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Orderpreservation of solution correspondence for parametric generalized variational inequalities on Banach lattices
Fixed Point Theory and Applicationsvolume 2015, Article number: 108 (2015)
Abstract
In this paper, several ordertheoretic fixed point theorems are proved on Banach lattices. As applications of these fixed point theorems, we study the orderpreservation of solution correspondence for parametric generalized variational inequalities. Furthermore, the existence of solutions to equilibrium problems is also examined. Our approach is ordertheoretic and all results obtained in this paper do not require the considered mappings to be continuous or semicontinuous with respect to any topology equipped on the underlying space.
Introduction
Let X be a Banach space whose (topological) dual is denoted by $X^{*}$, let C be a nonempty closed and convex subset of X, and let $\Gamma:C\rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be a setvalued mapping. In this paper, we consider the following generalized variational inequality problem, which is to find $\hat{x}\in C$ such that there exists $\phi\in\Gamma(\hat{x})$ with
Let us refer to this problem succinctly as $\operatorname{GVI}(C,\Gamma)$. If there is at least one solution to it, then we say that $\operatorname{GVI}(C,\Gamma)$ is solvable.
For studying the solvability of $\operatorname{GVI}(C,\Gamma)$, Nishimura and Ok [1] use the ordertheoretic methods to obtain several existence theorems of maximal solutions to $\operatorname{GVI}(C,\Gamma)$ on Hilbert lattices. Along this line, Li and Ok [2] explore the orderpreserving of generalized metric projection operator and study the existence of maximum (minimum) solutions to $\operatorname{GVI}(C,\Gamma)$ on Banach lattices, where the considered mappings are required to have topped (bottomed) values. In contrast to many previous literature works, the approach of Nishimura, Ok and Li is ordertheoretic, and hence their results do not require Γ to be continuous or semicontinuous with respect to any topology equipped on the underlying space. For more details, see, for instance, [3–5] and the references therein.
Besides studying the solvability of $\operatorname{GVI}(C,\Gamma)$, it is also an important subject to examine the behavior of solutions to parametric $\operatorname{GVI}(C,\Gamma)$. In traditional research, the authors were often concerned about the continuity or semicontinuity of the solution correspondence (see, e.g., [6–11]). However, in the ordertheoretic setup, one would be concerned, instead, with the orderpreservation of this solution correspondence. To the best of our knowledge, there are few papers on this subject except for the work of Nishimura and Ok [1]. However, Nishimura and Ok only consider one parameter in their research, where Γ is disturbed by the parameter θ and the underlying sets are Hilbert lattices.
Motivated and inspired by Nishimura, Ok and Li, we have the principal objective in this paper to explore the orderpreservation of solution correspondence for parametric $\operatorname{GVI}(C,\Gamma)$, where two parameters are considered and the underlying sets are Banach lattices. The content of this paper can be summarized as follows. Section 2 is devoted to some basic concepts related to posets as well as some useful lemmas. In Section 3, several ordertheoretic fixed point theorems are proved on Banach lattices. These fixed point theorems are the main tools for the following analysis. In Section 4, we first examine the existence of maximal and minimal solutions to $\operatorname{GVI}(C,\Gamma)$ without requiring the involved mapping to have topped (bottomed) values. Then we explore the upper orderpreservation and lower orderpreservation of solution correspondence for parametric $\operatorname{GVI}(C,\Gamma)$ on Banach lattices. Section 5 is devoted to some further discussions on equilibrium problems, which include variational inequalities as special cases. Since there are no (generalized) metric projection operators in this case, some other techniques are introduced in this section.
Preliminaries
In this section, we recall some basic concepts in a poset as well as several useful lemmas. For more details, please refer to [1, 12–14].
Some concepts in poset
A poset is an ordered pair $(X,\succcurlyeq)$, where X is a nonempty set and ≽ a partial order on X. For each $x\in (X,\succcurlyeq)$, we define $x^{\uparrow}=\{y\in(X,\succcurlyeq ):y\succcurlyeq x \}$ and $x_{\downarrow}=\{y\in(X,\succcurlyeq): x\succcurlyeq y \}$. In turn, for any nonempty subset S, we define $S^{\uparrow}=\bigcup\{x^{\uparrow}:x\in S\}$ and $S^{\downarrow}=\bigcup\{ x^{\downarrow}:x\in S\}$. We say that an element x of $(X,\succcurlyeq )$ is a ≽upper bound for S if $x\succcurlyeq S$, that is, $x\succcurlyeq y$ for each $y\in S$. The notation $S\succcurlyeq x$ is similarly understood. We say that S is ≽bounded from above if $x\succcurlyeq S$ for some $x\in (X,\succcurlyeq)$, and ≽bounded from below if $S\succcurlyeq x$ for some $x\in(X,\succcurlyeq)$. In turn, S is said to be ≽bounded if it is ≽bounded from above and below. Particularly, if $x\in S$ and x is a ≽upper bound for S, then we say that x is the ≽maximum in S. The ≽minimum element of S is similarly defined. We say that x is ≽maximal element of S if $x\in S$ and $y\succcurlyeq x$ does not hold for any $y\in S\setminus\{x\}$. Similarly, x is said to be the ≽minimal element of S if $x\in S$ and $x\succcurlyeq y$ does not hold for any $y\in S\setminus\{x\}$. A nonempty subset S of X is said to be a ≽chain in X if either $x\succcurlyeq y$ or $y\succcurlyeq x$ holds for each $x,y\in S$.
The ≽supremum of S is the ≽minimum of the set of all ≽upper bounds for S and is denoted by $\bigvee_{X}S$. The ≽infimum of S, which is denoted by $\bigwedge_{X}S$, is defined similarly. As is conventional, we denote $\bigvee_{X}\{x,y\}$ as $x\vee y$ and $\bigwedge_{X}\{x,y\}$ as $x\wedge y$ for any x, y in $(X,\succcurlyeq)$. If $x\vee y$ and $x\wedge y$ exist for every x and y in $(X,\succcurlyeq)$, then we say that $(X,\succcurlyeq)$ is a lattice, and if $\bigvee_{X}S$ and $\bigwedge_{X}S$ exist for every nonempty (≽bounded) $S\subseteq(X,\succcurlyeq)$, we say that $(X,\succcurlyeq )$ is a (Dedekind) complete lattice. If Y is a nonempty subset of $(X,\succcurlyeq)$ which contains $\bigvee_{X}\{x,y\}$ and $\bigwedge_{X}\{x,y\}$ for every $x,y\in Y$, then it is said to be a ≽sublattice of $(X,\succcurlyeq)$. In turn, if Y contains $\bigvee_{X}S$ and $\bigwedge_{X}S$ for every nonempty $S\subseteq Y$, then it is said to be a subcomplete ≽sublattice of $(X,\succcurlyeq)$.
Orderpreservation for correspondences
For any lattices $(X,\succcurlyeq_{X})$ and $(Y,\succcurlyeq_{Y})$, we say that a map $F:X\rightarrow Y$ is orderpreserving if $x\succcurlyeq_{X}y$ implies $F(x)\succcurlyeq_{Y}F(y)$ for every $x,y\in X$. In turn, if $\Gamma:X\rightarrow2^{Y}$ is a setvalued correspondence, we say that Γ is upper orderpreserving if $x\succcurlyeq_{X}y$ implies that $\Gamma(y)=\emptyset$, or for every $y'\in\Gamma(y)$ there is $x'\in\Gamma(x)$ such that $x'\succcurlyeq_{Y}y'$. Upper orderreversing maps are defined dually. Similarly, Γ is lower orderpreserving if $x\succcurlyeq_{X}y$ implies that $\Gamma(x)=\emptyset$, or for every $x'\in\Gamma(x)$ there is $y'\in\Gamma(y)$ such that $x'\succcurlyeq_{Y}y'$. Γ is orderpreserving if it is both upper and lower orderpreserving. If $(X,\succcurlyeq_{X})$ and $(Y,\succcurlyeq_{Y})$ are subposets of a given poset $(Z,\succcurlyeq )$, then we use the phrase ≽preserving instead of orderpreserving.
Banach lattice
A Riesz space is a lattice $(X,\succcurlyeq)$ where X is a (real) linear space whose linear structure is compatible with the partial order ≽ in the sense that $\alpha\mathrm{id}_{X}+z$ is a ≽preserving selfmap on X for every $z\in X$ and real number $\alpha>0$. The positive cone of X is $X_{+}:=\{x\in X: x\succcurlyeq\mathbf{0} \}$, where 0 denotes the origin of X.
A Riesz space $(X,\succcurlyeq)$ is called a normed Riesz space if X is a normed linear space whose norm $\\cdot\$ is a compatible with the partial order ≽ in the sense that $\x\\geqslant \y\$ holds for every $x,y\in X$ with $x\succcurlyeqy$, where $z=(z\vee\mathbf{0})+(z\vee\mathbf{0})$ for every $z\in X$.
We say $(X,\succcurlyeq)$ is a Banach lattice, that is, $(X,\succcurlyeq)$ is an ordered Riesz space with X being a Banach space. If X is a Hilbert space here, then $(X,\succcurlyeq)$ is referred to as a Hilbert lattice.
Lemma 2.1
Let $(X,\succcurlyeq)$ be a Banach lattice, then the following results hold:

(i)
The lattice operations ∨ and ∧ are continuous.

(ii)
The positive cone $X_{+}$ is closed.

(iii)
For any $y\in X$, $\{z\in X: z\succcurlyeq y \}=y+X_{+}$, $\{ z\in X: y\succcurlyeq z \}=yX_{+}$ and $\{z\in X: y_{2}\succcurlyeq z\succcurlyeq y_{1} \}=(y_{1}+X_{+})\cap (y_{2}X_{+})$.
Definition 2.1
(see [2])
Let $(X,\succcurlyeq)$ be a Banach lattice. The dual of ≽ is the partial order ≽^{∗} on $X^{*}$ defined as follows:
It is well known that $(X^{*},\succcurlyeq^{*})$ is a Banach lattice, which is called the dual of $(X,\succcurlyeq)$. As usual, we denote the positive cone of $(X^{*},\succcurlyeq^{*})$ by $X^{*}_{+}$.
Some results related to generalized metric projection operator on Banach space
In 1996, Alber [12] introduced the following generalized metric projection operator on a Banach space. Let X be a Banach space whose dual is denoted by $X^{*}$. Denote the operator norm on $X^{*}$ by $\ \cdot\_{*}$, and consider the map $V:X^{*}\times X\rightarrow R$ defined by
It is easy to check that this map is nonnegativevalued. For any nonempty, closed and convex subset C of X, the generalized metric projection operator onto C is the correspondence $\pi _{C}:X^{*}\rightarrow2^{C}\setminus\{\emptyset\}$ defined by
When X is a Hilbert space, we can regard any ϕ in $X^{*}$ as lying in X. In this case, $V(\phi,x)=\\phix\$ for each $(\phi ,x)\in X^{*}\times X$, which, in turn, implies that $\pi_{C}$ is the standard metric projection operator onto C. In [13], Li derived many important properties of $\pi_{C}$. Some of them are listed as follows; for more details, see, for instance, [12, 13].
Lemma 2.2
(see [13])
Let X be a reflexive Banach space with dual space $X^{*}$ and C be a nonempty, closed and convex subset of X. We have the following results.

(i)
Define $J_{X}:X\rightarrow2^{X^{*}}$ which is called normalized duality mapping by
$$ J_{X}(x)=\bigl\{ \phi(x)\in X^{*}:\bigl\langle \phi(x),x \bigr\rangle =\bigl\Vert \phi(x)\bigr\Vert _{*}\ x\=\x\^{2}= \bigl\Vert \phi(x)\bigr\Vert _{*}^{2} \bigr\} . $$(2.4)If X is also smooth and strictly convex, then $J_{X}$ is singledvalued and $\pi_{C}\circ J_{X}=\mathrm{id}_{C}$.

(ii)
If X is also smooth, then for any given $\phi\in X^{*}$, $x\in\pi_{C}(\phi)$ if and only if $\langle\phiJ(x),xy\rangle\geq0$ for every $y\in C$. This is the variational characterization of generalized metric projection operator.

(iii)
The operator $\pi_{C}:X^{*}\rightarrow C$ is singlevalued if and only if X is strictly convex.

(iv)
If X is also smooth and strictly convex, then the generalized projection operator $\pi_{C}: X^{*}\rightarrow C$ is continuous.
Based on the result (ii) of Lemma 2.2, we can establish the equivalence between $\operatorname{GVI}(C,\Gamma)$ and a fixed point problem as follows.
Lemma 2.3
(see [13])
Let X be a reflexive and smooth Banach space and $\lambda: X\rightarrow R_{++}$ be any function. Let C be a nonempty closed and convex subset of X, and let $\Gamma :C\rightarrow2^{X^{*}}$ be any correspondence. Then $x^{*}$ is a solution to ${\operatorname{GVI}}(C,\Gamma)$ if and only if $x^{*}$ is a fixed point of the correspondence $\pi_{C}\circ(J_{X}\lambda\Gamma)$, that is, $x^{*}\in\pi_{C}(J_{X}(x^{*})\lambda(x^{*})\Gamma(x^{*}))$.
In addition, we also need the following results.
Definition 2.2
(see [13])
Let X be a Banach lattice. A ≽sublattice of S of X is said to be regular if $\\cdot\$ is submodular on S with respect to ≽, that is, $\x\vee y\^{2}+\x\wedge y\^{2}\leq\ x\^{2}+\y\^{2}$ for every $x,y\in S$.
Lemma 2.4
(see [13])
Let X be a reflexive Banach lattice and C be a regular ≽sublattice of X. Then $\pi_{C}$ is orderpreserving.
Lemma 2.5
(Zorn’s lemma)
Let $(P,\succcurlyeq)$ be a poset, if every chain of P has an ≽upper bound in P, then P contains at least one maximal element.
Lemma 2.6
(Dual version of Zorn’s lemma)
Let $(P,\succcurlyeq)$ be a poset, if every chain of P has a ≽lower bound in P, then P contains at least one minimal element.
Ordertheoretic fixed point theorems on Banach lattices
In this section, we first use Zorn’s lemma to give an ordertheoretic fixed point theorem, which extends the result of Nishimura and Ok [1] from Hilbert lattices to Banach lattices.
Theorem 3.1
Let $(X,\succcurlyeq)$ be a Banach lattice and C be a subcomplete ≽sublattice of X. If $f:C\rightarrow2^{C}\setminus\{\emptyset\}$ is upper ≽preserving and compactvalued, then f has a maximal fixed point, which is the maximal element of the set of fixed points of f.
Proof
Let
Since C is a subcomplete ≽sublattice, we have $\bigwedge_{X}C\in C$, from which we get $\bigwedge_{X}C\in A$, that is, A is nonempty. Next, we show that A is inductive, that is, every chain in A has its upper bound in A.
To this end, take an arbitrary chain $S\subseteq A$. Then, for any $x\in S$, there is $\omega(x)\in f(x)$ such that $\omega (x)\succcurlyeq x$. Since for every $x\in S$, $\bigvee_{X}S\succcurlyeq x$ and f is upper ≽preserving, it follows that for each $x\in S$, there is $\mu(x)\in f(\bigvee_{X}S)$ such that $\mu (x)\succcurlyeq\omega(x)\succcurlyeq x$. Hence, $S\subseteq f(\bigvee_{X}S)^{\downarrow}$. We claim that $\{x^{\uparrow}\cap f(\bigvee_{X}S):x\in S \}$ has the finite intersection property. Indeed, let T be a nonempty finite subset of S. Since S is a ≽chain, T is also a ≽chain. Hence there is $\bar{x}\in T$ such that $\bar{x}\succcurlyeq T$. Since $\bar{x}\in T\subseteq S\subseteq f(\bigvee_{X}S)^{\downarrow}$, we have $y\succcurlyeq\bar {x}$ for some $y\in f(\bigvee_{X}S)$. By transitivity of ≽, then, $y\in\bigcap\{x^{\uparrow}\cap f(\bigvee_{X}S): x\in T \}$.
From Lemma 2.1, it follows that the positive cone $X_{+}$ of a Banach lattice is closed and $x^{\uparrow}=x+X_{+}$. Thus, $\{x^{\uparrow}\cap f(\bigvee_{X}S): x\in S \}$ is a collection of closed subsets of $f(\bigvee_{X}S)$. Again, since f is compactvalued, $f(\bigvee_{X}S)$ is a compact subset of C. By the above analysis, we get $\{ x^{\uparrow}\cap f(\bigvee_{X}S): x\in S \}\neq\emptyset$, which implies that there is a ≽upper bound for S in $f(\bigvee_{X}S)$. Denote this ≽upper bound by ω, then $\omega\succcurlyeq\bigvee_{X}S$. On the other hand, since $\omega\in f(\bigvee_{X}S)$, we have $\bigvee_{X}S\in A$. Therefore, A is inductive.
From Zorn’s lemma, there is a ≽ maximal element $\hat{x}$ in A. By the definition of A, there is $\hat{y}\in f(\hat{x})$ such that $\hat{y}\succcurlyeq\hat{x}$. In addition, since f is upper ≽preserving and $\hat{y}\in f(\hat{x})$, there is $\hat {z}\in f(\hat{y})$ such that $\hat{z}\succcurlyeq\hat{y}$. It follows that $\hat{y}\in A$. As $\hat{y}\succcurlyeq\hat{x}$ and $\hat{x}$ is a maximal element in A, thus $\hat{x}=\hat{y}$. Therefore, $\hat{x}\in f(\hat{x})$. Denote by $\operatorname{Fix}(f)$ the set of fixed points of f, then it is easy to check that $\operatorname{Fix}(f)\subseteq A$. Hence, f has a maximal fixed point. □
In [1], Nishimura and Ok prove that if C is a closed and ≽bounded ≽sublattice of a separable Hilbert lattice, then C is a subcomplete ≽sublattice of X. Observing that the above result also holds in a Banach lattice, we get the following corollary from Theorem 3.1 immediately.
Corollary 3.1
Let $(X,\succcurlyeq)$ be a separable Banach lattice and C be a closed and ≽bound ≽sublattice of X. If $f:C\rightarrow2^{C}\setminus\{ \emptyset\}$ is upper ≽preserving and compactvalued, then f has a maximal fixed point.
Employing the idea of Fujimoto [3] and applying the dual version of Zorn’s lemma (Lemma 2.6), we can derive the following ordertheoretic fixed point theorem for lower orderpreserving setvalued mappings.
Theorem 3.2
Let $(X,\succcurlyeq)$ be a Banach lattice and C be a subcomplete ≽sublattice of X. If $f:C\rightarrow2^{C}\setminus\{\emptyset\}$ is lower ≽preserving and compactvalued, then f has a minimal fixed point.
By a similar argument to that in Corollary 3.1, we have the following result for Theorem 3.2.
Corollary 3.2
Let $(X,\succcurlyeq)$ be a separable Banach lattice and C be a closed and ≽bound ≽sublattice of X. If $f:C\rightarrow2^{C}\setminus\{ \emptyset\}$ is lower ≽preserving and compactvalued, then f has a minimal fixed point.
Remark 3.1
If $(X,\succcurlyeq)$ is a Hilbert lattice, then Theorem 3.1 is reduced to a similar result introduced by Nishimura and Ok [1]. However, our results are more general. Firstly, the underlying set that we considered is a Banach lattice; secondly, besides the case of upper ≽preserving setvalued mappings, we also examine the fixed point theorems for lower ≽preserving setvalued mappings; finally, our results do not require the domain C to be convex, and it is easy to construct this kind of examples. Take $C=[1,2]\cup[3,4]$ and define a setvalued mapping $f:C\rightarrow2^{C}\setminus\{\emptyset\}$ by
Obviously, C is not convex, but f has infinitely many fixed points.
Orderpreservation of solution correspondence for parametric generalized variational inequalities
Throughout this section, let $(X,\succcurlyeq)$ be a Banach lattice, and let Θ, Ω be posets, whose partial order relations are denoted by $\succcurlyeq_{\Theta}$ and $\succcurlyeq_{\Omega}$, respectively. In addition, we can define the product of $(\Omega ,\succcurlyeq_{\Omega})$ and $(\Theta,\succcurlyeq_{\Theta})$ as follows. Let $\Omega\times\Theta$ be the product of Ω and Θ, and the partial order on $\Omega\times\Theta$ is defined by
where $(\omega_{1},\theta_{1})$, $(\omega_{2},\theta_{2})$ belong to $\Omega\times\Theta$.
Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ and $\Gamma :X\times\Theta\rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be setvalued mappings. For given $(\omega,\theta)\in\Omega\times\Theta$, the parametric generalized variational inequality considered here is to find $x^{*}\in C(\omega)$ such that there exists $\varphi^{*}\in \Gamma(x^{*},\theta)$ with
Denote this problem by ${\operatorname{GVI}}(C(\omega),\Gamma(\cdot,\theta))$. Generally speaking, the set of solutions to ${\operatorname{GVI}}(C(\omega),\Gamma (\cdot,\theta))$ is perturbed by parameters ω and θ, that is, the set of solutions to ${\operatorname{GVI}}(C(\omega),\Gamma(\cdot,\theta ))$ may vary with these parameters. Therefore, we can define a solution correspondence $\Lambda:\Omega\times\Theta\rightarrow2^{X}$ by
Next, we examine the orderpreservation properties of this solution correspondence Λ.
Theorem 4.1
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Omega ,\succcurlyeq_{\Omega})$, $(\Theta,\succcurlyeq_{\Theta})$ be posets. Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ and $\Gamma: X\times \Theta\rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be setvalued correspondences that satisfy the following conditions:

(i)
$C(\omega)$ is a regular, closed and convex subcomplete $\succcurlyeq_{X}$sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed and convex subcomplete $\succcurlyeq _{X}$sublattice Y of X such that $C(\omega)\in Y$ for any $\omega \in\Omega$;

(ii)
$\Gamma(x,\cdot)$ is upper orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma(\cdot,\theta)$ is upper orderpreserving and compactvalued for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $(\omega,\theta)\in\Omega\times\Theta$, $\Lambda (\omega,\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is upper orderpreserving.
Proof
(a) Define a setvalued mapping $f:X\times\Omega\times\Theta\rightarrow2^{X}$ by setting
From Lemma 2.3, we know that the set of solutions to $\operatorname{GVI}(C(\omega ),\Gamma(\cdot,\theta))$ coincide with the set of fixed points of $f(\cdot,\omega,\theta)$ for each $(\omega,\theta)\in\Omega\times\Theta $. Therefore, we only need to prove that $f(\cdot,\omega,\theta)$ has a fixed point for each $(\omega,\theta)\in\Omega\times\Theta$. To this end, we first show that $f(\cdot,\omega,\theta)$ is upper $\succcurlyeq _{X}$preserving for any given $(\omega,\theta)$. In fact, since X is a reflexive Banach lattice and $C(\omega)$ is a regular ≽sublattice for each $\omega\in\Omega$ by assumption (i), $\pi _{C(\omega)}$ is orderpreserving for every $\omega\in\Omega$ by Lemma 2.4. Again, as $J_{X}\lambda\Gamma(\cdot,\theta)$ is upper orderpreserving for each $\theta\in\Theta$, it follows that $f(\cdot ,\omega,\theta)$ is upper $\succcurlyeq_{X}$preserving for any $(\omega ,\theta)\in\Omega\times\Theta$. On the other hand, since $J_{X}\lambda \Gamma(\cdot,\theta)$ is compactvalued for each $\theta\in\Theta$ by assumption (ii) and $\pi_{C(\omega)}$ is continuous for any $\omega\in \Omega$ by Lemma 2.1, we conclude that $f(\cdot,\omega,\theta)$ is compactvalued for any $(\omega,\theta)\in\Omega\times\Theta$. Since $C(\omega)$ is a subcomplete ≽sublattice of X, it follows from Theorem 3.1 that $f(\cdot,\omega,\theta)$ has a fixed point for any $(\omega,\theta)\in\Omega\times\Theta$, that is, $\Lambda (\omega,\theta)\neq\emptyset$ for each $(\omega,\theta)\in\Omega\times \Theta$.
(b) In this part, we examine the upper orderpreservation of Λ. Firstly, note the following facts. Recalling the proof of (a), we know that $f(\cdot,\omega,\theta)$ is upper $\succcurlyeq _{X}$preserving for each $(\omega,\theta)\in\Omega\times\Theta$. In fact, $f(x,\cdot,\cdot)$ is also upper orderpreserving for each $x\in X$. To see this, take arbitrary $x\in X$ and any $(\omega_{1},\theta _{1}), (\omega_{2},\theta_{2})\in\Omega\times\Theta$ with $(\omega _{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta}(\omega_{2},\theta _{2})$. Picking any $x(\omega_{2},\theta_{2})\in f(x,\omega_{2},\theta _{2})$, from the definition of f, there exists $y(\theta_{2})\in \Gamma(x,\theta_{2})$ such that $x(\omega_{2},\theta_{2})=\pi_{C(\omega _{2})}(J_{X}(x)\lambda(x)y(\theta_{2}))$. Note that $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta }(\omega_{2},\theta_{2})$ implies $\theta_{1}\succcurlyeq_{\Theta}\theta _{2}$ and that $\Gamma(x,\cdot)$ is upper orderreversing for each $x\in X$, there exists $y(\theta_{1})\in\Gamma(x,\theta_{1})$ such that $y(\theta_{2})\succcurlyeq_{X^{*}} y(\theta_{1})$, which implies that $J_{X}(x)\lambda(x)y(\theta_{1})\succcurlyeq_{X^{*}} J_{X}(x)\lambda(x)y(\theta_{2})$. Again, since $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta }(\omega_{2},\theta_{2})$ implies $\omega_{1}\succcurlyeq_{\Omega}\omega _{2}$ and $\pi_{C(\cdot)}(x)$ is orderpreserving by condition (iii), we have
Denote $\pi_{C(\omega_{1})}(J_{X}(x)\lambda(x)y(\theta_{1}))$ by $x(\omega_{1},\theta_{1})$, then (4.5) is reduced to $x(\omega _{1},\theta_{1})\succcurlyeq_{X} x(\omega_{2},\theta_{2})$. It is obvious that $x(\omega_{1},\theta_{1})\in f(x,\omega_{1},\theta_{1})$. Therefore, $f(x,\cdot,\cdot)$ is upper orderpreserving for any $x\in X$.
For proving the upper orderpreservation of Λ, take any $(\omega _{1},\theta_{1}), (\omega_{2},\theta_{2})\in\Omega\times\Theta$ with $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta} (\omega _{2},\theta_{2})$, and pick any $x(\omega_{2},\theta_{2})\in\Lambda (\omega_{2},\theta_{2})$. Our aim is to find an $x(\omega_{1},\theta _{1})\in\Lambda(\omega_{1},\theta_{1})$ such that $x(\omega_{1},\theta _{1})\succcurlyeq_{X} x(\omega_{2},\theta_{2})$. The rest of this proof is divided into four steps.
Step 1. Construct a correspondence as follows.
Since $x(\omega_{2},\theta_{2})\in C(\omega_{2})$ and $C(\omega _{2})\subseteq Y $ by condition (i), we have $x(\omega_{2},\theta _{2})\in Y\cap x(\omega_{2}, \theta_{2})^{\uparrow}$, which implies that $Y\cap x(\omega_{2},\theta_{2})^{\uparrow}\neq\emptyset$. Hence, we can define a setvalued correspondence $g:Y\cap x(\omega _{2},\theta_{2})^{\uparrow}\rightarrow2^{Y\cap x(\omega_{2},\theta _{2})^{\uparrow}}$ by
Step 2. Show that g is well defined, that is, $f(x,\omega _{1},\theta_{1})\cap C(\omega_{1})\cap x(\omega_{2},\theta _{2})^{\uparrow}\neq\emptyset$ for any $x\in x(\omega_{2},\theta _{2})^{\uparrow}$.
Since $x(\omega_{2},\theta_{2})\in\Lambda(\omega_{2},\theta_{2})$, we have $x(\omega_{2},\theta_{2})\in f(x(\omega_{2},\theta_{2}),\omega _{2},\theta_{2})$. As $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times \Theta} (\omega_{2},\theta_{2})$ and $f(x(\omega_{2},\theta_{2}),\cdot ,\cdot)$ is upper orderpreserving, there exists $y\in f(x(\omega _{2},\theta_{2}),\omega_{1},\theta_{1})$ such that $y\succcurlyeq_{X} x(\omega_{2},\theta_{2})$. Again, since $f(\cdot,\omega_{1},\theta _{1})$ is upper $\succcurlyeq_{X}$preserving, and hence for any $x\succcurlyeq_{X} x(\omega_{2},\theta_{2})$, there exists $z\in f(x,\omega_{1},\theta_{1})$ such that $z\succcurlyeq_{X} y$. From the transitivity of $\succcurlyeq_{X}$, we have $z\succcurlyeq_{X} x(\omega _{2},\theta_{2})$. Hence, $z\in f(x,\omega_{1},\theta_{1})\cap C(\omega_{1})\cap x(\omega_{2},\theta_{2})^{\uparrow}$, which implies that $f(x,\omega_{1},\theta_{1})\cap C(\omega_{1})\cap x(\omega_{2},\theta_{2})^{\uparrow}\neq\emptyset$ for any $x\succcurlyeq _{X} x(\omega_{2},\theta_{2})$.
Step 3. Show that $Y\cap x(\omega_{2},\theta_{2})^{\uparrow}$ is a subcomplete $\succcurlyeq_{X}$sublattice of X.
Take any $a,b\in Y\cap x(\omega_{2},\theta_{2})^{\uparrow}$, then we have $a,b\in Y$ and $a,b \in x(\omega_{2},\theta_{2})^{\uparrow}$, which implies that $a\succcurlyeq x(\omega_{2},\theta_{2})$ and $b\succcurlyeq x(\omega _{2},\theta_{2})$. Thus, we get $a\vee b\succcurlyeq a\wedge b\succcurlyeq x(\omega_{2},\theta_{2})$, that is, $a\vee b, a\wedge b\in x(\omega_{2},\theta_{2})^{\uparrow}$. Since Y is a sublattice of X, we have $a\vee b, a\wedge b\in Y$. So $a\vee b, a\wedge b\in x(\omega_{2},\theta_{2})^{\uparrow}\cap Y$, which implies that $Y\cap x(\omega_{2},\theta_{2})^{\uparrow}$ is a sublattice of X. In a similar manner, take any nonempty subset V of $Y\cap x(\omega_{2},\theta_{2})^{\uparrow}$, and then we can prove $\bigvee_{X}V, \bigwedge_{X}V\in Y\cap x(\omega_{2},\theta _{2})^{\uparrow}$. From the definition of subcomplete $\succcurlyeq _{X}$sublattice, we conclude that $Y\cap x(\omega_{2},\theta _{2})^{\uparrow}$ is a subcomplete $\succcurlyeq_{X}$sublattice of X.
Step 4. Prove that g is compactvalued and upper $\succcurlyeq _{X}$preserving.
Recalling the proof of (a), we know that $f(\cdot,\omega_{1},\theta _{1})$ is compactvalued for each $(\omega_{1},\theta_{1})\in\Omega \times\Theta$. Since the positive cone $X_{+}$ of X is closed and noting that $x(\omega_{2},\theta_{2})^{\uparrow}=x(\omega_{2},\theta _{2})+X_{+}$, we have $x(\omega_{2},\theta_{2})^{\uparrow}$ is closed. Moreover, $C(\omega_{1})$ is also closed by assumption (i). Hence, we claim that g is compactvalued.
Next, we prove that g is upper $\succcurlyeq_{X}$preserving. Actually, take any $x,x'\in x(\omega_{2},\theta_{2})^{\uparrow}$ with $x\succcurlyeq_{X} x'$, and pick any $y'\in g(x')$. By definition of g, we have $y'\in f(x',\omega_{1},\theta_{1})$. Since $f(\cdot,\omega_{1},\theta_{1})$ is upper $\succcurlyeq _{X}$preserving, there exists $y\in f(x,\omega_{1},\theta_{1})$ such that $y\succcurlyeq_{X} y'$. Noting that $y'\in g(x')$ implies $y'\succcurlyeq_{X} x(\omega_{2},\theta_{2})$, which leads to $y\succcurlyeq_{X} x(\omega_{2},\theta_{2})$. That is, there exists $y\in g(x)$ such that $y\succcurlyeq_{X} y'$. Hence, g is upper $\succcurlyeq_{X}$preserving.
Applying Theorem 3.1, g has a fixed point, that is, there exists $x(\omega_{1},\theta_{1})\in x(\omega_{2},\theta_{2})^{\uparrow}$ such that $x(\omega_{1},\theta_{1})\in f(x(\omega_{1},\theta_{1}),\omega _{1},\theta_{1})$, that is, $x(\omega_{1},\theta_{1})\in\Lambda(\omega _{1},\theta_{1})$. Therefore, Λ is upper orderpreserving. □
By a similar argument to that in Theorem 4.1 and applying Corollary 3.1, we obtain the following result.
Corollary 4.1
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth, strictly convex and separable Banach lattice, and let $(\Omega,\succcurlyeq_{\Omega})$, $(\Theta,\succcurlyeq_{\Theta})$ be posets. Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ and $\Gamma: X\times\Theta\rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be setvalued correspondences that satisfy the following conditions:

(i)
$C(\omega)$ is a regular, closed, convex and ≽bounded ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed, convex and ≽bounded ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in\Omega$;

(ii)
$\Gamma(x,\cdot)$ is upper orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma(\cdot,\theta)$ is upper orderpreserving and compactvalued for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $(\omega,\theta)\in\Omega\times\Theta$, $\Lambda (\omega,\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is upper orderpreserving.
When $C(\omega)=C$ for each $\omega\in\Omega$ but Γ is perturbed by θ, we deduce the following result from Theorem 3.1.
Corollary 4.2
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Theta ,\succcurlyeq_{\Theta})$ be a poset. Let $\Gamma: X\times\Theta \rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be a setvalued correspondence. Assume that the following conditions hold:

(i)
C is a regular, closed and convex subcomplete ≽sublattice of X;

(ii)
$\Gamma(x,\cdot)$ is upper orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma(\cdot,\theta)$ is upper orderpreserving and compactvalued for each $\theta\in\Theta$.
Then

(a)
for each $\theta\in\Theta$, $\Lambda(\theta)\neq \emptyset$;

(b)
the solution correspondence Λ is upper orderpreserving.
When Γ is fixed but C is perturbed by ω, we can obtain the following result from Theorem 3.1.
Corollary 4.3
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Omega ,\succcurlyeq_{\Omega})$ be a poset. Let $C:\Omega\rightarrow 2^{X}\setminus\{\emptyset\}$ and $\Gamma: X\rightarrow 2^{X^{*}}\setminus\{\emptyset\}$ be a setvalued correspondences. Assume that the following conditions hold:

(i)
$C(\omega)$ is a regular, closed and convex subcomplete ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed and convex subcomplete ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in \Omega$;

(ii)
there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma$ is upper orderpreserving and compactvalued;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $\omega\in\Omega$, $\Lambda(\omega)\neq\emptyset$;

(b)
the solution correspondence Λ is upper orderpreserving.
When Γ is a singlevalued mapping, then we have the following result.
Corollary 4.4
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Omega ,\succcurlyeq_{\Omega})$, $(\Theta,\succcurlyeq_{\Theta})$ be posets. Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ be a setvalued mapping and $\Gamma: X\times\Theta\rightarrow X^{*}$ be a singlevalued correspondence. Assume that the following conditions hold:

(i)
$C(\omega)$ is a regular, closed and convex subcomplete ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed and convex subcomplete ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in \Omega$;

(ii)
$\Gamma(x,\cdot)$ is orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda \Gamma(\cdot,\theta)$ is orderpreserving for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $(\omega,\theta)\in\Omega\times\Theta$, $\Lambda (\omega,\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is upper orderpreserving.
Next, we consider the lower orderpreservation of Λ. To this end, the ordertheoretic fixed point theorems for lower ≽preserving correspondence are applied, for instance, Theorem 3.2 and Corollary 3.2.
Theorem 4.2
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Omega ,\succcurlyeq_{\Omega})$, $(\Theta,\succcurlyeq_{\Theta})$ be posets. Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ and $\Gamma: X\times \Theta\rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be setvalued correspondences that satisfy the following conditions:

(i)
$C(\omega)$ is a regular, closed and convex subcomplete ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed and convex subcomplete ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in \Omega$;

(ii)
$\Gamma(x,\cdot)$ is lower orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma(\cdot,\theta)$ is lower orderpreserving and compactvalued for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $(\omega,\theta)\in\Omega\times\Theta$, $\Lambda (\omega,\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is lower orderpreserving.
Proof
(a) In a similar way to the proof of Theorem 4.1, define a setvalued mapping $f:X\times\Omega\times\Theta \rightarrow2^{X}$ by setting
From Lemma 2.3, we only need to prove that $f(\cdot,\omega,\theta)$ has a fixed point for each $(\omega,\theta)\in\Omega\times\Theta$. In contrast to Theorem 4.1, we need to show that $f(\cdot,\omega,\theta)$ is lower $\succcurlyeq_{X}$preserving for any given $(\omega,\theta )\in\Omega\times\Theta$. In the proof of Theorem 4.1, we have showed that $\pi_{C(\omega)}$ is orderpreserving for every $\omega\in\Omega $. Again, since $J_{X}\lambda\Gamma(\cdot,\theta)$ is lower orderpreserving for each $\theta\in\Theta$, it follows that $f(\cdot ,\omega,\theta)$ is lower $\succcurlyeq_{X}$preserving for any $(\omega ,\theta)\in\Omega\times\Theta$. By the same argument as that in Theorem 4.1, we conclude that $f(\cdot,\omega,\theta)$ is compactvalued for any $(\omega,\theta)\in\Omega\times\Theta$ and $C(\omega)$ is a subcomplete ≽sublattice of X. From Theorem 3.2, $f(\cdot,\omega,\theta)$ has a fixed point for any $(\omega,\theta)\in\Omega\times\Theta$, which implies that $\Lambda (\omega,\theta)\neq\emptyset$ for each $(\omega,\theta)\in\Omega\times \Theta$.
(b) For proving the lower orderpreservation of Λ, we first prove that $f(x,\cdot,\cdot)$ is lower orderpreserving for each $x\in X$. To see this, take arbitrary $x\in X$ and any $(\omega _{1},\theta_{1}), (\omega_{2},\theta_{2})\in\Omega\times\Theta$ with $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta}(\omega _{2},\theta_{2})$. Picking any $x(\omega_{1},\theta_{1})\in f(x,\omega _{1},\theta_{1})$, from the definition of f, there exists $y(\theta _{1})\in\Gamma(x,\theta_{1})$ such that $x(\omega_{1},\theta_{1})=\pi _{C(\omega_{1})}(J_{X}(x)\lambda(x)y(\theta_{1}))$. Note that $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta }(\omega_{2},\theta_{2})$ implies $\theta_{1}\succcurlyeq_{\Theta}\theta _{2}$ and $\Gamma(x,\cdot)$ is lower orderreversing for each $x\in X$, there exists $y(\theta_{2})\in\Gamma(x,\theta_{2})$ such that $y(\theta _{2})\succcurlyeq_{X^{*}} y(\theta_{1})$, which implies that $J_{X}(x)\lambda(x)y(\theta_{1})\succcurlyeq_{X^{*}} J_{X}(x)\lambda (x)y(\theta_{2})$. Again, since $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times\Theta }(\omega_{2},\theta_{2})$ implies $\omega_{1}\succcurlyeq_{\Omega}\omega _{2}$ and $\pi_{C(\cdot)}(x)$ is orderpreserving by condition (iii), we have
Denote $\pi_{C(\omega_{2})}(J_{X}(x)\lambda(x)y(\theta_{2}))$ by $x(\omega_{2},\theta_{2})$, then (4.8) is reduced to $x(\omega _{1},\theta_{1})\succcurlyeq_{X} x(\omega_{2},\theta_{2})$. It is obvious that $x(\omega_{2},\theta_{2})\in f(x,\omega_{2},\theta_{2})$. Therefore, $f(x,\cdot,\cdot)$ is lower orderpreserving for every $x\in X$.
Based on the above analysis, let us prove the lower orderpreservation of Λ, take any $(\omega_{1},\theta_{1}), (\omega_{2},\theta _{2})\in\Omega\times\Theta$ with $(\omega_{1},\theta_{1})\succcurlyeq _{\Omega\times\Theta} (\omega_{2},\theta_{2})$, and pick any $x(\omega _{1},\theta_{1})\in\Lambda(\omega_{1},\theta_{1})$. Our aim is to find $x(\omega_{2},\theta_{2})\in\Lambda(\omega_{2},\theta_{2})$ such that $x(\omega_{1},\theta_{1})\succcurlyeq_{X} x(\omega_{2},\theta_{2})$. Similar to the proof of Theorem 4.1, we break the rest of proof into four steps.
Step 1. Construct a correspondence as follows.
Since $x(\omega_{1},\theta_{1})\in C(\omega_{1})$ and $C(\omega _{1})\subseteq Y $, by condition (i), we have $x(\omega_{1},\theta _{1})\in Y\cap x(\omega_{1},\theta_{1})^{\downarrow}$, which implies that $Y\cap x(\omega_{1},\theta_{1})^{\downarrow}\neq\emptyset$. Hence, we can define a setvalued correspondence $g:Y\cap x(\omega _{1},\theta_{1})^{\downarrow}\rightarrow2^{Y\cap x(\omega_{1},\theta _{1})^{\downarrow}}$ by
Step 2. Show that g is well defined, that is, $f(x,\omega _{2},\theta_{2})\cap C(\omega_{2})\cap x(\omega_{1},\theta _{1})^{\downarrow}\neq\emptyset$ for any $x\in x(\omega_{2},\theta _{2})^{\downarrow}$.
Since $x(\omega_{1},\theta_{1})\in\Lambda(\omega_{1},\theta_{1})$, we have $x(\omega_{1},\theta_{1})\in f(x(\omega_{1},\theta_{1}),\omega _{1},\theta_{1})$. As $(\omega_{1},\theta_{1})\succcurlyeq_{\Omega\times \Theta} (\omega_{2},\theta_{2})$ and $f(x(\omega_{1},\theta_{1}),\cdot ,\cdot)$ is lower orderpreserving, there exists $y\in f(x(\omega _{1},\theta_{1}),\omega_{2},\theta_{2})$ such that $x(\omega_{1},\theta _{1})\succcurlyeq_{X}y$. Again, since $f(\cdot,\omega_{1},\theta_{1})$ is lower $\succcurlyeq_{X}$preserving, and hence for any $x(\omega _{1},\theta_{1})\succcurlyeq_{X}x $, there exists $z\in f(x,\omega _{2},\theta_{2})$ such that $y\succcurlyeq_{X} z$. From the transitivity of $\succcurlyeq_{X}$, we have $x(\omega_{1},\theta _{1})\succcurlyeq_{X}z $. Hence, $z\in f(x,\omega_{2},\theta _{2})\cap C(\omega_{2})\cap x(\omega_{1},\theta_{1})^{\downarrow }$, which implies that $f(x,\omega_{2},\theta_{2})\cap C(\omega _{2})\cap x(\omega_{1},\theta_{1})^{\downarrow}\neq\emptyset$ for any $x(\omega_{1},\theta_{1})\succcurlyeq_{X}x$.
Step 3. In the same way as Theorem 4.1, one can show that $Y\cap x(\omega_{1},\theta_{1})^{\downarrow}$ is a subcomplete $\succcurlyeq_{X}$sublattice of X.
Step 4. Prove that g is compactvalued and lower $\succcurlyeq _{X}$preserving.
Note that the positive cone $X_{+}$ of X is closed and $x(\omega _{1},\theta_{1})^{\downarrow}=x(\omega_{1},\theta_{1})X_{+}$, then by a similar argument to that of Theorem 4.1, we claim that g is compactvalued.
For proving that g is lower $\succcurlyeq_{X}$preserving, take any $x,x'\in x(\omega_{1},\theta_{1})^{\downarrow}$ with $x\succcurlyeq _{X} x'$, and pick any $y\in g(x)$. By definition of g, we have $y\in f(x,\omega_{1},\theta_{1})$. Since $f(\cdot,\omega_{1},\theta _{1})$ is lower $\succcurlyeq_{X}$preserving, there exists $y'\in f(x',\omega_{1},\theta_{1})$ such that $y\succcurlyeq_{X} y'$. Note that $y\in g(x)$ implies $x(\omega_{1},\theta_{1})\succcurlyeq_{X} y$, which leads to $x(\omega_{1},\theta_{1})\succcurlyeq_{X} y'$. That is, there exists $y'\in g(x')$ such that $y\succcurlyeq_{X} y'$. Hence, g is lower $\succcurlyeq_{X}$preserving.
Applying Theorem 3.2, g has a fixed point, that is, there exists $x(\omega_{2},\theta_{2})\in x(\omega_{1},\theta_{1})^{\downarrow}$ such that $x(\omega_{2},\theta_{2})\in f(x(\omega_{2},\theta_{2}),\omega _{2},\theta_{2})$, that is, $x(\omega_{2},\theta_{2})\in\Lambda(\omega _{2},\theta_{2})$. That is, Λ is lower orderpreserving. □
From Corollary 3.2, we can deduce the following result from Theorem 4.2.
Corollary 4.5
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth, strictly convex and separable Banach lattice, and let $(\Omega,\succcurlyeq_{\Omega})$, $(\Theta,\succcurlyeq_{\Theta})$ be posets. Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ and $\Gamma: X\times\Theta\rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be setvalued correspondences that satisfy the following conditions:

(i)
$C(\omega)$ is a regular, closed, convex and ≽bounded ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed, convex and ≽bounded ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in\Omega$;

(ii)
$\Gamma(x,\cdot)$ is lower orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma(\cdot,\theta)$ is lower orderpreserving and compactvalued for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $(\omega,\theta)\in\Omega\times\Theta$, $\Lambda (\omega,\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is lower orderpreserving.
When $C(\omega)=C$ for each $\omega\in\Omega$ but Γ is perturbed by θ, then we get the following result.
Corollary 4.6
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Theta ,\succcurlyeq_{\Theta})$ be a poset. Let $\Gamma: X\times\Theta \rightarrow2^{X^{*}}\setminus\{\emptyset\}$ be a setvalued correspondence. Assume that the following conditions hold:

(i)
C is a regular, closed and convex subcomplete ≽sublattice of X;

(ii)
$\Gamma(x,\cdot)$ is lower orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma(\cdot,\theta)$ is lower orderpreserving and compactvalued for each $\theta\in\Theta$.
Then

(a)
for each $\theta\in\Theta$, $\Lambda(\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is lower orderpreserving.
When Γ is fixed but C is perturbed by ω, then we can get the following result.
Corollary 4.7
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Omega ,\succcurlyeq_{\Omega})$ be a poset. Let $C:\Omega\rightarrow 2^{X}\setminus\{\emptyset\}$ and $\Gamma: X \rightarrow 2^{X^{*}}\setminus\{\emptyset\}$ be setvalued correspondences that satisfy the following conditions:

(i)
$C(\omega)$ is a regular, closed and convex subcomplete ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed and convex subcomplete ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in \Omega$;

(ii)
there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda\Gamma$ is lower orderpreserving and compactvalued for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $\omega\in\Omega$, $\Lambda(\omega)\neq\emptyset$;

(b)
the solution correspondence Λ is lower orderpreserving.
When Γ is a singlevalued mapping, then we have the following result.
Corollary 4.8
Let $(X,\succcurlyeq_{X})$ be a reflexive, smooth and strictly convex Banach lattice, and let $(\Omega ,\succcurlyeq_{\Omega})$, $(\Theta,\succcurlyeq_{\Theta})$ be posets. Let $C:\Omega\rightarrow2^{X}\setminus\{\emptyset\}$ and $\Gamma: X\times \Theta\rightarrow X^{*}$ be setvalued correspondences that satisfy the following conditions:

(i)
$C(\omega)$ is a regular, closed and convex subcomplete ≽sublattice of X for any $\omega\in\Omega$, and there exists a regular, closed and convex subcomplete ≽sublattice Y of X such that $C(\omega)\in Y$ for any $\omega\in \Omega$;

(ii)
$\Gamma(x,\cdot)$ is orderreversing for each $x\in X$, and there exists a map $\lambda:X\rightarrow R_{++}$ such that $J_{X}\lambda \Gamma(\cdot,\theta)$ is orderpreserving and compactvalued for each $\theta\in\Theta$;

(iii)
$\pi_{C(\cdot)}(x):X^{*}\times\Omega\rightarrow X$ is orderpreserving for any $x\in X$.
Then

(a)
for each $(\omega,\theta)\in\Omega\times\Theta$, $\Lambda (\omega,\theta)\neq\emptyset$;

(b)
the solution correspondence Λ is lower orderpreserving.
Remark 4.1
Since a Hilbert lattice is reflexive, smooth and strictly convex, all of the above results also hold on a Hilbert lattice.
Remark 4.2
Our approach is ordertheoretic and all results obtained in this section do not require the involved mapping Γ to be continuous or semicontinuous.
Further discussions on discontinuous equilibrium problems
Let $(X,\succcurlyeq)$ be a Banach lattice and C be a subset of X. Let $f: C\times C\rightarrow R$ be a bifunction. In this section, we consider the following equilibrium problem, which is to find $\hat {x}\in C$ such that
Denote this problem by $\operatorname{EP}(C,f)$. It is well known that $\operatorname{EP}(C,f)$ include variational inequality problems, complementary problems and Nash equilibrium problems etc. as special cases (see, e.g., [15, 16]). For instance, if T is a singlevalued mapping from C to $X^{*}$ and let $f(x,y)=\langle T(x),yx \rangle$, then $\operatorname{EP}(C,f)$ is reduced to the following variational inequality problem, which is to find $\hat{x}\in C$ such that
For studying the solvability of $\operatorname{EP}(C,f)$, various methods have been developed, for instance, topological fixed point theorems, KKM theorems, Ekeland’s variational principle and so on (see, e.g., [17–20]). For applying these methods, the involved mappings are always required to be continuous or semicontinuous. However, if the continuity of involved mappings is unknown, these methods may fail. Therefore, it is necessary to explore some other techniques to overcome this burdensome issue. In the sequel, we use ordertheoretic fixed point theorems to study the discontinuous equilibrium problems.
Theorem 5.1
Let $(X,\succcurlyeq)$ be a Banach lattice and let C be a subcomplete ≽sublattice of X. Let $f: C\times C\rightarrow R$ be a bifunction. Assume that the following conditions hold:

(i)
$f(x,x)\geqslant0$ for any $(x,x)\in C\times C$;

(ii)
the setvalued mapping $\Phi: C\rightarrow2^{C}$ defined by setting
$$\Phi(x)=\bigl\{ y\in C: f(x,y)< 0 \bigr\} $$is upper ≽preserving and compactvalued.
Then $\operatorname{EP}(C,f)$ is solvable.
Proof
We claim that there exists $\hat{x}\in E$ such that $\Phi(\hat{x})=\emptyset$. Arguing by contradiction, assume that $\Phi(x)\neq\emptyset$ for all $x\in E$, then Φ is a setvalued mapping from C to $2^{C}\setminus\{\emptyset\}$. Since Φ is upper ≽preserving and compactvalued and C is a subcomplete ≽sublattice of X, it follows from Theorem 3.1 that Φ has a fixed point. Denote this fixed point by $\bar {x}$, then $\bar{x}\in\Phi(\bar{x})$, which implies that $f(\bar {x},\bar{x})<0$. Contradiction! Hence, there exists $\hat{x}\in C$ such that $\Phi(\hat{x})=\emptyset$, that is, $f(\hat{x},y)\geqslant0$ for every $y\in C$. □
Observing that assumption (ii) in Theorem 5.1 is not convenient for applications, we introduce some original conditions on f as follows.
Theorem 5.2
Let $(X,\succcurlyeq)$ be a Banach lattice and let C be a subcomplete ≽sublattice of X. Let $f: C\times C\rightarrow R$ be a bifunction. Assume that the following conditions hold:

(i)
$f(x,x)\geqslant0$ for any $(x,x)\in C\times C$;

(ii)
$f(\cdot,y)$ is ≽reversing for each $y\in C$ and the set $\{y\in C: f(x,y)<0 \}$ is compact for each $x\in C$.
Then $\operatorname{EP}(C,f)$ is solvable.
Proof
We claim that there exists $\hat{x}\in C$ such that $\{y\in C:f(\hat{x},y)<0\}=\emptyset$. Arguing by contradiction, assume that $\{y\in C:f(x,y)<0\}\neq\emptyset$ for all $x\in C$, then we can define a setvalued mapping $\Phi:C\rightarrow2^{C}\setminus\{\emptyset \}$ by setting
Next, we show that Φ is upper ≽preserving. To see this, take any $x_{1},x_{2}\in C$ with $x_{1}\succcurlyeq x_{2}$ and pick any $y_{2}\in\Phi(x_{2})$, then $f(x_{2},y_{2})<0$. Since $f(\cdot ,y)$ is ≽reversing for each $y\in C$, we have $f(x_{2},y_{2})\geqslant f(x_{1},y_{2})$. Choose $y_{1}=y_{2}$, then $y_{1}\in C$ and $f(x_{1},y_{1})<0$, which implies that $y_{1}\in\Phi (x_{1})$ with $y_{1}\succcurlyeq y_{2}$. Hence, Φ is upper ≽preserving. Again, as C is a subcomplete ≽sublattice of X, it follows from Theorem 3.1 that Φ has a fixed point. Denote this fixed point by $\bar{x}$, then $\bar{x}\in\Phi(\bar{x})$, which implies that $f(\bar{x},\bar{x})<0$. Contradiction! Hence, there exists $\hat{x}\in C$ such that $\Phi(\hat{x})=\emptyset$, that is, $f(\hat{x},y)\geqslant0$ for every $y\in C$. □
Applying Theorem 3.2 and Corollary 3.2, we can also consider the case when Φ is lower ≽preserving and $f(\cdot ,y)$ is ≽preserving, respectively.
Theorem 5.3
Let $(X,\succcurlyeq)$ be a Banach lattice and let C be a subcomplete ≽sublattice of X. Let $f: C\times C\rightarrow R$ be a bifunction. Assume that the following conditions hold:

(i)
$f(x,x)\geqslant0$ for any $(x,x)\in C\times C$;

(ii)
the setvalued mapping $\Phi: C\rightarrow2^{C}$ defined by setting
$$\Phi(x)=\bigl\{ y\in C: f(x,y)< 0 \bigr\} $$is lower ≽preserving and compactvalued.
Then $\operatorname{EP}(C,f)$ is solvable.
Theorem 5.4
Let $(X,\succcurlyeq)$ be a Banach lattice and let C be a subcomplete ≽sublattice of X. Let $f: C\times C\rightarrow R$ be a bifunction. Assume that the following conditions hold:

(i)
$f(x,x)\geqslant0$ for any $(x,x)\in C\times C$;

(ii)
$f(\cdot,y)$ is ≽preserving for each $y\in C$ and the set $\{y\in C: f(x,y)<0 \}$ is compact for each $x\in C$.
Then $\operatorname{EP}(C,f)$ is solvable.
At the end of this paper, we give an example, which shows that the conditions of Theorem 5.2 can be satisfied easily. Of course, the examples for Theorem 5.4 can be analogously considered.
Example 5.1
Let $(X,\succcurlyeq)=(R,\geqslant)$ and $C=[0,2]\subseteq R$. Denote by E the set $\{(x,y)\in[0,2]\times [0,2]:xy\geqslant1\}$. Define a mapping $f:[0,2]\times[0,2]\rightarrow R$ by
It is easy to check that f satisfies all the condition of Theorem 5.2, and hence we conclude that there must exists a solution to $\operatorname{EP}([0,2],f)$. In fact, taking $\hat{x}=\frac{1}{2}$, we have
Moreover, observing the above example, we can see that f is discontinuous.
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Acknowledgements
The authors sincerely thank the anonymous reviewers for their valuable suggestions, which improved the presentation of this paper. The second author was supported financially by the National Natural Science Foundation of China (11071109). This work was also supported by the National Natural Science Foundation of China (11401296), Jiangsu Provincial Natural Science Foundation of China (BK20141008) and Natural Science Fund for colleges and universities in Jiangsu Province (14KJB110007).
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Keywords
 ordertheoretic fixed point
 orderpreservation
 parametric generalized variational inequalities
 equilibrium problems
 Banach lattices