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Wellposedness for lexicographic vector quasiequilibrium problems with lexicographic equilibrium constraints
Journal of Inequalities and Applications volume 2015, Article number: 163 (2015)
Abstract
We consider the wellposedness for lexicographic vector equilibrium problems and optimization problems with lexicographic equilibrium constraints in metric spaces. Sufficient conditions for a family of such problems to be (uniquely) wellposed at the reference point are established. Numerous examples are provided to explain that all the assumptions we impose are very relaxed and cannot be dropped.
1 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, saddle point/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 the most considerable attention of researchers; see, e.g., [2, 3]. A relatively new but rapidly growing topic is the stability of solutions, including semicontinuity properties in the sense of Berge and Hausdorff; see, e.g., [4, 5] and the Hölder/Lipschitz continuity of solution mappings; see, e.g., [6–10].
On the other hand, wellposedness of optimizationrelated problems can be defined in two ways. The first and oldest is Hadamard wellposedness [11], which means existence, uniqueness, and continuous dependence of the optimal solution and optimal value from perturbed data. The second is Tikhonov wellposedness [12], which means the existence and uniqueness of the solution and convergence of each minimizing sequence to the solution. Wellposedness properties have been intensively studied and the two classical wellposedness notions have been extended and blended. Recently, the Tikhonov notion has been more interested. The major reason is its vital role in numerical methods. Any algorithm can generate only an approximating sequence of solutions. Hence, this sequence is applicable only if the problem under consideration is wellposed. For parametric problems, wellposedness is closely related to stability. Up to now, there have been many works dealing with wellposedness of optimizationrelated problems as mathematical programming [13, 14], constrained minimization [15, 16] variational inequalities [17–19], Nash equilibria [20], and equilibrium problems [21].
On the other hand, many papers appeared dealing with bilevel problems such as mathematical programming with equilibrium constraints [22], optimization problems with variational inequality constraints [20], optimization problems with Nash equilibrium constraints [20], optimization problems with equilibrium constraints [23, 24], etc. The increasing importance of these bilevel problems in mathematical applications in engineering and economics is recognized. For instance, the multileaderfollower game in economics is a bilevel problem, since each leader has to solve a Stackelberg game formulated as a mathematical program with equilibrium constraints. Recently, Anh et al. in [25] considered the bilevel equilibrium and optimization problems with equilibrium constraints. They proposed a relaxed level closedness and use it together with pseudocontinuity assumptions to establish sufficient conditions for the wellposedness and unique wellposedness.
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., [26–30]. For instance, Chadli et al. in [31] obtained conditions for the existence of solutions of a sequential equilibrium problem via a viscosity argument under quite strong conditions. Bianchi et al. in [32] 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. However, for equilibrium problems, the main emphasis has been on the issue of solvability/existence. To the best of our knowledge, very recently, Anh et al. in [26] studied the wellposedness for lexicographic vector equilibrium problems in metric spaces and gave the sufficient conditions for a family of such problems to be wellposed and uniquely wellposed at the considered point. Furthermore, they derived several results on wellposedness for a class of variational inequalities.
Motivated by the work reported above, this paper aims to consider the lexicographic vector equilibrium problems and optimization problems with lexicographic equilibrium constraints in metric spaces and establishes necessary and/or sufficient conditions for such problems to be wellposed and uniquely wellposed at the considered point assumed always that the mentioned solutions exist.
The layout of the paper is as follows. In Section 2, we propose the lexicographic vector equilibrium problems and optimization problems with lexicographic equilibrium constraints in metric spaces under our consideration and recall notions and preliminaries needed in the sequel. In Section 3, we study the wellposedness of the lexicographic vector equilibrium problems with lexicographic equilibrium constraints in metric spaces. Section 4 is devoted to the wellposedness of optimization problems with lexicographic equilibrium constraints.
2 Preliminaries
We first recall the concept of lexicographic cone in finite dimensional spaces and models of equilibrium problems with the order induced by such a cone. The lexicographic cone of \(\mathbb{R}^{n}\), denoted \(C_{l}\), is the collection of zero and all vectors in \(\mathbb{R}^{n}\) with the first nonzero coordinate being positive, i.e.,
This cone is convex and pointed, and it induces the total order as follows:
We also 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 see that \(\operatorname{int} C_{1} \subsetneq C_{l} \subsetneq C_{1}\), while
Throughout this paper, if not otherwise specified, X and Λ denote the metric spaces. Let \(f:=(f_{1}, f_{2},\ldots, f_{n}) : X\times X\times\Lambda\rightarrow\mathbb{R}^{n}\) and \(K_{i} : X\times\Lambda\rightarrow2^{X}\), \(i=1,2\). The lexicographic vector quasiequilibrium problem consists of, for each \(\lambda\in\Lambda\),
 (\(\mathrm{LQEP}_{\lambda}\)):

finding \(\bar{x} \in K_{1}(\bar{x},\lambda)\) such that
$$f(\bar{x}, y, \lambda)\geq_{l} 0 ,\quad \forall y \in K_{2} (\bar{x},\lambda). $$
Remark 2.1

(i)
When \(f:= f_{1} : X\times X\times\Lambda\rightarrow\mathbb{R}\), the (\(\mathrm{LQEP}_{\lambda}\)) collapses to the parametric quasiequilibrium problem (QEP) considered by Anh et al. [25].

(ii)
When \(K_{i}(\bar{x},\lambda) = K(\lambda)\), for all \(i =1,2\), that is, \(K_{i}\) does not depend on \(\bar{x}\), the (\(\mathrm{LQEP}_{\lambda}\)) reduces to the lexicographic vector equilibrium problem (\(\mathrm{LEP}_{\lambda}\)) considered by Anh et al. [26].
Instead of writing \(\{(\mathrm{LQEP}_{\lambda})  \lambda\in\Lambda \}\) for the family of lexicographic vector quasiequilibrium problem, i.e., the lexicographic parametric problem, we will simply write (LQEP) in the sequel. Let \(S_{f}:\Lambda\rightarrow2^{X}\) be the solution map of (LQEP).
Following the line of investigating εsolutions to vector optimization problems initiated by Loridan [33], we consider the following approximate problem: for each \(\varepsilon\in[0,\infty)\),
 (\(\mathrm{LQEP}_{\lambda,\varepsilon}\)):

find \(\bar{x} \in K_{1}(\bar{x}, \lambda)\) such that
$$f(\bar{x}, y, \lambda) + \varepsilon e \geq_{l} 0, \quad \forall y \in K_{2}(\bar{x}, \lambda), $$
where \(e=(0,0,\ldots, 0,1) \in\mathbb{R}^{n}\). The solution set of (\(\mathrm{LQEP}_{\lambda,\varepsilon}\)) is denoted by \(\tilde{S}_{f}(\lambda,\varepsilon)\).
Let \(Y=X \times\Lambda\) and \(F=:(F_{1}, F_{2}, \ldots, F_{n}) :Y \times Y\rightarrow\mathbb{R}^{n} \) be given. The lexicographic vector equilibrium problem with lexicographic equilibrium constraints under question is
 (LVQEPLEC):

finding \(\bar{\mathbf{y}} \in\operatorname{gr} S_{f}\) such that
$$F(\bar{\mathbf{y}}, \mathbf{y}) \geq_{l} 0, \quad \forall\mathbf {y} \in\operatorname{gr} S_{f}, $$
where \(\operatorname{gr} S_{f}\) denotes the graph of \(S_{f}\), i.e., \(\operatorname{gr} S_{f} := \{ (x,\lambda)  x \in S_{f}(\lambda)\}\). We denote the solution set of (LVQEPLEC) by \(\mathbf{S}_{F}\). Next we consider for each \(\xi\in[0,\infty)\), the following approximate problem of (LVQEPLEC):
 (\(\mathrm{LVQEPLEC}_{\xi}\)):

find \(\bar {\mathbf{y}} \in\operatorname{gr} S_{f}\) such that
$$F(\bar{\mathbf{y}}, \mathbf{y})+\xi e \geq_{l} 0, \quad \forall \mathbf {y} \in\operatorname{gr} S_{f}. $$
For the function \(g : X \times\Lambda\rightarrow\bar{\mathbb {R}}\), where \(\bar{\mathbb{R}}=(\infty,\infty]\), the optimization problem with lexicographic equilibrium constraints is the problem of, for each \(\lambda\in\Lambda\),
 (OPLEC):

finding \(\bar{\mathbf{x}} := (\bar {x},\lambda) \in\operatorname{gr}S_{f} \) such that
$$ g(\bar{\mathbf{x}}) = \min\bigl\{ g(\mathbf{x})  \mathbf{x} := (x,\lambda) \in \operatorname{gr}S_{f}\bigr\} . $$
Let \(\mathrm{S}_{g} : \Lambda\rightarrow2^{X\times\Lambda}\) be the solution map for (OPLEC); that is,
Remark 2.2
When \(f:= f_{1} : X\times X\times\Lambda\rightarrow\mathbb {R}\), the (OPLEC) collapses to the optimization problem with equilibrium constraints (OPEC) considered by Anh et al. [25].
We next give the concept of an approximating sequence, wellposedness, and unique wellposedness for (LQEP), (LVQEPLEC), and (OPLEC).
Definition 2.3
A sequence \(\{x_{n}\}\) is an approximating sequence of (LQEP) corresponding to a sequence \(\{ \lambda_{n}\}\subset\Lambda\) converging to \(\bar{\lambda}\) if there is a sequence \(\{\varepsilon_{n}\} \subset(0,\infty)\) converging to 0 such that \(x_{n} \in\tilde{S}_{f}(\lambda_{n}, \varepsilon_{n})\) for all n.
Definition 2.4
A sequence \(\{\mathbf{x}_{n} \} := \{ (x_{n}, \lambda _{n}) \} \subseteq Y:= X \times\Lambda\) is termed an approximating sequence for (LVQEPLEC) iff there exists \(\varepsilon_{n} \downarrow0\) such that

(i)
\(F(\mathbf{x}_{n},\mathbf{y}) + \varepsilon_{n}e \geq_{l} 0\), for all \(\mathbf{y}:= (y, \lambda) \in S_{f}(\lambda) \times\Lambda\);

(ii)
\(\{ x_{n}\}\) is an approximating sequence for (LQEP) corresponding to \(\{ \lambda_{n} \}\).
Definition 2.5
A sequence \(\{\mathbf{x}_{n} \} := \{(x_{n}, \lambda _{n})\} \subseteq Y := X \times\Lambda\) is called an approximating (or minimizing) sequence for (OPLEC) iff there exists \(\varepsilon _{n} \downarrow0\) such that

(i)
\(g(\mathbf{x}_{n}) \leq g(\mathbf{y}) + \varepsilon_{n}\), for all \(\mathbf{y}:= (y, \lambda) \in S_{f}(\lambda) \times\Lambda\);

(ii)
\(\{x_{n}\}\) is an approximating sequence for (LQEP) corresponding to \(\{\lambda_{n}\}\).
Definition 2.6
Problem (LVQEPLEC) or (OPLEC) is called wellposed at \(\bar{\lambda}\) iff

(i)
it has solutions;

(ii)
for any approximating sequence \(\{ \mathbf{x}_{n} \}:= \{(x_{n}, \lambda_{n})\}\) for (LVQEPLEC), where \(\lambda_{n} \rightarrow\bar {\lambda}\), has a subsequence converging to a solution.
Definition 2.7
Problem (LVQEPLEC) or (OPLEC) is called uniquely wellposed at \(\bar{\lambda}\) iff

(i)
it has a unique solution \(\bar{\mathbf{x}} := (\bar{x}, \bar {\lambda})\);

(ii)
every approximating sequence \(\{ \mathbf{x}_{n} \}:= \{(x_{n}, \lambda_{n})\}\) for (LVQEPLEC) or (OPLEC), where \(\lambda_{n} \rightarrow \bar{\lambda}\), converges to \(\bar{\mathbf{x}}\).
Now we recall the continuitylike properties which will be used for our analysis.
Definition 2.8
[34]
Let \(Q:X\rightrightarrows Y\) be a setvalued mapping between two metric spaces.

(i)
Q is upper semicontinuous (usc) at \(\bar{x}\) if, for any open set \(U\supseteq Q(\bar{x})\), there is a neighborhood N of \(\bar{x}\) such that \(Q(N)\subseteq U\).

(ii)
Q is lower semicontinuous (lsc) at \(\bar{x}\) if, for any open subset U of Y with \(Q(\bar{x})\cap U\neq\emptyset\), there is a neighborhood N of \(\bar{x}\) such that \(Q(x)\cap U \neq \emptyset\) for all \(x \in N\).

(iii)
Q is closed at \(\bar{x}\) if, for any sequences \(\{x_{k}\} \) and \(\{y_{k}\}\) with \(x_{k}\rightarrow\bar{x}\) and \(y_{k} \rightarrow\bar{y}\) and \(y_{k} \in Q(x_{k})\), we have \(\bar{y} \in Q(\bar{x})\).
Lemma 2.9

(i)
If Q is usc at \(\bar{x}\) and \(Q(\bar{x})\) is compact, then for any sequence \(\{x_{n}\}\) converging to \(\bar{x}\), every sequence \(\{y_{n}\}\) with \(y_{n} \in Q(x_{n})\) has a subsequence converging to some point in \(Q(\bar{x})\). If, in addition, \(Q(\bar{x}) = \{\bar{y}\}\) is a singleton, then such a sequence \(\{y_{n}\}\) must converge to \(\bar{y}\).

(ii)
Q is lsc at \(\bar{x}\) if and only if, for any sequence \(\{ x_{n}\}\) with \(x_{n}\rightarrow\bar{x}\) and any point \(y \in Q(\bar{x})\), there is a sequence \(\{y_{n}\}\) with \(y_{n} \in Q(x_{n})\) converging to y.
Definition 2.10
Let g be an extended realvalued function on a metric space X and ε be a real number.

(i)
g is upper εlevel closed at \(\bar{x} \in X\) if, for any sequence \(\{x_{n}\}\) satisfying
$$x_{n}\rightarrow\bar{x} \quad \text{and} \quad g(x_{n})\geq \varepsilon\quad \text{for all } n, $$\(g(\bar{x})\geq\varepsilon\).

(ii)
g is strongly upper εlevel closed at \(\bar {x} \in X\) if, for any sequences \(\{x_{n}\}\) in X and \(\{r_{n}\}\subset [0,\infty)\) satisfying
$$x_{n}\rightarrow\bar{x},\qquad r_{n} \rightarrow0 \quad \text{and}\quad g(x_{n}) + r_{n} \geq\varepsilon \quad \text{for all } n, $$\(g(\bar{x}) \geq\varepsilon\).
Definition 2.11
Let X be a topological space and \(f : X\rightarrow\bar{\mathbb{R}}\).

(i)
f is called upper pseudocontinuous at \(x_{0} \in X\) iff for any point x and sequence \(\{x_{n}\} \) in X such that
$$f(x_{0}) < f(x)\quad \text{and}\quad x_{n} \rightarrow x_{0}, $$\(\limsup_{n\rightarrow\infty} f(x_{n}) < f(x)\).

(ii)
f is called lower pseudocontinuous at \(x_{0} \in X\) iff for any point x and sequence \(\{x_{n}\} \) in X such that
$$f(x) < f(x_{0})\quad \text{and}\quad x_{n} \rightarrow x_{0}, $$\(f(x) < \liminf_{n\rightarrow\infty} f(x_{n}) \).

(iii)
f is termed pseudocontinuous at \(x_{0} \in X\) iff it is both lower and upper pseudocontinuous at this point.
Remark 2.12
The class of the upper pseudocontinuous functions strictly contains that of the usc functions; see [16].
Let A, B be two subsets of a metric space X. The Hausdorff distance between A and B is defined as follows:
where \(H^{*}(A,B) = \sup_{a \in A}d(a,B)\), and \(d(x,A) = \inf_{y \in A}d(x,y)\).
3 Lexicographic vector equilibrium problems with lexicographic equilibrium constraints (LVQEPLEC)
In this section, we shall establish necessary and/or sufficient conditions for (LVQEPLEC) to be (uniquely) wellposed at the reference point \(\bar{\lambda} \in \Lambda\). To simplify the presentation, in the sequel, the results will be formulated for the case \(n=2\).
For any positive numbers ϵ and ξ, as above, \(\tilde {S}_{f}(\lambda, \varepsilon)\) and \(\tilde{S}_{F}(\xi)\) are defined by the solution sets of (\(\mathrm{LQEP}_{\lambda,\epsilon}\)) and (\(\mathrm{LVQEPLEC}_{\xi}\)), respectively; that is,
and
For positive ξ and ε, the corresponding approximate solution set of (LVQEPLEC) is defined by
The setvalued mapping \(Z_{f}:X \times\Lambda\rightarrow 2^{X}\) next defined will play an important role our analysis
where \(Z_{1,f} : \Lambda\rightarrow2^{X}\) denotes the solution mapping of the scalar equilibrium problem determined by the realvalued function \(f_{1}\); that is,
Then the problem (\(\mathrm{LQEP}_{\lambda,\varepsilon}\)) can be equivalently stated as follows:
 (\(\mathrm{LQEP}_{\lambda,\varepsilon}\)):

find \(\bar{x} \in K_{1}(\bar{x},\lambda)\) such that
$$ \left \{ \begin{array}{l} f_{1} (\bar{x}, y, \lambda)\geq0, \quad \forall y \in K_{2}(\bar{x},\lambda); \\ f_{2}(\bar{x},z,\lambda)+\varepsilon\geq0 ,\quad \forall z\in Z_{f}(\bar{x},\lambda). \end{array} \right . $$
Next, let the setvalued map \(Z_{F}: X \times\Lambda\rightarrow 2^{X}\) be defined by
where \(Z_{1,F}(\lambda):= \{ \bar{\mathbf{y}} = (\bar{y},\bar {\lambda}) \in\operatorname{gr}S_{f}  F_{1}(\bar{\mathbf{y}},\mathbf {y}')\geq0, \forall\mathbf{y}' \in\operatorname{gr}S_{f}\}\). Then the problem (\(\mathrm{LVQEPLEC}_{\xi}\)) can be equivalently stated as follows:
 (\(\mathrm{LVQEPLEC}_{\xi}\)):

find \(\bar{\mathbf{y}} \in\operatorname{gr}S_{f}\) such that
$$ \left \{ \begin{array}{l} F_{1} (\bar{\mathbf{y}}, \mathbf{y})\geq0, \quad \forall \mathbf{y} \in\operatorname{gr}S_{f}; \\ F_{2}(\bar{\mathbf{y}},\mathbf{y}')+\varepsilon\geq0, \quad \forall\mathbf{y}' \in Z_{F}(\bar{\mathbf{y}}). \end{array} \right . $$
Thus, for any positive numbers ξ and ε, \(\Gamma(\xi ,\varepsilon)\) is equivalent to
Lemma 3.1
Let \(\{x_{n}\}\) converging to \(\bar{x} \in Z_{1,f}(\bar{\lambda})\) be an approximating sequence of (\(\mathrm{LQEP}_{\bar{\lambda}}\)) corresponding to a sequence \(\lambda _{n}\rightarrow\bar{\lambda}\) and assume that \(Z_{f}\) is lsc at \((\bar{x},\bar{\lambda})\) and \(f_{2}\) is strongly upper 0level closed on \(\{\bar{x}\} \times Z_{f}(\bar{x},\bar{\lambda})\times\{\bar {\lambda}\}\). Then \(\bar{x} \in S_{f}(\bar{\lambda})\).
Proof
Suppose to the contrary that \(\bar{x} \notin S_{f}(\bar{\lambda})\). Then there exists \(\bar{z} \in Z_{f}(\bar{x},\bar{\lambda})\) such that \(f_{2}(\bar{x}, \bar{z}, \bar{\lambda})<0\). For each n, we conclude with the lower semicontinuity of \(Z_{f}\) at \((\bar{x},\bar {\lambda})\) and Lemma 2.9(ii) there exists \(z_{n} \in Z_{f}(x_{n},\lambda_{n})\) such that \(z_{n} \rightarrow\bar{z}\). Since \(\{ x_{n}\}\) is an approximating sequence of (\(\mathrm{LQEP}_{\bar{\lambda}}\)) corresponding to a sequence \(\lambda_{n} \), there is a sequence \(\{\varepsilon_{n}\} \subset(0,\infty)\) converging to 0 such that \(x_{n} \in\tilde{S}_{f}(\lambda_{n}, \varepsilon_{n})\) for all n. This implies that
This together with the strongly upper 0level closedness of \(f_{2}\) at \((\bar{x}, \bar{z}, \bar {\lambda})\) implies that
This yields a contradiction; we have \(\bar{x} \in S_{f}(\bar{\lambda })=\tilde{S}_{f}(\bar{\lambda},0)\). □
Theorem 3.2
Assume that X be compact and

(i)
in \(X \times\Lambda\), \(K_{1}\) is closed and \(K_{2}\) is lsc;

(ii)
\(Z_{f}\) is lsc on \(Z_{1,f}(\bar{\lambda}) \times\{\bar{\lambda }\}\);

(iii)
\(f_{1}\) is upper 0level closed on \(K_{1}(\bar{x},\bar{\lambda}) \times K_{2}(\bar{x}, \bar{\lambda}) \times\{\bar{\lambda}\}\);

(iv)
\(f_{2}\) is strongly upper 0level closed on \(K_{1}(\bar{x},\bar {\lambda}) \times K_{2}(\bar{x}, \bar{\lambda}) \times\{\bar {\lambda}\}\);

(v)
\(F_{1}(\cdot,\mathbf{y})\) is upper 0level closed at \((\bar {x},\bar{\lambda})\), for all \(\mathbf{y} \in X \times\Lambda\);

(vi)
\(F_{2}(\cdot,\mathbf{y})\) is strongly upper 0level closed at \((\bar{x}, \bar{\lambda})\), for all \(\mathbf{y} \in X \times \Lambda\).
Then (LVQEPLEC) is wellposed at \(\bar{\lambda}\). Furthermore, if \(S_{f} : \Lambda\rightarrow X\) is singlevalued and (LVQEPLEC) admits a unique solution \(\bar{\mathbf{x}}\), then (LVQEPLEC) is uniquely wellposed.
Proof
Step I: We first prove that \(Z_{1,f}\) is closed at \(\bar{\lambda}\). Suppose to the contrary that there are two sequences \(\{\lambda_{n} \}\) and \(\{x_{n}\}\) satisfying \(\lambda_{n} \rightarrow\bar{\lambda}\) and \(x_{n} \rightarrow\bar {x}\) with \(x_{n} \in Z_{1,f}(\lambda_{n})\) and \(\bar{x} \notin Z_{1,f}(\bar{\lambda})\). Since \(K_{1}\) is closed in \(X\times\Lambda \) and \(x_{n} \in K_{1}(x_{n}, \lambda_{n})\) for all n, we conclude that \(\bar{x} \in K_{1}(\bar{x},\bar{\lambda})\). Then there exists \(\bar{y} \in K_{2}(\bar{x} , \bar{\lambda})\) satisfying \(f_{1}(\bar{x}, \bar{y}, \bar{\lambda})<0\). The lower semicontinuity of \(K_{2}\) at \((\bar{x},\bar{\lambda})\) ensures that, for each n, there is \(y_{n} \in K_{2}(x_{n},\lambda_{n})\) such that \(y_{n} \rightarrow\bar{y}\) as \(n \rightarrow\infty\). Since \(x_{n} \in Z_{1,f}(\lambda_{n})\), it follows that
This together with the upper 0level closedness of \(f_{1}\) implies that
which yields a contradiction and, hence, \(Z_{1,f}\) is closed at \(\bar {\lambda}\).
Step II: Next, we show that \(\tilde{S}_{f}(\cdot,\cdot)\) is usc at \((\bar{\lambda},0)\). Indeed, if it were otherwise, then there is an open set \(U \supseteq\tilde{S}_{f}(\bar{\lambda},0)\) such that for all neighborhood \(N(\bar{\lambda},0)\) of \((\bar {\lambda},0)\),
In particular, for each \(\{\lambda_{n}\}\) and \(\{\epsilon_{n}\} \) satisfying \(\lambda_{n} \rightarrow\bar{\lambda}\) and \(\epsilon_{n} \rightarrow0\), there exists \(x_{n} \in\tilde{S}_{f}(\lambda _{n},\epsilon_{n})\) such that \(x_{n} \notin U\) for all n. Since X is compact, we can assume that \(\{ x_{n} \}\) converges to some \(\bar{x} \notin U\). By the closedness of \(Z_{1,f}\) at \(\bar{\lambda}\), one has \(\bar{x} \in Z_{1,f}(\bar{\lambda})\). Applying Lemma 3.1, we conclude that
which gives \(\bar{x} \in U\). This yields a contradiction. Therefore the map \(\tilde{S}_{f}\) is usc at \((\bar{\lambda},0)\).
Step III: We have to prove that \(\tilde{S}_{f}(\bar {\lambda}, 0)\) is compact by checking its closedness. Take an arbitrary sequence \(\{ x_{n}\}\) in \(S(\bar{\lambda})= \tilde{S}_{f}(\bar{\lambda}, 0)\) converging to \(\bar{x}\). Setting \(\lambda_{n} := \bar{\lambda}\) for all n, we have \(\lambda_{n} \rightarrow\bar{\lambda}\) and \(x_{n}\in Z_{1,f}(\lambda_{n})\) for all n. This together with the closedness of \(Z_{1,f}\) at \(\bar{\lambda}\) implies that \(\bar{x} \in Z_{1,f}(\bar {\lambda})\). Note that \(\{x_{n}\}\) is, of course, an approximating sequence of (\(\mathrm{LQEP}_{\bar{\lambda}}\)) corresponding to \(\{\lambda_{n}\} \). Then Lemma 3.1 again implies that \(\bar{x} \in S_{f}(\bar {\lambda})= \tilde{S}_{f}(\bar{\lambda}, 0)\), and hence \(S_{f}(\bar {\lambda})\) is compact; that is, \(\tilde{S}_{f}(\bar{\lambda}, 0)\) is compact.
Step IV: Finally, we prove that (LVQEPLEC) is wellposed at \(\bar{\lambda}\). To this end, let \(\{\mathbf{x}_{n} \}:= \{(x_{n}, \lambda_{n})\}\), where \(\lambda_{n} \rightarrow\bar{\lambda}\), be any approximating sequence for (LVQEPLEC). Hence, by Definition 2.4, \(\{ x_{n}\}\) is an approximating sequence for (LQEP) corresponding to \(\{ \lambda_{n}\}\). Then there exists a real sequence \(\{\varepsilon_{n}\} \downarrow0 \) such that
Applying Lemma 2.9(i), there exists a subsequence \(\{ x_{n_{k}}\}\) of \(\{x_{n}\}\) converging to some \(\bar{x} \in\tilde {S}_{f}(\bar{\lambda}, 0)\), and hence
Now we check that \(\bar{\mathbf{x}} := (\bar{x},\bar{\lambda})\) is a solution of (LVQEPLEC). Since \(\{\mathbf{x}_{n}\}\) is an approximating sequence, there exists \(\{\varepsilon_{n}\}\downarrow0\) such that \(F_{1}(\mathbf{x}_{n}, \mathbf{y})\geq0\) and \(F_{2}(\mathbf{x}_{n}, \mathbf{y})+\varepsilon_{n} \geq0\) for all \(\mathbf{y} \in\operatorname{gr}S_{f}\). The upper 0level closedness of \(F_{1}\) and the strongly upper 0level closedness of \(F_{2}\) implies that \(F_{1}(\bar{\mathbf{x}},\mathbf{y}) \geq0\) and \(F_{2}(\bar{\mathbf{x}},\mathbf{y}) \geq0\) for all \(\mathbf{y} \in \operatorname{gr} S_{f}\), i.e., \(\bar{\mathbf{x}}\) is a solution. Thus, (LVQEPLEC) is wellposed at \(\bar{\lambda}\).
Furthermore, suppose that \(S_{f} : \Lambda\rightarrow X\) is singlevalued and (LVQEPLEC) admits a unique solution \(\bar{\mathbf{x}}\). We have to show that (LVQEPLEC) is uniquely wellposed. Let \(\{\mathbf {x}_{n}\}\) be an approximating sequence for (LVQEPLEC). By the same argument as in the preceding part, there is a subsequence converging to \(\bar{\mathbf{x}}\). If \(\{ \mathbf{x}_{n}\}\) did not converge to \(\bar{\mathbf{x}}\), there would be an open set U containing \(\bar{\mathbf{x}}\) such that some subsequence was outside U. By the above argument, this subsequence has a subsequence convergent to \(\bar {\mathbf{x}}\), an impossibility. □
The following examples show that none of the assumptions in Theorem 3.2 can be dropped.
Example 3.3
(The compactness of X cannot be dropped)
Let \(X=\mathbb{R}\), \(\Lambda=[0,1]\), \(K_{1}(x,\lambda) = K_{2}(x,\lambda) =[\lambda ,+\infty)\),
and
It is clear that in \(X\times\Lambda\), \(K_{1}\) is closed and \(K_{2}\) is lsc. One can check that \(Z_{1,f}(\lambda)=[\frac{1}{\lambda},+\infty)\). Thus \(Z_{f}\) is lsc. Furthermore, (iii)(vi) hold as f and F are continuous in \(X \times X \times\Lambda\) and \((X\times\Lambda) \times (X\times\Lambda)\), respectively. The solution set of (LVQEPLEC) is gr \(S_{f}\). But \(S_{f}(0)=\{0\}\) and \(S_{f}(\lambda)=[\frac{1}{\lambda},\infty )\) for all \(\lambda\in(0,1]\), \(\operatorname{gr}S_{f}=\{(0,0)\} \cup\{ ([\frac{1}{\lambda},\infty),\lambda)  \lambda\in(0,1]\}\). Hence, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=n\), \(\lambda _{n}=\frac{1}{n}\) for all \(n \in\mathbb{N}\). We see that \(\mathbf {x}_{n}:=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC). It is clear that \(\{\mathbf{x}_{n}\}\) has no convergent subsequence. The reason is that X is not compact. We note further that \(S_{f}(\cdot)\) is neither usc nor lsc at 0, even under the continuity assumptions of \(K_{1}\), \(K_{2}\), and f.
Example 3.4
(The closedness of \(K_{1}\) is essential)
Let \(X=[1,1]\), \(\Lambda=[0,1]\), \(K_{1}(x,\lambda)= K_{2}(x, \lambda)=(0,1]\),
and
It is not hard to see that X is compact, \(K_{2}\) is lsc in \(X\times \Lambda\). One can check that \(Z_{1,f}(\lambda)= (0,1]\) and
Thus \(Z_{f}\) is lsc, (ii)(vi) are satisfied (by the continuity of f and F). We see also that the solution set of (LVQEPLEC) is \(\operatorname{gr}S_{f}\). But \(S_{f}(\lambda)=(0,1] \) for all \(\lambda\in[0,1]\), i.e., \(\operatorname{gr} S_{f}=\{ (x, \lambda) x \in(0,1], \lambda\in [0,1]\}\).
Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=\frac {1}{n}\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb{N}\). Then \(\mathbf{x}_{n}:=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\{\mathbf{x}_{n}\}\) converges to \(\mathbf {x}:=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
Example 3.5
(The lower semicontinuity of \(K_{2}\) cannot be dispensed)
Let \(X = \Lambda= [0, 2]\),
and
One can check that \(K_{1}\) is closed but \(K_{2}\) is not lsc at \(\bar {\lambda}=0\) and
Thus (ii)(vi) hold. One can check that
Moreover, the solution set of (LVQEPLEC) coincides with \(\operatorname{gr} S_{f}\). But
i.e., \(\operatorname{gr} S_{f} :=(2,0) \cup\{ (1, \lambda)  \lambda\in(0,2]\}\). Hence, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=1\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb {N}\). We see that \(\mathbf{x}_{n} :=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\{\mathbf{x}_{n}\}\) converges to \(\mathbf{x}:=(1,0)\). But x does not belong to the solution set of (LVQEPLEC).
Example 3.6
(The lower semicontinuity of \(Z_{f}\) cannot be dropped)
Let \(X=\Lambda=[0,1]\) (compact), \(K_{1}(x,\lambda)=[0,1]\) closed, \(K_{2}=[0,1]\) lsc,
and
One can check that
and, for each \((x,\lambda) \in\operatorname{gr}S_{1,f}\),
\(Z_{f}\) is not lsc at \((0,1)\) because by taking \(\{(\lambda_{n} = \frac{1}{n}, x_{n} = 1)\}\rightarrow(0,1)\), we have \(Z_{f}(x_{n},\lambda_{n}) = \{1\}\) for all n, while \(Z_{f}(0,1) = [0,1]\). Assumptions (i), (iii)(vi) are obviously satisfied. Finally, we observe that (LVQEPLEC) is not wellposed at \(\bar{\lambda}\) by calculating the solution mapping S explicitly as follows:
i.e., \(\operatorname{gr}S_{f}:= (0,0) \cup\{ (x,\lambda)  x=0,1, \lambda\in(0,1]\}\). Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=1\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb {N}\). We see that \(\mathbf{x}_{n} :=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\{\mathbf{x}_{n}\}\) converges to \(\mathbf{x}:=(1,0)\). But x does not belong to the solution set of (LVQEPLEC).
Example 3.7
(Upper 0level closedness of \(f_{1}\))
Let \(X = \Lambda= [0,1]\) (compact), \(K_{1}(x,\lambda) = K_{2}(x,\lambda) =[0,1]\) (continuous and closed), \(\bar{\lambda}= 0\),
and
One can check that
i.e., \(\operatorname{gr} S_{f}:= (1,0) \cup\{(0, \lambda) \lambda\in(0,1]\}\). Hence, all the assumptions except (iii) hold true. However, (LVQEPLEC) is not wellposed at \(\bar{\lambda}\). Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb {N}\). We see that \(\mathbf{x}_{n} :=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\{\mathbf{x}_{n}\}\) converges to \(\mathbf{x}:=(0,0)\). But x does not belong to the solution set of (LVQEPLEC). Finally, we show that assumption (iii) is not satisfied. Indeed, take \(\{x_{n}\}\) and \(\{ \lambda_{n}\}\) as above and \(\{y_{n} = 1\}\), we have \((x_{n}, y_{n}, \lambda _{n})\rightarrow(0,1,0)\) and \(f_{1}(x_{n},y_{n},\lambda_{n}) = 1 > 0\) for all n, while \(f_{1}(0,1,0) = 1 < 0\).
Example 3.8
(Strong upper 0level closedness of \(f_{2}\))
Let X, Λ, \(K_{1}\), \(K_{2}\), \(\bar{\lambda}\), and F be as in Example 3.7,
One can check that
i.e., \(\operatorname{gr} S_{f} :=(1,0) \cup\{ (x, \lambda)  x=0,1, \lambda\in(0,1]\}\). We can conclude that all the assumptions of Theorem 3.2 except (iv) are satisfied. Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb {N}\). We see that \(\mathbf{x}_{n} :=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\mathbf{x}_{n}\) converges to \(\mathbf {x}:=(0,0)\). But x does not belong to the solution set of (LVQEPLEC). Finally, we show that assumption (iv) is not satisfied. Indeed, take sequences \(x_{n} = 0\), \(y_{n} = 1\), \(\lambda_{n} = \frac{1}{n} \) and \(\varepsilon_{n} = \frac{1}{n}\), we have \(\{ (x_{n}, y_{n}, \lambda_{n}, \varepsilon_{n})\}\) and \(f_{2}(x_{n}, y_{n}, \lambda_{n})+\varepsilon_{n} > 0\) for all n, while \(f_{2}(0,1,0) = 1 < 0\).
Example 3.9
(Upper 0level closedness of \(F_{1}\))
Let \(X= \Lambda=[0,1]\), \(K_{1}(x, \lambda)=[0,1]\) closed, \(K_{2}(x, \lambda)=[0,1]\) lsc,
and
Then assumptions (i)(vi) and (vi) are satisfied. We have \(\operatorname{gr} S_{f}:= [0,1] \), \(\lambda\in[0,1]\). The solution set of (LVQEPLEC) is \((1,0)\cup\{ (x,\lambda) x=0,1, \lambda\in(0,1]\}\). We can conclude that all the assumptions of Theorem 3.2 except (v) are satisfied. Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb {N}\). We see that \(\mathbf{x}_{n} :=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\mathbf{x}_{n}\) converges to \(\mathbf {x}:=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
Example 3.10
(Strong upper 0level closedness of \(F_{2}\))
Let X, Λ, \(K_{1}\), \(K_{2}\), \(\bar{\lambda}\), and f be as in Example 3.9 and
One can check that
i.e., \(\operatorname{gr} S_{f}:= \{ (x,\lambda)  x \in[0,1] , \lambda\in[0,1]\}\). The solution set of (LVQEPLEC) is \((1,0)\cup\{ (x,\lambda) x=0,1, \lambda\in(0,1]\}\). We can conclude that all the assumptions of Theorem 3.2 except (vi) are satisfied. Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\) for all \(n \in\mathbb {N}\). We see that \(\mathbf{x}_{n} :=(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\mathbf{x}_{n}\) converges to \(\mathbf {x}:=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
Theorem 3.11
Let X and Λ be two metric spaces. Then:

(i)
If (LVQEPLEC) is uniquely wellposed at \(\bar{\lambda}\), then \(\operatorname{diam} \Gamma(\xi,\varepsilon) \downarrow0\) as \((\xi,\varepsilon)\downarrow(0,0)\).

(ii)
Conversely, suppose that X and Λ are complete, assumptions (i)(vi) in Theorem 3.2 hold and \(\operatorname{diam} \Gamma(\xi,\varepsilon) \downarrow0\) as \(\xi\downarrow0\) and \(\varepsilon\downarrow0\). Then (LVQEPLEC) is uniquely wellposed at \(\bar{\lambda}\).
Proof
(1) Suppose that (LVQEPLEC) be uniquely wellposed at \(\bar{\lambda }\). Then (LVQEPLEC) has a unique solution \(\bar{\mathbf{x}} := (\bar {x}, \bar{\lambda})\) for some \(\bar{x}\in X\). Assume to the contrary that \(\operatorname{diam}\Gamma(\xi_{n},\varepsilon_{n})\) does not converge to 0 as \(n\rightarrow\infty\). This lead to the existence of a number \(r > 0\) such that for any \(k \in\mathbb{N}\), there exists \(n_{k} \geq k\) with
This implies that, for each k, there exist \((x_{n_{k}}^{1}, \lambda _{n_{k}}^{1}), (x_{n_{k}}^{2}, \lambda_{n_{k}}^{2}) \in\Gamma(\xi _{n_{k}},\varepsilon_{n_{k}}) \) such that
Since \(\{(x_{n_{k}}^{1}, \lambda_{n_{k}}^{1})\}\) and \(\{(x_{n_{k}}^{2}, \lambda _{n_{k}}^{2})\}\) are approximating sequences for (LVQEPLEC), it follows from (3.2) that \(0 = d(\bar{\mathbf{x}}, \bar{\mathbf{x}}) > r/2\). Then we arrive at a contradiction.
(2) Let \(\{\mathbf{x}_{n}\} := \{ (x_{n}, \lambda_{n})\}\) be an approximating sequence of (LVQEPLEC) with \(\lambda_{n} \rightarrow \bar{\lambda}\) as \(n\rightarrow\infty\). Then there exists \(\xi_{n} \downarrow0\) such that
Furthermore, there is a sequence \(\varepsilon_{n} \downarrow0\) such that
Hence we have \(\mathbf{x}_{n} \in \Gamma(\xi_{n},\varepsilon_{n})\) for all n. By choosing subsequences if necessary, we can assume that both sequences \(\{\xi_{n}\}\) and \(\{\varepsilon_{n}\}\) are nonincreasing. Thus,
From this observation and \(\operatorname{diam}\Gamma(\xi_{n},\varepsilon _{n})\downarrow0\) as \(n\rightarrow\infty\), one can directly check that \(\{\mathbf{x}_{n}\}\) is a Cauchy sequence in \(X\times\Lambda\). The completeness of \(X\times\Lambda\) implies that \(\mathbf{x}_{n} \rightarrow\bar{\mathbf{x}} := (\bar{x}, \bar {\lambda})\) as \(n\rightarrow\infty\). By (3.3), we have
for all \(\mathbf{y}:= (y, \lambda) \in\operatorname{gr}S_{f}\). This together with the upper 0level closedness of \(F_{1}\) and the strongly upper 0level closedness of \(F_{2}\) implies that
i.e., \(\bar{\mathbf{x}}\) is a solution of (LVQEPLEC). Finally, we show that \(\bar{\mathbf{x}}:=(\bar{x}, \bar{\lambda})\) is the only solution to (LVQEPLEC). Suppose to the contrary that \(\mathbf{x}'\) is another solution to (LVQEPLEC), i.e., \(\mathbf{x}'\neq\bar{\mathbf {x}}\). It is clear that they both belong to \(\Gamma(\xi,\varepsilon)\) for any \(\xi,\varepsilon> 0\). Then it follows that
which gives a contradiction. Thus, (LVQEPLEC) is uniquely wellposed at \(\bar{\lambda}\). □
To weaken the assumption of unique wellposedness in Theorem 3.11, we are going to use the notions of measures of noncompactness in a metric space X. We recall that a subset A of a metric space X is εdiscrete iff \(d(x, y) \geq\varepsilon\) for all \(x,y \in A\) with \(x \neq y\).
Definition 3.12
Let M be a nonempty subset of a metric space X.

(i)
The Kuratowski measure of M is
$$ \mu(M)=\inf \Biggl\{ \varepsilon>0 \Big M\subseteq\bigcup _{k=1}^{n} M_{k} \text{ and } \operatorname{diam} M_{k} \leq\varepsilon, k=1,\ldots ,n, \exists n \in \mathbb{N} \Biggr\} . $$ 
(ii)
The Hausdorff measure of M is
$$ \eta(M)=\inf \Biggl\{ \varepsilon>0 \Big M\subseteq\bigcup _{k=1}^{n} B(x_{k},\varepsilon), x_{k} \in X \text{ for some } n \in\mathbb {N} \Biggr\} . $$ 
(iii)
The Istrǎtescu measure of M is
$$ \iota(M)=\inf \{ \varepsilon>0  M \text{ have no infinite } \varepsilon \text{discrete subset} \}. $$
Daneš [36] obtained the following inequalities:
The measures μ, η, and ι share many properties and we will use γ in the sequel to denote either one of them. γ is a regular measure (see [37, 38]), i.e., it enjoys the following properties:

(i)
\(\gamma(M) = +\infty\) if and only if the set M is unbounded;

(ii)
\(\gamma(M) = \gamma(\operatorname{cl}M)\);

(iii)
from \(\gamma(M) = 0\) it follows that M is a totally bounded set;

(iv)
if X is a complete space and if \(\{A_{n}\}\) is a sequence of closed subsets of X such that \(A_{n+1} \subseteq A_{n}\) for each \(n \in \mathbb{N}\) and \(\lim_{n \rightarrow+\infty}\gamma(A_{n}) = 0\), then \(K := \bigcap_{n \in\mathbb{N}}A_{n}\) is a nonempty compact set and
$$\lim_{n \rightarrow+\infty}H(A_{n},K) = 0, $$where H is the Hausdorff metric;

(v)
from \(M \subseteq N\) it follows that \(\gamma(M) \leq\gamma(N)\).
In terms of a measure \(\gamma\in\{\mu,\eta,\iota\}\) of noncompactness we have the following result.
Theorem 3.13
Let X and Λ be metric spaces.

(i)
If (LVQEPLEC) is wellposed at \(\bar{\lambda}\), then \(\gamma (\Gamma(\xi,\varepsilon))\downarrow0\) as \(\xi\downarrow0\) and \(\varepsilon\downarrow0\).

(ii)
Conversely, suppose that \(\gamma(\Gamma(\xi,\varepsilon ))\downarrow0\) as \(\xi\downarrow0\) and \(\varepsilon\downarrow0\), and the following conditions hold:

(a)
X and Λ are complete;

(b)
\(K_{1}\) is closed and \(K_{2}\) is lsc at \((\bar{x},\bar{\lambda})\);

(c)
\(Z_{f}\) is lsc on \((X \times\Lambda) \cap\operatorname{gr}Z_{1,f}\);

(d)
\(f_{1}\) is upper 0level closed on \(K_{1}(\bar{x},\bar{\lambda }) \times K_{2}(\bar{x},\bar{\lambda})\times\{\bar{\lambda} \}\);

(e)
\(f_{2}\) is upper alevel closed on \(K_{1}(\bar{x},\bar{\lambda }) \times K_{2}(\bar{x},\bar{\lambda})\times\{\bar{\lambda}\}\) and \(a<0\);

(f)
\(F_{1}(\cdot, \mathbf{y})\) is upper 0level closed at \((\bar{x}, \bar{\lambda})\), for all \(\mathbf{y} \in X\times\Lambda\);

(g)
\(F_{2}(\cdot, \mathbf{y})\) is upper blevel closed at \((\bar {x},\bar{\lambda})\), for all \(\mathbf{y} \in X\times\Lambda\) and \(b <0\).

(a)
Then (LVQEPLEC) is wellposed at \(\bar{\lambda}\).
Proof
By (3.4) the proof is similar for the three mentioned measures of noncompactness. We discuss only the case \(\gamma=\mu\), the Kuratowski measure.
(1) Suppose that (LVQEPLEC) be wellposed. For each \(\xi> 0\) and \(\varepsilon> 0\), the solution set S of (LVQEPLEC) clearly satisfies the relation \(\mathbf{S}\subseteq\Gamma(\xi,\varepsilon)\). Hence,
Let \(\{\mathbf{x}_{n}\}:=\{(x_{n},\lambda_{n})\}\) be arbitrary sequence in S. Then, of course, \(\{\mathbf{x}_{n}\}\) is an approximating sequence of (LVQEPLEC). Thus, it has a subsequence converging to a point in S. Therefore, S is compact, and hence \(\mu(\mathbf{S}) = 0\). Now for any \(\delta> 0\), there are \(M_{1}^{\delta}, M_{2}^{\delta}, \ldots, M_{n}^{\delta}\) for some \(n\in\mathbb{N}\) such that
Next, for each \(k=1,\ldots,n\), define the set
Now, we show that \(\Gamma(\xi,\varepsilon) \subseteq\bigcup_{k=1}^{n} N_{k}^{\delta}\). Let \(\mathbf{x} \in\Gamma(\xi ,\varepsilon)\). Due to \(\mathbf{S} \subseteq\bigcup_{k=1}^{n} M_{k}^{\delta}\), one has
Then there is \(\bar{k} \in\{1,2,\ldots,n\}\) such that
which gives \(\mathbf{x} \in N_{\bar{k}}^{\delta}\). Therefore, we obtain the claim
Furthermore, we see that \(\operatorname{diam} N_{k}^{\delta}\leq\delta+ 2H(\Gamma(\xi,\varepsilon ), \mathbf{S})\). Indeed, for any \(\mathbf{y},\mathbf{y}' \in N_{k}^{\delta}\), \(\mathbf {m},\mathbf{m}' \in M_{k}^{\delta}\),
which gives
and we arrive at \(\operatorname{diam} N_{k}^{\delta}\leq\delta+ 2H(\Gamma(\xi,\varepsilon ), \mathbf{S})\). The definition of μ implies that
Therefore, we can conclude that
To check that \(H(\Gamma(\xi,\varepsilon),\mathbf{S}) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\) by contradiction, assume the existence of \(\rho>0\), \((\xi_{n}, \varepsilon_{n}) \downarrow(0, 0)\), and \(\mathbf{x}_{n} \in\Gamma(\xi_{n}, \varepsilon_{n})\) such that \(d(\mathbf{x}_{n},\mathbf{S}) \geq\rho\) for all \(n \in\mathbb{N}\). Since \(\{\mathbf{x}_{n}\}\) is an approximating sequence, one has a subsequence convergent to some point of S, which is impossible. Hence \(\mu(\Gamma (\xi,\varepsilon))\downarrow0\) as \(\xi\downarrow0\) and \(\varepsilon\downarrow0\).
(2) Assume that \(\mu(\Gamma(\xi,\varepsilon))\downarrow0\) as \(\xi\downarrow0\) and \(\varepsilon\downarrow0\). We claim that \(\Gamma(\xi,\varepsilon)\) is closed for all \(\xi,\epsilon>0\). Let the sequence \(\{\mathbf{x}_{n} \} := \{ (x_{n},\lambda_{n})\} \) in \(\Gamma(\xi,\varepsilon)\) with \(\mathbf{x}_{n} \rightarrow\mathbf {x}: = (x,\lambda)\). Then, for all \(\mathbf{y} \in\operatorname{gr}S_{f}\), \(\mathbf{y}' \in Z_{F}(\mathbf{x}_{n})\), \(y_{n} \in K_{2}(x_{n},\lambda_{n})\), and all \(z_{n} \in Z_{f}(x_{n},\lambda_{n})\), we have
and
As \(K_{1}\) is closed at \((x,\lambda)\), one has \(x \in K_{1}(x,\lambda)\). By the upper 0level closedness of \(F_{1}\) and upperξlevel closedness of \(F_{2}\), one obtains
Next, we show by a contrapositive argument that
Suppose that there exist \(y \in K_{2}(x,\lambda)\) and \(z \in Z_{f}(x,\lambda)\) such that
Since \(K_{2}\) is lsc at \((x,\lambda)\) and \(Z_{f}\) is lsc at \((x,\lambda )\), there are two sequences \(\{y_{n}\}\) and \(\{z_{n}\}\) such that \(y_{n} \in K_{2}(x_{n},\lambda_{n})\) and \(z_{n} \in Z_{f}(x_{n},\lambda_{n})\) and
By (d) and (e), there is \(n_{0} \in\mathbb{N}\) such that
which leads to a contradiction. As a result, \(\mathbf{x} \in\Gamma (\xi,\varepsilon)\) and this set is closed. Next, we observe further that
The properties of μ implies that S is compact and \(H(\Gamma(\xi,\varepsilon),\mathbf{S}) \downarrow0\) as \((\xi ,\epsilon) \downarrow(0,0)\). Let \(\{\mathbf{x}_{n}\} := \{(x_{n},\lambda _{n})\}\) be an approximating sequence. There is \((\xi_{n},\epsilon_{n}) \downarrow(0,0)\) such that, for all \(\mathbf{y} \in\operatorname{gr}S_{f}\), \(y' \in Z_{F}(\mathbf{x}_{n})\), \(y_{n} \in K_{2}(x_{n},\lambda_{n})\), and all \(z_{n} \in Z_{f}(x_{n}\lambda_{n})\),
Therefore \((x_{n},\lambda_{n}) \in\Gamma(\xi_{n},\epsilon_{n})\). Consequently,
By the compactness of S, there is a subsequence of \(\{ \mathbf{x}_{n}\}\) converging to a point of S. Hence (LVQEPLEC) is wellposed. This completes the proof. □
The following examples show that all assumptions of Theorem 3.13(ii) are essential.
Example 3.14
(The closedness of \(K_{1}\) is essential)
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.4. It is easy to check that \(Z_{f}\) is lsc and X is complete and \(K_{2}\) is lsc in \(X \times\Lambda\). Assumptions (ii)(c)(ii)(f) are fulfilled since f and F are continuous in \(X\times X \times\Lambda \) and \((X\times\Lambda) \times(X,\Lambda)\), respectively. Moreover, \(\Gamma(\xi, \varepsilon) \subseteq[1, 1] \times[0, 1]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) \leq\gamma([1, 1] \times [0, 1]) = 0\). It is easy to see that the solution set of (LVQEPLEC) coincides with \(\operatorname{gr} S_{f}\). But \(S_{f}(\lambda)=(0,1] \) for all \(\lambda\in[0,1]\), i.e., \(\operatorname{gr} S_{f}=\{ (x, \lambda) x \in(0,1], \lambda\in [0,1]\}\). With the same arguments as in Example 3.4, (LVQEPLEC) is not wellposed. The reason is that \(K_{1}\) is not closed at \((0, 0)\).
Example 3.15
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.5. Then X is complete, \(K_{1}\) is closed in \(X\times\Lambda\), and (ii)(b) and (ii)(f) hold. \(\Gamma(\xi,\varepsilon)\subseteq[0,2] \times[0, 2]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) = 0\). Furthermore, the solution set of (LVQEPLEC) is \(\operatorname{gr} S_{f}\). But
i.e., \(\operatorname{gr} S_{f} :=(2,0) \cup\{ (1, \lambda)  \lambda\in(0,2]\}\). Thus, (LVQEPLEC) is not wellposed. The reason is that \(K_{2}\) is not lsc in \(X \times\Lambda\).
Example 3.16
(The lower semicontinuity of \(Z_{f}\) cannot be dropped)
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.6. One can check that
and, for each \((x,\lambda) \in\operatorname{gr}S_{1,f}\),
\(Z_{f}\) is not lsc at \((0,1)\) because by taking \((\lambda_{n} = \frac {1}{n}, x_{n} = 1)\rightarrow(0,1)\), we have \(Z_{f}(x_{n},\lambda_{n}) = \{1\}\) for all n, while \(Z_{f}(0,1) = [0,1]\). Assumptions (i), (iii)(iv) are obviously satisfied. Furthermore, \(\Gamma(\xi,\varepsilon)\subseteq[0,1] \times[0, 1]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) = 0\). Finally, we observe that (LVQEPLEC) is not wellposed at \(\bar{\lambda }\) by calculating the solution mapping \(S_{f}\) explicitly as follows:
i.e., \(\operatorname{gr}S_{f}:= (0,0) \cup\{ (x,\lambda)  x=0,1, \lambda\in(0,1]\}\). Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=1\), \(\lambda_{n}=\frac{1}{n}\). We see that \(\mathbf {x}_{n} =(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\{\mathbf{x}_{n}\}\) converges to \(\mathbf{x}=(1,0)\). But x does not belong to the solution set of (LVQEPLEC).
Example 3.17
(Upper 0level closedness of \(f_{1}\))
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.7. One can check that
i.e., \(\operatorname{gr} S_{f}:= (1,0) \cup\{(0, \lambda) \lambda\in(0,1]\}\) Hence, all the assumptions except (iii) hold true. Moreover, \(\Gamma (\xi,\varepsilon)\subseteq[0,1] \times[0, 1]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) = 0\). However, (LVQEPLEC) is not wellposed at \(\bar{\lambda}\). Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\). We see that \(\mathbf {x}_{n} =(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\{\mathbf{x}_{n}\}\) converges to \(\mathbf{x}=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
Finally, we show that assumption (iii) is not satisfied. Indeed, take \(\{x_{n}\}\) and \(\{\lambda_{n}\}\) as above and \(y_{n} = 1\), we have \((x_{n}, y_{n}, \lambda_{n}) \rightarrow(0,1,0)\) and \(f_{1}(x_{n},y_{n},\lambda_{n}) = 1 > 0\) for all n, while \(f_{1}(0,1,0) = 1 < 0\).
Example 3.18
(Strong upper 0level closedness of \(f_{2}\))
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.8. One can check that
i.e., \(\operatorname{gr} S_{f} :=(1,0) \cup\{ (x, \lambda)  x=0,1, \lambda\in(0,1]\}\). We can conclude that all the assumptions of Theorem 3.2 except (iv) are satisfied. In addition, \(\Gamma(\xi,\varepsilon)\subseteq[0,1] \times[0, 1]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) = 0\). Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\). We see that \(\mathbf {x}_{n} =(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\mathbf{x}_{n}\) converges to \(\mathbf {x}=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
Finally, we show that assumption (iv) is not satisfied. Indeed, take sequences \(x_{n} = 0\), \(y_{n} = 1\), \(\lambda_{n} = \frac{1}{n} \), and \(\varepsilon_{n} = \frac{1}{n}\), we have \(\{ (x_{n}, y_{n}, \lambda_{n}, \varepsilon_{n})\}\) and \(f_{2}(x_{n}, y_{n}, \lambda_{n})+\varepsilon_{n} > 0\) for all n, while \(f_{2}(0,1,0) = 1 < 0\).
Example 3.19
(Upper 0level closedness of \(F_{1}\))
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.9. Then assumptions (i)(vi) and (vi) are satisfied. Moreover, \(\Gamma (\xi,\varepsilon)\subseteq[0,1] \times[0, 1]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) = 0\). We have \(\operatorname{gr} S_{f}:= [0,1] \), \(\lambda\in[0,1]\). The solution set of (LVQEPLEC) is \((1,0)\cup\{ (x,\lambda) x=0,1, \lambda\in(0,1]\}\). We can conclude that all the assumptions of Theorem 3.2 except (vii) are satisfied. Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\). We see that \(\mathbf {x}_{n} =(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\mathbf{x}_{n}\) converges to \(\mathbf {x}=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
Example 3.20
(Strong upper 0level closedness of \(F_{2}\))
Let X, Λ, \(K_{1}\), \(K_{2}\), f, and F be as in Example 3.10. One can check that
i.e., \(\operatorname{gr} S_{f}:= \{ (x,\lambda)  x \in[0,1] , \lambda\in[0,1]\}\). The solution set of (LVQEPLEC) is \((1,0)\cup\{ (x,\lambda) x=0,1, \lambda\in(0,1]\}\). We can conclude that all the assumptions of Theorem 3.2 except (vii) are satisfied. By the way, \(\Gamma(\xi,\varepsilon)\subseteq[0,1] \times[0, 1]\), and hence \(\gamma(\Gamma(\xi,\varepsilon)) = 0\). Therefore, (LVQEPLEC) is not wellposed. Indeed, let \(x_{n}=0\), \(\lambda_{n}=\frac{1}{n}\). We see that \(\mathbf {x}_{n} =(x_{n}, \lambda_{n})\) is a solution of (LVQEPLEC) and \(\mathbf{x}_{n}\) converges to \(\mathbf {x}=(0,0)\). But x does not belong to the solution set of (LVQEPLEC).
4 Optimization problem with lexicographic equilibrium constraints (OPLEC)
We prove first a sufficient condition for the wellposedness in topological settings.
Theorem 4.1
Assume that X is compact and

(i)
in \(X \times\Lambda\), \(K_{1}\) is closed and \(K_{2}\) is lsc;

(ii)
\(Z_{f}\) is lsc on \(Z_{1,f}(\bar{\lambda}) \times\{\bar{\lambda }\} \);

(iii)
\(f_{1}\) is upper 0level closed on \(K_{1}(\bar{x},\bar{\lambda}) \times K_{2}(\bar{x}, \bar{\lambda}) \times\{\bar{\lambda}\}\);

(iv)
\(f_{2}\) is strongly upper 0level closed at \(K_{1}(\bar{x},\bar {\lambda}) \times K_{2}(\bar{x}, \bar{\lambda}) \times\{\lambda\}\);

(v)
g is lower pseudocontinuous in \((x,\bar{\lambda})\).
Then (OPLEC) is wellposed at \(\bar{\lambda}\). Furthermore, if \(S_{f}(\lambda)\) is a singleton, for all \(\lambda\in\Lambda\), and (OPLEC) possesses a unique solution, then this problem is uniquely wellposed at \(\bar{\lambda}\).
Proof
Let the function \(F=:(F_{1}, F_{2}) : Y \times Y\rightarrow\mathbb{R}^{2} \) be given by
that is, for all \((x,\lambda_{1}),(y,\lambda_{2}) \in Y := X\times \Lambda\),
Hence, we have \(F_{2}(\cdot,\mathbf{y})\) is strongly upper 0level closed on \(X \times\Lambda\), for all \(\mathbf{y} \in X \times \Lambda\). To apply Theorem 3.2, we need to check only that \(F_{1}(\cdot ,\mathbf{y})\) is upper 0level closed on \(X \times\Lambda\) for all \(\mathbf{y} \in X \times\Lambda\). For each fixed point \(\mathbf{y} := (y, \lambda) \in X \times\Lambda \), let \(\{\mathbf{x}_{n}\} := \{(x_{n},\lambda_{n})\} \) be any sequence in \(X \times\Lambda\) converging to \(\mathbf{x} := (x,\bar{\lambda})\) and \(F_{1}(\mathbf{x}_{n},\mathbf{y}) \geq0\); that is, we obtain
We will show that \(F_{1}(\mathbf{x}, \mathbf{y} ) \geq0\); that is, we have to prove that \(g(y,\lambda) \geq g(x,\bar{\lambda})\). Suppose, on the contrary, that \(g(y,\lambda) < g(x,\bar{\lambda})\). \(\mathbf{y} \in X \times \Lambda\). The lower pseudocontinuity of g at \((x,\bar{\lambda })\) implies that
Thus there are \(t \in\mathbb{R}\) and \(n_{0} \in\mathbb{N}\) such that, for all \(n \geq n_{0}\),
which gives a contradiction with (4.1). Applying Theorem 3.2, we have (LVQEPLEC) generated by the function F is wellposed at \(\bar {\lambda}\). Consequently, (OPLEC) is wellposed at \(\bar{\lambda}\). The assertion on unique wellposedness is easy to demonstrate. This completes the proof. □
For \(\xi,\varepsilon>0\), the approximate solution set of (OPLEC) is defined by
where \(e=(0,1) \in\mathbb{R}^{2}\).
Theorem 4.2
Let X and Λ be two metric spaces. Then the following assertions hold:

(i)
If (OPLEC) is uniquely wellposed, then \(\operatorname{diam}M(\xi, \varepsilon) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\).

(ii)
Conversely, assume that \(\operatorname{diam}M(\xi, \varepsilon) \downarrow0\) as \((\xi, \varepsilon) \downarrow (0, 0)\), and that the following conditions hold:

(a)
X and Λ are complete;

(b)
\(K_{1}\) is closed and \(K_{2}\) is lsc at \((\bar{x}, \bar{\lambda})\);

(c)
\(Z_{f}\) is lsc on \(Z_{1,f}(\bar{\lambda}) \times\bar{\lambda}\);

(d)
\(f_{1}\) is upper 0level closed at \(K_{1}(\bar{x},\bar{\lambda}) \times K_{2}(\bar{x}, \bar{\lambda}) \times\{\bar{\lambda}\}\);

(e)
\(f_{2}\) is strongly upper 0level closed on \(K_{1}(\bar{x},\bar {\lambda}) \times K_{2}(\bar{x}, \bar{\lambda}) \times\{\bar {\lambda}\}\);

(f)
g be lower pseudocontinuous at \((\bar{x}, \bar{\lambda})\).

(a)
Then (OPLEC) is uniquely wellposed at \(\bar{\lambda}\).
Proof
(1) Suppose that (OPLEC) is uniquely wellposed. Assume, on the contrary, that there are a sequence \((\xi_{n},\epsilon_{n}) \downarrow 0\), \(n_{0} \in\mathbb{N}\), and \(r > 0\) such that
Then, for each \(n\geq n_{0}\), there exist \((x^{1}_{n},\lambda^{1}_{n})\) and \((x^{2}_{n},\lambda^{2}_{n})\) in \(M(\xi_{n},\epsilon_{n})\) such that
Since \(\{(x^{1}_{n},\lambda^{1}_{n})\}\) and \(\{(x^{2}_{n},\lambda^{2}_{n})\}\) are approximating sequences for (OPLEC), they have to converge to the same unique solution and hence we arrive at a contradiction.
(2) Let \(\{\mathbf{x}_{n}\} := \{(x_{n},\lambda_{n})\}\) be any approximating sequence for (OPLEC). Then there is a sequence \((\xi _{n},\epsilon_{n}) \downarrow(0,0)\), as \(n\rightarrow\infty\), such that, for all \(n\in\mathbb{N}\),
and
Consequently, we can obtain, for all \(n\in\mathbb{N}\),
This means that \(\mathbf{x}_{n} := (x_{n},\lambda_{n}) \in M(\xi _{n},\epsilon_{n})\), and hence \(\{\mathbf{x}_{n}\}\) is a Cauchy sequence \(X \times\Lambda\). The completeness of \(X\times\Lambda\) implies that \(\{\mathbf{x}_{n}\}\) converges to a point \(\bar{\mathbf{x}} := (\bar{x},\bar{\lambda})\). Since \(K_{1}\) is closed at \((\bar{x},\bar {\lambda})\) and \(x_{n} \in K_{1}(x_{n},\lambda_{n})\), one has \(\bar{x} \in K_{1}(\bar{x},\bar{\lambda})\). Using the same argument as for Theorem 3.2, one sees that \(\bar{\mathbf{x}}\) solves (OPLEC). Next, we will show that (OPLEC) has a unique solution. If (OPLEC) has two distinct solutions \((\bar{x}_{1},\bar{\lambda}_{1})\) and \((\bar {x}_{2},\bar{\lambda}_{2})\), they must belong to \(M(\xi,\epsilon)\) for all \(\xi,\epsilon> 0\). This yields the contradiction that
This completes the proof. □
For the wellposedness of (OPLEC) in terms of measures of noncompactness we have the following result. Let us consider only the case of the Hausdorff measure η; we get the corresponding results for the case μ and ι.
Theorem 4.3

(i)
If (OPLEC) is wellposed at \(\bar{\lambda}\), then \(\eta(M(\xi ,\varepsilon)) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\).

(ii)
Conversely, suppose that \(\eta(M(\xi, \varepsilon)) \downarrow 0\) as \((\xi, \varepsilon) \downarrow(0, 0)\), and the following conditions hold:

(a)
X and Λ are complete;

(b)
\(K_{1}\) is closed and \(K_{2}\) is lsc on \(X \times\Lambda\);

(c)
\(Z_{f}\) is lsc on \(Z_{1,f}(\bar{\lambda}) \times\bar{\lambda}\);

(d)
\(f_{1}\) is upper bupper level closed in \(K_{1}(X,\Lambda) \times K_{2}(X,\Lambda) \times\Lambda\), for all \(b <0\);

(e)
\(f_{2}\) is strongly upper bupper level closed in \(K_{1}(X,\Lambda ) \times K_{2}(X,\Lambda) \times\Lambda\), for all \(b <0\);

(f)
g is lsc in \(X \times\Lambda\).

(a)
Then (OPLEC) is wellposed at \(\bar{\lambda}\).
Proof
(1) Suppose that (OPLEC) is wellposed at \(\bar{\lambda}\). For all \(\xi,\varepsilon> 0\), the solution set \(\mathbf{S}_{g}(\bar{\lambda })\) of (OPLEC) satisfies obviously the containment \(\mathbf{S}_{g}(\bar {\lambda})\subseteq M(\xi,\varepsilon)\). Consequently, we have
Any sequence \(\{\mathbf{x}_{n}\}\) in \(\mathbf{S}_{g}(\bar{\lambda})\) is an approximating sequence of (OPLEC) and has a subsequence convergent to some point of \(\mathbf{S}_{g}(\bar{\lambda})\). So, \(\mathbf{S}_{g}(\bar{\lambda})\) is compact. Thus, there exist \(y_{1}, y_{2}, \ldots, y_{n} \in Y:= X \times\Lambda\) such that
Next, we claim that
Let \(y \in M(\xi,\varepsilon)\) and suppose that \(y \notin\bigcup_{k=1}^{n} B(y_{k},\varepsilon+H(M(\xi,\varepsilon),\mathbf {S}_{g}(\bar{\lambda})))\). This implies that
which gives
Since \(\mathbf{S}_{g}(\bar{\lambda})\) is compact, there is \(b \in \mathbf{S}_{g}(\bar{\lambda})\) such that
which leads to a contradiction with (4.2), and hence (4.3) holds. Consequently, we have
Hence, we obtain \(H(M(\xi,\varepsilon),\mathbf{S}_{g}(\bar{\lambda })) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\). Indeed, suppose that there exist a real number \(\rho> 0\), a sequence \((\xi _{n}, \varepsilon_{n}) \downarrow(0, 0)\) and \(\mathbf{x}_{n} \in M(\xi_{n}, \varepsilon_{n})\) such that
Being an approximating sequence for (OPLEC), \(\{\mathbf{x}_{n}\}\) has a subsequence convergent to some point of \(\mathbf{S}_{g}(\bar{\lambda})\), by which one arrives at a contradiction with \(\rho> 0\). We conclude that \(\eta(M(\xi,\varepsilon)) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\).
(2) Assume that \(\eta(M(\xi, \varepsilon)) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\). We first check that \(M(\xi, \varepsilon)\) is closed for all \(\xi, \varepsilon>0\). Let \(\mathbf{m}_{n} := (m_{n}, \lambda'_{n}) \in M(\xi, \varepsilon)\) with \(\mathbf{m}_{n} \rightarrow\mathbf{m} := (m, \lambda')\). Hence,
Since \(K_{1}\) is closed at \((m, \lambda')\), \(m \in K_{1}(m, \lambda')\). By the semicontinuity of g at \((m, \lambda')\), we have
Furthermore, we claim that
Indeed, if there exist \(z \in K_{2}(m, \lambda')\) and \(y \in Z_{f}(m,\lambda')\) such that
then there is \(z_{n} \in K_{2}(m_{n}, \lambda'_{n})\), \(y_{n} \in Z_{f}(m_{n},\lambda '_{n})\) such that
as \(K_{2}\) is lsc at \((m, \lambda')\). By (c) and (d), there is \(n_{0} \in\mathbb{N}\) such that
for all \(n \geq n_{0}\), which is a contradiction. Hence, \(M(\xi, \varepsilon)\) is closed. Note further that \(\mathbf {S}_{g}(\bar{\lambda}) =\bigcap_{\xi>0,\varepsilon>0} M(\xi, \varepsilon)\) and \(\eta(M(\xi, \varepsilon)) \downarrow 0\) as \((\xi, \varepsilon) \downarrow(0, 0)\). Therefore, by the earliermentioned properties of η, \(\mathbf{S}_{g}(\bar{\lambda })\) is compact and \(H(M(\xi,\varepsilon),\mathbf{S}_{g}(\bar{\lambda})) \downarrow0\) as \((\xi, \varepsilon) \downarrow(0, 0)\).
Finally, we prove that (OPLEC) is wellposed at \(\bar{\lambda}\). Let \(\{\mathbf{x}_{n}\} := \{(x_{n}, \lambda_{n})\}\) be an approximating sequence, i.e., there exists \((\xi_{n}, \varepsilon_{n}) \downarrow (0, 0)\) such that
Consequently, \((x_{n}, \lambda_{n}) \in M(\xi_{n}, \varepsilon_{n})\). So,
By the compactness of \(\mathbf{S}_{g}(\bar{\lambda})\), there is a subsequence of \(\{\mathbf{x}_{n}\}\) convergent to some point of \(\mathbf{S}_{g}(\bar{\lambda})\). Thus, (OPLEC) is wellposed. □
5 Conclusions
In this paper, we obtain the wellposedness for lexicographic vector equilibrium problems and optimization problems with lexicographic equilibrium constraints in metric spaces. Sufficient conditions for a family of such problems to be (uniquely) wellposed at the reference point are established. Numerous examples are provided to explain that all the assumptions we impose are very relaxed and cannot be dropped. The results presented in this paper extend and improve some known results.
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Acknowledgements
This work was partially supported by the Thailand Research Fund, Grant No. PHD/0035/2553 and Naresuan University. The authors would like to thank the referees for their remarks and suggestions, which helped to improve the paper.
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Wangkeeree, R., Bantaojai, T. & Yimmuang, P. Wellposedness for lexicographic vector quasiequilibrium problems with lexicographic equilibrium constraints. J Inequal Appl 2015, 163 (2015). https://doi.org/10.1186/s1366001506695
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DOI: https://doi.org/10.1186/s1366001506695