Semicontinuity and closedness of parametric generalized lexicographic quasiequilibrium problems

Abstract

This paper is mainly concerned with the upper semicontinuity, closedness, and the lower semicontinuity of the set-valued solution mapping for a parametric lexicographic equilibrium problem where both two constraint maps and the objective bifunction depend on both the decision variable and the parameters. The sufficient conditions for the upper semicontinuity, closedness, and the lower semicontinuity of the solution map are established. Many examples are provided to ensure the essentialness of the imposed assumptions.

Introduction

Equilibrium problems first considered by Blum and Oettli [1] have been playing an important role in optimization theory with many striking applications particularly in transportation, mechanics, economics, etc. Equilibrium models incorporate many other important problems such as: optimization problems, variational inequalities, complementarity problems, saddlepoint/minimax problems, and fixed points. Equilibrium problems with scalar and vector objective functions have been widely studied. The crucial issue of solvability (the existence of solutions) has attracted most considerable attention of researchers; see, e.g., [2–5].

With regard to vector equilibrium problems, most of the existing results correspond to the case when the order is induced by a closed convex cone in a vector space. Thus, they cannot be applied to lexicographic cones, which are neither closed nor open. These cones have been extensively investigated in the framework of vector optimization; see, e.g., [6–13]. For instance, Konnov and Ali [12] studied sequential problems, especially exploiting its relation with regularization methods. Bianchi et al. in [7] analyzed lexicographic equilibrium problems on a topological Hausdorff vector space, and their relationship with some other vector equilibrium problems. They obtained the existence results for the tangled lexicographic problem via the study of a related sequential problem.

As a unified model of vector optimization problems, vector variational inequality problems, variational inclusion problems and vector complementarity problems, vector equilibrium problems have been intensively studied. The stability analysis of the solution mapping for these problems is an important topic in vector optimization theory. Recently, a great deal of research has been devoted to the semicontinuity of the solution mapping for a parametric vector equilibrium problem. Based on the assumption of the (strong) C-inclusion property of a function, Anh and Khanh [14] obtained the upper and lower semicontinuity of the solution set map of parametric multivalued (strong) vector quasiequilibrium problems. Anh and Khanh [15] obtained the semicontinuity of a class of parametric quasiequilibrium problems by a generalized concavity assumption and a closedness of the level set of functions. Wangkeeree et al. [16] established the continuity of the efficient solution mappings to a parametric generalized strong vector equilibrium problem involving a set-valued mapping under the Holder relation assumption. Recently, Wangkeeree et al. [17] obtained the sufficient conditions for the lower semicontinuity of an approximate solution mapping for a parametric generalized vector equilibrium problem involving set-valued mappings. By using a scalarization method, they obtained the lower semicontinuity of an approximate solution mapping for such a problem without the assumptions of monotonicity and compactness. For other qualitative stability results on parametric generalized vector equilibrium problems, see [14–20] and the references therein.

It is well known that partial order plays an important role in vector optimization theory. The vector optimization problems in the previous references are studied in the partial order induced by a closed or open cone. But in some situations, the cone is neither open nor closed, such as the lexicographic cone. On the other hand, since the lexicographic order induced by the lexicographic cone is a total order, it can refine the optimal solution points to make it smaller in the theory of vector optimization. Thus, it is valuable to investigate the vector optimization problems in the lexicographic order. To the best of our knowledge, the first lower stability results of the solution set map based on the density of the solution set mapping for a parametric lexicographic vector equilibrium problem have been established by Shi-miao et al. [21]. Recently, Anh et al. [22] established the sufficient conditions for the upper semicontinuity, closedness, and continuity of the solution maps for a parametric lexicographic equilibrium problem. However, to the best of our knowledge, there is no work to study the stability analysis for a parametric lexicographic equilibrium problem where both two constraint maps and the objective bifunction depend on both the decision variable and the parameters. We observe that quasiequilibrium models are the important general models including as special cases quasivariational inequalities, complementarity problems, vector minimization problems, Nash equilibria, fixed-point and coincidence-point problems, traffic networks, etc. A quasioptimization problem is more general than an optimization one as constraint sets depend on the decision variable as well.

Motivated by the mentioned works, this paper is devoted to the study of closedness upper and lower of the solution map for a parametric lexicographic equilibrium problem where both two constraint maps and the objective bifunction depend on both the decision variable and the parameters. The sufficient conditions for the upper semicontinuity, closedness, and the lower semicontinuity of the solution map are established. Many examples are provided to ensure the essentialness of the imposed assumptions.

The paper is organized as follows. In Section 2, we first introduce the parametric lexicographic equilibrium problem where both two constraint maps and the objective bifunction depend on both the decision variable and the parameters, and we recall some basic definitions on semicontinuity of set-valued maps. Section 3 establishes the sufficient conditions for the upper semicontinuity and closedness of the solution map. Many examples are provided to ensure the essentialness of the imposed assumptions. Section 4 establishes the sufficient conditions for the lower semicontinuity of the solution map. Furthermore, we give also many examples ensuring the essentialness of the imposed assumptions.

Preliminaries

Throughout this paper, if not otherwise specified, let X and Λ be Hausdorff topological vector spaces. Let $A\subseteq X$ be nonempty. Let $K_{1}, K_{2}: A \times\Lambda \rightarrow 2^{X}$ be two multivalued constraint maps and $f: =(f_{1},f_{2},\ldots,f_{n}): A \times A \times\Lambda\rightarrow \mathbb {R}^{n}$ a vector-valued function where, for each $i\in I_{n}:=\{1,2,\ldots,n\}$, $f_{i} : A \times A \times\Lambda\rightarrow \mathbb {R}$ is a real valued function. We assume that, for every $x\in X$ and $i\in I_{n}$, $f_{i}(x,x,\lambda)=0$, i.e., $f_{i}$ is an equilibrium function. Set $\Bbb {R} = (-\infty, +\infty)$, $\Bbb {R}_{+} = [0, +\infty)$, $\Bbb {R}_{-} = -\Bbb {R}_{+}$ and $\bar{\mathbb {R}}:=\mathbb {R}\cup\{+\infty\}$. For a subset A of X, intA, clA and bdA stand for the interior, closure, and boundary of A, respectively. For any given $\alpha\in \Bbb {R}$, the upper α-level set and the lower α-level set of the function $f:X\rightarrow \bar{\mathbb {R}}$ are denoted, respectively, by

$$\operatorname{lev}_{\geq\alpha}f:=\bigl\{ x\in X| f(x)\geq\alpha\bigr\} $$

and

$$\operatorname{lev}_{\leq\alpha}f:=\bigl\{ x\in X| f(x)\leq\alpha\bigr\} . $$

Recall that the lexicographic cone of $\mathbb {R}^{n}$, denoted by $C_{L}$, is defined as

$$C_{L}:=\{0\}\cup\bigl\{ x\in \mathbb {R}^{n} | \exists i\in I_{n}: x_{i}>0, \forall j< i, x_{j} =0\bigr\} . $$

We observe that it is neither closed nor open. Indeed, when comparing with the cone $C_{1}:=\{x\in \mathbb {R}^{n} | x_{1} \geq0\}$, we have

$$\operatorname {int}C_{1} \subsetneq C_{L}\subsetneq C_{1}, \qquad \operatorname {int}C_{L}=\operatorname {int}C_{1} \quad\mbox{and}\quad \operatorname{cl}C_{L} = C_{1}. $$

However, it is worth noticing that the lexicographic cone is convex, pointed, and total (‘total’ means that $C_{L}\cup(-C_{L})=\mathbb {R}^{n}$). The lexicographic order, $\geq_{L}$, in $C_{L}$ is defined by

$$x\geq_{L} y\quad \Longleftrightarrow\quad x -y \in C_{L}. $$

This is a total (called also linear) order, i.e., any pair of elements is comparable. In [23], it was shown that, for a fixed orthogonal base, the lexicographic order is the unique total order. We will see later that this causes difficulties in studies of many topics related to ordering cones.

Next, we shall introduce and study a problem where both the two constraint maps and the bifunction depend on parameters. For a given $\lambda\in\Lambda$, the parametric generalized lexicographic quasiequilibrium problem, denoted by $\operatorname{GLQEP}_{\lambda}$, is

$$(\operatorname{GLQEP}_{\lambda}) \left \{ \textstyle\begin{array}{@{}l} \mbox{finding }\bar{x} \in K_{1}(\bar{x}, \lambda)\mbox{ such that, for all }y\in K_{2}(\bar{x}, \lambda), \\ f(\bar{x},y,\lambda)\geq_{L} 0. \end{array}\displaystyle \right . $$

Remark 2.1

When $K_{1}= K_{2}:= K : \Lambda \rightarrow 2^{X}$, the $(\operatorname{GLQEP}_{\lambda})$ collapses to the lexicographic vector quasiequilibrium problem $(\operatorname{LEP}_{\lambda})$: for each $\lambda \in\Lambda$,

$$(\operatorname{LEP}_{\lambda}) \left \{ \textstyle\begin{array}{@{}l} \mbox{finding } \bar{x} \in K(\lambda) \mbox{ such that }\\ f(\bar{x}, y, \lambda)\geq_{L} 0, \forall y \in K(\lambda). \end{array}\displaystyle \right . $$

The stability analysis of the set-valued solution mapping for $(\operatorname {LEP}_{\lambda})$ are studied in Anh et al. [6] and Shi-miao et al. [21].

Let the set-valued mappings $E : \Lambda \rightarrow 2^{X}$ and $S_{f_{1}}: \Lambda \rightarrow 2^{X}$ be defined by

$$E(\lambda) = \bigl\{ x\in A : x\in K_{1}(x,\lambda) \bigr\} $$

and

$$S_{f_{1}}(\lambda) = \bigl\{ x\in E(\lambda) : f_{1}(x,y, \lambda) \geq0 , \forall y\in K_{2}(x,\lambda) \bigr\} . $$

Furthermore, let a mapping $Z: S_{f_{1}}(\lambda)\times\Lambda \rightarrow2^{X}$ be given by

$$Z(x,\lambda):=\bigl\{ y\in K_{2}(x, \lambda) \mid f_{1}(x,y, \lambda)=0\bigr\} . $$

For the sake of simplicity, we consider the case $n=2$, since the general case is similar. Then $\operatorname{GLQEP}_{\lambda}$ collapses to: find $\bar{x} \in K_{1}(\bar{x}, \lambda)$ such that

$$\textstyle\begin{cases} f_{1}(\bar{x}, y, \lambda) \geq0, & \forall y\in K_{2}(\bar{x}, \lambda ),\\ f_{2}(\bar{x}, z, \lambda) \geq0, & \forall z\in Z(\bar{x},\lambda). \end{cases} $$

Thus, $\operatorname{GLQEP}_{\lambda}$ can be rewritten as

$$ \mbox{find }\bar{x} \in S_{f_{1}}(\lambda)\mbox{ such that }f_{2}(\bar{x}, y,\lambda)\geq0, \mbox{ for all }y\in Z(\bar{x}, \lambda). $$

(2.1)

The solution mapping for $\operatorname{GLQEP}_{\lambda}$ is denoted by $S_{f}$. We denote the whole family of problems, say of $\operatorname {GLQEP}_{\lambda}$, for $\lambda\in\Lambda$, by $(\operatorname{GLQEP}_{\lambda})_{\lambda\in\Lambda}$. We first observe some basic facts about lexicographic equilibrium problems. The lexicographic cone $C_{L}$ contains clearly all pointed closed and convex cones C included in the closed half space $\{x\in\Bbb {R}^{n} : x_{1}\geq0\}$. Then, for an ordering cone C, we consider some kinds of parametric equilibrium problems: the parametric generalized quasiequilibrium problem [23], denoted by $\operatorname{GQEP}_{\lambda}$, is

$$(\operatorname{GQEP}_{\lambda}) \left \{ \textstyle\begin{array}{@{}l} \mbox{finding }\bar{x} \in K_{1}(\bar{x}, \lambda)\mbox{ such that, for all }y\in K_{2}(\bar{x}, \lambda), \\ f(\bar{x},y,\lambda)\in C. \end{array}\displaystyle \right . $$

The solution mapping for $\operatorname{GQEP}_{\lambda}$ is denoted by $S_{GQEP}$. Therefore, for any pointed closed and convex cones C included in the closed half space $\{x\in\Bbb {R}^{n} : x_{1}\geq0\}$, we can get the following fact: $S_{GQEP} \subseteq C_{L}$. Hence, the existence results of solutions for $GLQEP$ can be obtained by the nonemptiness of $S_{GQEP}$. Next, we need to recall some well-known definitions.

Definition 2.2

[24]

Let $\{A_{n}\}$ be a sequence of subsets of X. Then

• (i) the upper limit or outer limit of the sequence $\{ A_{n}\}$ is a subset of X given by

  $$\limsup_{n\rightarrow \infty}A_{n} = \Bigl\{ x\in X \big| \liminf _{n\rightarrow \infty} \operatorname{dist} (x,A_{n}) = 0\Bigr\} ; $$

• (ii) the lower limit or inner limit of the sequence $\{ A_{n}\}$ is a subset of X given by

  $$\liminf_{n\rightarrow \infty}A_{n} = \Bigl\{ x\in X \big| \lim _{n\rightarrow \infty} \operatorname{dist} (x,A_{n}) = 0\Bigr\} ; $$

• (iii) if $\limsup_{n\rightarrow \infty}A_{n} = \liminf_{n\rightarrow \infty}A_{n}$, then we say that the limit of $\{ A_{n}\}$ exist and

  $$\lim_{n\rightarrow \infty}A_{n} = \limsup_{n\rightarrow \infty}A_{n} = \liminf_{n\rightarrow \infty}A_{n}. $$

Consequently, we have the following result.

Proposition 2.3

Let $\{A_{n}\}$ be a sequence of subsets of X. Then

• (i) $\limsup_{n\rightarrow \infty}A_{n} = \{x\in A | x_{n_{k}}\in A_{n_{k}} : x_{n_{k}} \rightarrow x \}$;

• (ii) $\liminf_{n\rightarrow \infty}A_{n} = \{x\in A | x_{n}\in A_{n} : x_{n} \rightarrow x \}$.

Definition 2.4

[25]

Let X and Y be Hausdorff topological vector spaces and $S: X\rightarrow2^{Y}$ a given set-valued map.

• (i) S is said to be upper semicontinuous (usc, for short) at $x_{0} \in X$ iff for any open set $V\subset Y$, where $S(x_{0})\subset V$, there exists a neighborhood $U\subset X$ of $x_{0}$ such that

  $$S(x) \subset V, \quad \forall x\in U. $$

  The map $S(\cdot)$ is said to be u.s.c. on X if it is u.s.c. at every $x\in X$.

• (ii) S is said to be lower semicontinuous (lsc, for short) at $x_{0} \in X$ iff for any open set $V\subset Y$ such that $S(x_{0}) \cap V \neq\emptyset$, there exists a neighborhood $U\subset X$ of $x_{0}$ such that

  $$S(x) \cap V \neq\emptyset, \quad \forall x\in U. $$

  The map $S(\cdot)$ is said to be l.s.c. on X if $S(\cdot)$ is l.s.c. at every $x\in X$.

• (iii) S is said to be closed at $x_{0}$ if from $(x_{n},y_{n})$ in the graph $\operatorname{gr} S:=\{(x,y)\in X\times Y \mid y\in S(x)\}$ of S and tends to $(x_{0},y_{0})$ it follows that $(x_{0},y_{0})\in \operatorname{gr} S$.

We will often use the well-known fact: if $S(x)$ is compact, then S is usc at x if and only if for any sequence $\{x_{n}\}$ in X converging to x and $y_{n}\in Q(x_{n})$, there is a subsequence of $\{y_{n}\}$ converging to a point $y\in Q(x)$. Next we give equivalent forms of the lower semicontinuity of S.

For a set-valued map $Q: X\rightarrow 2^{Y}$ between two linear spaces, Q is called concave [15] on a convex subset $A\subseteq X$ if, for each $x_{1},x_{2}\in A$ and $t\in[0,1]$,

$$Q\bigl( (1-t)x_{1} + t x_{2}\bigr) \subseteq tQ(x_{1}) + (1-t)Q(x_{2}). $$

Lemma 2.5

Let $S: X\rightarrow2^{Y}$ be a given set-valued map. The following are equivalent:

• (i) S is lsc at $x_{0}$;

• (ii) if $\{x_{n}\}$ is any sequence such that $x_{n} \rightarrow x_{0}$ and $V\subset Y$ an open subset such that $S(x_{0}) \cap V \neq\emptyset$, then

  $$\exists N \geq1 : S(x_{n})\cap V \neq\emptyset,\quad \forall n\geq N; $$

• (iii) if $\{x_{n}\}$ is a sequence such that $x_{n} \rightarrow x_{0}$ and $y_{0} \in S(x_{0})$ arbitrary, then there is a sequence $\{y_{n}\}$ with $y_{n}\in S(x_{n})$ such that $y_{n} \rightarrow y_{0}$ as $n\rightarrow \infty$.

From Proposition 2.3 and Lemma 2.5 we can obtain the following lemma immediately.

Lemma 2.6

Let $S: X\rightarrow2^{Y}$ be a given set-valued map. Then S is lsc at $x_{0}$ iff for any sequence $\{x_{n}\} \subseteq X$ converging to $x_{0}$,

$$S(x_{0}) \subset\liminf_{n\rightarrow \infty}S(x_{n}). $$

The following relaxed continuity properties are also needed and can be found in [26].

Definition 2.7

([26])

Let X be a topological space and $g :X\rightarrow\bar{\mathbb {R}}$ be a function on X.

• (i) g is said to be (sequentially) upper pseudocontinuous at $x_{0}\in X$ if for any sequence $\{x_{n}\}$ in X converging to $x_{0}$ and for each $x\in X$ such that $g(x) > g(x_{0})$,

  $$g(x)>\limsup_{n\rightarrow \infty} g(x_{n}). $$

• (ii) g is called (sequentially) lower pseudocontinuous at $x_{0}\in X$ if for any sequence $\{x_{n}\}$ in X converging to $x_{0}$ and for each $x\in X$ such that $g(x) < g(x_{0})$,

  $$g(x)< \liminf_{n\rightarrow \infty} g(x_{n}). $$

• (iii) g is pseudocontinuous at $x_{0}\in X$ if it is both lower and upper pseudocontinuous at this point.

The class of the pseudocontinuous functions strictly contains that of the semicontinuous functions as shown by the following.

Example 2.8

The function $g: \mathbb {R}\rightarrow \mathbb {R}$ defined by

$$g(x)= \textstyle\begin{cases} 2,&\mbox{if } x>0,\\ 0,&\mbox{if } x=0,\\ -2,&\mbox{if } x< 0, \end{cases} $$

is pseudocontinuous, but neither upper nor lower semicontinuous at 0.

Lemma 2.9

([16])

Let X be a topological space. Then $g:X\rightarrow \overline{\mathbb {R}}$ is pseudocontinuous in X if and only if, for all sequences $\{x_{n}\}$ and $\{y_{n}\}$ in X such that $x_{n}\rightarrow x$ and $y_{n}\rightarrow y$ as $n\rightarrow \infty$ and $g(y) < g(x)$,

$$\limsup_{n\rightarrow \infty} g(y_{n})< \liminf_{n\rightarrow \infty} g(x_{n}). $$

The following important definition can be found in [15].

Definition 2.10

Let $g: X\times Z \rightarrow \Bbb {R}$ and $\Delta\subset\Bbb {R}$, where $\operatorname{int} \Delta\neq\emptyset$. g is called generalized Δ-concave in a convex set $A \subset Z$, if for each $x\in X$ and $z_{1}, z_{2}\in A$ satisfying $g(x,z_{1})\in\Delta$ and $g(x,z_{2})\in\operatorname{int}\Delta$, it follows that, for all $t\in(0,1)$,

$$g\bigl(x, (1-t)z_{1}+ tz_{2}\bigr) \in\operatorname{int} \Delta. $$

The upper semicontinuouity and closedness of $S_{f}$

In this section, we discuss the upper semicontinuity and closedness of the solution mapping $S_{f}$. Since there have been a number of contributions to existence issues, focusing on stability we always assume that $S_{f_{1}}(\lambda)$ and $S_{f}(\lambda)$ are nonempty for all λ in a neighborhood of the considered point λ̄. First of all, we shall establish the upper semicontinuity and closedness of the solution mapping $S_{f_{1}}$.

Lemma 3.1

For $(\operatorname{GLQEP}_{\lambda})_{\lambda\in\Lambda}$ assume that

• (i) E is usc at λ̄ and $E(\bar{\lambda})$ is compact;

• (ii) $K_{2}$ is lsc in $K_{1}(A, \Lambda) \times\{\bar{\lambda}\}$;

• (iii) $\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$ is closed in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$ for $i=1,2$.

Then the solution map $S_{f_{1}}$ is both usc and closed at λ̄.

Proof

We first prove that the solution map $S_{f_{1}}$ is usc at λ̄. Suppose on the contrary that there exists an open set $U\supseteq S_{f_{1}}(\bar{\lambda})$ such that for any neighborhood $N(\bar{\lambda})$ of λ̄, there exists $\lambda\in N(\bar{\lambda})$ such that $S_{f_{1}}(\lambda) \nsubseteq U$. In particular, for each $n\in\Bbb {N}$, there exist sequences $\{\lambda_{n}\} \subseteq\Lambda$ converging to λ̄ and $\{ x_{n}\} \subseteq S_{f_{1}}(\lambda_{n}) \subseteq E(\lambda_{n})$ with $x_{n} \notin U$. By the upper semicontinuity of E and the compactness of $E(\bar{\lambda})$, one can assume that $x_{n} \rightarrow x_{0}$, for some $x_{0} \in E(\bar{\lambda})$. Next, we claim that $x_{0} \in S_{f_{1}}(\bar{\lambda})$. Again suppose on the contrary that there exists $y_{0} \in K_{2}(x_{0}, \bar{\lambda})$ such that $f_{1}(x_{0},y_{0},\bar{\lambda})<0.$ The lower semicontinuity of $K_{2}$ at $(x_{0}, \bar{\lambda})$, by Lemma 2.5, implies that there exists a sequence $\{y_{n}\}$ in $K_{2}(x_{n}, \lambda_{n})$ such that $y_{n} \rightarrow y_{0}$ as $n\rightarrow \infty$. For each $n\in\Bbb {N}$, since $x_{n} \in S_{f_{1}}(\lambda_{n})$, we have

$$f_{1}(x_{n},y_{n},\lambda_{n})\geq0. $$

It follows from the closedness of $\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$ that $f_{1}(x_{0},y_{0},\bar{\lambda})\geq0$, which leads to a contradiction. Therefore, $x_{0}\in S_{f_{1}}(\bar{\lambda})\subseteq U$, again a contradiction, since $x_{n}\notin U$ for all n. Thus, $S_{f_{1}}$ is usc at λ̄.

Next, we prove that $S_{f_{1}}$ is closed at λ̄. We suppose on the contrary that $S_{f_{1}}$ is not closed at λ̄, i.e., there a sequence $\{\lambda_{n}\}$ converging to λ̄ and $\{x_{n}\}\subseteq S_{f_{1}}(\lambda_{n})$ with $x_{n} \rightarrow x_{0}$ but $x_{0} \notin S_{f_{1}}(\bar{ \lambda})$. The same argument as above ensures that $x_{0}\in S_{f_{1}}(\bar{ \lambda })$, which gives a contradiction. Therefore, we can conclude that $S_{f_{1}}$ is closed at λ̄. □

Now, we are in the position to discuss the upper semicontinuity and closedness of the solution mapping $S_{f}$.

Theorem 3.2

For $(\operatorname{GLQEP}_{\lambda})_{\lambda\in\Lambda}$ assume that

• (i) E is usc at λ̄ and $E(\bar{\lambda})$ is compact;

• (ii) $K_{2}$ is lsc in $K_{1}(A, \Lambda) \times\{\bar{\lambda}\}$;

• (iii) $\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$ is closed in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$ for $i=1,2$;

• (iv) Z is lsc in $S_{f_{1}}(\bar{\lambda})\times\{\bar{\lambda}\}$.

Then the solution map $S_{f}$ is both usc and closed at λ̄.

Proof

We first claim that the solution map $S_{f}$ is usc at λ̄. Suppose there exist an open set $U\supseteq S_{f}(\bar{\lambda})$, $\{\lambda_{n}\}\rightarrow\bar{\lambda}$, and $\{x_{n}\}\subseteq S_{f}(\lambda_{n})$ such that $x_{n}\notin U$ for all n. By the upper semicontinuity of $S_{f_{1}}$ at λ̄ and the compactness of $S_{f_{1}}(\bar{\lambda})$, without loss of generality we can assume that $x_{n} \rightarrow x_{0}$ as $n\rightarrow \infty$ for some $x_{0}\in S_{f_{1}}(\bar{\lambda})$. If $x_{0}\notin S_{f}(\bar{\lambda})$, there exists $y_{0}\in Z(x_{0},\bar{\lambda})$ such that $f_{2}(x_{0},y_{0},\bar{\lambda})<0$. The lower semicontinuity of Z in turn yields $y_{n}\in Z(x_{n},\lambda_{n})$ tending to $y_{0}$. Notice that for each $n\in\Bbb {N}$, $f_{2}(x_{n},y_{n},\lambda_{n})\geq0$. This together with the closedness of $\operatorname{lev}_{\geq0}f_{2}(\cdot,\cdot, \bar{\lambda})$ in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$ implies that $f_{2}(x_{0},y_{0},\bar{\lambda}) \geq0$, which gives a contradiction. If $x_{0}\in S_{f}(\bar{\lambda})\subseteq U$, one has another contradiction, since $x_{n}\notin U$ for all n. Thus, $S_{f}$ is usc at λ̄.

Now we prove that $S_{f}$ is closed at λ̄. Suppose on the contrary that there exists a sequence $\{(\lambda_{n}, x_{n})\}$ converging to $(\bar{\lambda}, x_{0})$ with $x_{n}\in S_{f}(\lambda_{n})$ but $x_{0}\notin S_{f}(\bar{\lambda})$. Then $f_{2}(x_{0},y_{0},\bar{\lambda})<0$ for some $y_{0}\in Z(x_{0},\bar{\lambda})$. Due to the lower semicontinuity of Z, there is $y_{n}\in Z(x_{n},\lambda_{n})$ such that $y_{n}\rightarrow y_{0}$. Since $x_{n}\in S_{f}(\lambda_{n})$, $f_{2}(x_{n},y_{n},\lambda_{n})\geq0$. By the closedness of the set $\operatorname{lev}_{\geq0}f_{2}$, $f_{2}(x_{0},y_{0},\bar{\lambda})\geq0$, which is impossible since $f_{2}(x_{0},y_{0},\bar{\lambda})<0$. Therefore, $S_{f}$ is closed at λ̄. □

Corollary 3.3

For GLQEP, suppose that the conditions (i), (ii), and (iv) given in Theorem  3.2 are satisfied. Further, for each $i = 1,2$, assume that $f_{i}$ is upper pseudocontinuous in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$. Then the solution map $S_{f}$ is both usc and closed at λ̄.

Proof

It is suffice to derive the condition (iii) that given in Theorem 3.2. For $i = 1,2$, suppose $\{ (x_{n},y_{n},\lambda _{n})\}$ is any sequence in $\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$ such that $(x_{n},y_{n},\lambda_{n})\rightarrow (\bar{x},\bar{y}, \bar{\lambda})$ as $n\rightarrow \infty$. Assume that on the contrary that $(\bar{x},\bar{y}, \bar{\lambda})\notin \operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$, which implies that $f_{i}(\bar{x},\bar{y}, \bar{\lambda})<0 = f_{i}(\bar{x},\bar{x}, \bar{\lambda})$. The upper pseudocontinuity of $f_{i}$ at $(\bar{x},\bar{y}, \bar{\lambda})$ implies that

$$0=f_{i}(\bar{x},\bar{x}, \bar{\lambda}) > \limsup_{n\rightarrow \infty} f_{i}(x_{n},y_{n},\lambda_{n})\geq0, $$

which gives a contradiction. Hence, we can conclude that $(\bar{x},\bar{y}, \bar{\lambda}) \in \operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$. Now, the closedness of $\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})$ is proved. Applying Theorem 3.2, we get the desired result. □

The following examples show that all assumptions imposed in Theorem 3.2 are very essential and cannot be relaxed.

Example 3.4

(The upper semicontinuity and compactness in (i) are crucial)

Let $A = X = \mathbb {R}$, $\Lambda= [0,1]$, $\bar{\lambda}= 0$. Define the mappings $K_{1}$, $K_{2}$, and f by

$$K_{1}(x, \lambda) = (\lambda, 1 + \lambda] \quad\mbox{and}\quad K_{2}(x, \lambda) = (0, 1] $$

and

$$f(x,y,\lambda) = \bigl(x(x-y)^{2}(1+ \lambda), 2^{\lambda+ xy}x(x -y) \bigr). $$

Then we have

$$E(0) = (0, 1] \quad\mbox{and}\quad E(\lambda) = (\lambda, 1+ \lambda], \quad \forall\lambda\in(0,1]. $$

Hence, $E(0)$ is not compact and E is not usc. Indeed, we choose an open set $U := (0, 3/2) \supseteq E(0) = (0, 1]$. We observe that, for any $\varepsilon> 0$, we can choose $\lambda'= -\varepsilon/2 \in N(0, \varepsilon)$ such that

$$E\bigl(\lambda'\bigr) = ( -\varepsilon/2, \varepsilon/2 ] \nsubseteq U. $$

Clearly, the conditions (ii) and (iii) are all satisfied. Easy calculations yield

$$S_{f_{1}}(\lambda)= \textstyle\begin{cases} (0,1], & \mbox{if } \lambda=0, \\ (\lambda, 1 + \lambda], & \mbox{if } \lambda\neq0, \end{cases} $$

and $Z(x,\lambda)=\{x\}$. Hence, assumption (iv) is satisfied. Direct computations give

$$S_{f}(\lambda) = \textstyle\begin{cases} (0,1], & \mbox{if } \lambda=0, \\ (\lambda, 1 + \lambda], & \mbox{if } \lambda\neq0. \end{cases} $$

It is evident that $S_{f}$ is neither usc nor closed at $\bar{\lambda}= 0$. This is caused by the fact that E is neither upper semicontinuous nor compact-valued at $\bar{\lambda}=0$.

Example 3.5

(The lower semicontinuity of $K_{2}$ in $K_{1}(A, \Lambda) \times\{\bar{\lambda}\}$ is essential)

Let $A = X = \mathbb {R}$, $\Lambda= [0,1]$, $\bar{\lambda}= 0$. Define the mappings $K_{1}$, $K_{2}$, and f by

$$K_{1}(x, \lambda) = K_{2}(x, \lambda) = \textstyle\begin{cases} [-1, 0], & \mbox{if } \lambda= 0, \\ {[}-\frac{1}{2}, 0], & \mbox{if } \lambda\neq0, \end{cases} $$

and

$$f(x,y,\lambda) = \bigl( (1 +\lambda) (y - x), 2^{\lambda y}(y -x) \bigr). $$

Then we have

$$E(\lambda) = \textstyle\begin{cases} [-1, 0], & \mbox{if } \lambda= 0, \\ {[}-\frac{1}{2}, 0], & \mbox{if } \lambda\neq0, \end{cases} $$

which shows that E is usc at 0 and $E(0)$ is compact, that is, (i) is satisfied. Clearly, the condition (iii) in Theorem 3.2 is satisfied. Furthermore, easy calculations yield

$$S_{f_{1}}(\lambda) = \textstyle\begin{cases} \{-1\}, & \mbox{if } \lambda=0, \\ \{-\frac{1}{2}\} , & \mbox{if } \lambda\neq0, \end{cases} $$

and $Z(x, \lambda) = \{x\}$, which is lsc in $S_{f_{1}}(\bar{\lambda})\times\{\bar{\lambda}\}$. Direct calculation gives

$$S_{f}(\lambda) = \textstyle\begin{cases} \{-1\}, & \mbox{if } \lambda=0, \\ \{-\frac{1}{2}\} , & \mbox{if } \lambda\neq0. \end{cases} $$

It is evident that $S_{f}$ is neither usc nor closed at $\bar{\lambda}= 0$. This is caused by the fact that $K_{2}$ is not lsc at $\bar{\lambda}=0$. Indeed, we observe that $(0,0) \in K_{1}(A, \Lambda) \times\{0\}$ and $\{ (\frac{1}{n}, \frac{1}{n} )\} \rightarrow (0,0)$. We choose $y := -1 \in K_{2}(0,0) = [-1, 0]$ such that there is not any sequence $\{y_{n}\}$ in $K_{2}(\frac{1}{n}, \frac{1}{n}) = [-\frac{1}{2}, 0]$ converging to y.

Example 3.6

(The lower semicontinuity of Z in $S_{f_{1}}(\bar{\lambda})\times\{\bar{\lambda}\}$ are crucial)

Let $A = X = \mathbb {R}$, $\Lambda= [0,1]$, $\bar{\lambda}= 0$. Define the mappings $K_{1}$, $K_{2}$, and f by

$$K_{1}(x, \lambda) = K_{2}(x, \lambda) = [0,1] $$

and

$$f(x,y,\lambda) = \bigl(\lambda(x - y), 2^{\lambda y}(y -x) \bigr). $$

Hence Conditions (i), (ii), and (iii) clearly hold. By direct calculations, we can get

$$\begin{aligned}& S_{f_{1}}(\lambda) = \textstyle\begin{cases} [0,1], & \mbox{if } \lambda= 0, \\ \{1\}, & \mbox{if } \lambda\neq0, \end{cases}\displaystyle \\& Z(x, \lambda) = \textstyle\begin{cases} [0,1], & \mbox{if } \lambda= 0, \\ \{x\}, & \mbox{if } \lambda\neq0. \end{cases}\displaystyle \end{aligned}$$

Hence Z is not lsc in $[0,1]\times\{ 0 \}$. Further

$$S_{f}(\lambda) = \textstyle\begin{cases} \{0\}, & \mbox{if } \lambda= 0, \\ \{1\}, & \mbox{if } \lambda\neq0. \end{cases} $$

It is evident that $S_{f}$ is neither usc nor closed at $\bar{\lambda}= 0$.

The lower semicontinuouity of $S_{f}$

For investigation the lower semicontinuity of the solution mapping $S_{f}$, as an auxiliary problem we consider, for a given $\lambda\in \Lambda$, an auxiliary parametric generalized lexicographic quasiequilibrium problem, denoted by $\operatorname{AGLQEP}_{\lambda}$:

$$(\operatorname{AGLQEP}_{\lambda}) \left \{ \textstyle\begin{array}{@{}l} \mbox{finding }\bar{x} \in K_{1}(\bar{x}, \lambda)\mbox{ such that} \\ f_{1}(\bar{x},y,\lambda) > 0, \mbox{ for all }y\in K_{2}(\bar{x}, \lambda). \end{array}\displaystyle \right . $$

Let the set-valued mappings $E : \Lambda \rightarrow 2^{X}$ and $S_{AGQEP}: \Lambda \rightarrow 2^{X}$ be defined by

$$E(\lambda) = \bigl\{ x\in A : x\in K_{1}(x,\lambda) \bigr\} , $$

and the solution mapping for $\operatorname{AGLQEP}_{\lambda}$ is denoted by $S_{AGLQEP}(\lambda)$, i.e.

$$S_{AGQEP}(\lambda) = \bigl\{ x\in E(\lambda) : f_{1}(x,y, \lambda) > 0 , \forall y\in K_{2}(x,\lambda) \bigr\} . $$

First, we establish the lower semicontinuity of the solution mapping $S_{AGQEP}$.

Lemma 4.1

For AGLQEP, assume that the following conditions are satisfied:

• (i) E is lsc at λ̄;

• (ii) $K_{2}$ is usc and compact-valued in $K_{1}(A, \Lambda) \times\{\bar{\lambda}\}$;

• (iii) $\operatorname{lev}_{\leq0}f_{1}(\cdot,\cdot, \bar{\lambda})$ is closed in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$.

Then the solution map $S_{AGQEP}$ is lsc at λ̄.

Proof

Suppose to the contrary that $S_{AGQEP}$ is not lsc at λ̄, i.e., there a sequence $\{\lambda_{n}\} \subseteq \Lambda$ with $\lambda_{n} \rightarrow \bar{\lambda}$ and there exists $\bar{x}\in S_{AGQEP}(\bar{\lambda}) \subseteq E(\bar{\lambda})$ such that,

$$ \mbox{for all sequence } \{y_{n}\}\subseteq S_{AGQEP}( \lambda_{n}) \subseteq E(\lambda_{n} ), \quad y_{n}\nrightarrow\bar{x}. $$

(4.1)

Since E is lsc at λ̄, there exists a sequence $\{x_{n}\}$ in $E(\lambda_{n})$ with $x_{n} \rightarrow \bar{x}$ as $n\rightarrow \infty$. It follows from (4.1) that there exists $\{ x_{n_{k}} \}$ of $\{x_{n}\}$ such that $x_{n_{k}} \notin S_{AGQEP_{1}}( \lambda_{n_{k}})$ for all $k\in\Bbb {N}$. This implies that there $y_{n_{k}} \in K_{2}(x_{n_{k}}, \lambda_{n_{k}} )$ satisfying

$$f_{1}(x_{n_{k}}, y_{n_{k}}, \lambda_{n_{k}} ) \leq0, \quad\mbox{for each } k\in \Bbb {N}. $$

As $K_{2}$ is usc at $(\bar{x}, \bar{\lambda})$ and $K_{2}(\bar{x}, \bar {\lambda})$ is compact, there exists $\bar{y}\in K_{2}(\bar{x}, \bar {\lambda})$ such that

$$y_{n_{k}} \rightarrow \bar{y} \quad\mbox{as } k\rightarrow \infty\ (\mbox{taking a subsequence if necessary}). $$

It follows from the closedness of $\operatorname{lev}_{\leq0}f_{1}(\cdot,\cdot, \bar{\lambda})$ that $f_{1}(\bar{x}, \bar{y}, \bar{\lambda})\leq0$, which is impossible since $\bar{x} \in S_{AGQEP}(\bar{\lambda})$. The proof is completed. □

Now, we establish the lower semicontinuity of the solution mapping $S_{f}$.

Theorem 4.2

For $(\operatorname{GLQEP})$ let the following conditions be satisfied:

• (i) E is lsc at λ̄ and $E(\bar{\lambda })$ is convex;

• (ii) $K_{2}(\cdot , \bar{\lambda})$ is usc and compact-valued in $K_{1}(A, \Lambda) \times\{\bar{\lambda}\}$; $K_{2}(\cdot , \bar{\lambda})$ is concave in $E(\bar{\lambda})$;

• (iii) $\operatorname{lev}_{\leq0}f_{1}(\cdot,\cdot, \bar{\lambda})$ is closed in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$;

• (iv) $f_{1}(\cdot, \cdot , \bar{\lambda})$ is generalized $\Bbb {R}_{+}$-concave in $E(\bar{\lambda})\times K_{2}(A, \bar{\lambda})$.

Then the solution map $S_{f}$ is lsc at λ̄.

Proof

First, we notice that $S_{f}(\bar{\lambda}) \subseteq \operatorname{cl}S_{AGLQEP}(\bar{\lambda})$. Indeed, let $\bar{x} \in S_{f}(\bar {\lambda})$ be arbitrary. For any $\bar{x}_{A} \in S_{AGLQEP}(\bar {\lambda })$ and $t\in(0,1)$, define $x_{t} = (1-t)\bar{x} + t\bar{x}_{A}$. Clearly, $x_{t} \rightarrow \bar{x}$ as $t \downarrow0$ and by the virtue of the convexity of $E(\bar{\lambda})$, $x_{t} \in K_{1}(x_{t}, \bar{ \lambda })$. Further, for all $y\in K_{2}(x_{t}, \bar{\lambda})$, the concavity of $K_{2}(\cdot , \bar{\lambda})$ implies that there exist $\bar{y} \in K_{2}(\bar {x}, \bar{\lambda})$ and $\bar{y}_{A} \in K_{2}(\bar{x}_{A}, \bar {\lambda})$ such that $y= (1-t)\bar{y} + t\bar{y}_{A}$. It is clear that $f_{1}(\bar{x}, \bar{y},\bar{\lambda}) \geq0$ and $f_{1}(\bar{x}_{A}, \bar {y}_{A}, \bar{\lambda}) > 0$. It follows from the generalized $\Bbb {R}_{+}$-concavity of $f_{1}(\cdot, \cdot, \bar{\lambda})$ that $f_{1}(x_{t}, y, \bar{\lambda}) > 0$, i.e., $x_{t} \in S_{AGQEP}(\bar {\lambda})$. Therefore, we conclude that $\bar{x} \in\operatorname{cl}S_{AGQEP}(\bar {\lambda })$, which shows that $S_{f}(\bar{\lambda}) \subseteq\operatorname {cl}S_{AGQEP}(\bar{\lambda})$. Next, for any sequence $\{\lambda_{n}\}$ satisfying $\lambda_{n} \rightarrow \bar{\lambda}$ as $n\rightarrow \infty$, by the lower semicontinuity of $S_{AGQEP}$ at λ̄ given in Lemma 4.1, we have

$$S_{f}(\bar{\lambda}) \subseteq \operatorname{cl}S_{AGQEP}( \bar{\lambda}) \subseteq\operatorname{cl}\liminf_{n\rightarrow \infty} S_{AGQEP}(\lambda_{n}) \subseteq \operatorname{cl}\liminf _{n\rightarrow \infty} S_{f}(\lambda_{n}), $$

which gives the lower semicontinuity of $S_{f}$ at λ̄ since Lemma 2.6. The proof is completed. □

Corollary 4.3

For GLQEP, suppose that the conditions (i), (ii), and (iv) given in Theorem  4.2 are satisfied. Further, assume that $f_{1}$ is lower pseudocontinuous in $K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}$. Then the solution map $S_{f}$ is lsc at λ̄.

Proof

It is suffice to derive the condition (iii) that imposed in Theorem 4.2. Let $\{(x_{n},y_{n},\lambda_{n})\}$ be any sequence in $\operatorname{lev}_{\leq0}f_{1}(\cdot,\cdot, \bar{\lambda})$ such that $(x_{n},y_{n},\lambda_{n})\rightarrow (\bar{x},\bar{y}, \bar{\lambda})$ as $n\rightarrow \infty$. Assume that on the contrary that $(\bar{x},\bar{y}, \bar{\lambda})\notin \operatorname{lev}_{\leq0}f_{1}$, which implies that $f_{1}(\bar{x},\bar{y}, \bar{\lambda}) > 0 = f_{1}(\bar{x},\bar{x}, \bar{\lambda})$. It follows from the lower pseudocontinuity of $f_{1}$ at $(\bar{x},\bar{y}, \bar{\lambda})$ that

$$0 = f_{1}(\bar{x},\bar{x}, \bar{\lambda}) < \liminf_{n\rightarrow \infty} f_{1}(x_{n},y_{n},\lambda_{n})\leq0, $$

which gives a contradiction. Hence, we can conclude that $(\bar{x},\bar{y}, \bar{\lambda}) \in \operatorname{lev}_{\leq0}f_{1}$. Now, the closedness of $\operatorname{lev}_{\leq0}f_{1}(\cdot,\cdot, \bar{\lambda})$ is proved. Applying Theorem 4.2 we obtain the desired result. □

The following example illustrates that the lower semicontinuity assumption for the set E cannot be relaxed in Theorem 4.2.

Example 4.4

(The lower semicontinuity of E at λ̄ is crucial)

Let $A = X = \mathbb {R}$, $\Lambda= [0,1]$, $\bar{\lambda}= 0$. Define the mappings $K_{1}$, $K_{2}$, and f by

$$\begin{aligned}& K_{1}(x, \lambda) = \textstyle\begin{cases} [-1,1], & \mbox{if } \lambda= 0, \\ {[}-1,0] \cup\{1 \}, & \mbox{if } \lambda\neq0, \end{cases}\displaystyle \qquad K_{2}(x, \lambda) = [-1, 0], \end{aligned}$$

and

$$f(x,y,\lambda) = \bigl((1+ \lambda) (x - y), 2^{\lambda y}(x-2y) \bigr). $$

Hence conditions (ii)-(iv) clearly hold. However, E is not lsc at $\bar{\lambda}= 0$. Indeed, we choose a sequence $\{1/n\} \subseteq\Lambda$ such that $1/n \rightarrow 0$ and $1/2 \in E(0) = [-1,1]$. We can see that, for all sequences $\{y_{n}\} \subseteq E(1/n) := [-1,0] \cup\{1 \}$, $y_{n} \nrightarrow 1/2$ as $n\rightarrow \infty$. By direct calculations, we can get

$$S_{AGQEP}(\lambda) = S_{f}(\lambda) = \textstyle\begin{cases} (0,1], & \mbox{if } \lambda= 0, \\ \{1\}, & \mbox{if } \lambda\neq0. \end{cases} $$

Hence $S_{f}$ is not lsc at $\bar{\lambda}= 0$, indeed, we choose $\lambda_{n} = \frac{1}{n} \rightarrow 0$ and $\frac{1}{2} \in S_{f}(0)$ but we cannot find a sequence in $S_{f}(\frac{1}{n})$ which converges to $\frac {1}{2}$.

The next example indicates the essential role of the upper semicontinuity assumption for the set $K_{2}$ in Theorem 4.2.

Example 4.5

(The upper semicontinuity of $K_{2}$ is crucial)

Let $A = X = \mathbb {R}$, $\Lambda= [0,1]$, $\bar{\lambda}= 0$. Define the mappings $K_{1}$, $K_{2}$, and f by

$$K_{1}(x, \lambda) = [0,1],\qquad K_{2}(x, \lambda) = \textstyle\begin{cases} \{-\frac{1}{2} \} \cup[0,1], & \mbox{if } \lambda= 0, \\ {[}-\frac{2}{3},\frac{2}{3}], & \mbox{if } \lambda\neq0, \end{cases} $$

and

$$f(x,y,\lambda) = (x+y, x-y). $$

It is clear that the upper semicontinuity of $K_{2}$ is not satisfied. Indeed, for each $n\in\Bbb {N}$, we choose

$$(x_{n},\lambda_{n}) = \biggl(\frac{1}{n}, \frac{1}{n} \biggr) \quad\mbox{and}\quad y_{n} = - \frac{2}{3} \in K_{2}(x_{n},\lambda_{n}). $$

It is obvious that there is not any subsequence of $\{y_{n}\}$ converging to an element in $\{-\frac{1}{2} \} \cup[0,1]:= K(0,0)$. However, all conditions (i), (iii)-(v) of Theorem 4.2 are satisfied. By direct calculations, we have

$$S_{AGQEP}(\lambda) = S_{f}(\lambda) = \textstyle\begin{cases} (\frac{1}{2},1], & \mbox{if } \lambda= 0, \\ (\frac{2}{3},1], & \mbox{if } \lambda\neq0. \end{cases} $$

Hence $S_{f}$ is not lsc at $\bar{\lambda}= 0$. The cause is that (ii) is not fulfilled.

Conclusion

We presented the upper semicontinuity, closedness, and the lower semicontinuity of the set-valued solution mapping for a parametric lexicographic equilibrium problem where both two constraint maps and the objective bifunction depend on both the decision variable and the parameters. The sufficient conditions for the upper semicontinuity, closedness, and the lower semicontinuity of the solution map are established. Many examples are provided to ensure the essentialness of the imposed assumptions.