1 Introduction

He and Liu^[1] and He et al.^[2] first introduced and studied the inverse variational inequalities. Yang^[3] considered the dynamic power price problem and formulated such problem as an inverse variational inequality. He et al.^[4] studied a normative control problem and characterized such problem as an inverse variational inequality. They also provided a proximal point based algorithm for solving the inverse variational inequality and showed an application to a bipartite market equilibrium problem. Inverse variational inequalities have a wide range of applications in different fields, such as electrical power network management, traffic network equilibrium problems, economic equilibrium problems, normative flow control problems, and equilibrium state control problems. Therefore, various theoretical results, numerical algorithms, and applications for inverse variational inequalities have been studied by many authors (see, for example, Refs.[4]-[15] and the references therein).

Recently, in order to study the road pricing problem, Aussel et al.^[12] introduced an inverse quasi-variational inequality problem. They considered the gap functions and error bounds for an inverse quasi-variational inequality problem which can be considered as a mixture of the quasi-variational inequalities and the inverse variational inequalities. They showed global error bound results for the inverse quasi-variational inequality by using the residual gap function. They also gave error bounds for the inverse quasi-variational inequality with the residual gap function, the regularized gap function, and the $D$-gap function. We note that in the results of Ref.^[12], the assumption that the solution set of the inverse quasi-variational inequality is nonempty plays a key role. As pointed out by Aussel et al.^[12], the discipline of the inverse (quasi) variational inequality was still not fully investigated and much work was expected to be done. We note that the inverse (quasi) variational inequality problem is a special case of the general (quasi) variational inequality problems. However, it is worthwhile to study these problems independently since, as we shall see, they enjoy stronger results than those currently known for the general (quasi) variational inequality problems. Therefore, it is important to explore some new conditions to ensure the existence of solutions to the inverse (quasi) variational inequality problems. The first purpose of this paper is to show two new existence theorems of solutions to inverse variational inequality and inverse quasi-variational inequality problems involving the information on the inverse mapping $f^{-1}$.

Moreover, we know that the stability analysis of the solution mappings to variational inequalities and equilibrium problems is an important topic in the optimization theory. Recently, semicontinuity, especially lower semicontinuity, of solution mappings to parametric vector optimization problems and parametric vector equilibrium problems has been studied by several authors (see, for example, Refs.[16]-[29] and the references therein). However, to the best of the authors' knowledge, upper semicontinuity and lower semicontinuity of solution mappings to parametric inverse variational inequality problems have not been explored so far. Therefore, it is interesting to discuss upper semicontinuity and lower semicontinuity of solution mappings to parametric inverse variational inequality problems. The second purpose of this paper is to establish upper semicontinuity and lower semicontinuity of solution mapping and approximate solution mapping to parametric inverse variational inequality problems concerned with the information on $f^{-1}$.

The rest of this paper is organized as follows. In Section 2, some necessary notations and lemmas are given. In Section 3, we establish two new existence theorems of solutions for inverse variational and quasi-variational inequality problems using the Fan-Knaster-Kuratowski-Mazurkiewicz (KKM) theorem and the Kakutani-Fan-Glicksberg fixed point theorem, respectively. We also give some sufficient conditions to guarantee lower semicontinuity of ${f^{-1}}(\cdot)$. In Section 4, we discuss upper semicontinuity and lower semicontinuity of solution mapping and approximate solution mapping to parametric inverse variational inequality problems. Finally, we give an application to a road pricing problem in which the traffic management authority attempts to control the environment impact within a certain range in Section 5.

2 Preliminaries

Throughout this paper, let $W$ and $\Lambda$ be two normed vector spaces. Let $\Omega $ be a nonempty subset of ${{\mathbb{R}}^n}$ and $f$ be a mapping from ${{\mathbb{R}}^n}$ into ${{\mathbb{R}}^n}$. Assume that

For a given set $A \subseteq {{\mathbb{R}}^n}$, ${\rm{co}}(A)$ denotes the convex hull of $A$.

Let $K$ be a nonempty subset of ${{\mathbb{R}}^n}$ and $\Phi :{{\mathbb{R}}^n} \to {2^K}$ be a set-valued mapping. The inverse quasi-variational inequality is to find a vector ${x^*} \in {{\mathbb{R}}^n}$ such that

which was introduced and studied by Aussel et al.^[12], where T represents transpose.

When $\Phi (x)=\Omega$ for all $x\in {\mathbb{R}}^n$, the inverse quasi-variational inequality reduces to the inverse variational inequality: find a vector ${x^*} \in {{\mathbb{R}}^n}$ such that

which was considered by several authors^{[1, 3-10]}.

Let $\Omega :{{\mathbb{R}}^n} \to {2^{{\mathbb{R}}^n} }$ be a set-valued mapping and $f:{{\mathbb{R}}^n} \times W \to {{\mathbb{R}}^n}$ be a mapping. For $({u, \lambda }) \in W \times \Lambda $, we consider the following parametric inverse variational inequality: find a vector ${x^*} \in {{\mathbb{R}}^n}$ such that

For $({u, \lambda }) \in W \times \Lambda $, let ${S}({u, \lambda })$ denote the set of all solutions to the parametric inverse variational inequality, i.e.,

For $({\varepsilon, u, \lambda }) \in {\mathbb{R}}_+ ^1 \times W \times \Lambda $, let ${E}({\varepsilon, u, \lambda })$ denote the set of all approximate solutions to the parametric inverse variational inequality, i.e.,

Definition 1^[30] Let $K$ be a nonempty subset of a topological vector space $X$. A set-valued mapping $F:K \to {2^X}$ is said to be a KKM mapping$, $ if for any finite subset $\left\{ {{y_1}, {y_2}, \cdots, {y_n}} \right\}$ of $K, $ one has

Let $X$ be a real linear space. For $x, y \in X$, let us denote by $\left[{x, y} \right]: =\{z = ({1-t}) x + ty:t \in [0, 1]\}$ the closed line segment with the endpoints $x$ and $y$.

Definition 2^[31] Let $X$ and $Y$ be two real linear spaces. We say that a mapping $A:D \subseteq X \to Y$ is of type ql if for every $x, y \in D$ and every $z \in \left[{x, y} \right] \cap D, $ one has $A (z) \in \left[{A (x), A (y)} \right]$.

Remark 1 If a mapping $A$ is linear, then $A$ is of type ql (see Proposition 3.3 of Ref.[31]).

Definition 3 Let $D$ be a nonempty convex subset of ${\mathbb{R}}^n$. A mapping $f:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ is said to be

(ⅰ)${\mathbb{R}}_+ ^n$-convex on $D$ if for any ${x_1}, {x_2} \in D$ and for any $t \in \left[{0, 1} \right], $ one has

(ⅱ) natural quasi ${\mathbb{R}}_+ ^n$-convex on $D$^[32] if for any ${x_1}, {x_2} \in D$ and for any $t \in \left[{0, 1} \right], $ there exists $\lambda \in \left[{0, 1} \right]$ such that

(ⅲ) strictly natural quasi ${\mathbb{R}}_+ ^n$-convex on $D$ if for any ${x_1}, {x_2} \in D$ with ${x_1} \ne {x_2}$ and for any $t \in ({0, 1}), $ there exists $\lambda\in \left[{0, 1}\right]$ such that

Remark 2 If $f$ is ${\mathbb{R}}_+ ^n$-convex, then $f$ is natural quasi ${\mathbb{R}}_+ ^n$-convex (see Lemma 2.1 of Ref.[32]). The class of natural quasi $R_+^n$-convex mappings is strictly larger than that of ${\mathbb{R}}_+ ^n$-convex mappings (see Remark 2.1 of Ref.[32]). Moreover, we can see that if $f$ is of type ql, then $f$ is natural quasi ${\mathbb{R}}_+ ^n$-convex.

Definition 4 Let $K$ be a nonempty subset of ${\mathbb{R}}^n$. A mapping $f:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ is said to be monotone on $K$ if

Definition 5 Let $T$ and $T_1$ be two topological spaces. A set-valued mapping $G:T \to {2^{{T_1}}}$ is said to be

(ⅰ) upper semicontinuous ($u.s.c.$) at ${u_0} \in T$ if for any neighborhood $V$ of $G ({{u_0}}), $ there exists a neighborhood $U ({{u_0}})$ of ${u_0}$ such that for every $u \in U ({{u_0}}), $ $G (u) \subseteq V; $

(ⅱ) lower semicontinuous ($l.s.c.$) at ${u_0} \in T$ if for any $x \in G ({{u_0}})$ and any neighborhood $V$ of $x, $ there exists a neighborhood $U ({{u_0}})$ of ${u_0}$ such that for every $u \in U ({{u_0}}), $ $G (u) \cap V \ne \varnothing$.

We say that $G$ is u.s.c. and l.s.c. on $T, $ if it is u.s.c. and l.s.c. at each $u \in T, $ respectively. $G$ is said to be continuous on $T, $ if it is both u.s.c. and l.s.c. on $T$.

Lemma 1^[33] Let $G:W\to 2^{\Lambda}\backslash\varnothing$. Then$, $ $G$ is l.s.c.at ${u_0} \in W $ if and only if for any sequence $\left\{ {{u_n}} \right\} \subseteq W $ with ${u_n} \to {u_0}$ and for any ${x_0} \in G ({{u_0}}), $ there exists ${x_n} \in G ({{u_n}})$ such that ${x_n} \to {x_0}$.

Lemma 2^[34] Let $G:W\to 2^{\Lambda}\backslash\varnothing$. If $G ({{u_0}})$ is compact$, $ then $G$ is u.s.c.at ${u_0} \in W $ if and only if for any sequence $\left\{ {{u_n}} \right\} \subseteq W $ with ${u_n} \to {u_0}$ and for any ${x_n} \in G ({{u_n}}), $ there exists ${x_0} \in G ({{u_0}})$ and a subsequence $\left\{ {{x_{{n_k}}}} \right\}$ of $\left\{ {{x_n}} \right\}$ such that ${x_{{n_k}}} \to {x_0}$.

Let $T_1$, $T_2$, and $T_3$ be three topological spaces, and let $F:{T_1} \to {2^{{T_2}}}$ and $G:{T_2} \to {2^{{T_3}}}$ be two set-valued mappings. The set-valued mapping $M:{T_1} \to {2^{{T_3}}}$ can be defined as follows:

Lemma 3^[33] (ⅰ)If $F$ is u.s.c.on $T_1$ and $G$ is u.s.c.on $T_2, $ then $M$ is u.s.c.on $T_1$.

(ⅱ) If $F$ is l.s.c.on $T_1$ and $G$ is l.s.c.on $T_2, $ then $M$ is l.s.c.on $T_1$.

Lemma 4^[30] Let $K$ be a nonempty subset of a Hausdorff topological vector space $X, $ and let $F:K \to {2^X}\backslash\varnothing$ be a KKM mapping with closed values. If there exists ${x_0} \in K$ such that $F ({{x_0}})$ is compact$, $ then $\bigcap\limits_{y \in K} {F (y)} \ne \varnothing $.

Lemma 5^[35-36] Let $K$ be a nonempty compact convex subset of a locally convex Hausdorff topological vector space $X, $ and let $F:K \to {2^K}\backslash\varnothing$ be an u.s.c. set-valued mapping with nonempty compact convex values. Then$, $ there exists ${x_0} \in K$ such that ${x_0} \in F ({x_0})$.

Lemma 6 Let $g:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ be a mapping and $K$ be a nonempty subset of $g ({{{\mathbb{R}}^n}})$. Assume that ${g^{-1}}(K)$ is compact and $g$ is continuous on ${{\mathbb{R}}^n}$. Then$, $ ${g^{-1}}(\cdot)$ is u.s.c. on $K$.

Proof Suppose, on the contrary, that ${g^{-1}}(\cdot)$ is not u.s.c. at ${y_0} \in K$. Then, there exists a neighborhood ${W_0}$ of ${g^{-1}}({y_0})$ and a sequence $\left\{ {{y_n}} \right\} \subseteq K$ with ${y_n} \to {y_0}$ such that ${g^{-1}}({{y_n}}) \not\subset {W_0}$ for all $n \in \mathbb{N}.$ Then, for each $n \in \mathbb{N}$, there exists

(1)

such that

(2)

It follows from (1) that ${x_n} \in {g^{-1}}({{y_n}}) \subseteq {g^{-1}}(K)$. Since ${g^{-1}}(K)$ is compact, without loss of generality, we can assume that ${x_n} \to {x_0} \in {g^{-1}}(K)$, and so $g ({{x_n}}) \to g ({{x_0}})$. Noting that ${y_n} = g ({{x_n}}) \to {y_0}$, we have ${y_0} = g ({{x_0}})$, that is, ${x_0} \in {g^{-1}}({{y_0}})$. Then, ${x_n} \to {x_0} \in {g^{-1}}({{y_0}}) \subseteq {W_0}$, which contradicts (2). This completes the proof.

Lemma 7 Assume that $\Omega$ is convex$, $ $f$ is monotone on ${f^{-1}}(\Omega), $ and ${f^{-1}}(\cdot)$ is l.s.c.on $\Omega $. Then$, $ the following two statements are equivalent$:$

(ⅰ) ${x^*}$ is a solution to inverse variational inequalities$, $ that is$, $

(ⅱ) ${x^*}$ is a solution to the following associated inverse variational inequalities$:$ find ${x^*} \in {{\mathbb{R}}^n}$ such that

Proof $({\rm{i}}) \Rightarrow ({\rm{ii}})$. Let ${x^*}$ be a solution to the inverse variational inequalities. Then,

For each $y \in \Omega$ and for each $x \in {f^{-1}}(y)$, we have $y = f (x)$. Since $f$ is monotone on ${f^{-1}}(\Omega)$, one has ${({f (x)-f ({{x^*}})})^{\rm T}}({x-{x^*}}) \ge 0$ and so ${({y-f ({{x^*}})})^{\rm T}}({x-{x^*}}) \ge 0.$ This shows that

$({\rm{ii}}) \Rightarrow ({\rm{i}})$. Let ${x^*}$ be a solution to the associated inverse variational inequalities. Then,

(3)

Let $\left\{ {{t_n}} \right\} \subseteq ({0, 1})$ with ${t_n} \to 0$. For each $y \in \Omega$, let ${z_n} = ({1-{t_n}}) f ({{x^*}}) + {t_n}y$ for $n \in \mathbb{N}$. Then, ${z_n} \in \Omega $ and ${z_n} \to f ({{x^*}})$. Since ${x^*} \in {f^{-1}}({f ({{x^*}})})$ and ${f^{-1}}(\cdot)$ is l.s.c.at ${f ({{x^*}})}$, by Lemma 1, there exists ${x_n} \in {f^{-1}}({{z_n}})$ such that ${x_n} \to {x^*}$. It follows from (3) that ${({{z_n}-f ({{x^*}})})^{\rm T}}{x_n} \ge 0.$ Noting that ${z_n} = ({1-{t_n}}) f ({{x^*}}) + {t_n}y$ and $\left\{ {{t_n}} \right\} \subseteq ({0, 1})$, we have ${({y-f ({{x^*}})})^{\rm T}}{x_n} \ge 0.$ It follows from ${x_n} \to {x^*}$ that ${({y-f ({{x^*}})})^{\rm T}}{x^*} \ge 0.$ This completes the proof.

3 Existence theorems

In this section, we establish two existence theorems for inverse variational and quasivariational inequality problems, respectively.

Theorem 1 Assume that ${f^{-1}}(\Omega)$ is bounded convex$, $ $\Omega$ is closed$, $ $\Omega \subseteq f ({{\mathbb{R}}^n}), $ and ${f^{-1}}(\Omega) \subseteq {\mathbb{R}}_+ ^n$. If $f$ is continuous on ${{\mathbb{R}}^n}$ and natural quasi ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}(\Omega), $ then the inverse variational inequality has a solution.

Proof Let $D = {f^{-1}}(\Omega) \subseteq {\mathbb{R}}_+ ^n$. Noting that $\Omega \subseteq f ({R ^n})$, it is easy to see that the inverse variational inequality is equivalent to the following problem: find ${x^*} \in D$ such that ${({f (x)-f ({{x^*}})})^{\rm T}}{x^*} \ge 0$ for all $x \in D.$ For each $y \in D$, define

It is clear that $y \in G (y)$. Then$, $ $G (y)$ is nonempty. Since $f$ is continuous, it is easy to see that $G (y)$ is closed.

We claim that $G:D \to {2^D}$ is a KKM mapping. In fact, by contradiction, suppose that there exist $\left\{ {{z_1}, {z_2}, \cdots, {z_k}} \right\} \subseteq D$ and $z \in {\rm{co}}({\left\{ {{z_1}, {z_2}, \cdots, {z_k}} \right\}})$ such that $z \notin G ({{z_i}})$ for $i = 1, 2, \cdots, k.$ Then,

(4)

Since $z \in {\rm{co}}({\left\{ {{z_1}, {z_2}, \cdots, {z_k}} \right\}})$, there exist ${\lambda _i} \ge 0, i = 1, 2, \cdots, k$ with $\sum\limits_{i = 1}^k {{\lambda _i}} = 1$ such that $z = \sum\limits_{i = 1}^k {{\lambda _i}{z_i}} $. Since $f$ is natural quasi ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}(\Omega)$, there exist ${\alpha _i} \ge 0, i = 1, 2, \cdots, k$ with $\sum\limits_{i = 1}^k {{\alpha _i}} = 1$ such that

By (4), we have

(5)

Noting that $z \in D \subseteq {\mathbb{R}}_+ ^n$ and $z = \sum\limits_{i = 1}^k {{\lambda _i}{z_i}} $, we have

which contradicts (5). It follows from Lemma 4 that $\bigcap\limits_{y \in D} {G (y)} \ne \varnothing$. Let ${x^*} \in \bigcap\limits_{y \in D} {G (y)} $. Then, for any $y \in D$, one has ${({f (y)-f ({{x^*}})})^{\rm T}}{x^*} \ge 0.$ This implies that ${x^*}$ is a solution to the inverse variational inequality. This completes the proof.

Remark 3 It is worth mentioning that the assumptions of Theorem 1 involve the information on the inverse mapping $f^{-1}$, which are different from the previous ones. Moreover, the classical fixed point method is no longer valid for Theorem 1. In fact, it is well known that the inverse variational inequality is equivalent to the following problem: find ${x^*} \in D=f^{-1}(\Omega)$ such that

(6)

where $\Omega$ is convex closed, and $P_\Omega$ is a projection from ${\mathbb{R}}^n$ to $\Omega$. Obviously, (6) can be rewritten as $x^*=F (x^*)$, where

It is easy to see that $F$ is a continuous mapping from $D$ to ${\mathbb{R}}^n$. Since $F (D)\not\subset D$, the Brouwer fixed point is not valid to ensure the existence of the fixed point of $F$. Moreover, we can write (6) as $x^*\in A (x^*)$, where

It is clear that $A$ is a set-valued mapping from $D$ onto $2^D$. For any given $x\in D$, since $f^{-1}(x)$ is not convex in general, it is not easy to show the convexity of $A (x)$. Thus, the Kakutani fixed point theorem is not valid to guarantee the existence of the fixed point of $A$.

Remark 4 (ⅰ) It is easy to see that if $\Omega$ is convex and $f$ is linear, then ${f^{-1}}(\Omega)$ is convex.

(ⅱ) If $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a linear continuous injective and surjective mapping, then the open mapping theorem shows that $f^{-1}:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is linear continuous. Moreover, if $\Omega$ is bounded, then ${f^{-1}}(\Omega)$ is bounded.

Now, we give an example to illustrate Theorem 1.

Example 1 Let $f:{{\mathbb{R}}_+ ^2} \to {{\mathbb{R}}^2}$ be defined as follows:

Then, it is easy to check that $f$ is continuous on ${{\mathbb{R}}_+ ^2}$ and ${\mathbb{R}}_+ ^2$-convex on ${{\mathbb{R}}_+ ^2}$. Let $\Omega=\left[{1, 2} \right] \times \left[{1, 3} \right] \cap f ({{{{\mathbb{R}}_+ ^2}}})$ and

Clearly, ${f^{-1}}(\Omega) \subseteq \Psi \subseteq {\mathbb{R}}_+ ^2$, ${f^{-1}}(\Omega)$ is bounded, and $\Psi $ is nonempty compact convex.

We first show that $\Omega$ is closed. In fact, let $\left\{ {{z_n}} \right\} \subseteq \Omega $ with ${z_n} \to {z_0}$. Obviously, ${z_0} \in \left[{1, 2} \right] \times \left[{1, 3} \right]$. Since $\left\{ {{z_n}} \right\} \subseteq \Omega $, there exists ${x_n} \in {f^{-1}}(\Omega)$ such that ${z_n} = f ({{x_n}})$. Noting that ${x_n} \in {f^{-1}}(\Omega) \subseteq \Psi $ and $\Psi $ is compact, without loss of generality, we can assume that ${x_n} \to {x_0} \in \Psi $. Since $f ({{x_n}}) \to f ({{x_0}})$, ${z_n} = f ({{x_n}})$, and ${z_n} \to {z_0}$, we know that ${z_0} = f ({{x_0}})$. Thus, ${z_0} \in f ({{{\mathbb{R}}_+ ^2}})$ and so ${z_0} \in \Omega$.

Next, we claim that ${f^{-1}}(\Omega)$ is convex. In fact, let ${x_1}, {x_2} \in {f^{-1}}(\Omega)$ and $t \in \left[{0, 1} \right]$. Noting that ${f^{-1}}(\Omega) \subseteq \Psi $ and $\Psi $ is convex, we know that $t{x_1} + ({1-t}){x_2} \in \Psi $ and so

It is clear that $f ({t{x_1} + ({1-t}){x_2}}) \in f ({{{\mathbb{R}}_+ ^2}})$ and so $f ({t{x_1} + ({1-t}){x_2}}) \in \Omega $, which implies that $t{x_1} + ({1-t}){x_2} \in {f^{-1}}(\Omega)$. This shows that all the conditions of Theorem 1 can be satisfied.

Theorem 2 Let ${f^{-1}}(K)$ be bounded convex and $K \subseteq f ({{{\mathbb{R}}^n}})$ be compact. Assume that

(ⅰ) $f$ is continuous on ${{\mathbb{R}}^n}$ and natural quasi ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}(K); $

(ⅱ) $f$ is monotone on ${f^{-1}}(K), $ and ${f^{-1}}(\cdot)$ is l.s.c. on $K; $

(ⅲ) $\Phi $ is continuous on ${{\mathbb{R}}^n}$ and for each $u \in {{\mathbb{R}}^n}, $ $\Phi (u)$ is convex closed$, $ and ${f^{-1}}({\Phi (u)})$ is bounded and convex with ${f^{-1}}({\Phi (u)}) \subseteq {\mathbb{R}}_+ ^n$.

Then$, $ the inverse quasi-variational inequality has a solution.

Proof For each $v \in {f^{-1}}(K)$, define

It suffices to prove that the set-valued mapping $S:{f^{-1}}(K) \to {2^{{f^{-1}}(K)}}$ has a fixed point. By Theorem 1, we know that $S (v)$ is nonempty. Since $f$ is continuous and $\Phi (v)$ is closed, it is easy to see that $S (v)$ is closed.

Now, we claim that $S (v)$ is convex. In fact, let

Then, it follows from Lemma 7 that $S (v) = Q (v)$. It suffices to prove that $Q (v)$ is convex. For any ${x_1}, {x_2} \in Q (v)$ and for any $\lambda \in \left[{0, 1} \right]$, it is easy to obtain

Since ${x_1}, {x_2} \in Q (v)$, for each $ y \in \Phi (v)$ and for each $ \alpha \in {f^{-1}}(y)$, one has

(7)

Since $f$ is natural quasi ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}(K)$, there exists $\eta \in \left[{0, 1} \right]$ such that

(8)

It follows from (7) that

(9)

By (8) and (9), we have

and so $\lambda {x_1} + ({1-\lambda }){x_2} \in Q (v)$.

Next, we show that $S$ is u.s.c. on ${f^{-1}}(K) $. Suppose, on the contrary, that $S$ is not u.s.c. at ${v_0} \in {f^{-1}}(K)$. Then, there exists a neighborhood ${W_0}$ of $S ({{v_0}})$, for any neighborhood $U ({{v _0}})$ of $v_0$, there exists $v' \in U ({{v_0}})$ such that $S ({v'})\not \subset {W_0}$. Hence, there exists a sequence $\left\{ {{v_n}} \right\}$ with ${v_n} \to {v_0}$ such that $S ({{v_n}})\not \subset {W_0}$ for $n=1, 2, \cdots.$ This shows that there exist

(10)

such that

(11)

It follows from Lemma 6 that ${f^{-1}}(\cdot)$ is u.s.c. on $K$. By Lemma 3 and the assumptions, we can see that the set-valued mapping ${f^{-1}}({\Phi (\cdot)})$ is u.s.c. on ${f^{-1}}(K)$ with compact values. Noting that ${x_n} \in {f^{-1}}({\Phi ({{v_n}})})$ and Lemma 2, there exists ${x_0} \in {f^{-1}}({\Phi ({{v_0}})})$ and a subsequence $\left\{ {{x_{{n_k}}}} \right\}$ of $\left\{ {{x_n}} \right\}$ such that ${x_{{n_k}}} \to {x_0}$. Without loss of generality, we can assume that ${x_n} \to {x_0}$. Suppose that ${x_0} \notin S ({{v_0}})$. Then, there exists ${y_0} \in \Phi ({{v_0}})$ such that

(12)

Since $\Phi (\cdot)$ is l.s.c.at ${v _0}$ and ${y_0} \in \Phi ({{v_0}})$, by Lemma 1, there exists ${y_n} \in \Phi ({{v_n}})$ such that ${y_n} \to {y_0}$. It follows from (10) that ${({{y_n}-f ({{x_n}})})^{\rm T}}{x_n} \ge 0.$ Since ${y_n} \to {y_0}$ and ${x_n} \to {x_0}$, we have ${({{y_0}-f ({{x_0}})})^{\rm T}}{x_0} \ge 0, $ which contradicts (12). Hence, ${x_0} \in S ({{v_0}})$. We can see that ${x_n} \to {x_0}\in {W_0}$, which contradicts (11). Therefore, for each $v \in {f^{-1}}(K)$, $S (v)$ is nonempty convex closed, and $S$ is u.s.c. on ${f^{-1}}(K) $. By Lemma 5, $S$ has a fixed point. This completes the proof.

Now, we give an example to illustrate Theorem 2.

Example 2 Let $f:{{\mathbb{R}}_+ ^2} \to {{\mathbb{R}}^2}$ be defined as follows:

We can see that $f$ is continuous on ${{\mathbb{R}}_+ ^2}$ and ${\mathbb{R}}_+ ^2$-convex on ${{\mathbb{R}}_+ ^2}$, $f$ is monotone on ${{\mathbb{R}}_+ ^2}$, and ${f^{-1}}(\cdot)$ is l.s.c. on $f ({{\mathbb{R}}_+ ^2})$. Let $K = \left[{0, 3} \right] \times \left[{0, 3} \right] \cap f ({{\mathbb{R}}_+ ^2})$ and

Then, it is clear that ${f^{-1}}(K) \subseteq \Delta $, ${f^{-1}}(K)$ is bounded, and $\Delta $ is nonempty compact convex. Similar to Example 1, we can prove that $K$ is closed, and ${f^{-1}}(K)$ is convex.

Let $\Phi :{{\mathbb{R}}^2} \to {2^K}$ be defined as follows:

Then, it is easy to see that $\Phi $ is continuous on ${{\mathbb{R}}^2}$. Moreover, for each $u \in {{\mathbb{R}}^2}$, $\Phi (u)$ is closed, and ${f^{-1}}({\Phi (u)})$ is bounded and convex with $\Phi (u) \subseteq f ({{\mathbb{R}}_+ ^2})$. This shows that all the conditions of Theorem 2 can be satisfied.

We note that lower semicontinuity of ${f^{-1}}(\cdot)$ plays a key role in Theorem 2. Thus, it is important to give some sufficient conditions to guarantee lower semicontinuity of ${f^{-1}}(\cdot)$. To this end, we give the following propositions.

Proposition 1 Let $X$ be a normed vector space. If $f:X \to R$ is linear$, $ then ${f^{-1}}(\cdot)$ is l.s.c.on $R$.

Proof Suppose, on the contrary, that ${f^{-1}}(\cdot)$ is not l.s.c.at ${y_0} \in R$. Then, there exists ${x_0} \in {f^{-1}}({{y_0}})$, a neighborhood ${W_0}$ of $0 \in X$, and a sequence $\left\{ {{y_n}} \right\}$ with ${y_n} \to {y_0}$ such that

(13)

By (13), it is clear that ${y_n} \ne {y_0}$ for any $n \in \mathbb{N}$. Since ${y_n} \in R$ and ${y_n} \to {y_0}$, without loss of generality, we can assume that ${y_n} < {y_{n + 1}} < {y_0}$ for any $n \in \mathbb{N}$. Let ${x_1} \in {f^{-1}}({{y_1}})$ and so ${y_1} = f ({{x_1}})$. Since ${y_1} < {y_{n + 1}} < {y_0}$ for any $n \in \mathbb{N}$ and ${y_n} \to {y_0}$, we can find ${y_{{n_0}}}$ and ${t_0} \in ({0, 1})$ small enough such that

Then,

and so ${t_0}{x_1} + ({1-{t_0}}){x_0} \in {f^{-1}}({{y_{{n_0}}}})$. We can obtain

which contradicts (13). This completes the proof.

Proposition 2 Let $f:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ be continuous on ${{\mathbb{R}}^n}$ and ${y_0} \in f ({{{\mathbb{R}}^n}})$. Assume that for any $x \in {f^{-1}}({{y_0}}), $ there exists a constant $\delta > 0$ such that ${y_0} \in {\mathop{\rm int}} ({f ({x + \delta B})})$ and $f$ is injective on $x + \delta B, $ where $B$ denotes the closed unit ball of ${\mathbb{R}}^n $. Then$, $ ${f^{-1}}(\cdot)$ is l.s.c.at ${y_0}$.

Proof For any $x \in {f^{-1}}({{y_0}})$, by the assumptions, there exists $\delta > 0$ such that $f$ is injective on $x + \delta B$ and ${y_0} \in {\mathop{\rm int}} ({f ({x + \delta B})})$. Let

It is clear that for any $y \in f ({x + \delta B})$, $G (y) \ne \varnothing $. We claim that for any $y \in f ({x + \delta B})$, $G (y)$ is singleton. In fact, let ${\alpha _1}, {\alpha _2} \in G (y)$ with ${\alpha _1} \ne {\alpha _2}$. Then, ${\alpha _1}, {\alpha _2} \in x + \delta B$, and $f ({{\alpha _1}}) = f ({{\alpha _2}}) = y$. Since $f$ is injective on $x + \delta B$, we get a contradiction.

Next, we show that $G:f ({x + \delta B}) \to x + \delta B$ is continuous at ${y_0} \in {\mathop{\rm int}} ({f ({x + \delta B})})$. In fact, if not, then there exists a neighborhood ${W_0}$ of $G ({{y_0}})$ and a sequence $\left\{ {{y_n}} \right\} \subseteq f ({x + \delta B})$ with ${y_n} \to {y_0}$ such that

(14)

Let ${x_n} = G ({{y_n}})$. Then, ${x_n} \in x + \delta B$. Since $B$ is the closed unit ball of ${\mathbb{R}}^n $, it is easy to see that $x + \delta B$ is compact. Without loss of generality, we can assume that ${x_n} \to {x_0} \in x + \delta B$. It follows from the continuity of $f$ that ${y_n} = f ({{x_n}}) \to f ({{x_0}})$. Noting that ${y_n} \to {y_0}$, we know that $f ({{x_0}}) = {y_0}$ and so ${x_0} = G ({{y_0}})$. Moreover, we have

which contradicts (14).

For any neighborhood ${U}$ of $0 \in {\mathbb{R}}^n$, let ${U_x} = ({x + U}) \cap ({x + \delta B})$. It is clear that ${U_x}$ is a neighborhood of $x = G ({{y_0}})$. Since $G:f ({x + \delta B}) \to x + \delta B$ is continuous at ${y_0}$, there exists a neighborhood $U ({{y_0}})$ of $y_0$ such that

Let $N ({{y_0}}) = U ({{y_0}}) \cap {\mathop{\rm int}} ({f ({x + \delta B})})$. Then, it is clear that $N ({{y_0}})$ is a neighborhood of ${y_0}$ and so

Therefore, ${f^{-1}}(\cdot)$ is l.s.c. at ${y_0}$. This completes the proof.

Remark 5 It is clear that $f (x) = {x^2}$ is not injective on ${\mathbb{R}}$, but $f (x) = {x^2}$ is injective on $\left\{ {x \in {\mathbb{R}}:x \ge 0} \right\}$ and $\left\{ {x \in {\mathbb{R}}:x \leq 0} \right\}$. Moreover, it is easy to check that $g (x) = \cos x$ is not injective on ${\mathbb{R}}$, but $g (x) = \cos x$ is injective on $\left[{k\pi, ({k + 1})\pi } \right]$, where $k$ is an integer.

Proposition 3 Let $f:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ be continuous on ${{\mathbb{R}}^n}$. If $f$ is of type ql$, $ injective$, $ and surjective$, $ then $f^{-1}:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ is continuous on ${\mathbb{R}}^n$.

Proof Since $f:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ is injective and surjective, we can see that for any $y \in {{\mathbb{R}}^n}$, ${f^{-1}}(y)$ is singleton. It follows from Theorem 3.2.4 of Ref.[37] that $f$ is an open mapping and so ${f^{-1}}:{{\mathbb{R}}^n} \to {{\mathbb{R}}^n}$ is continuous on ${{\mathbb{R}}^n}$. This completes the proof.

4 Semicontinuity of solution mappings

In this section, we establish the upper semicontinuity and the lower semicontinuity of $S ({ \cdot, \cdot })$ and $E ({ \cdot, \cdot, \cdot })$. Let $u \in W$ and $A$ be a nonempty subset of ${{\mathbb{R}}^n}$. Set

We always assume that $\Omega (\lambda) \subseteq f ({{{\mathbb{R}}^n}, u})$ for any $({u, \lambda }) \in W \times \Lambda $. Then, for any $y \in \Omega (\lambda)$,

Theorem 3 Let $({{\varepsilon _0}, {u_0}, {\lambda _0}}) \in {\mathbb{R}}_+ ^1 \times W \times \Lambda $. Assume that $\Omega (\cdot)$ is continuous at ${\lambda _0}, $ $\Omega ({{\lambda _0}})$ is compact$, $ $f ({ \cdot, \cdot })$ is continuous on ${{\mathbb{R}}^n} \times W, $ and ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$ is bounded. Then$, $

(ⅰ) $S ({ \cdot, \cdot })$ is u.s.c. at $({{u_0}, {\lambda _0}}); $

(ⅱ) $E ({ \cdot, \cdot, \cdot })$ is u.s.c. at $({{\varepsilon _0}, {u_0}, {\lambda _0}})$.

Proof It suffices to prove (ii). Suppose, on the contrary, that ${E}({ \cdot, \cdot, \cdot })$ is not u.s.c.at $({{\varepsilon _0}, {u_0}, {\lambda _0}})$. Then, there exists a neighborhood ${W_0}$ of ${E}({{\varepsilon _0}, {u_0}, {\lambda _0}})$ and a sequence $\left\{ {({{\varepsilon _n}, {u_n}, {\lambda _n}})} \right\}$ with $({{\varepsilon _n}, {u_n}, {\lambda _n}}) \to ({{\varepsilon _0}, {u_0}, {\lambda _0}})$ such that ${E}({{\varepsilon _n}, {u_n}, {\lambda _n}}) \not\subset {W_0}$ for $n \in \mathbb{N}.$ Then, there exist

(15)

such that

(16)

It follows from (15) that $f ({{x_n}, {u_n}}) \in \Omega ({{\lambda _n}})$, and so ${x_n} \in {f^{-1}}({\Omega ({{\lambda _n}}), {u_n}})$. By Lemmas 3 and 6, we can see that ${f^{-1}}({\Omega (\cdot), \cdot })$ is u.s.c. at $({{\lambda _0}, {u_0}})$. It is easy to see that ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$ is compact. It follows from Lemma 2 that there exists ${x_0} \in {f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$ and a subsequence $\left\{ {{x_{{n_k}}}} \right\}$ of $\left\{ {{x_n}} \right\}$ such that ${x_{{n_k}}} \to {x_0}$. Without loss of generality, we can assume that ${x_n} \to {x_0}$.

Suppose that ${x_0} \notin {E}({{\varepsilon _0}, {u_0}, {\lambda _0}})$. Then, there exists ${y_0} \in \Omega ({{\lambda _0}})$ such that

(17)

Since $\Omega (\cdot)$ is l.s.c. at ${\lambda _0}$ and ${y_0} \in \Omega ({{\lambda _0}})$, by Lemma 1, there exists ${y_n} \in \Omega ({{\lambda _n}})$ such that ${y_n} \to {y_0}$. It follows from (15) that ${({{y_n}-f ({{x_n}, {u_n}})})^{\rm T}}{x_n} + {\varepsilon _n} \ge 0$, and so ${({{y_0}-f ({{x_0}, {u_0}})})^{\rm T}}{x_0} + {\varepsilon _0} \ge 0$, which contradicts (17). Therefore, ${x_0} \in {E}({{\varepsilon _0}, {u_0}, {\lambda _0}})$. We can see that ${x_n} \to {x_0} \in {W_0}$, which contradicts (16). Therefore, $E ({ \cdot, \cdot, \cdot })$ is u.s.c. at $({{\varepsilon _0}, {u_0}, {\lambda _0}})$. This completes the proof.

Theorem 4 Let $({{u_0}, {\lambda _0}}) \in W \times \Lambda $. Assume that

(ⅰ) $\Omega (\cdot)$ is continuous at ${\lambda _0}$ with nonempty convex values$, $ $\Omega ({{\lambda _0}})$ is compact, and $f ({ \cdot, \cdot })$ is continuous on ${{\mathbb{R}}^n} \times W; $

(ⅱ) for any $({u, \lambda }) \in W \times \Lambda, $ $f ({ \cdot, u})$ is monotone on ${f^{-1}}({\Omega (\lambda), u}), $ and ${f^{-1}}({ \cdot, u})$ is l.s.c. on ${\Omega (\lambda)}, $ and $\bigcup\limits_{({u, \lambda }) \in W \times \Lambda } {{f^{-1}}({\Omega (\lambda), u})} $ is bounded$; $

(ⅲ) $f ({ \cdot, {u_0}})$ is strictly natural quasi ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}}), $ and ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}}) \subseteq {\mathop{\rm int}} {\mathbb{R}}_+ ^n$ is bounded convex.

Then$, $ ${S}({ \cdot, \cdot })$ is l.s.c.at $({{u_0}, {\lambda _0}})$.

Proof Suppose, on the contrary, that $S ({\cdot, \cdot })$ is not l.s.c.at $({{u_0}, {\lambda _0}})$. Then, there exists ${x_0} \in S ({{u_0}, {\lambda _0}})$, a neighborhood ${W_0}$ of $0 \in {\mathbb{R}}^n$, and a sequence $\left\{ {({{u_n}, {\lambda _n}})} \right\}$ with $({{u_n}, {\lambda _n}}) \to ({{u_0}, {\lambda _0}})$ such that

(18)

There are two cases to be considered.

Case 1 $S ({{u_0}, {\lambda _0}})$ is singleton. Let

(19)

It follows from (19) that $f ({{x_n}, {u_n}}) \in \Omega ({{\lambda _n}})$, and so ${x_n} \in {f^{-1}}({\Omega ({{\lambda _n}}), {u_n}})$. By Lemmas 3 and 6, we can see that ${f^{-1}}({\Omega (\cdot), \cdot })$ is u.s.c. at $({{\lambda _0}, {u_0}})$. It is easy to see that ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$ is compact. It follows from Lemma 2 that there exists ${\bar{x}} \in {f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$ and a subsequence $\left\{ {{x_{{n_k}}}} \right\}$ of $\left\{ {{x_n}} \right\}$ such that ${x_{{n_k}}} \to {\bar{x}}$. Without loss of generality, we can assume that ${x_n} \to {\bar{x}}$. Similar to the proof of Theorem 3, we can prove that ${\bar{x}} \in S ({{u_0}, {\lambda _0}})$. Noting the fact that $S ({{u_0}, {\lambda _0}})$ is singleton, we have $\bar{x} = {x_0}$ and so ${x_n} \to \bar{x} = {x_0}$. Thus, ${x_n} \in {x_0} + {W_0}$ when $n$ is large enough. This together with (19) implies that $({{x_0} + {W_0}}) \cap S ({{u_n}, \lambda {}_n}) \not= \varnothing$ when $n$ is large enough, which contradicts (18).

Case 2 $S ({{u_0}, {\lambda _0}})$ is not singleton. Then, there exists $x' \in S ({{u_0}, {\lambda _0}})$ such that $x' \ne {x_0}$. Since $x', {x_0} \in S ({{u_0}, {\lambda _0}})$, by Lemma 7, we have

(20)

(21)

Since $f ({ \cdot, {u_0}})$ is strictly natural quasi ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$, for any $t \in ({0, 1})$, there exists $\lambda \in \left[{0, 1} \right]$ such that

(22)

It follows from (20) and (21) that

(23)

Noting that $x \in {f^{-1}}({y, {u_0}}) \subseteq {f^{-1}}({\Omega ({{\lambda _0}}), {u_0}}) \subseteq {\mathop{\rm int}} {\mathbb{R}}_+ ^n$, by (22) and (23), we have

(24)

Let $x (t): = tx' + ({1-t}){x_0}$. Then, it is clear that $x (t) \in {f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$. For the above $W_0$, there exists a neighborhood $W_1$ of $0 \in {\mathbb{R}}^n$ such that ${W_1} + {W_1} \subseteq {W_0}$. Obviously, there exists ${t_0} \in ({0, 1})$ such that $x ({{t_0}}) \in {x_0} + {W_1}$. Thus,

(25)

Since ${f^{-1}}({ \cdot, u_0})$ is l.s.c.on ${\Omega (\lambda _0)}$, and $\Omega (\cdot)$ is l.s.c.at ${\lambda _0}$, by Lemma 3, we can see that ${f^{-1}}({\Omega (\cdot), \cdot })$ is l.s.c. at $({{\lambda _0}, {u_0}})$. Noting that $x ({{t_0}}) \in {f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$, it follows from Lemma 1 that there exists $x{'_n} \in {f^{-1}}({\Omega ({{\lambda _n}}), {u_n}})$ such that ${x'_n} \to {x ({{t_0}})}$ and so $x{'_n} \in x ({{t_0}}) + {W_1}$ when $n$ is large enough. Noting (18) and (25), we have $x{'_n} \notin S ({{u_n}, {\lambda _n}})$. Then, it follows from Lemma 7 that there exist ${y_n} \in \Omega ({{\lambda _n}})$ and ${\overline{x}_n} \in {f^{-1}}({{y_n}, {u_n}})$ such that

(26)

Since ${y_n} \in \Omega ({{\lambda _n}})$, it follows from Lemma 2 that there exists ${y_0} \in \Omega ({{\lambda _0}})$ and a subsequence $\left\{ {{y_{{n_k}}}} \right\}$ of $\left\{ {{y_n}} \right\}$ such that ${y_{{n_k}}} \to {y_0}$. Without loss of generality, we can assume that ${y_n} \to {y_0}$. Since

and

is bounded, without loss of generality, we can assume that ${\overline{x}_n} \to \overline{x} \in {\mathbb{R}}^n$, and so $f ({{\overline{x}_n}, {u_n}}) \to f ({\overline{x}, {u_0}})$. Noting that ${y_n} = f ({{\overline{x}_n}, {u_n}}) \to {y_0}$, we have ${y_0} = f ({\overline{x}, {u_0}})$, that is, $\overline{x} \in {f^{-1}}({{y_0}, {u_0}})$. It follows from (26) that ${({{y_0}-f ({x ({{t_0}}), {u_0}})})^{\rm T}}\overline{x} \le 0$, which contradicts (24). Thus, ${S}({ \cdot, \cdot })$ is l.s.c. at $({{u_0}, {\lambda _0}})$. This completes the proof.

Theorem 5 Let $({{\varepsilon _0}, {u_0}, {\lambda _0}}) \in {\mathbb{R}}_+ ^0 \times W \times \Lambda $. Assume that

(ⅰ) $\Omega (\cdot)$ is continuous at ${\lambda _0}, $ $\Omega ({{\lambda _0}})$ is compact convex$, $ and $f ({ \cdot, \cdot })$ is continuous on ${{\mathbb{R}}^n} \times W; $

(ⅱ) $f ({ \cdot, u_0})$ is monotone on ${f^{-1}}({\Omega (\lambda _0), u_0}), $ and ${f^{-1}}({ \cdot, u_0})$ is l.s.c. on ${\Omega (\lambda _0)}; $

(ⅲ) $f ({ \cdot, {u_0}})$ is ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}}), $ and ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}}) \subseteq {\mathbb{R}}_+ ^n$ is bounded convex. Then$, $ ${E}({ \cdot, \cdot, \cdot })$ is l.s.c.at $({{\varepsilon _0}, {u_0}, {\lambda _0}})$.

Proof Suppose, on the contrary, that ${E}({ \cdot, \cdot, \cdot })$ is not l.s.c.at $({{\varepsilon _0}, {u_0}, {\lambda _0}})$. Then, there exists ${x_0} \in {E}({{\varepsilon _0}, {u_0}, {\lambda _0}})$, a neighborhood ${W_0}$ of $0 \in X, $ and a sequence $\left\{ {({{\varepsilon _n}, {u_n}, {\lambda _n}})} \right\}$ with $({{\varepsilon _n}, {u_n}, {\lambda _n}}) \to ({{\varepsilon _0}, {u_0}, {\lambda _0}})$ such that

(27)

For the above $W_0$, there exists a neighborhood $W_1$ of $0 \in X$ such that

(28)

Define a set-valued mapping $Q:{\mathbb{R}}_+ ^1 \to {2^{{{\mathbb{R}}^n}}}$ by

We claim that ${Q}(\cdot)$ is l.s.c. on ${\mathbb{R}}_+ ^0$. Suppose, on the contrary, that there exists ${\varepsilon _0} \in {R}_ + ^0$ such that ${Q}(\cdot)$ is not l.s.c. at ${\varepsilon _0}$. Then, there exists ${x_0} \in {Q}({{\varepsilon _0}})$, a neighborhood ${W_0}$ of $0 \in X, $ and a sequence $\left\{ {{\varepsilon _n}} \right\}$ with ${\varepsilon _n} \to {\varepsilon _0}$ such that

(29)

It is easy to see that, if $0 \le \alpha \le \beta $, then ${Q}(\alpha) \subseteq {Q}(\beta)$. Suppose that ${\varepsilon _0} \le {\varepsilon _n}$. Then, ${x_0} \in {Q}({{\varepsilon _0}}) \subseteq {Q}({{\varepsilon _n}})$, which contradicts (29). Thus, we know that ${\varepsilon _0} > {\varepsilon _n}$ for any $n \in \mathbb{N}$. It follows from Theorem 1 that $Q (0) = E ({0, {u_0}, {\lambda _0}}) = S ({{u_0}, {\lambda _0}}) \ne \varnothing $. We choose $x' \in {Q}(0)$. It follows from ${\varepsilon _n} \to {\varepsilon _0}$ that there exists ${\varepsilon _{n_0}}$ such that

(30)

Now, we claim that

(31)

In fact, since ${x_0} \in {Q_f}({{\varepsilon _0}})$ and $x' \in {Q_f}(0)$, by Lemma 7, we have

(32)

(33)

Since $f ({ \cdot, {u_0}})$ is ${\mathbb{R}}_+ ^n$-convex on ${f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$, we have

(34)

It follows from (32) and (33) that

(35)

Noting that $x \in {f^{-1}}({y, {u_0}}) \subseteq {f^{-1}}({\Omega (\lambda), {u_0}}) \subseteq {\mathbb{R}}_+ ^n$, by (34) and (35), we have

Then, it follows from Lemma 7 that (31) holds. By (30) and (31), we have

which contradicts (29). Therefore, ${Q}(\cdot)$ is l.s.c. on ${\mathbb{R}}_ + ^0$.

For the above ${x_0} \in {E}({{\varepsilon _0}, {u_0}, {\lambda _0}}) = Q ({{\varepsilon _0}})$ and for the above $W_1$, there exists a neighborhood $U ({{\varepsilon _0}})$ of ${\varepsilon _0}$ such that

Choose $\overline{\varepsilon } \in U ({{\varepsilon _0}})$ with $0 < \overline{\varepsilon } < {\varepsilon _0}$. Then,

Let

(36)

Then,

(37)

Since ${f^{-1}}({ \cdot, u_0})$ is l.s.c.on ${\Omega (\lambda _0)}$ and $\Omega (\cdot)$ is l.s.c.at ${\lambda _0}$, by Lemma 3, we can see that ${f^{-1}}({\Omega (\cdot), \cdot })$ is l.s.c.at $({{\lambda _0}, {u_0}})$. Noting that $\overline{x} \in {f^{-1}}({\Omega ({{\lambda _0}}), {u_0}})$, it follows from Lemma 1 that there exists ${\overline{x}_n} \in {f^{-1}}({\Omega ({{\lambda _n}}), {u_n}})$ such that ${\overline{x}_n} \to \overline{x}$. Then, ${\overline{x}_n} \in \overline{x} + {W_1}$ when $n$ is large enough.

We claim that there exists ${\overline{x}_{{n_0}}} \in \overline{x} + {W_1}$ such that ${\overline{x}_{{n_0}}} \in E ({{\varepsilon _{{n_0}}}, {u_{{n_0}}}, {\lambda _{{n_0}}}})$. In fact, if not, then for any $n \in \mathbb{N}$ with ${\overline{x}_n} \in \overline{x} + {W_1}$, we have ${\overline{x}_n} \notin E ({{\varepsilon _n}, {u_n}, {\lambda _n}})$. Thus, there exists ${y_n} \in \Omega ({{\lambda _n}})$ such that

(38)

Since ${y_n} \in \Omega ({{\lambda _n}})$, it follows from Lemma 2 that there exists ${y_0} \in \Omega ({{\lambda _0}})$ and a subsequence $\left\{ {{y_{{n_k}}}} \right\}$ of $\left\{ {{y_n}} \right\}$ such that ${y_{{n_k}}} \to {y_0}$. Without loss of generality, we can assume that ${y_n} \to {y_0}$. It follows from (38) that ${({{y_0}-f ({\overline{x}, {u_0}})})^{\rm T}}\overline{x} + {\varepsilon _0} \le 0$ and so

which contradicts with (37). Therefore, there exists ${\overline{x}_{{n_0}}} \in \overline{x} + {W_1}$ such that ${\overline{x}_{{n_0}}} \in E ({{\varepsilon _{{n_0}}}, {u_{{n_0}}}, {\lambda _{{n_0}}}})$. It follows from (28) and (36) that

and so

which contradicts (27). Therefore, ${E}({ \cdot, \cdot, \cdot })$ is l.s.c. at $({{\varepsilon _0}, {u_0}, {\lambda _0}})$. This completes the proof.

5 Application

In this section, we give an application of Theorem 1 to a road pricing problem in which the traffic management authority attempts to control the environment impact within a certain range.

As discussed by Aussel et al.^[12], we consider the road pricing model in which the environment impact problem due to traffic flow is taken into account to fix the road taxes. For the sake of brevity, we discuss here the simple traffic network. In particular, note the following:

(ⅰ) $n$ is the number of lines or edges in the network;

(ⅱ) $e (u)\in {\mathbb{R}}^n$ is the vector of environment impact due to the traffic flow $u$ on the $n$ lines.

We consider the problem from the traffic management authority's point of view. Assume that the goal of the traffic management authority is to control the environment impact within a certain range. Suppose that there exists a predetermined environment impact function $e (u)$ due to some amount of traffic flow $u$ by the previous statistics of data. Under this perspective, initially, we can fix the lower and upper bounds for $e (u)$, but after a certain increment in the traffic flow $u$, it is not easy to maintain the same fixed bounds for $e (u)$. Therefore, we can assume that there are two functions $a (u)$ and $b (u)$ of the traffic flow $u$ such that the environment impact function $e (u)$ satisfying $a (u) \preceq e (u) \preceq b (u)$, where the following notations are used for the partial orders $(\preceq, \prec)$ with respect to ${\mathbb{R}}^n_+$: for all $x, y \in {\mathbb{R}}^n$, $x \preceq y$ iff $y-x \in {\mathbb{R}}^n_+$ and $x \prec y$ iff $y-x \in {\mathbb{R}}^n_+ \backslash \{0\}.$ Let

The goal of the traffic management authority is to control the environment impact $e (u)\in K (u)$. It is known that the tax adjustment is an efficient means of regulating the traffic flow. Since the environment impact depends on the traffic flow, in order to control the environment impact, the traffic management authority can control the traffic flow by adjusting taxes on the selected lines of the network. Let $x\in {\mathbb{R}}^n$ denote the vector of taxes on the $n$ lines, and let $q (x) \in {\mathbb{R}}^n$ denote the vector of flow on the $n$ lines. Then, it is easy to see that $e (q (x))$ is the vector of environment impact due to the flow $q (x)$. Let

Therefore, the goal of the traffic management authority is to maintain $e (q (x))\in K (x)$ by adjusting the tax $x$ on the lines of network. Thus, the tax $x^*\in {\mathbb{R}}^n$ is a solution to this road pricing problem if $a (q (x^*)) \preceq e (q (x^*)) \preceq b (q (x^*))$, and the following conditions can be satisfied^[12]:

Obviously, $x^*=0$ means "no tax", and $0 \prec x^*$ stands for "positive tax". Moreover, $x^*\prec 0$ can be considered as any kind of reward by the traffic management authority to the user.

It is easy to see that the above problem can be formulated as the following inverse quasi-variational inequality: find a vector $x^*\in {\mathbb{R}}^n$ such that $e (q (x^*))\in K (x^*)$ and

(39)

In many real situations, we can assume that the functions $a (q (x))$ and $b (q (x))$ are bounded. Then, there exist two vectors $a\in {\mathbb{R}}^n$ and $b\in {\mathbb{R}}^n$ with $a \preceq b$ such that $K (x)=\left\{\xi \in {\mathbb{R}}^n: a \preceq \xi \preceq b\right\}$ for all $x\in {\mathbb{R}}^n$. Let $\Delta =\left\{\xi \in {\mathbb{R}}^n: a \preceq \xi \preceq b\right\}$. We can see that $\Delta = ({a + {\mathbb{R}}_+ ^n}) \cap ({b-{\mathbb{R}}_+ ^n})$ is convex and closed. Then, it is easy to see that (39) reduces to the following inverse variational inequality: find a vector $x^*\in {\mathbb{R}}^n$ such that $e (q (x^*))\in \Delta$ and

(40)

Let $\Omega=-\Delta $ and $f (\cdot)=-e ({q (\cdot)})$. Then, (40) is equivalent to finding a vector $x^*\in {\mathbb{R}}^n$ such that $f (x^*)\in \Omega$ and

Moreover, we can also assume that $\Delta \subseteq e ({q ({{{\mathbb{R}}^n}})})$, and the functions $e (u)$ and $q (x)$ are both of type ql and continuous. Thus, we can show that all conditions of Theorem 1 are satisfied. In fact, since $e (u)$ and $q (x)$ are both of type ql, by Proposition 3.9 of Ref.[31], we can see that the composite function $e ({q (\cdot)})$ is also of type ql and $f (\cdot) =-e ({q (\cdot)})$ is also of type ql. Since $\Delta$ is convex and closed, we can see that $\Omega =-\Delta $ is convex and closed. It is clear that $f (\cdot) =-e ({q (\cdot)})$ is continuous and $\Omega \subseteq f ({{{\mathbb{R}}^n}})$. In some practical cases, the taxes can be considered nonnegative and finite. Thus, we can assume that the vector $x$ of taxes on the $n$ lines is in a bounded subset $A \subseteq {\mathbb{R}}_+ ^n$ and so ${f^{-1}}(\Omega) \subseteq A \subseteq {\mathbb{R}}_+ ^n$ is bounded. Since $f (\cdot)$ is of type ql and $\Omega$ is convex, it is easy to see that ${f^{-1}}(\Omega)$ is convex. Noting that $f (\cdot)$ is of type ql, by Remark 2, we can see that $f (\cdot)$ is natural quasi ${\mathbb{R}}_+ ^n$-convex. Therefore, all conditions of Theorem 1 are satisfied. In particular, if the functions $e (u)$ and $q (x)$ are both linear continuous, then it follows from Remark 1 that $e (u)$ and $q (x)$ are both of type ql and continuous. Thus, when $e (u)$ and $q (x)$ are both linear continuous, all conditions of Theorem 1 are also satisfied.

Acknowledgements The authors are grateful to the editor and the referees for their valuable comments and suggestions.