which converges weakly to some $u^*.$ Now, we prove that $\{u_n\}$ converges weakly to $u^*.$ Suppose the contrary, then there exists a weak neighborhood $V^w$ of $u^*$ and a subsequence $\{u_{\varphi (n)}\}$ of $\{u_n\}$, such that $u_{\varphi (n)}\not \in V^w$ for all $n\ge 0.$ By virtue of the weak relative compactness of $\{u_n\},$ there exists a subsequence $\{u_{\varphi (\psi (n))}\}$ of $\{u_{\varphi (n)}\}$ which converges weakly to some $v^*.$ Obviously, for all $n\in {\mathbb {N}},$ we have $\varphi (\psi (\tau (n)))\ge \tau (n)$, so that $u_{\tau (n)}\le u_{\varphi (\psi (\tau (n)))}.$ Letting $n\rightarrow +\infty ,$ we conclude that $u^*\le v^*.$ Similarly, $\varphi (\psi (n))\le \tau (\varphi (\psi (n)))$ for any integer n implies $v^*\le u^*$. Hence, $u^*=v^*.$ This implies that for n large enough, we have $u_{\varphi (\psi (n))}\in V^w,$ which is a contradiction. This proves that $\{u_n\}$ converges weakly to $u^*.$ The proof of the second statement is similar to that of the first one and so will be omitted. $\square$

Remark 3.1

• 1. It should be noted that Lemma 3.1 is a sharpening of Lemma 2.4. In Lemma 2.4, the cone P is required to be normal.

• 2. It is remarkable to find that the result of Theorem 2.3 remains true without the assumption of normality on cones if we use Lemma 3.1 instead of Lemma 2.4 in the proof of this theorem.

Lemma 3.2

Let E be an OBS with a generating cone P. Let $A:E\rightarrow E$ be an increasing operator satisfying the condition $({\mathcal {H}}2)$. Then:

• (i) There are $u,v\in E$, such that $u\le Au$ and $Av\le v.$

• (ii) A has at most one fixed point in E.

Proof

To prove (i), pick up any $\varphi \in P^\prime .$ In view of Lemma 2.1, there are $u\le 0$ and $v\ge 0$, such that $\frac{1}{1+c}A(0)\ge u$ and $\frac{1}{1-c}A(0)\le v.$ Hence, by (${\mathcal {H}}2$), we have:

$$\begin{aligned} \left\langle \varphi , Au-u\right\rangle= & {} \left\langle \varphi , (A(0)-u)-(A(0)-Au)\right\rangle \\\ge & {} c\left\langle \varphi , u\right\rangle -c\left\langle \varphi , u\right\rangle =0, \end{aligned}$$

and

$$\begin{aligned} \left\langle \varphi , v-Av\right\rangle= & {} \left\langle \varphi , v-A(0)\right\rangle -\left\langle \varphi , Av-A(0)\right\rangle \\\ge & {} c\left\langle \varphi , v\right\rangle -c \left\langle \varphi , v\right\rangle =0. \end{aligned}$$

Invoking Theorem 2.2, we conclude that $u\le Au$ and $Av\le v.$ To prove (ii), suppose that there exist $u,v\in E$, such that $Au=u$ and $Av=v.$ Since P is generating, then, by Lemma 2.1, there exists $w\in E$, such that $u\le w$ and $v\le w.$ Pick up any $\varphi \in P^\prime .$ Using a straightforward mathematical induction, we can prove that for each integer $n\ge 0$, $A^nw$ is comparable to $A^nv$ and $A^nu.$ This leads to:

$$\begin{aligned} \vert \left\langle \varphi , A^nu -A^nw\right\rangle \vert \le c^n\left\langle \varphi , w-u\right\rangle , \end{aligned}$$

(3.1)

and

$$\begin{aligned} \vert \left\langle \varphi , A^nv-A^n w\right\rangle \vert \le c^n\left\langle \varphi , w-v\right\rangle . \end{aligned}$$

(3.2)

Linking (3.1) and (3.2), we arrive:

$$\begin{aligned} \vert \left\langle \varphi , u-v\right\rangle \vert\le & {} \vert \left\langle \varphi , A^nu-A^nv\right\rangle \vert \\\le & {} \vert \left\langle \varphi , A^nu-A^nw\right\rangle \vert +\vert \left\langle \varphi , A^nw-A^nv\right\rangle \vert \\\le & {} c^n\left( \left\langle \varphi , w-u\right\rangle +\left\langle \varphi , w-v\right\rangle \right) ; \end{aligned}$$

upon letting $n\rightarrow +\infty$, we see that $\left\langle \varphi , u-v\right\rangle =0.$ Using again Corollary 2.1, we obtain $u=v$. $\square$

Now, we are in a position to state the first main result of this section.

Theorem 3.1

Let E be an OBS with a generating cone P. Let $A:E\rightarrow E$ be an increasing operator satisfying the conditions $({\mathcal {P}}(n_0))$ and $({\mathcal {H}}2).$ Then, A has unique fixed point in E.

Proof

In view of Lemma 3.2, there exist $u_0,v_0\in E$, such that $u_0\le Au_0$ and $Av_0\le v_0.$ Let $u_n=Au_{n-1}$ and $v_n=Av_{n-1}$ for $n\ge 1.$ Since A is increasing, then:

$$\begin{aligned} u_0\le u_1\le \,\cdots \,\le u_{n}\le \,\cdots \,v_n \le \cdots v_1\le v_0. \end{aligned}$$

(3.3)

Let $S=\{u_0,u_1,\ldots ,u_n,\ldots \}$. Clearly, for any integer $k\ge 1$, we have:

$$\begin{aligned} A^{k}(S)\cup \{u_0,u_1,\ldots ,u_{k-1}\}=S. \end{aligned}$$

From our hypotheses,, we know that S is weakly relatively compact. In light of Lemma 3.1, this shows that $\{u_n\}$ converges weakly to some $u^*\in E.$ By monotonicity arguments, we deduce that $u_n\le u^*$ for all $n\ge 0.$ Accordingly, for any $\phi \in P^\prime ,$ we have:

$$\begin{aligned} \vert \left\langle \phi , Au^*-u^*\right\rangle \vert\le & {} \vert \left\langle \phi , Au^*-Au_n\right\rangle \vert +\vert \left\langle \phi , u_{n+1}-u^*\right\rangle \vert \\\le & {} c\vert \left\langle \phi , u^*-u_n\right\rangle \vert +\vert \left\langle \phi , u_{n+1}-u^*\right\rangle \vert . \end{aligned}$$

Letting $n\rightarrow +\infty ,$ we conclude that $\left\langle \phi , Au^*-u^*\right\rangle =0.$ Referring to Corollary 2.1, we see that $Au^*=u^*.$ The uniqueness follows from Lemma 3.2. $\square$

A similar result for non-monotonic mappings may be stated as follows.

Theorem 3.2

Let E be a weakly sequentially complete OBS with a normal cone P, and let $A:E\rightarrow E$ be a mapping. Assume that for any $\varphi \in P^\prime ,$ there exists $c=c(\varphi )\in [0,1)$, such that for all $u,v\in E$ with $v\le u$, we have:

$$\begin{aligned} -c\left\langle \varphi , u-v\right\rangle \le \left\langle \varphi ,Au-Av\right\rangle \le c\left\langle \varphi ,u-v\right\rangle . \end{aligned}$$

(3.4)

Then, A has a unique fixed point $x^*$ in E. Moreover, for each $x\in E,$ the sequence $\{A^nx\}$ converges weakly to $x^*.$

Proof

For any fixed $\varphi \in P^\prime ,$ we define the mapping $P_\varphi$ on E by:

$$\begin{aligned} P_\varphi (x)=\vert \left\langle \varphi , x\right\rangle \vert ,\quad x\in E. \end{aligned}$$

It is readily verified that for each $\varphi \in P^\prime ,$ $P_\varphi$ is a seminorm in E. We show that the family of seminorms $(P_\varphi )_{\varphi \in P^\prime }$ separates points. To see this, let $x\in E$ such that $P_\varphi (x)=0$ for any positive functional $\varphi \ge 0.$ By Corollary 2.1 we therefore obtain $x=0.$ Thus, we have proved that the family of seminorms $(P_\varphi )_{\varphi \in P^\prime }$ produces a Hausdorff locally convex topology $\tau$ on E. Now, we perform that $(E,\tau )$ is sequentially complete. In order to achieve this, we consider a Cauchy sequence $\{x_n\}$ in $(E,\tau ).$ Then, for every positive functional $\varphi \in P^\prime ,$ we have $P_\varphi (x_n-x_m)=\vert \left\langle \varphi , x_n-x_m\right\rangle \vert \rightarrow 0$ as $n,m\rightarrow +\infty .$ Now, let $\phi \in E^\prime .$ Since P is normal then $E^\prime =P^\prime -P^\prime .$ Hence, there exist $\varphi _1,\varphi _2\in P^\prime$ such that $\phi =\varphi _1-\varphi _2.$ Thus,

$$\begin{aligned} \left\langle \phi , x_n-x_m\right\rangle =\left\langle \varphi _1, x_n-x_m\right\rangle -\left\langle \varphi _2, x_n-x_m\right\rangle \rightarrow 0 \end{aligned}$$

as $n,m\rightarrow +\infty$. This proves that $\{x_n\}$ is a weak Cauchy sequence, and so (since E is weakly sequentially complete) there exists $x^*\in E$, such that $x_n{\mathop {\rightarrow }\limits ^{w}} x^*.$ This implies that $x_n{\mathop {\rightarrow }\limits ^{\tau }} x^*.$ On the other hand, From (3.4), it follows that for every $\varphi \in P^\prime$ and every $x,y\in E$, we have:

$$\begin{aligned} P_\varphi (Ax-Ay)\le c(\varphi ) P_\varphi (x-y). \end{aligned}$$

Invoking the Cain–Nashed fixed point theorem [7, Theorem 2.2], we infer that A has a unique fixed point theorem in E. Moreover, for every $x\in E$, we have $A^nx{\mathop {\rightarrow }\limits ^{\tau }} x^*$ and so $A^nx{\mathop {\rightarrow }\limits ^{w}} x^*$. $\square$

Remark 3.2

In Theorems 3.1 and 3.2, no continuity assumption is required.

Before making a formal statement of the next main result, let us fix some notation. Let E be an OBS with a cone P and let $\psi$ be a mapping from P to P satisfying:

• (i) $\psi$ is increasing and $\sum _{k=0}^\infty \Vert \psi ^k(x)\Vert <+\infty$ for each fixed $x\in P,$ where $\psi ^1=\psi ,\,\psi ^{n+1}=\psi ^n(\psi (x)),\,n=1,2\ldots$

• (ii) $P\subset (id-\psi )P$ and $\Vert \psi (x)\Vert \rightarrow 0$, when $\Vert x\Vert \rightarrow 0.$

Following [13], we define:

$$\begin{aligned} \Psi = \{\psi :P \rightarrow P\,\hbox { satisfies}\, \mathrm{(i)}{-}\mathrm{(ii)}\}. \end{aligned}$$

Remark 3.3

Let

• 1. $\psi _1(x)= \lambda x,\,\forall x\in P,$ where $0\le \lambda <1$ is a constant.

• 2. $\psi _2(x)= Lx,\,\forall x\in P,$ where L is a nonnegative linear operator satisfying $r(L)<1.$ In this case, we should point out that $id-L$ is invertible and $(id-L)^{-1}=\sum _{k=0}^\infty L^k.$

Then, $\psi _1,\psi _2\in \Psi$.

Before formulating our next fixed point result, we need the following key lemma.

Lemma 3.3

Let E be an OBS with a generating cone P and let $A:E\rightarrow E$ be an increasing operator. Assume that there exists $\psi \in \Psi$, such that for any $u,v\in E$ with $v\le u,$ we have:

$$\begin{aligned} Au-Av\le \psi (u-v). \end{aligned}$$

(3.5)

Then, the following statements hold.

• (i) There exist $u_0,v_0\in E$, such that $u_0\le Au_0$ and $Av_0\le v_0.$

• (ii) A has at most one fixed point.

Remark 3.4

We point out that the special case corresponding to $\psi =c\,id$ ($0\le c<1$) in Lemma 3.3 follows from Lemma 3.2.

Proof

In view of Lemma 2.1, there are $u\le 0$ and $v\ge 0$, such that $A(0)\ge u$ and $A(0)\le v$. Since $P\subset (id-\psi )P,$ then there exist $u_0\le 0$ and $v_0\ge 0$, such that $-u=-u_0-\psi (-u_0)$ and $v=v_0-\psi (v_0)$. Accordingly:

$$\begin{aligned} A(u_0)\ge A(0)-\psi (-u_0)\ge u-\psi (-u_0)=u_0, \end{aligned}$$

and

$$\begin{aligned} A(v_0)\le A(0)+\psi (v_0)\le v+\psi (v_0)=v_0. \end{aligned}$$

To prove (ii), suppose that there exist $u,v\in E$, such that $Au=u$ and $Av=v.$ Since P is generating, then, by Lemma 2.1, there exists $w\in E$, such that $u\le w$ and $v\le w.$ Pick up any $\varphi \in P^\prime .$ Using a simple mathematical induction, we can prove that for each integer $n\ge 0$, we have $A^nu\le A^nw$ and $A^nv\le A^nw.$ This together with (3.5) lead to:

$$\begin{aligned} A^nw -A^nu \le \psi ^n( w-u), \end{aligned}$$

(3.6)

and

$$\begin{aligned} A^nw-A^nv\le \psi ^n(w-v). \end{aligned}$$

(3.7)

Linking (3.6) and (3.7), we arrive at:

$$\begin{aligned} \vert \left\langle \varphi , u-v\right\rangle \vert\le & {} \vert \left\langle \varphi , A^nu-A^nv\right\rangle \vert \\\le & {} \vert \left\langle \varphi , A^nu-A^nw\right\rangle \vert +\vert \left\langle \varphi , A^nw-A^nv\right\rangle \vert \\\le & {} \left\langle \varphi , \psi ^n(w-u)\right\rangle +\left\langle \varphi ,\psi ^n( w-v)\right\rangle ; \end{aligned}$$

upon letting $n\rightarrow +\infty$, we see that $\left\langle \varphi , u-v\right\rangle =0.$ Using again Corollary 2.1, we obtain $u=v$. $\square$

Now, we are ready to state the following result.

Theorem 3.3

Let E be an OBS with a generating cone P. Let $A:E\rightarrow E$ be an increasing operator, such that there exists a positive bounded linear operator $L:E\rightarrow E$ with spectral radius $r(L)<1$ satisfying:

$$\begin{aligned} Au-Av\le L(u-v),\quad \hbox {for}\,\,u,v\in E\,\,\hbox {with}\,\, v\le u. \end{aligned}$$

(3.8)

Assume further that A satisfies ${\mathcal {P}}(n_0)$ for some integer $n_0\ge 1.$ Then, A has a unique fixed point in E.

Proof

In view of Lemma 3.3, there exist $u_0,v_0\in E$, such that $u_0\le Au_0$ and $Av_0\le v_0.$ Define the sequence $\{u_n\}$ by $u_n=Au_{n-1},\,n=1,2,\ldots$ It is straightforward to deduce that:

$$\begin{aligned} u_0\le u_1\le \,\cdots \,\le u_{n}\le \,\cdots \end{aligned}$$

(3.9)

Let $S=\{u_0,u_1,\ldots ,u_n,\ldots \}$. Clearly, for any integer $k\ge 1$, we have:

$$\begin{aligned} A^{k}(S)\cup \{u_0,u_1,\ldots ,u_{k-1}\}=S. \end{aligned}$$

From our hypotheses, we know that S is weakly relatively compact, and so, by Lemma 3.1, the sequence $\{u_{n}\}$ converges weakly to some $u^*.$ By monotonicity arguments, we deduce that $u_n\le u^*$ for all $n\ge 0.$ Accordingly, for any $\phi \in P^\prime ,$ we have:

$$\begin{aligned} \vert \left\langle \phi , Au^*-u^*\right\rangle \vert\le & {} \vert \left\langle \phi , Au^*-Au_n\right\rangle \vert +\vert \left\langle \phi , u_{n+1}-u^*\right\rangle \vert \\\le & {} \vert \left\langle \phi , L(u^*-u_n)\right\rangle \vert +\vert \left\langle \phi , u_{n+1}-u^*\right\rangle \vert . \end{aligned}$$

Letting $n\rightarrow +\infty ,$ we conclude (since L is weakly continuous) that $\left\langle \phi , Au^*-u^*\right\rangle =0.$ Referring to Corollary 2.1, we see that $Au^*=u^*.$ The uniqueness follows from Lemma 3.2. $\square$

Remark 3.5

The main novelty in Theorem 3.3, compared with results of a similar flavor (see, for example, [20, Theorem 49.3] and [13, Theorem 3.2]), is the fact that the cone P need not be normal. This permits the establishment of existence and uniqueness of solutions for boundary value problems in Banach spaces partially ordered by non-normal cones (see Sects. 5 and 6).

From Theorem 3.3, we can derive the following corollaries.

Corollary 3.1

Let E be an OBS with a generating cone P and let $\mu$ be a generalized measure of weak noncompactness on E. Let $A:E\rightarrow E$ be an increasing operator, such that there exists a positive bounded linear operator $L:E\rightarrow E$ with spectral radius $r(L)<1$ satisfying:

$$\begin{aligned} Au-Av\le L(u-v),\quad \hbox {for}\,\,u,v\in E\,\,\hbox {with}\,\, v\le u. \end{aligned}$$

(3.10)

Assume further that there exists an integer $n_0\ge 1$, such that $A^{n_0}(E)$ is bounded, and for any countable bounded set $\Omega \subset E$ with $\mu (\Omega )\ne 0$, we have:

$$\begin{aligned} \mu (A^{n_0}(\Omega ))<\mu (\Omega ). \end{aligned}$$

Then, A has a unique fixed point in E.

Corollary 3.2

Let E be an OBS with a generating cone P. Let $A:E\rightarrow E$ be an increasing operator, such that there exists a positive bounded linear operator $L:E\rightarrow E$ with spectral radius $r(L)<1$ satisfying:

$$\begin{aligned} Au-Av\le L(u-v),\quad \hbox {for}\,\,u,v\in E\,\,\hbox {with}\,\, v\le u. \end{aligned}$$

(3.11)

Assume further that there exists an integer $n_0\ge 1$, such that $A^{n_0}(E)$ is weakly relatively compact. Then, A has a unique fixed point in E.

Corollary 3.3

Let E be an OBS with a generating cone P and let $\mu$ be a generalized measure of weak noncompactness on E. Let $A:E\rightarrow E$ be an increasing operator, such that there exist a positive bounded linear operator $L:E\rightarrow E$ with spectral radius $r(L)<1$ and positive integers $m_0$ and $n_0$ with:

$$\begin{aligned} A^{n_0}u-A^{n_0}v\le L^{m_0}(u-v),\quad \hbox {for}\,\,u,v\in E\,\,\hbox {with}\,\, v\le u. \end{aligned}$$

(3.12)

Suppose that $A^{n_0}(E)$ is bounded and for any bounded countable set $\Omega \subset E$ with $\mu (\Omega )\ne 0$, we have:

$$\begin{aligned} \mu (A^{n_0}(\Omega ))<\mu (\Omega ). \end{aligned}$$

Then, A has a unique fixed point in E.

Proof

First notice that $r(L^{m_0})\le r(L)^{m_0}<1.$ Invoking Corollary 3.1, we deduce that $A^{n_0}$ has a unique fixed point $x^*\in E.$ However:

$$\begin{aligned} A^{n_0}(A(x^*))=A(A^{n_0}(x^*))=A(x^*), \end{aligned}$$

so $A(x^*)$ is also a fixed point of $A^{n_0}.$ Since the fixed point of $A^{n_0}$ is unique, it must be the case that $A(x^*)=x^*.$ Now, suppose $A(y^*)=y^*$, then $A^{n_0}(y^*)=y^*$ proving (again by uniqueness) that $y^*=x^*$. $\square$

Corollary 3.4

Let E be an OBS with a generating cone P and let $\mu$ be a generalized measure of weak noncompactness on E. Let $A:E\rightarrow E$ be a decreasing operator, such that there exists a positive bounded linear operator $L:E\rightarrow E$ with spectral radius $r(L)<1$ satisfying:

$$\begin{aligned} Au-Av\ge - L(u-v),\quad \hbox {for}\,\,u,v\in E\,\,\hbox {with}\,\, v\le u. \end{aligned}$$

(3.13)

Assume further that A(E) is bounded and for any bounded countable set $\Omega \subset E$ with $\mu (\Omega )\ne 0$, we have:

$$\begin{aligned} \mu (A(\Omega ))<\mu (\Omega ). \end{aligned}$$

Then, A has a unique fixed point in E.

Proof

Since A is decreasing then from (3.13), it follows that:

$$\begin{aligned} A^{2}u-A^{2}v\le L^{2}(u-v),\quad \hbox {for}\,\,u,v\in E\,\,\hbox {with}\,\, v\le u. \end{aligned}$$

(3.14)

A direct application of Corollary 3.3 yields the desired conclusion. $\square$

4 Coupled fixed point theorems

In this section, we establish some existence and uniqueness results for mixed monotone operators in Banach spaces. We point out that, in 1987, Guo and Lakshmikantham [15] first introduced the concept of mixed monotone mappings to study the existence and uniqueness of solutions to ordinary differential equations. Thereafter, many authors have studied the existence of coupled fixed points for mappings having mixed monotone property and obtained many important results, which were applied to integral equations, fractional differential equations, and matrix equations (see [6, 16, 18, 21] and the references therein).

Before making a formal statement of the main results of this section, we need to fix notations and recall standard definitions and related results. Throughout this section, E is a real Banach space which is partially ordered by a cone P.

Definition 4.1

Let $D\subset E$ and $A:D\times D\rightarrow E.$ We say that A has the mixed monotone property if, for any $x,y\in D:$

$$\begin{aligned} x_1,x_2\in D,\quad x_{1} \le x_{2} \,\,\hbox {implies}\,\, A\left( x_{1}, y\right) \le A\left( x_{2}, y\right) \end{aligned}$$

and

$$\begin{aligned} y_1,y_2\in D,\quad y_{1} \le y_{2}\,\,\hbox {implies}\,\,A\left( x, y_{1}\right) \ge A\left( x, y_{2}\right) . \end{aligned}$$

Definition 4.2

An element $(x^*, y^*) \in D\times D$ is called a coupled fixed point of $A:D\times D\rightarrow E$ if $A(x^*, y^*) = x^*$ and $A(y^*,x^*)= y^*.$ When $x^*=y^*,$ i.e., $x^*=A(x^*,x^*),$ we say that $x^*$ is a fixed point of A.

Definition 4.3

Let $D\subset E.$ We say that $A:D\times D\rightarrow E$ is (monotone) subcontinuous if, for any (monotone) sequences $(u_n)_{n}$ and $(v_n)_{n}$ in D that converge strongly to u and v, respectively, the sequence $(A(u_n,v_n))_{n}$ converges weakly to A(u, v).

For convenience purposes, we need to set up some additional notation. Let D be a non-empty subset of E and $A:D\times D\rightarrow D$ be a mapping.

• For all $(x,y)\in D\times D$, we define:

  $$\begin{aligned} A^0(x,y):=x \text { and } A^{k+1}(x,y):=A\Big (A^k(x,y),A^k(y,x)\Big ),\quad k=1,2,\ldots . \end{aligned}$$

• Let U, V be two subsets of D and $k\ge 1$ be an integer. We use $A^k(U,V)$ to denote the set:

  $$\begin{aligned} A^k(U,V)=\{A^k(x,y):x\in U,\,y\in V\}. \end{aligned}$$

• Let $k\ge 1$ be an integer and let $U=\{u_n\}$ and $V=\{v_n\}$ be two sequences of elements of D. We use the following notations:

  $$\begin{aligned} U\triangle V=\{(u_n,v_n):n\ge 0\} \end{aligned}$$

  and

  $$\begin{aligned} A^k(U\triangle V)=\{A^k(u_n,v_n):n\ge 0\}. \end{aligned}$$

Now, we are ready to state the following monotone iterative principle for operators with the mixed monotone property.

Theorem 4.1

Let E be an OBS ordered by a normal cone P and $u_0,v_0\in E$ with $u_0<v_0.$ Let $A:[u_0,v_0]\times [u_0,v_0]\rightarrow E$ be a monotone subcontinuous mapping having the mixed monotone property, such that:

$$\begin{aligned} u_0\le A(u_0,v_0),\quad A(v_0,u_0)\le v_0. \end{aligned}$$

(4.1)

Assume further that A satisfies the following condition:

${\mathcal {P}}^\prime (n_0)$: there exists an integer $n_0\ge 1$, such that for any monotone sequences $U=\{u_n\}$ and $V=\{v_n\}$ in $[u_0,v_0]$ and any finite subsets $M_{n_0},N_{n_0}$ of cardinal $n_0,$ we :

$$\begin{aligned} \left\{ \begin{array}{l} U= M_{n_0}\cup A^{n_0}(U\triangle V)\\ \\ V= N_{n_0}\cup A^{n_0}(V\triangle U) \end{array}\right. \,\, \Rightarrow \,\, U \,\hbox {and}\, V \,\hbox {are weakly relatively compact}. \end{aligned}$$

(4.2)

Then, A has a coupled fixed point $(x_*,x^*)$ in $[u_0,v_0]\times [u_0,v_0]$ which is minimal and maximal, in the sense that $x_*\le x \le x^*$ and $x_*\le y \le x^*$ for any coupled fixed point $(x,y)\in [u_0,v_0]\times [u_0,v_0]$ of A.

Moreover, the sequence of the successive approximations $(A^n(u_0,v_0), A^n(v_0,u_0))$ converges to $(x_*,x^*)$ .

Proof

Define the sequences $\{u_n\}$ and $\{v_n\}$ by:

$$\begin{aligned} u_n=A(u_{n-1},v_{n-1}),\quad v_n=A(v_{n-1},u_{n-1}),\quad n=1,2,\ldots . \end{aligned}$$

It is routine to verify that:

$$\begin{aligned} u_0\le u_1\le \cdots \le u_n\le \cdots \le v_n \le \cdots \le v_1\le v_0. \end{aligned}$$

Now, let $U=\{u_0,u_1,\ldots ,u_n,\ldots \}$ and $V=\{v_0,v_1,\ldots ,v_n,\ldots \}$.

It is readily verified that for each fixed integer $k\ge 1,$ we have:

$$\begin{aligned} A^k(u_n,v_n)=u_{n+k}\quad \hbox {and}\quad A^k(v_n,u_n)=v_{n+k},\,n=0,1,\ldots , \end{aligned}$$

so that:

$$\begin{aligned} U=\{u_0,u_1,\ldots ,u_{k-1}\}\cup A^{k}(U\triangle V), \end{aligned}$$

and

$$\begin{aligned} V=\{v_0,v_1,\ldots ,v_{k-1}\}\cup A^{k}(V\triangle U). \end{aligned}$$

From our assumptions, we know that U and V are weakly relatively compact. Referring to Lemma 2.4, we see that there exist $x_*,x^*\in E$, such that $\lim _{n \rightarrow \infty } u_n=x_*$ and $\lim _{n \rightarrow \infty } v_n=x^*.$ It follows from the monotone subcontinuity of A that:

$$\begin{aligned} u_{n+1}=A(u_n,v_n)\rightharpoonup A(x_*,x^*) \end{aligned}$$

and

$$\begin{aligned} v_{n+1}=A(v_n,u_n)\rightharpoonup A(x^*,x_*). \end{aligned}$$

Therefore, $x_*=A(x_*,x^*)$ and $x^*=A(x^*,x_*);$ that is, $(x_*,x^*)$ is a coupled fixed point of A. Now, we illuminate that the coupled fixed point $(x_*,x^*)$ is minimal and maximal (in the sense indicated above). To do this, let $(x,y)\in [u_0,v_0]\times [u_0,v_0]$ be any coupled fixed point of T. It follows from the mixed monotone property of A that:

$$\begin{aligned} u_1=A(u_0,v_0)\le x=A(x,y)\le v_1=A(v_0,u_0) \end{aligned}$$

and

$$\begin{aligned} u_1=A(u_0,v_0) \le y=A(y,x) \le v_1=A(v_0,u_0). \end{aligned}$$

By a straightforward mathematical induction, we obtain $u_n\le x\le v_n$ and $u_n\le y\le v_n$ for any integer $n\ge 0.$ Letting $n\rightarrow +\infty ,$ we conclude that $x_*\le x\le x^*$ and $x_*\le y\le x^*$. $\square$

Theorem 4.2

Let the conditions of Theorem 4.1 be satisfied. Suppose there exists $\psi \in \Psi$, such that for any $x,y\in E$ with $x\le y,$ we have:

$$\begin{aligned} A(y,x)-A(x,y)\le \psi (y-x). \end{aligned}$$

(4.3)

Then, A has exactly one fixed point $\overline{x}\in [u_0,v_0].$ Moreover, if we successively construct the sequences:

$$\begin{aligned} x_n=A(x_{n-1},y_{n-1}),\, y_n=A(y_{n-1},x_{n-1}),\quad n=1,2\ldots , \end{aligned}$$

(4.4)

for any initial $(x_0,y_0)\in [u_0,v_0]\times [u_0,v_0],$ then we have $x_n\rightarrow \overline{x}$ and $y_n\rightarrow \overline{x}$.

Proof

By (4.3), we know that for any integer $n\ge 1,$ we have:

$$\begin{aligned} 0\le v_n-u_n= A(v_{n-1},u_{n-1})-A(u_{n-1},v_{n-1})\le \psi (v_{n-1}-u_{n-1}), \end{aligned}$$

and so for any $\varphi \in P^\prime ,$ we have:

$$\begin{aligned} 0\le \varphi (v_n-u_n)\le \varphi (\psi ^n( v_{0}-u_{0}))\rightarrow 0\quad \hbox {as }\,n\rightarrow \infty . \end{aligned}$$

Hence, $\varphi (x^*-x_*)=0$ for any $\varphi \in P^\prime .$ Invoking Corollary 2.1, we conclude that $x^*=x_*.$ Let $\overline{x}=x_*=x^*,$ and then, $\overline{x}$ is a fixed point of A. By virtue of the minimal and maximal property of $(x_*,x^*)$, we can show easily that $\overline{x}$ is the unique fixed point of A in $[u_0,v_0].$ Let $(x_0,y_0)\in [u_0,v_0]\times [u_0,v_0],$ be given and (4.4) be constructed. Similar to the proof of Theorem 4.1, we get $u_n\le x_n\le v_n$ and $u_n\le y_n\le v_n$ for $n=0,1,\ldots .$ It follows, therefore, that $x_n\rightarrow \overline{x}$ and $y_n\rightarrow \overline{x}$. $\square$

Remark 4.1

Theorem 4.2 extends [16, Corollary 2.1.5]. For example, in the case where $\psi =\lambda \,id,$ it may happen that $0<\lambda <1$ and $\delta \lambda \ge 1$ where $\delta$ is the normal constant of P. In such a case, the mapping A is “contractive” in order without being “contractive” in norm.

As a convenient specialization of Theorem 4.1, we obtain the following result.

Corollary 4.1

Let E be an OBS ordered by a normal cone P and let $\mu$ be a generalized measure of weak noncompactness on E. Let $A:[u_0,v_0]\times [u_0,v_0]\rightarrow E$ be a monotone subcontinuous mapping having the mixed monotone property. Suppose that there exists an integer $n_0\ge 1$, such that for any bounded monotone sequences $U=\{u_n\}$ and $V=\{v_n\}$ in $[u_0,v_0]$ with $\max \{\mu (U),\mu (V)\} >0,$ we have:

$$\begin{aligned} \mu (A^{n_0}(U\triangle V))<\max \{\mu (U),\mu (V)\}. \end{aligned}$$

(4.5)

If, moreover, the condition (4.1) is satisfied, then A has a coupled fixed point $(x_*,x^*)$ in $[u_0,v_0]\times [u_0,v_0]$ which is minimal and maximal, in the sense that $x_*\le x \le x^*$ and $x_*\le y \le x^*$ for any coupled fixed point $(x,y) \in [u_0,v_0]\times [u_0,v_0]$ of A. In addition, the sequence of the successive approximations $(A^n(u_0,v_0),A^n(v_0,u_0))$ converges to $(x_*,x^*)$.

Proof

It is enough to show that the condition ${\mathcal {P}}^\prime (n_0)$ is satisfied. To do this, take any two monotone sequences $U=\{u_n\}$ and $V=\{v_n\}$ in $[u_0,v_0]$ satisfying:

$$\begin{aligned} U=M_{n_0}\cup A^{n_0}(U\triangle V)\quad \hbox {and}\quad V=N_{n_0}\cup A^{n_0}(V\triangle U ), \end{aligned}$$

where $M_{n_0}$ and $N_{n_0}$ are finite sets of $[u_0,v_0].$ Employing the properties of a generalized measure of weak noncompactness, we obtain:

$$\begin{aligned} \mu (U)= \mu (A^{n_0}(U\triangle V ))\quad \hbox {and}\quad \mu (V)= \mu (A^{n_0}(V\triangle U )). \end{aligned}$$

(4.6)

Assume that either $\mu (U)>0$ or $\mu (V)>0$. By (4.5), we have:

$$\begin{aligned} \mu (A^{n_0}(U\triangle V ))< \max \{\mu (U),\mu (V)\}\, \text { and } \mu (A^{n_0}(V\triangle U )) < \max \{\mu (U),\mu (V)\}. \end{aligned}$$

(4.7)

Combining (4.6) and (4.7), we arrive at:

$$\begin{aligned} \mu (U)< \max \{\mu (U),\mu (V)\}\, \text { and } \mu (V) < \max \{\mu (U),\mu (V)\}, \end{aligned}$$

which is a contradiction. Consequently, U and V are weakly relatively compact.

Remark 4.2

• Corollary 4.1 extends [18, Theorem 3.3.1], [9, Theorem 1], and [8, Theorem 1].

• It is noteworthy that the condition of the boundedness of $A([u_0,v_0]\times [u_0,v_0])$ is fulfilled whenever the cone P is normal.

Let E be an OBS with a cone P. We equip $E\times E$ with the norm $\Vert (x,y)\Vert =\sup \{\Vert x\Vert ,\Vert y\Vert \}.$ Put:

$$\begin{aligned} \tilde{P}=\{(x,y)\in E\times E:x\ge \theta ,y\le \theta \}. \end{aligned}$$

It is clear that $\tilde{P}$ is a cone in $E\times E$ and defines a partial ordering in $E\times E$ by:

$$\begin{aligned} (x_1,y_1)\le (x_2,y_2)\quad \hbox {iff}\quad x_1\le x_2\quad \hbox {and}\quad y_1\ge y_2. \end{aligned}$$

The following lemma is probably well known. For the sake of completeness, we give here a sketch of its proof.

Lemma 4.1

• (i) If P is normal then, so is $\tilde{P}.$

• (ii) If P is generating then, so is $\tilde{P}.$

Proof

To prove (i), take any $(x_1,y_1)$ and $(x_2,y_2)$ elements of $E\times E$, such that $\theta \le (x_1,y_1)\le (x_2,y_2).$ Then, $\theta \le x_1\le x_2$ and $y_2\le y_1\le \theta .$ By the normality of P, we have $\Vert x_1\Vert \le N\Vert x_2\Vert$ and $\Vert y_1\Vert \le N\Vert y_2\Vert .$ Thus, $\Vert (x_1,y_1)\Vert \le N\Vert (x_2,y_2)\Vert .$ To prove (ii), take $(x,y)\in E\times E.$ Since P is generating then, by Lemma 2.1, there are $u\ge \theta$ and $v\le \theta$, such that $x\le u$ and $y\ge v.$ Thus, $(u,v)\ge \theta$ and $(x,y)\le (u,v).$ This implies, by virtue of Lemma 2.1, that $\tilde{P}$ is generating. $\square$

Now, let $D\subset E$ and $A:D\times D\rightarrow E$ be a mixed monotone operator. We define $\tilde{A}:D\times D\rightarrow E\times E$ by:

$$\begin{aligned} \tilde{A}(u,v)=(A(u,v),A(v,u)), \end{aligned}$$

where $(u,v)\in D\times D.$ Assume that $(x_1,y_1)\le (x_2,y_2).$ Then:

$$\begin{aligned} A(x_2,y_2)\ge A(x_1,y_2)\ge A(x_1,y_1) \end{aligned}$$

and

$$\begin{aligned} A(y_2,x_2)\le A(y_1,x_2)\le A(y_1,x_1). \end{aligned}$$

It follows that $\tilde{A}(x_2,y_2)\ge \tilde{A}(x_1,y_1)$ and $\tilde{A}$ is increasing. Furthermore, it is easily seen that $(x^*,y^*)\in D\times D$ is a fixed point of $\tilde{A}$ if and only if $(x^*,y^*)$ is a coupled fixed point of A. In view of this observation, a standard tactic in addressing the problem of the existence of coupled fixed point of A is to seek for a fixed point of the operator $\tilde{A}$ by taking advantage of the wide variety of fixed point results developed for increasing operators.

Lemma 4.2

[10, Proposition 6.4] Let E and F be Banach spaces, and let $E\times F$ be the product space endowed with one of the equivalent norms:

$$\begin{aligned} \Vert (x,y)\Vert _\infty =\max \{\Vert x\Vert ,\Vert y\Vert \} \end{aligned}$$

or

$$\begin{aligned} \Vert (x,y)\Vert _p=\left( \Vert x\Vert ^p+\Vert y\Vert ^p\right) ^{1/p},\,\, 1\le p<\infty . \end{aligned}$$

Then, for any bounded subset M of E and any bounded subset N of F, we have:

$$\begin{aligned} \beta (M\times N)\le \rho _p(\beta (M),\beta (N)), \end{aligned}$$

(4.8)

where $\rho _p(r,s)=(r^p+s^p)^{1/p}$ for $1\le p<\infty$, $\rho _\infty (r,s)=\max \{r,s\}$ and $\beta (\cdot )$ is the De Blasi measure of weak noncompactness.

Proof

We give the proof for the convenience of the reader. Let M be a bounded subset of E and N be a bounded subset of F. Let $r>\beta (M)$ and $s>\beta (N).$ Then, there exist a weakly compact subset $K_1$ of E and a weakly compact subset $K_2$ of F, such that $M\subset K_1+rB$ and $N\subset K_2+sB.$ It is an easy matter to check that $M\times N\subset K_1\times K_2+\rho _p(r,s)B.$ This implies that:

$$\begin{aligned} \beta (M\times N)\le \rho _p(r,s). \end{aligned}$$

Letting $r\rightarrow \beta (M)$ and $s\rightarrow \beta (N)$, we get the desired result. $\square$

Now, we consider coupled fixed points for operators which have not necessarily the mixed monotone property.

Theorem 4.3

Let E be a weakly sequentially complete OBS with a normal cone P and let $A:E\times E\rightarrow E$ be a mapping. Assume that for any $\varphi \in P^\prime ,$ there exists $c=c(\varphi )\in [0,1)$, such that for all $x_1,x_2,y_1,y_2\in E$ with $x_1\ge x_2$ and $y_1\le y_2$, we have:

$$\begin{aligned} -c(\varphi )\left\langle \varphi , x_1-x_2\right\rangle \le \left\langle \varphi ,A(x_1,y_1)-A(x_2,y_2)\right\rangle \le c(\varphi )\left\langle \varphi ,x_1-x_2\right\rangle . \end{aligned}$$

(4.9)

Then, A has a unique coupled fixed point $(x^*,y^*)\in E\times E,$ and for each initial point $(u_0,v_0)\in E\times E,$ the sequence $\{(u_n,v_n)\}$ defined by

$$\begin{aligned} (u_n,v_n)=\tilde{A}(u_{n-1},v_{n-1}),\quad n=1,2,\ldots , \end{aligned}$$

converges weakly to $(x^*,y^*)$.

Proof

First, notice that (4.9) is equivalent :

$$\begin{aligned} -c(\varphi )\left\langle \varphi , y_2-y_1\right\rangle \le \left\langle \varphi ,A(y_2,x_2)-A(y_1,x_1)\right\rangle \le c(\varphi )\left\langle \varphi ,y_2-y_1\right\rangle , \end{aligned}$$

(4.10)

where $x_1,x_2,y_1,y_2,\,\,x_1\ge x_2,$ and $y_1\le y_2$.

Let $\phi \in \tilde{P}^\prime .$ Then, for all $(x,y)\ge 0,$ we have $\phi (x,y)\ge 0.$ Let $\varphi _1$ and $\varphi _2$ be two functionals defined on E by $\varphi _1(x)=\phi (x,0)$ and $\varphi _2(x)=\phi (0,-x)$ for each $x\in E.$ Clearly, $\varphi _1,\varphi _2\in P^\prime$ and $\phi (x,y)=\varphi _1(x)-\varphi _2(y).$ Moreover:

$$\begin{aligned} \left\langle \phi ,\tilde{A}(x_1,y_1)-\tilde{A}(x_2,y_2)\right\rangle= & {} \left\langle \varphi _1, A(x_1,y_1)-A(x_2,y_2)\right\rangle \\&-\left\langle \varphi _2, A(y_1,x_1)-A(y_2,x_2)\right\rangle \\\le & {} \max \{c(\varphi _1),c(\varphi _2)\}\left\langle \phi ,(x_1-x_2,y_1-y_2)\right\rangle . \end{aligned}$$

By Theorem 3.2, $\tilde{A}$ has a unique fixed point $(x^*,y^*)\in E\times E$ and for each initial point $(u_0,v_0)\in E\times E,$ the sequence $\{(u_n,v_n)\}$ defined by

$$\begin{aligned} (u_n,v_n)=\tilde{A}(u_{n-1},v_{n-1}),\quad n=1,2,\ldots , \end{aligned}$$

converges weakly to $(x^*,y^*).$ This achieves the proof. $\square$

Theorem 4.4

Let E be an OBS with a generating cone P and $L:E\rightarrow E$ be a positive linear operator whose spectral radius $r(L)<1.$ Suppose that $A:E\times E\rightarrow E$ satisfies:

$$\begin{aligned} -L(x_1-x_2)\le A(x_1,y_1)-A(x_2,y_2)\le L(x_1-x_2), \end{aligned}$$

(4.11)

where $x_1,x_2,y_1,y_2\in E,$ $x_1\ge x_2$ and $y_1\le y_2.$ Assume further that there is a positive integer $n_0$, such that $A^{n_0}(E)$ is bounded and for any bounded countable subset D of $E\times E$ with $\beta (D)>0,$ we have:

$$\begin{aligned} \beta (A^{n_0}(D))<\beta (D). \end{aligned}$$

(4.12)

Then, A has a unique coupled fixed point $(x^*,y^*)\in E\times E,$ and for each initial point $(u_0,v_0)\in E\times E,$ the sequence $\{(u_n,v_n)\}$ defined by

$$\begin{aligned} (u_n,v_n)=\tilde{A}(u_{n-1},v_{n-1}),\quad n=1,2,\ldots , \end{aligned}$$

converges weakly to $(x^*,y^*).$

Proof

First, we note that (4.11) is equivalent to

$$\begin{aligned} L(y_2-y_1)\ge A(y_2,x_2)-A(y_1,x_1)\ge -L(y_2-y_1), \end{aligned}$$

(4.13)

where $x_1,x_2,y_1,y_2\in E,\,x_1\ge x_2,$ and $y_1\le y_2.$ Define $\tilde{L}:E\times E\rightarrow E\times E$ by

$$\begin{aligned} \tilde{L}(x,y)=(Lx,Ly),\quad x,y\in E. \end{aligned}$$

It is clear that $\tilde{L}$ is a positive linear operator and $r(\tilde{L})=r(L)<1.$ Now (4.11) and (4.13) imply

$$\begin{aligned} \tilde{A}(x_1,y_1)-\tilde{A}(x_2,y_2)\le \tilde{L}(x_1-x_2,y_1-y_2), \end{aligned}$$

where $x_1,x_2,y_1,y_2\in E,\,x_1\ge x_2,$ and $y_1\le y_2.$ On the other hand, let $D\subset E\times E$ be a countable bounded set and denote $D^\prime =\{(y,x):(x,y)\in D\}.$ Then, $\beta (D)=\beta (D^\prime ).$ By induction, it is easy to see that:

$$\begin{aligned} \tilde{A}^n(x,y)=(A^n(x,y),A^n(y,x)),\quad n=1,2\ldots , \end{aligned}$$

for $(x,y)\in D.$ Hence, for any integer $n\ge 1$:

$$\begin{aligned} \tilde{A}^n(D)\subset A^n(D)\times A^n(D^\prime ). \end{aligned}$$

If $\beta (D)>0,$ then:

$$\begin{aligned} \beta (\tilde{A}^{n_0}(D))\le \beta (A^{n_0}(D)\times A^{n_0}(D^\prime )) \le \max \{\beta (A^{n_0}(D)), \beta (A^{n_0}(D^\prime ))\}<\beta (D). \end{aligned}$$

By Corollary 3.1, $\tilde{A}$ has a unique fixed point $(x^*,y^*)\in E\times E,$ and for each initial point $(u_0,v_0)\in E\times E,$ the sequence $\{(u_n,v_n)\}$ defined by:

$$\begin{aligned} (u_n,v_n)=\tilde{A}(u_{n-1},v_{n-1}),\quad n=1,2\ldots \end{aligned}$$

converges to $(x^*,y^*).$ This proves the assertion. $\square$

5 Application to second-order differential equations

This section is concerned with the existence and uniqueness of solution of a boundary value problem for nonlinear second-order ordinary differential equation of the type :

$$\begin{aligned} x^{\prime \prime } =f(t,x)\,\,\hbox {for a.a.}\,\,t\in I=[0,1],\quad x(0)=x(1)=0. \end{aligned}$$

(5.1)

We are interested in solutions $x:I\rightarrow {\mathbb {R}}$ of (5.1) in the Sobolev space $W^{2,1}(I)$ of all absolutely continuous real-valued functions x, such that $x^\prime$ is also absolutely continuous and $x^{\prime \prime }\in L^1(I).$

Equation (5.1) will be studied under the following assumptions :

• (C1) The mapping $f:I\times {\mathbb {R}}\rightarrow {\mathbb {R}}$ satisfies Carathéodory conditions; that is, for every $u\in {\mathbb {R}},$ the mapping $t\in I \mapsto f(t,u)$ is measurable and for a.a. $t\in I,$ the mapping $u\in {\mathbb {R}}\mapsto f(t,u)$ is continuous.

• (C2) There exists $h \in L^1 (0, 1)$, such that for a.e. $t \in [0,1]$ and all $u\in {\mathbb {R}}$, we have $\vert f(t,u)\vert \le h(t).$

• (C3) There exists $\mu >0$, such that for all $t\in I$ and all $v\ge u$, we have:

  $$\begin{aligned} 0\le f(t,u)-f(t,v)\le \mu (v-u). \end{aligned}$$

Remark 5.1

We point out that (C1) implies that any composition $t\in I \mapsto f(t,x(t))$ is measurable whenever $x\in {\mathcal {C}}(I).$ We refer the readers to [3] for more information on measurability of compositions.

Solutions of (5.1) coincide then with fixed points of the usual operator $A:{\mathcal {C}}^1(I)\rightarrow {\mathcal {C}}^1(I)$ defined by:

$$\begin{aligned} Ax(t)=\int _0^1G(t,s)f(s,x(s)) \mathrm{d}s \quad (t\in I,\, x\in {\mathcal {C}}^1(I)), \end{aligned}$$

where ${\mathcal {C}}^1(I)$ is the Banach space of continuously differentiable functions $x:I\rightarrow {\mathbb {R}}$ equipped with the norm:

$$\begin{aligned} \Vert x\Vert _{{\mathcal {C}}^1}=\max _{t\in I}\vert x(t)\vert + \max _{t\in I}\vert x^\prime (t)\vert , \end{aligned}$$

and ordered by the generating (non-normal) cone:

$$\begin{aligned} P=\{x\in {\mathcal {C}}^1(I):x(t)\ge 0\,\hbox { for all}\,t\in I\}. \end{aligned}$$

The function G is the Green’s function corresponding to problem (5.1), which is given by:

$$\begin{aligned} G(t,s)=\left\{ \begin{array}{ll} (t-1)s&{}\quad \hbox {if}\,\,0\le s\le t\le 1,\\ \\ (s-1)t&{}\quad \hbox {if}\,\,0\le t\le s\le 1. \end{array}\right. \end{aligned}$$

It can easily be checked that:

$$\begin{aligned} \int _0^1\vert G(t,s)\vert \mathrm{d}s =\frac{t(1-t)}{2}\le \frac{t}{2}\ \hbox {and} \ \int _0^1\left| \frac{\partial G}{\partial t}(t,s)\right| \mathrm{d}s =t^2-t+\frac{1}{2} \le \frac{1}{2}\quad \hbox {for}\;t\in I. \end{aligned}$$

Let:

$$\begin{aligned}&M=\sup \{\left| G(t,s)\right| :t,s\in I\},\,N =\sup \left\{ \left| \frac{\partial G(t,s)}{\partial t}\right| :t,s\in I\right\} ; \\&r_0=M\int _0^1h(s)\mathrm{d}s,\, r_1=N\int _0^1h(s)\mathrm{d}s. \end{aligned}$$

For any $x\in {\mathcal {C}}^1(I)$, the function

$$\begin{aligned} t\mapsto \int _0^1G(t,s)f(s,x(s))\mathrm{d}s \end{aligned}$$

satisfies the inequalities:

$$\begin{aligned} \vert u^{\prime \prime }(t)\vert =\vert f(t,x(t))\vert \le h(t),\,\vert u^\prime (t)\vert \le r_1,\,\vert u(t)\vert \le r_0, \end{aligned}$$

and consequently, by the mean value theorem:

$$\begin{aligned} \vert u^\prime (t)-u^\prime (\tau )\vert \le \left| \int _\tau ^th(s)\mathrm{d}s\right| \quad \hbox {and}\quad \vert u(t)-u(\tau )\vert \le r_1\vert t-\tau \vert \end{aligned}$$

for $t,\tau \in I.$ Invoking the Arzela–Ascoli theorem [17, Theorem 1.2.7], we conclude that $A({\mathcal {C}}^1(I))$ is a relatively compact subset of ${\mathcal {C}}^1(I).$

Furthermore, it is readily verified that A is increasing, and for any $x,y\in {\mathcal {C}}^1(I)$ with $x\le y$, we have $Ay-Ax\le L(y-x),$ where:

$$\begin{aligned} (Lz)(t)=\int _0^1-\mu G(t,s) z(s)\mathrm{d}s. \end{aligned}$$

Then:

$$\begin{aligned} \vert Lz(t)\vert \le \mu \int _0^1\vert G(t,s)\vert \vert z(s)\vert \mathrm{d}s \le \frac{\mu }{2}t\Vert z\Vert _{{\mathcal {C}}^1}, \end{aligned}$$

and hence:

$$\begin{aligned} \vert L^2z(t)\vert \le \mu \int _0^1\vert G(t,s)\vert \vert Lz(s)\vert \mathrm{d}s \le \left( \frac{\mu ^2}{2}\int _0^1s\vert G(t,s)\vert \mathrm{d}s\right) \Vert z\Vert _{{\mathcal {C}}^1} \le \frac{\mu ^2}{2^2}\frac{t^2}{2} \Vert z\Vert _{{\mathcal {C}}^1}. \end{aligned}$$

By a straightforward mathematical induction, we obtain:

$$\begin{aligned} \vert L^nz(t)\vert \le \frac{\mu ^n}{2^n}\frac{t^n}{n!} \Vert z\Vert _{{\mathcal {C}}^1}. \end{aligned}$$

(5.2)

Accordingly:

$$\begin{aligned} \vert (L^nz)^\prime (t)\vert \le \mu \int _0^1\left| \frac{\partial G}{\partial t} (t,s) \right| \vert L^{n-1}z(s)\vert \mathrm{d}s\le \frac{\mu }{2}\int _0^1 \vert L^{n-1}z(s)\vert \mathrm{d}s \le \frac{\mu ^n}{2^n}\frac{1}{n!} \Vert z\Vert _{{\mathcal {C}}^1}. \end{aligned}$$

(5.3)

Linking (5.2) and (5.3), we arrive at:

$$\begin{aligned} \Vert L^nz\Vert _{{\mathcal {C}}^1}\le \frac{\mu ^n}{2^{n-1}}\frac{1}{n!}\Vert z\Vert _{{\mathcal {C}}^1}. \end{aligned}$$

Easy computations lead to:

$$\begin{aligned} \Vert L^n\Vert ^{\frac{1}{n}}_{{\mathcal {C}}^1} \le \left( \frac{1}{2}\right) ^{\frac{1}{n}}\frac{\mu }{2}\left( \frac{1}{n!}\right) ^{\frac{1}{n}} \rightarrow 0,\quad \hbox {as }\,\,n\rightarrow +\infty , \end{aligned}$$

so that:

$$\begin{aligned} r(L)=\lim _{n\rightarrow +\infty }\Vert L^n\Vert ^{\frac{1}{n}}_{{\mathcal {C}}^1}=0<1. \end{aligned}$$

Consequently, A fulfills all the conditions of Corollary 3.2 and, therefore, has a unique fixed point in ${\mathcal {C}}^1(I),$ which is turn a unique solution of (5.1) in $W^{2,1}(I).$ We are therefore in a position to state the following existence and uniqueness result.

Theorem 5.1

Assume that (C1),(C2), and (C3) hold. Then, Eq. (5.1) has a unique solution in $W^{2,1}(I).$

6 Application to a system of integral equations

In this section, we study the existence and the uniqueness of solution for the following system of integral equations:

$$\begin{aligned} \left\{ \begin{array}{l} u(t) =\displaystyle \int _{0}^{t} f(s,u(s),v(s))\mathrm{d}s,\quad t\in I \\ \\ v(t) =\displaystyle \int _{0}^{t} f(s,v(s),u(s))\mathrm{d}s, \end{array}\right. \end{aligned}$$

(6.1)

where $I=[0,a]$ ($a>0$) is a compact interval of ${\mathbb {R}}$ and E is an OBS partially ordered by a normal cone P. Let ${\mathcal {C}}(I, E)$ denote the Banach space of all continuous functions $u:I\rightarrow E$, equipped with the standard norm $\left\| u \right\| =\sup _{t\in I}\left\| u(t) \right\| _E$ and partially ordered by the following order relation:

$$\begin{aligned} u,v\in {\mathcal {C}}(I, E),\, u\le v \Leftrightarrow u(t)\le v(t),\quad \forall t\in I. \end{aligned}$$

Our main purpose in the immediate sequel is to show the existence of a unique solution to Eq. (6.1). Before doing so, it is appropriate to clarify the definition of solution we will consider.

Definition 6.1

By a solution of (6.1), we mean a couple $(u,v)\in {\mathcal {C}}(I, E)\times {\mathcal {C}}(I, E)$ which satisfies (6.1).

We also recall the definition of coupled lower-upper solutions of (6.1).

Definition 6.2

An element $(u_0,v_0)\in {\mathcal {C}}(I, E)\times {\mathcal {C}}(I, E)$ is called a coupled lower–upper solution of the integral equation (6.1) if:

$$\begin{aligned} u_0(t)\le & {} \displaystyle \int _{0}^{t} f(s,u_0(s),v_0(s))\mathrm{d}s,\quad \forall t\in I , \nonumber \\ v_0(t)\ge & {} \displaystyle \int _{0}^{t} f(s,v_0(s),u_0(s))\mathrm{d}s,\quad \forall t\in I . \end{aligned}$$

(6.2)

The system (6.1) will be considered under the following set of assumptions:

$(H_0)$:

The system (6.1) has a coupled lower–upper solution $(u_0,v_0)$, such that $u_0(t)\le v_0(t),$ for all $t\in I.$

$(H_1)$:

The mapping $f:I\times E\times E\rightarrow E$ satisfies Carathéodory conditions; that is, for every $(u,v)\in E\times E,$ the mapping $t\in I \mapsto f(t,u,v)$ is measurable and for a.a. $t\in I,$ the mapping $(u,v)\in E\times E\mapsto f(t,u,v)$ is continuous.

$(H_2)$:

For any $z,w\in [u_0,v_0]:$

$$\begin{aligned} z_1,z_2\in [u_0,v_0],\quad z_{1} \le z_{2} \,\,\hbox {implies} \,\, f\left( t, z_{1}, w(t)\right) \le f\left( t, z_{2}, w(t)\right) \end{aligned}$$

and

$$\begin{aligned} w_1,w_2\in [u_0,v_0],\quad w_{1} \le w_{2}\,\,\hbox {implies} \,\,f\left( t,z(t), w_{1}\right) \ge f\left( t, z(t), w_{2}\right) . \end{aligned}$$

$(H_3)$:

There is a constant $k\ge 0$, such that, for any bounded subsets $\Gamma ,\Lambda$ of E if either $\beta (\Gamma )$ or $\beta (\Lambda )$ is greater than 0, then:

$$\begin{aligned} \beta \big (f(I, \Gamma , \Lambda )\big ) \le k \max \{\beta (\Gamma ),\beta (\Lambda )\}. \end{aligned}$$

$(H_4)$:

There is a constant $\lambda \ge 0$, such that for any $z,z^\prime \in [u_0,v_0]$ with $z\le z^\prime$ and for any $t\in I$, we have:

$$\begin{aligned} f(t,z^\prime (t),z(t))-f(t,z(t),z^\prime (t))\le \lambda (z^\prime (t)-z(t)). \end{aligned}$$

$(H_5)$:

There exists a nonnegative function $h\in L^1(I)$, such that for any $z,z^\prime \in [u_0,v_0]$ and for any $t\in I,$ we have:

$$\begin{aligned} \Vert f(t,z(t),z^\prime (t))\Vert _E\le h(t). \end{aligned}$$

To allow the abstract formulation of our problem, we consider the operator T defined by:

$$\begin{aligned} T(u,v)(t):=\int _{0}^{t} f(s,u(s),v(s))\mathrm{d}s,\quad t\in I,\quad u,v\in {\mathcal {C}}(I,E). \end{aligned}$$

Remark 6.1

• Keeping in mind $(H_2),$ it is straightforward to see that T has the mixed monotone property.

• From hypotheses $(H_0)$ and $(H_2)$, we know that:

  $$\begin{aligned} \forall u,v\in [u_0,v_0],\,\,\, u_0 \le T(u_0,v_0) \le T(u,v)\le T(v_0,u_0) \le v_0. \end{aligned}$$

  Therefore:

  $$\begin{aligned} T\big ([u_0,v_0]\times [u_0,v_0]\big )\subseteq [u_0,v_0]. \end{aligned}$$

• If E is reflexive, then the condition $(H_3)$ is automatically fulfilled. This is due to the fact that a subset of a reflexive Banach space is weakly relatively compact if and only if it is norm bounded.

It is plainly visible that (u, v) is a solution of (6.1) if and only if (u, v) is a coupled fixed point of T. With this in mind, we shall show that the operator T satisfies all conditions of Theorem 4.2. This will be achieved in a series of lemmas.

Lemma 6.1

The operator T is continuous on $[u_0,v_0]\times [u_0,v_0]$.

Proof

Let $((u_n,v_n))_n$ be a sequence in $[u_0,v_0]\times [u_0,v_0]$ that converges to some (u, v). By $(\mathbf{H1}),$ we have:

$$\begin{aligned} \displaystyle \lim _{n \rightarrow \infty } f(s,u_n(s),v_n(s)) = f(s,u(s),v(s)) \end{aligned}$$

$\text{ for } \text{ a.e. } s\in [0,a]$. On the other hand, standard computations yield:

$$\begin{aligned} \Vert T (u_n,v_n)- T(u,v)\Vert _\infty \le \int _{0}^{a} \Vert f(s,u_n(s),v_n(s)) - f(s,u(s),v(s))\Vert _E \mathrm{d}s. \end{aligned}$$

Using the dominated convergence Theorem, we obtain:

$$\begin{aligned} \displaystyle \lim _{n \rightarrow \infty }\Vert T(u_n,v_n)- T(u,v)\Vert _\infty = 0. \end{aligned}$$

This completes the proof. $\square$

Lemma 6.2

For any two subsets U, V of $[u_0,v_0]$, $T^n(U, V)$ is equicontinuous for each integer $n\ge 1.$

Proof

Let $0\le \tau \le t\le a$, $u\in U$, and $v\in V.$ Then:

$$\begin{aligned} \left\| T(u,v)(t)- T(u,v)(\tau )\right\|= & {} \left\| \int _{0}^{t}f \big (s,u(s),v(s))\mathrm{d}s-\int _{0}^{\tau }f(s,u(s),v(s)\big )\mathrm{d}s \right\| \\= & {} \left\| \int _{\tau }^{t}f\big (s,u(s),v(s)\big )\mathrm{d}s \right\| \ \\\le & {} \int _{\tau }^{t}\left\| f\big (s,u(s),v(s)\big ) \right\| \mathrm{d}s\\\le & {} \int _{\tau }^{t}h(s)\mathrm{d}s. \end{aligned}$$

Hence:

$$\begin{aligned} \left\| T(u,v)(t)- T(u,v)(\tau )\right\| \rightarrow 0\quad \hbox {as }\,\, t\rightarrow \tau , \end{aligned}$$

uniformly in (u, v). This proves the equicontinuity of the set T(U, V). Using a straightforward mathematical induction, we can show that $T^n(U, V)$ is equicontinuous for any integer $n\ge 1$. $\square$

Lemma 6.3

For any $t\in I$ and any integer $n\in {\mathbb {N}},$ we have:

$$\begin{aligned} \beta \big (T^n(U, V)(t)\big )\le \frac{(kt)^n}{n!} \max \{\beta (U([0,t])),\beta (V([0,t]))\}. \end{aligned}$$

Proof

The proof runs along the same lines as [1, p. 15], so we omit it. $\square$

Lemma 6.4

The operator T satisfies the condition ${\mathcal {P}}^\prime (n_0)$ for some integer $n_0\ge 1$.

Proof

Since $\lim _{n\rightarrow +\infty } \frac{(ka)^n}{n!}=0,$ we may choose an integer $n_0$ as large as we please, so that $\frac{(ka)^{n_0}}{{n_0}!}<1.$ Now, let $U,\, V$ to be two subsets of $[u_0,v_0]$ satisfying:

$$\begin{aligned} \left\{ \begin{array}{l} U \subseteq A\cup T^{n_0}(U\Delta V )\\ \\ V \subseteq B\cup T^{n_0}(V\Delta U ) \end{array}\right. \end{aligned}$$

(6.3)

with A and B some arbitrary finite sets of $[u_0,v_0].$ In view of Lemma 6.2, $T^{n}(U\Delta V )$ and $T^{n}(V\Delta U )$ are equicontinuous for any integer $n\ge 1.$ By virtue of Lemmas 6.3 and 2.2, we conclude that:

$$\begin{aligned} \beta _{{\mathcal {C}}} \big (T^n(U\Delta V)\big )\le \beta _{{\mathcal {C}}} \big (T^n(U, V)\big ) \le \frac{(ka)^n}{n!} \max \{\beta _{{\mathcal {C}}}(U),\beta _{{\mathcal {C}}}(V)\}, \end{aligned}$$

(6.4)

for any integer $n\ge 1$. Suppose that either $\beta _{{\mathcal {C}}} (U)$ or $\beta _{{\mathcal {C}}} (V)$ is greater than 0. Linking (6.4) and (6.3), we arrive at:

$$\begin{aligned} \max \{\beta _{{\mathcal {C}}}(U),\beta _{{\mathcal {C}}}(V)\} \le \beta _{{\mathcal {C}}}\big (T^{n_0}(U\Delta V)\big )\le & {} \frac{(ka)^{n_0}}{n_0!} \max \{\beta _{{\mathcal {C}}}(U),\beta _{{\mathcal {C}}}(V)\}\\< & {} \max \{\beta _{{\mathcal {C}}}(U),\beta _{{\mathcal {C}}}(V)\}, \end{aligned}$$

which is a contradiction. Therefore, $\beta _{{\mathcal {C}}}(U)=\beta _{{\mathcal {C}}}(V)=0$, and so, U and V are weakly relatively compact. $\square$

Lemma 6.5

There is a positive bounded linear operator L on C([0, a]; E) with spectral radius $r(L)=0$, such that for any $z,z^\prime \in [u_0,v_0]$ with $z\le z^\prime$, we have:

$$\begin{aligned} T(z^\prime ,z)-T(z,z^\prime )\le L (z^\prime -z). \end{aligned}$$

(6.5)

Proof

Let L be the positive bounded linear operator defined on C([0, a]; E) by:

$$\begin{aligned} Lu(t)=\int _ 0^t\lambda u(s)\mathrm{d}s,\quad u\in C([0,a]; E),\,t\in [0,a]. \end{aligned}$$

Using a simple mathematical induction, we can prove that:

$$\begin{aligned} L^nu(t)=\int _ 0^t\lambda ^n\frac{(t-s)^{n-1}}{(n-1)!} u(s)\mathrm{d}s,\quad n=1,2,\ldots , \end{aligned}$$

so that $\Vert L^n\Vert \le \frac{\lambda ^n}{(n-1)!}$. As a result:

$$\begin{aligned} r(L)=\lim _{n\rightarrow +\infty } \Vert L^n\Vert ^{\frac{1}{n}}\le \lambda \left( \frac{1}{(n-1)!}\right) ^{\frac{1}{n}}\rightarrow 0\quad \hbox {as }\,\, n\rightarrow +\infty . \end{aligned}$$

Therefore, $r(L)=0.$ On the other hand, by Hypothesis ($H_4$), we have:

$$\begin{aligned} (T(z^\prime ,z)-T(z,z^\prime ))(t)= & {} \int _0^t\left( f(s,z^\prime (s),z(s))-f(s,z(s),z^\prime (s))\right) \mathrm{d}s\\\le & {} \int _0^t\lambda \left( z^\prime (s)-z(s)\right) \mathrm{d}s\\= & {} L(z^\prime -z)(t), \end{aligned}$$

for any $z,z^\prime \in [u_0,v_0]$ with $z\le z^\prime$ and for any $t\in I$.

Consequently, for any $z,z^\prime \in [u_0,v_0]$ with $z\le z^\prime$, we have:

$$\begin{aligned} T(z^\prime ,z)-T(z,z^\prime )\le L (z^\prime -z). \end{aligned}$$

(6.6)

$\square$

Theorem 6.1

Assume that the hypotheses $(H_0)$–$(H_5)$ hold. Then, the problem (6.1) has a unique solution (u, v) in $[u_0,v_0]\times [u_0,v_0],$ with $u=v.$

Proof

The result follows from Theorem 4.2 on the basis of Lemmas 6.1, 6.2, 6.3, 6.4, and 6.5.