In this section, we denote by \(\operatorname{Sol}(F_{1})\), \(\operatorname{Sol}(F_{2})\), and \(\operatorname{Sol}(\widetilde{F})\) the sets of solutions for (GSVQEP 1), (GSVQEP 2), and (SEVP), respectively.
Firstly, we consider a local property for conevalued mapping which is c.u.c.
Lemma 3.1
Let
\(x_{0}\in K\). Suppose that
\(C(x_{0})\)
has a bounded and closed base and
C
is c.u.c. at
\(x_{0}\). Then, for any
\(0<\varepsilon <l\), there exists a neighborhood
U
of
\(x_{0}\)
such that
$$ \bigl[C(x)+C(x_{0})\bigr]\cap B_{l}\subseteq C(x_{0})+B_{\varepsilon }, \quad \forall x\in U\cap K. $$
Proof
The proof is divided into four steps.
Step 1. For any \(0<\varepsilon <l\) and \(\xi \in C(x_{0})^{ \Delta }\), there exist \(r>\varepsilon \) and \(c>0\) such that
$$ \xi (v)>c,\quad \forall v\in \bigl[C(x_{0})+B_{\varepsilon } \bigr]\cap S_{r}. $$
(1)
Since \(C(x_{0})\) has a bounded and closed base, it follows from Lemma 2.2 that \(C(x_{0})^{\Delta }\neq \emptyset \). Thus, for any \(\xi \in C(x_{0})^{\Delta }\subseteq Z^{*}\), there exists \(a>0\) such that
$$ \xi (v)\geq a\v\, \quad \forall v\in C(x_{0}). $$
(2)
Let \(b=\max \{a,\\xi \_{*}\}+1\), where \(\\xi \_{*}\) denotes the norm of the linear operator ξ. For any \(v\in Z\), it follows from \(b>\\xi \_{*}\) and \(\xi \in Z^{*}\) that
$$ \bigl\vert \xi (v) \bigr\vert \leq \\xi \_{*} \cdot \v\< b\v\. $$
(3)
Set \(c=bl+1\) and \(r = (c+a\varepsilon +b\varepsilon )/a\). We can see that \(a(r\varepsilon )b\varepsilon = c >0\) and \(r>\varepsilon \). For any \(v_{0}\in [C(x_{0})+B_{\varepsilon }]\cap S_{r}\), there exist \(v_{1}\in C(x_{0})\) and \(v_{2}\in B_{\varepsilon }\) such that \(v_{0}=v_{1}+v_{2}\). It follows from \(\v_{0}\=r\) and \(v_{2}\in B _{\varepsilon }\) (i.e., \(\v_{2}\<\varepsilon \)) that
$$ \v_{1}\=\v_{0}v_{2}\\geq \v_{0}\\v_{2}\> r\varepsilon >0. $$
(4)
Since \(v_{1}\in C(x_{0})\), applying (2) and (4), we have
$$ \xi (v_{1})\geq a\v_{1}\ > a(r \varepsilon )>0. $$
(5)
We conclude from (3) and \(v_{2}\in B_{\varepsilon }\) that \(\xi (v_{2})< b\v_{2}\< b\varepsilon \). This together with (5) implies that
$$ \xi (v_{0})=\xi (v_{1}+v_{2})>a(r\varepsilon )b\varepsilon = c. $$
By the arbitrariness of \(v_{0}\), we obtain that (1) is true.
Step 2. \(\operatorname{cone}\{[C(x_{0})+B_{\varepsilon }]\cap S_{r} \}\subseteq \operatorname{cone}\{x\in C(x_{0})+B_{\varepsilon }: \xi (x)=c\} \subseteq \operatorname{cone}\{[C(x_{0})+B_{\varepsilon }]\cap S_{l}\}\).
Let \(C_{1}=\operatorname{cone}\{[C(x_{0})+B_{\varepsilon }]\cap S_{r}\}\) and \(C_{2}=\operatorname{cone}\{x\in C(x_{0})+B_{\varepsilon }: \xi (x)=c\}\). It is easy to see that \(C_{1}\) and \(C_{2}\) are both cones. We claim that \(\{[C(x_{0})+B_{\varepsilon }]\cap S_{r}\}\subseteq C_{2}\). This together with the definition of \(C_{1}\) implies that \(C_{1}\subseteq C_{2}\). In fact, for any \(u_{1}\in [C(x_{0})+B_{\varepsilon }]\cap S _{r}\), it follows from (1) and \(u_{1}\in S_{r}\) that \(\xi (u_{1})>c>0\). Then there exists \(0<\lambda _{1} = c/\xi (u_{1}) <1\) such that
$$ \xi (\lambda _{1}u_{1}) = \lambda _{1}\xi (u_{1}) = c. $$
(6)
On the other hand, we can see that \(u_{1}\in C(x_{0})+B_{\varepsilon }\) and \(0\in C(x_{0})+B_{\varepsilon }\). This together with the convexity of \(C(x_{0})+B_{\varepsilon }\) and \(0< \lambda _{1} < 1\) implies that
$$ \lambda _{1}u_{1}\in C(x_{0})+B_{\varepsilon }. $$
(7)
Applying (6), (7) and the definition of \(C_{2}\), we can see that \(\lambda _{1}u_{1}\in C_{2}\). It is obvious that \(u_{1}\in C_{2}\). By the arbitrariness of \(u_{1}\), we can get that \(\{[C(x_{0})+B_{\varepsilon }]\cap S_{r}\}\subseteq C_{2}\). Hence, \(C_{1}\subseteq C_{2}\).
Let \(C_{3}=\operatorname{cone}\{[C(x_{0})+B_{\varepsilon }]\cap S_{l}\}\). Obviously, \(C_{2}\) and \(C_{3}\) are both cones. If we can prove that \(\{x\in C(x_{0})+B_{\varepsilon }: \xi (x)=c\}\subseteq C_{3}\), then we can easily get \(C_{2}\subseteq C_{3}\) by the definition of \(C_{2}\). In fact, for each \(u_{2}\in C(x_{0})+B_{\varepsilon }\) satisfying \(\xi (u_{2})=c\), we claim that \(\u_{2}\\geq l\). If it is not true, then we have \(\xi (u_{2})< b\u_{2}\<bl<c\) from (3), which contradicts \(\xi (u_{2})=c\). Therefore, \(\u_{2}\\geq l\) which implies that there exists \(0<\lambda _{2} \leq 1\) such that
$$ \\lambda _{2}u_{2}\=l. $$
(8)
On the other hand, \(u_{2}\in C(x_{0})+B_{\varepsilon }\) and \(0\in C(x_{0})+B_{\varepsilon }\). This together with the convexity of \(C(x_{0})+B_{\varepsilon }\) and \(0< \lambda _{2} < 1\) implies that
$$ \lambda _{2}u_{2}\in C(x_{0})+B_{\varepsilon }. $$
(9)
Applying (8), (9), and the definition of \(C_{3}\), it is easy to obtain that \(\lambda _{2}u_{2}\in C_{3}\). Obviously, \(u_{2}\in C_{3}\). By the arbitrariness of \(u_{2}\), we have \(\{x\in C(x_{0})+B_{\varepsilon }: \xi (x)=c\}\subseteq C_{3}\), which implies that \(C_{2}\subseteq C_{3}\).
Step 3. There exists a neighborhood U of \(x_{0}\) such that \(C(x)\cap S_{r}\subseteq [C(x_{0})+B_{\varepsilon }]\cap S_{r}\) for all \(x\in U\cap K\).
Since C is c.u.c. at \(x_{0}\), for the preceding positive numbers ε, l, c, and r, there exists a neighborhood U of \(x_{0}\) such that, for any \(x\in U\cap K\),
$$ C(x)\cap \operatorname{cl}(B_{r})\subseteq C(x_{0})\cap \operatorname{cl}(B _{r})+B_{\varepsilon }\subseteq C(x_{0})+B_{\varepsilon }. $$
By taking intersection with the sphere surface \(S_{r}\) in both sides of the above formula, we can get
$$ C(x)\cap S_{r}\subseteq \bigl[C(x_{0})+B_{\varepsilon } \bigr]\cap S_{r},\quad \forall x\in U\cap K. $$
(10)
Step 4. For any \(x\in U\cap K\), we have \([C(x)+C(x_{0})] \cap B_{l} \subseteq C(x_{0})+B_{\varepsilon }\).
Applying (10), we can see that
$$ C(x)=\operatorname{cone}\bigl\{ C(x)\cap S_{r}\bigr\} \subseteq \operatorname{cone}\bigl\{ \bigl[C(x_{0})+B _{\varepsilon }\bigr]\cap S_{r}\bigr\} =C_{1}. $$
This together with \(C_{1} \subseteq C_{2}\) implies that \(C(x)\subseteq C_{2}\). Furthermore, if we can prove that \(C_{2}\) is convex, then we have
$$ C(x)+C(x_{0})\subseteq C_{2}+C_{2}=C_{2}, \quad \forall x\in U\cap X. $$
This together with \(C_{2} \subseteq C_{3}\) implies that
$$ C(x)+C(x_{0})\subseteq C_{3},\quad \forall x\in U\cap X. $$
(11)
On the other hand, for any \(u\in C_{3}\cap B_{l}\) with \(u\neq 0\), we have \(0<\u\<l\). By the definition of \(C_{3}\), there exists \(\lambda >0\) such that \(\lambda u\in [C(x_{0})+B_{\varepsilon }] \cap S_{l}\). Thus we can see that \(\lambda >1\). This together with \(\lambda u\in C(x_{0})+B_{\varepsilon }\) implies that \(u\in C(x_{0})+B _{\varepsilon }\). Then, by the arbitrariness of u, we get
$$ C_{3}\cap B_{l}\subseteq C(x_{0})+B_{\varepsilon }. $$
(12)
We conclude from (11) and (12) that, for any \(x\in U\cap K\),
$$ \bigl[C(x)+C(x_{0})\bigr]\cap B_{l}\subseteq C_{3}\cap B_{l}\subseteq C(x_{0})+B _{\varepsilon }. $$
Finally, we show that \(C_{2}\) is convex. Since \(C_{2}\) is a cone, we only need to prove that, for any \(w_{1},w_{2}\in C_{2}\setminus \{0\}\), one has \(w_{1}+w_{2}\in C_{2}\). According to the definition of \(C_{2}\), there exist \(\lambda _{1},\lambda _{2}>0\) such that \(\lambda _{1}w_{1},\lambda _{2}w_{2}\in C(x_{0})+B_{\varepsilon }\) and \(\xi (\lambda _{1}w_{1})=\xi (\lambda _{2}w_{2})=c\). Let \(\lambda = \lambda _{1}/(\lambda _{1}+\lambda _{2})\), which implies that \(0<\lambda <1\). By the linearity of ξ, we have
$$ \xi \bigl((1\lambda )\lambda _{1}w_{1}+ \lambda \lambda _{2}w_{2}\bigr)=(1\lambda )\xi (\lambda _{1}w_{1})+\lambda \xi (\lambda _{2}w_{2})=(1 \lambda )c+ \lambda c=c. $$
(13)
Noting that \(C(x_{0})+B_{\varepsilon }\) is convex, we can see that \((1\lambda )\lambda _{1}w_{1}+\lambda \lambda _{2}w_{2}\in C(x_{0})+B _{\varepsilon }\). This together with (13) implies that \((1\lambda )\lambda _{1}w_{1}+\lambda \lambda _{2}w_{2}\in C_{2}\). Hence,
$$ (1\lambda )\lambda _{1}w_{1}+\lambda \lambda _{2}w_{2}=\frac{\lambda _{1}\lambda _{2}}{\lambda _{1}+\lambda _{2}}w_{1}+ \frac{\lambda _{1} \lambda _{2}}{\lambda _{1}+\lambda _{2}}w_{2}\in C_{2}, $$
which implies that \(w_{1}+w_{2}\in C_{2}\). □
For the simplification of the proof for our main results in this paper, we also need the following lemma.
Lemma 3.2
Let
\((t_{0},x_{0},y_{0})\in E\times K\times K\). Assume that
F
is lower
\(\{C(x_{0})\}\)continuous at
\(x_{0}\), C
is c.u.c. at
\(x_{0}\), and
\(C(x_{0})\)
has a bounded and closed base. Then, for any net
\(\{(t_{\alpha },x_{\alpha },y_{\alpha })\}\to (t_{0},x_{0},y_{0})\)
with
\(F(t_{\alpha },x_{\alpha },y_{\alpha })\subseteq C(x_{\alpha })\), we have
\(F(t_{0},x_{0},y_{0})\subseteq C(x_{0})\).
Proof
Suppose to the contrary that there exists \(w_{0}\in F(t_{0},x_{0},y _{0})\) such that \(w_{0}\notin C(x_{0})\). Since \(C(x_{0})\) is closed, there exists \(\varepsilon >0\) such that
$$ [w_{0}+B_{\varepsilon }]\cap \bigl[C(x_{0})+B_{\varepsilon } \bigr]=\emptyset . $$
(14)
Applying the lower \(\{C(x_{0})\}\)continuity of F, there exists \(\alpha _{1}\) such that, for all \(\alpha >\alpha _{1}\),
$$\begin{aligned} w_{0} \in F(t_{0},x_{0},y_{0}) & \subseteq F(t_{\alpha },x_{\alpha },y _{\alpha })+B_{\varepsilon }+C(x_{0}) \\ &\subseteq C(x_{\alpha })+B_{\varepsilon }+C(x_{0}). \end{aligned}$$
This implies that
$$ [w_{0}+B_{\varepsilon }]\cap \bigl[C(x_{\alpha })+C(x_{0}) \bigr]\neq \emptyset ,\quad \forall \alpha >\alpha _{1}. $$
(15)
Take any real number \(l>\varepsilon \) such that \(w_{0}+B_{\varepsilon }\subseteq B_{l}\). Since C is c.u.c. at \(x_{0}\in K\) and \(C(x_{0})\) has a bounded closed base, it follows from Lemma 3.1 that there exists \(\alpha _{2}\) such that, for any \(\alpha >\alpha _{2}\),
$$ \bigl[C(x_{\alpha })+C(x_{0})\bigr]\cap B_{l}\subseteq C(x_{0})+B_{\varepsilon}. $$
(16)
Let \(\alpha _{0}>\alpha _{1}\) and \(\alpha _{0}>\alpha _{2}\). Then, for any \(\alpha >\alpha _{0}\), we conclude from (15) and \(w_{0}+B _{\varepsilon }\subseteq B_{l}\) that
$$ [w_{0}+B_{\varepsilon }]\cap \bigl[C(x_{\alpha })+C(x_{0}) \bigr]\cap B_{l}=[w _{0}+B_{\varepsilon }]\cap \bigl[C(x_{\alpha })+C(x_{0})\bigr]\neq \emptyset . $$
(17)
Combining (16) with (17), we can see that \([w_{0}+B_{\varepsilon }]\cap [C(x_{0})+B_{\varepsilon }]\neq \emptyset \), which contradicts (14). Hence, \(F(t_{0},x_{0},y_{0}) \subseteq C(x_{0})\). □
Remark 3.1
In Lemma 3.2, we use the assumptions that C is c.u.c. and F is lower \(\{C(x)\}\)continuous, which is very different from the ones that C is closed and F is l.s.c. frequently used in the previous literature.
Now, we are ready to present our main results.
Theorem 3.1
Let
\(F(t,x,x)\subseteq C(x)\)
for any
\((t,x)\in E\times K\). Assume that
\(\varOmega =\{x\in K:x\in Q(x)\}\)
is closed, \(Q^{1}(y)\)
is open in
K
for any
\(y\in K\), and the following conditions are satisfied:

(i)
C
is c.u.c. on
K, and for any
\(x\in K\), \(C(x)\)
has a bounded and closed base;

(ii)
T
is l.s.c. on
\(K\times K\)
with nonempty and convex values;

(iii)
for any
\(y\in K\), the mapping
\((t,x)\to F(t,x,y)\)
is lower
\(\{C(x)\}\)continuous on
\(E\times K\);

(iv)
for any
\(x\in K\), the mapping
\((t,y)\to F(t,x,y)\)
is upper properly
\(C(x)\)quasiconvex on
\(E\times K\);

(v)
there exist a nonempty and compact subset
M
of
K
and a nonempty, compact, and convex subset
N
of
K
such that, for each
\(x\in K\setminus M\), there exists some
\(y\in N\cap Q(x)\)
such that
\(F(t,x,y)\nsubseteq C(x)\)
for some
\(t\in T(x,x)\).
Then
\(\operatorname{Sol}(F_{1})\)
is nonempty.
Proof
Define two setvalued mappings \(A,H:K\rightrightarrows K\) as follows: for each \(x\in K\),
$$ A(x)=\bigl\{ y\in K:\exists t\in T(x,x), F(t,x,y)\nsubseteq C(x)\bigr\} $$
and
$$ H(x)= \textstyle\begin{cases} Q(x)\cap A(x), & x\in \varOmega , \\ Q(x), & x\in K\setminus \varOmega . \end{cases} $$
We assert that there exists \(\bar{x}\in K\) such that \(H(\bar{x})= \emptyset \). If \(\bar{x}\in K\setminus \varOmega \), then, by the definition of H, we have \(H(\bar{x})=Q(\bar{x})=\emptyset \), which contradicts the fact that \(Q(x)\neq \emptyset \) for all \(x\in K\). Hence \(\bar{x}\in \varOmega \) and \(Q(\bar{x})\cap A(\bar{x})=\emptyset \). This implies that \(\bar{x}\in Q(\bar{x})\), and for any \(y\in Q(\bar{x})\) and any \(t\in T(\bar{x},\bar{x})\), we have \(F(t,\bar{x},y)\subseteq C( \bar{x})\), and so x̄ is a solution of (GSVQEP 1). Therefore, it only suffices to prove that there exists \(\bar{x}\in K\) such that \(H(\bar{x})=\emptyset \).
The proof is divided into three steps.
Step 1. For any \(x\in K\), \(x\notin \operatorname{conv}(H(x))\).
In fact, we can firstly show that \(A(x)\) is convex. Indeed, for any \(y_{1},y_{2}\in A(x)\), there exist \(t_{1},t_{2}\in T(x,x)\) such that
$$ F(t_{i},x,y_{i})\nsubseteq C(x), \quad i={1,2}. $$
(18)
Set \(y_{\lambda }=\lambda y_{1}+(1\lambda )y_{2}\) and \(t_{\lambda }= \lambda t_{1}+(1\lambda )t_{2}\) for any \(0<\lambda <1\). Since K and \(T(x,x)\) are both convex, we have \(y_{\lambda }\in K\) and \(t_{\lambda }\in T(x,x)\). If \(F(t_{\lambda },x,y_{\lambda })\subseteq C(x)\), then, by the upper properly \(C(x)\)quasiconvexity of \(F(\cdot ,x,\cdot )\), we know that there exists \(i_{0}\in \{1,2\}\) such that
$$ F(t_{i_{0}},x,y_{i_{0}})\subseteq F(t_{\lambda },x,y_{\lambda })+C(x) \subseteq C(x)+C(x)=C(x). $$
This contradicts (18). Hence \(F(t_{\lambda },x,y_{\lambda })\nsubseteq C(x)\). It follows that \(y_{\lambda }\in A(x)\). Then \(A(x)\) is convex.
Next, we show that \(x\notin \operatorname{conv}(H(x))\). In fact, for any given \(x\in K\), since \(Q(x)\) and \(A(x)\) are convex, we can easily see that \(H(x)\) is also convex. Thus, we consider the next two cases. (1) If \(x\in \varOmega \), then the assumption \(F(t,x,x)\subseteq C(x)\) implies \(x\notin A(x)\). Hence \(x\notin \operatorname{conv}(A(x))\). This together with \(H(x)\subseteq A(x)\) implies that \(x\notin \operatorname{conv}(H(x))\). (2) If \(x\in K\setminus \varOmega \), then \(x\notin Q(x)\). This together with \(\operatorname{conv}(Q(x))=\operatorname{conv}(H(x))\) implies that \(x\notin \operatorname{conv}(H(x))\).
Step 2. For all \(y\in X\), \(H^{1}(y)\) is open in K.
Indeed, for each \(y\in K\), by the definitions of A and H, we have
$$ A^{1}(y)=\bigl\{ x\in K:\exists t\in T(x,x), F(t,x,y)\nsubseteq C(x) \bigr\} $$
and
$$\begin{aligned} H^{1}(y) = & \bigl[\varOmega \cap Q^{1}(y)\cap A^{1}(y)\bigr]\cup \bigl[(K\setminus \varOmega )\cap Q^{1}(y) \bigr] \\ = & Q^{1}(y)\cap \bigl[(K\setminus \varOmega )\cup A^{1}(y) \bigr]. \end{aligned}$$
Noting that Ω is closed and \(Q^{1}(y)\) is open in K, it suffices to show that \(A^{1}(y)\) is also open in K. Thus, we only need to show that \(K\setminus A^{1}(y)\) is closed. In fact, for any net \(\{x_{\alpha }\}\subseteq K\setminus A^{1}(y)\) with \(x_{\alpha }\to x_{0}\in K\), if \(x_{0}\in A^{1}(y)\), then there exists \(t_{0}\in T(x_{0},x_{0})\) such that
$$ F(t_{0},x_{0},y)\nsubseteq C(x_{0}). $$
(19)
We conclude from condition (ii) and Lemma 2.1(ii) that there exists a net \(\{t_{\alpha }\}\) such that \(t_{\alpha }\to t_{0}\) with \(t_{\alpha }\in T(x_{\alpha },x_{\alpha })\). Since \(x_{\alpha }\notin A^{1}(y)\), we can see that
$$ F(t,x_{\alpha },y)\subseteq C(x_{\alpha }), \quad \forall t\in T(x_{\alpha },x_{\alpha }). $$
This together with \(t_{\alpha }\in T(x_{\alpha },x_{\alpha })\) implies that \(F(t_{\alpha },x_{\alpha },y)\subseteq C(x_{\alpha })\). Then, by conditions (i), (iii) and Lemma 3.2, we have \(F(t_{0},x_{0},y) \subseteq C(x_{0})\), which contradicts (19). Hence \(x_{0}\notin K\setminus A^{1}(y)\).
Step 3. There exist a nonempty and compact subset M of K and a nonempty, compact, and convex subset N of K such that, for each \(x\in K\setminus M\), \(N\cap H(x)\neq \emptyset \).
Indeed, by condition (v), there exist a nonempty and compact subset M of K and a nonempty, compact, and convex subset N of K such that, for each \(x\in K\setminus M\), there exists \(y\in N\cap Q(x)\) such that \(F(t,x,y)\nsubseteq C(x)\) for some \(t\in T(x,x)\). Hence \(y\in Q(x)\cap A(x)\cap N\). It follows that \(y\in N\cap H(x)\).
Thus all the conditions of Lemma 2.3 are satisfied. We conclude from Lemma 2.3 that there exists \(\bar{x}\in X\) such that \(H(\bar{x})=\emptyset \). □
In the following, the closedness of the solution set \(\operatorname{Sol}(F_{1})\) is discussed.
Lemma 3.3
Suppose that the conditions in Theorem 3.1
are all satisfied. Assume that
Q
is l.s.c. on
K
and
F
is lower
\(\{C(x)\}\)continuous on
\(E\times K\times K\). Then
\(\operatorname{Sol}(F_{1})\)
is closed.
Proof
In fact, for any net \(\{x_{\alpha }\}\subseteq \operatorname{Sol}(F_{1})\) with \(x_{\alpha }\to x_{0}\in K\), we can see that \(x_{\alpha }\in Q(x_{ \alpha })\), which implies that \(x_{\alpha }\in \varOmega \). Since Ω is closed in K, we have \(x_{0}\in \varOmega \), i.e., \(x_{0}\in Q(x_{0})\). If \(x_{0}\notin \operatorname{Sol}(F_{1})\), then there exist some \(y_{0}\in Q(x_{0})\) and \(t_{0}\in T(x_{0},x_{0})\) such that
$$ F(t_{0},x_{0},y_{0})\nsubseteq C(x_{0}). $$
(20)
Since Q and T are l.s.c. on K and \(K\times K\), respectively, we conclude from Lemma 2.1(ii) that there exist \(y_{\alpha } \in Q(x_{\alpha })\) and \(t_{\alpha }\in T(x_{\alpha },x_{\alpha })\) such that \(y_{\alpha }\to y_{0}\) and \(t_{\alpha }\to t_{0}\). Applying the assumption that \(x_{\alpha }\in \operatorname{Sol}(F_{1})\), we can get \(F(t,x_{ \alpha },y)\subseteq C(x_{\alpha })\) for all \(y\in Q(x_{\alpha })\) and \(t\in T(x_{\alpha },x_{\alpha })\). Therefore,
$$ F(t_{\alpha },x_{\alpha },y_{\alpha })\subseteq C(x_{\alpha }). $$
(21)
By conditions (i) and (iii), we conclude from Lemma 3.2 and (21) that
$$ F(t_{0},x_{0},y_{0})\subseteq C(x_{0}), $$
which contradicts (20). Hence \(x_{0}\in \operatorname{Sol}(F_{1})\), which implies that \(\operatorname{Sol}(F_{1})\) is a closed subset of K. □
Remark 3.2
Theorem 3.1 is very different from Theorem 4.2 of Lin and Huang [18]. More specifically, (a) in order to establish the existence result of solutions for (GSVQEP 1), Lin and Huang used the upper semicontinuity condition for the conevalued mapping, while we used here the cosmically upper continuity condition for it; (b) in Theorem 4.2 of Lin and Huang[18], the setvalued mapping \(F(\cdot ,\cdot ,y)\) is assumed to be l.s.c. on \(E\times K\), while in Theorem 3.1 of this paper, it is assumed to be lower \(\{C(x)\}\)continuous on \(E\times K\).
Remark 3.3
Chen, Yang, and Yu [26] discussed another form of equilibrium problem with variable ordering structure. By applying the scalarization method and the conditions that the conevalued mappings \(C(\cdot )\) and \(W(\cdot ):=Z\setminus \operatorname{int}C(\cdot )\) are both u.s.c., they proved an existence theorem of solutions for a singlevalued weak vector quasiequilibrium problem with variable ordering structure. Different from Theorem 3.2 of Chen, Yang, and Yu [26], in Theorem 3.1 of this paper, by using the condition that C is c.u.c., we also obtain the existence result of solutions for a setvalued strong vector quasiequilibrium problem with variable ordering structure.
Example 3.1
Let \(Z=\mathbb{R}^{2}\), \(X=Y=\mathbb{R}\), \(K=[0,\pi /2]\), and \(E=[0,1]\). For each \(x,y\in K\) and \(t\in E\), let
$$\begin{aligned}& C(x)=\operatorname{cone}\, \operatorname{conv}\bigl\{ (1,0),(\cos {x},\sin {x}) \bigr\} , \\& T(x)=[0,1], \qquad Q(x)=[0,1], \\& h(x,y)=\bigl((xy)\cos {x},(xy)\sin {x}\bigr) \end{aligned}$$
and
$$ F(t,x,y)= \textstyle\begin{cases} \operatorname{cone}\{(\cos {tx},\sin {tx})\}+h(x,y), & x\leq y, \\ \operatorname{cone}\{(\cos {tx},\sin {tx})\}\cap \operatorname{cl}(B_{5}), & x> y. \end{cases} $$
For any given \(y_{0}\in (0,\pi /2)\) and \(t_{0}\in E\), if \(x_{0}=y_{0}\), then \(h(x_{0},y_{0})=0\), and so
$$ F(t_{0},x_{0},y_{0})=\operatorname{cone}\bigl\{ (\cos {t_{0}x_{0}},\sin {t_{0}x_{0}}) \bigr\} . $$
We can verify that \(F(\cdot ,\cdot ,y_{0})\) is not l.s.c. at the point \((t_{0},x_{0})\). In fact, for any neighborhood \(V\times U\) of \((t_{0},x_{0})\), there exists \((t_{0},\tilde{x})\in V\times U\) such that \(\tilde{x}>x_{0}=y_{0}\). It follows from the definition of F that
$$ F(t_{0},\tilde{x},y_{0})= \operatorname{cone}\bigl\{ (\cos {t_{0}\tilde{x}},\sin {t_{0} \tilde{x}})\bigr\} \cap \operatorname{cl}(B_{5})\subseteq \operatorname{cl}(B _{5}). $$
(22)
Clearly, there exist a point \(v_{0}\in F(t_{0},x_{0},y_{0})\) with \(\v_{0}\>5\) and a neighborhood W of \(v_{0}\) such that \(W\cap \operatorname{cl}(B_{5})=\emptyset \). This together with (22) implies that \(W\cap F(t_{0},\tilde{x},y_{0})=\emptyset \). It is easy to know that \(F(\cdot ,\cdot ,y_{0})\) is not l.s.c. at the point \((t_{0},x_{0})\). Therefore, the condition that F is l.s.c. on \(E\times K\times K\) is not satisfied in Theorem 4.2 of [18]. So, we are unable to use Theorem 4.2 of [18] to decide the existence of the solutions for (GSVQEP 1).
However, for this example, we can verify that all the conditions in Theorem 3.1 of this paper are satisfied.
Indeed, for each \(x\in K\), it is easy to see that \(C(x)\subseteq \mathbb{R}^{2}\) is a closed, convex, and pointed cone and Ω is closed as \(\varOmega =[0,1]\). For each \(y\in K\), \(Q^{1}(y)\) is open in K because \(Q^{1}(y)=K\) or \(Q^{1}(y)=\emptyset \). In addition, for any \((t,x)\in E\times K=[0,1]\times [0,\pi /2]\), by the definitions of C and F, we have
$$ F(t,x,x)\subseteq C(x). $$
(23)
In the following, we verify that conditions (i)–(v) of Theorem 3.1 are all satisfied. By the assumptions of C and T, it is easy to get that conditions (i) and (ii) hold. Thus we consider whether the mapping \((t,y)\to F(t,x,y)\) is upper properly \(C(x)\)quasiconvex on \(E\times K\) for any \(x\in K\). In fact, let \(x\in K\), \(\lambda \in (0,1)\), \((t_{1},y_{1}),(t_{2},y_{2})\in E\times K\), and \(y_{2}\geq y_{1}\). Set \(y_{\lambda }=\lambda y_{1}+(1\lambda )y_{2}\) and \(t_{\lambda }=\lambda t_{1}+(1\lambda )t_{2}\). It suffices to show that there exists \(i_{0}\in \{1,2\}\) such that \(F(t_{i_{0}},x,y_{i _{0}})\subseteq F(t_{\lambda },x,y_{\lambda })+C(x)\). By the definition of C, we have
$$ (0,0)\in \operatorname{cone}\bigl\{ (\cos {tx},\sin {tx})\bigr\} \subseteq \operatorname{cone}\, \operatorname{conv}\bigl\{ (1,0),(\cos {x},\sin {x})\bigr\} =C(x). $$
(24)
By the definition of F, we can see that \(h(x,y)\in C(x)\). This together with (24) implies that
$$ C(x)\subseteq F(t,x,y)+C(x). $$
(25)
If \(x> y_{1}\), from (24) and (25), we can see that
$$ F(t_{1},x,y_{1})\subseteq C(x)\subseteq F(t_{\lambda },x,y_{\lambda })+C(x). $$
If \(x\leq y_{1}\), then \(y_{2}\geq y_{\lambda }\geq y_{1}\), and so \(h(x,y_{\lambda })h(x,y_{1})=((y_{1}y_{\lambda })\cos {x},(y_{1}y _{\lambda })\sin {x})\in C(x)\). It follows that \(h(x,y_{1})+C(x) \subseteq h(x,y_{\lambda })+C(x)\). This together with (24) implies that
$$\begin{aligned} F(t_{1},x,y_{1}) &\subseteq h(x,y_{1})+C(x) \subseteq h(x,y_{\lambda })+C(x) \\ &\subseteq F(t_{\lambda },x,y_{\lambda })+C(x). \end{aligned}$$
Therefore, the mapping \((t,y)\to F(t,x,y)\) is upper properly \(C(x)\)quasiconvex on \(E\times K\) for any \(x\in K\).
Next, we show that \((t,x)\to F(t,x,y)\) is lower \(\{C(x)\}\)continuous on \(E\times K\) for any given \(y\in K\). Let \((t_{0},x_{0})\in E\times K\) and \(\varepsilon >0\). We consider the following two cases. a) Assume that \(x_{0} > y\). Since \(F(t,x,y)=\operatorname{cone}\{(\cos {tx},\sin {tx}) \}\cap \operatorname{cl}(B_{5})\) for any \(x> y\), it is obvious that \(F(\cdot ,\cdot ,y)\) is lower \(\{C(x_{0})\}\)continuous at \((t_{0},x_{0})\in E\times K\). b) Assume that \(x_{0} > y\). We firstly define a vectorvalued mapping \(g:K\to Z\) as follows:
$$ g(x)= \textstyle\begin{cases} h(x,y), & x\leq y, \\ (0,0), & x> y. \end{cases} $$
Since \(x_{0}\leq y\), we conclude from the continuity of \(h(\cdot , y)\) and \(h(y,y)=(0,0)\) that g is continuous on K. Thus there exists a neighborhood W of \(x_{0}\) such that, for all \(x\in U(x_{0})\), \(g(x_{0})\in g(x)+ B_{\varepsilon }\). Hence,
$$\begin{aligned} F(t_{0},x_{0},y) &=h(x_{0},y)+ \operatorname{cone}\bigl\{ (\cos {tx_{0}},\sin {tx_{0}}) \bigr\} =g(x_{0})+\operatorname{cone}\bigl\{ (\cos {tx_{0}}, \sin {tx_{0}})\bigr\} \\ &\subseteq g(x)+B_{\varepsilon }+C(x_{0})\subseteq F(t,x,y)+B_{ \varepsilon }+C(x_{0}). \end{aligned}$$
This implies that \((t,x)\to F(t,x,y)\) is lower \(\{C(x)\}\)continuous on \(E\times K\).
Since K is a nonempty, compact, and convex subset of X, it is easy to see that condition (v) in Theorem 3.1 is satisfied with \(M=N=K\).
Therefore, conditions (i)–(v) in Theorem 3.1 of this paper are all satisfied. Thus, by Theorem 3.1, (GSVQEP 1) has at least one solution in K.
In fact, we can also verify that \(\bar{x}=1\) is the solution of this example. For all \(t\in T(\bar{x})=[0,1]\) and for all \(y\in Q(\bar{x})=[0,1]\), we conclude from \(y\leq 1=\bar{x}\) and \(t\in [0,1]\) that
$$ F(t,\bar{x},y)\subseteq \operatorname{cone}\bigl\{ (\cos {t},\sin {t})\bigr\} \subseteq \operatorname{cone}\, \operatorname{conv}\bigl\{ (1,0),(\cos {1},\sin {1})\bigr\} =C(\bar{x}). $$
This indicates that \(\bar{x}=1\) solves (GSVQEP 1).
Next, the existence of solutions and the closedness of the solution set for (GSVQEP 2) are discussed.
Theorem 3.2
Let
X, Y, Z, E, K, C, T, Q, and
F
be the same as in Theorem 3.1. Suppose that
E
is compact and
Ω
is closed. If the following conditions are satisfied:

(i)
C
is c.u.c. on
K, and for all
\(x\in K\), \(C(x)\)
has a bounded and closed base;

(ii)
T
is u.s.c. on
\(K\times K\)
with nonempty and closed values;

(iii)
for each
\(y\in K\), the mapping
\((t,x)\to F(t,x,y)\)
is lower
\(\{C(x)\}\)continuous on
\(E\times K\);

(iv)
for any
\((t,x)\in E\times K\), the mapping
\(y\to F(t,x,y)\)
is upper properly
\(C(x)\)quasiconvex on
K;

(v)
there exist a nonempty and compact subset
M
of
K
and a nonempty, compact, and convex subset
N
of
K
such that, for each
\(x\in K\setminus M\), there exists some
\(y\in N\cap Q(x)\)
such that
\(F(t,x,y)\nsubseteq C(x)\)
for all
\(t\in T(x,x)\).
Then
\(\operatorname{Sol}(F_{2})\)
is nonempty.
Proof
For each \(x\in K\), let
$$ A(x)=\bigl\{ y\in K:\forall t\in T(x,x), F(t,x,y)\nsubseteq C(x)\bigr\} , $$
and
$$ H(x)= \textstyle\begin{cases} Q(x)\cap A(x), & x\in \varOmega , \\ Q(x), & x\in K\setminus \varOmega . \end{cases} $$
Similar to the analysis as Theorem 3.1, it suffices to be proved that there exists \(\bar{x}\in K\) such that \(H(\bar{x})=\emptyset \).
The proof is divided into three steps.
Step 1. For all \(x\in K\), \(x\notin \operatorname{conv}(H(x))\).
In fact, for every \(t\in T(x,x)\), by the similar arguments for the convexity of A in the proof of Theorem 3.1, we can show that \(A(x)\) is convex for each \(x\in K\). In the same way as in Theorem 3.1, we can show that \(x\notin \operatorname{conv}(H(x))\).
Step 2. For all \(y\in X\), \(H^{1}(y)\) is open in K.
By the definition of A and H, we can see that \(A^{1}(y)=\{x \in K:\forall t\in T(x,x), F(t,x,y)\nsubseteq C(x)\}\) and \(H^{1}(y) = Q^{1}(y)\cap [( K\setminus \varOmega )\cup A^{1}(y)]\). Similar to the proof of openness of \(H^{1}(y)\) in Theorem 3.1, it suffices to show that \(K\setminus A^{1}(y)\) is closed.
In fact, for any net \(\{x_{\alpha }\}\subseteq K\setminus A^{1}(y)\) with \(x_{\alpha }\to x_{0}\in K\), we have \(x_{\alpha }\notin A^{1}(y)\). Thus, for each α, there exists some \(t_{\alpha }\in T(x_{\alpha },x_{\alpha })\subseteq E\) such that
$$ F(t_{\alpha },x_{\alpha },y)\subseteq C(x_{\alpha }). $$
(26)
Since E is compact, we can see that T is u.s.c. with compact values. By Lemma 2.1(i), there exists a subnet of \(\{t_{ \alpha }\}\) which converges to \(t_{0}\in T(x_{0},x_{0})\). Without loss of generality, we assume that \(t_{\alpha }\to t_{0}\). It follows from (26) and Lemma 3.2 that \(F(t_{0},x_{0},y)\subseteq C(x _{0})\). Hence, \(x_{0}\in K\setminus A^{1}(y)\), which implies that \(K\setminus A^{1}(y)\) is closed.
Step 3. There exist a nonempty and compact subset M of K and a nonempty, convex, and compact subset N of K such that, for each \(x\in K\setminus M\), \(N\cap H(x)\neq \emptyset \).
Indeed, by condition (v), there exist a nonempty compact subset M of K and a nonempty compact convex subset N of K such that, for each \(x\in K\setminus M\), there exists some \(y\in N\cap Q(x)\) such that \(F(t,x,y)\nsubseteq C(x)\) for all \(t\in T(x,x)\). It follows that \(y\in Q(x)\cap A(x)\), and so \(y\in N\cap H(x)\).
Applying Lemma 2.3, there exists \(\bar{x}\in K\) such that \(H(\bar{x})=\emptyset \). □
Lemma 3.4
Let
\(\operatorname{Sol}(F_{2})\neq \emptyset \). Suppose that
Q
is l.s.c. on
K, T
is u.s.c. on
\(K\times K\)
with compact values, and
F
is lower
\(\{C(x)\}\)continuous on
\(E\times K\times K\). Then
\(\operatorname{Sol}(F_{2})\)
is closed.
Proof
In fact, for any net \(\{x_{\alpha }\}\subseteq \operatorname{Sol}(F_{2})\) with \(x_{\alpha }\to x_{0}\in K\), we need to prove that \(x_{0}\in \operatorname{Sol}(F _{2})\). If it is not true, then there exists \(y_{0}\in Q(x_{0})\) such that, for all \(t\in T(x_{0},x_{0})\),
$$ F(t,x_{0},y_{0})\nsubseteq C(x_{0}). $$
(27)
Since \(y_{0}\in Q(x_{0})\) and Q is l.s.c. on K, applying Lemma 2.1(ii), there exists \(y_{\alpha }\in Q(x_{\alpha })\) such that \(y_{\alpha }\to y_{0}\). We conclude from \(x_{\alpha }\in \operatorname{Sol}(F_{2})\) and \(y_{\alpha }\in Q(x_{\alpha })\) that there exists \(t_{\alpha }\in T(x _{\alpha },x_{\alpha })\) such that
$$ F(t_{\alpha },x_{\alpha },y_{\alpha })\subseteq C(x_{\alpha }). $$
(28)
Noting that T is u.s.c. with compact values, by Lemma 2.1(i), there exists a subnet of \(\{t_{\alpha }\}\) which converges to \(t_{0}\in T(x_{0},x_{0})\). Without loss of generality, let \(t_{\alpha }\to t_{0}\). It follows from (28) and Lemma 3.2 that \(F(t_{0},x_{0},y_{0})\subseteq C(x_{0})\), which contradicts (27). Thus \(x_{0}\in \operatorname{Sol}(F_{2})\). This implies that \(\operatorname{Sol}(F_{2})\) is closed. □
Remark 3.4
If K is a nonempty, compact, and convex subset of X, we can take \(M=N=K\). Then, condition (v) in Theorems 3.1 and 3.2 is satisfied trivially. Furthermore, if Q is u.s.c. with nonempty, closed, and convex values, then the set \(\varOmega =\{x\in K:x\in Q(x)\}\) is closed.
Remark 3.5
In the situation that \(Q(x)\equiv K\) for all \(x\in K\), Fu and Wang [29] obtained an existence theorem of solutions for (GSVQEP 2)—Theorem 3.1, which is very different from Theorem 3.2 of this paper. Above all the differences, the continuity condition on the conevalued mapping C is the biggest one. Indeed, the conevalued mapping C is assumed to be u.s.c. in Theorem 3.1 of Fu and Wang [29], while it is assumed to be c.u.c. in Theorem 3.2 of this paper.
By applying Theorem 3.1, we can get the following corollary.
Corollary 3.1
Let
X, Z, K, and
C
be same as in Theorem 3.1. Let
\(\widetilde{F}:K \times K\rightrightarrows Z\)
be a setvalued mapping. Suppose that the following conditions are satisfied:

(i)
C
is c.u.c. on
K, and for all
\(x\in K\), \(C(x)\)
has a bounded and closed base;

(ii)
for all
\(x\in K\), \(\widetilde{F}(x,x)\subseteq C(x)\);

(iii)
for each
\(x\in K\), the mapping
\(y\to \widetilde{F}(x,y)\)
is upper properly
\(C(x)\)quasiconvex on
K;

(iv)
for each
\(y\in K\), the mapping
\(x\to \widetilde{F}(x,y)\)
is lower
\(\{C(x)\}\)continuous on
K;

(v)
there exist a nonempty and compact subset
M
of
K
and a nonempty, compact, and convex subset
N
of
K
such that, for each
\(x\in K\setminus M\), there exists
\(y\in N\cap Q(x)\)
satisfying
\(\widetilde{F}(x,y)\nsubseteq C(x)\).
Then the solution set
\(\operatorname{Sol}(\widetilde{F})\)
is nonempty and closed.
Proof
Let \(Y=X\) and \(E=K\). For each \(x,y\in K\), let
$$ Q(x) = K,\qquad T(x,y)=\{x\}\quad \mbox{and}\quad F(t,x,y)=\widetilde{F}(x,y). $$
It is easy to verify that conditions (i)–(v) of Theorem 3.1 are all satisfied. Then, by Theorem 3.1, there exists \(\bar{x} \in K\) such that
$$ \widetilde{F}(\bar{x},y)\subseteq C(\bar{x}),\quad \forall y\in Q( \bar{x})= K. $$
Hence, x̄ is a solution of (SVEP).
Next, we show that \(\operatorname{Sol}(\widetilde{F})\) is closed. In fact, for any net \(\{x_{\alpha }\}\subseteq \operatorname{Sol}(\widetilde{F})\) with \(x_{\alpha }\to x _{0}\in K\), we can see that
$$ \widetilde{F}(x_{\alpha },y)\subseteq C(x_{\alpha }),\quad \forall y \in K. $$
Then, for each \(y\in X\), we conclude from (i), (iv), and Lemma 3.2 that \(\widetilde{F}(x_{0},y)\subseteq C(x_{0})\). This implies that \(x_{0}\in \operatorname{Sol}(\widetilde{F})\). Hence, \(\operatorname{Sol}(\widetilde{F})\) is closed. □