In this section, we shall investigate the existence results for GSVQEP and GQVLIP with monotonicity and without monotonicity. First, we present the following lemma which is of Minty’s type for GSVQEP.
Lemma 3.1 Let K be a nonempty and convex subset of X, let A:K\to {2}^{K} be a setvalued mapping such that for any x\in K, A(x) is a nonempty convex subset of K and let F:K\times K\to {2}^{y} be g.h.c. in the first argument, Cconvex in the second argument and Cstrongly pseudomonotone. Then the following problems are equivalent:

(i)
Find x\in K such that x\in A(x), F(x,y)\u2288C\setminus \{0\}, \mathrm{\forall}y\in A(x).

(ii)
Find x\in K such that x\in A(x), F(y,x)\subseteq C, \mathrm{\forall}y\in A(x).
Proof (i) → (ii) It is clear by the Cstrong pseudomonotonicity.
(ii) → (i) Let x\in K. For any y\in A(x) and \theta \in (0,1), we set {z}_{\theta}=\theta y+(1\theta )X. By the assumption (ii) and the convexity of A(x), we conclude that
x\in A(x),\phantom{\rule{1em}{0ex}}F({z}_{\theta},x)\subseteq C.
Since F is Cconvex in the second argument, we have
\begin{array}{rcl}0& \in & F({z}_{\theta},{z}_{\theta})\\ \subseteq & \theta F({z}_{\theta},y)+(1\theta )F({z}_{\theta},x)C\\ \subseteq & \theta F({z}_{\theta},y)C.\end{array}
Then we have F({z}_{\theta},y)\cap C\ne \mathrm{\varnothing}, because C is a convex cone. Since F is g.h.c. in the first argument, we have x\in A(x), F(x,y)\cap C\ne \mathrm{\varnothing}, \mathrm{\forall}y\in A(x). It implies that x\in A(x), F(x,y)\u2288C\setminus \{0\} for all y\in A(x). This completes the proof. □
In the following theorem, we present the existence result for GSVQEP by assuming the monotonicity of the function.
Theorem 3.2 Let K be a nonempty compact convex subset of X. Let A:K\to {2}^{K} be a setvalued mapping such that for any x\in K, A(x) is a nonempty convex subset of K and for each y\in K, {A}^{1}(y) is open in K. Let the set P:=\{x\in X\mid x\in A(x)\} be closed. Assume that F:K\times K\to {2}^{Y} is Cstrongly pseudomonotone, g.h.c. in the first argument, Cconvex and l.s.c. in the second argument. Then GSVQEP has a solution.
Proof For any x\in K, we define the setvalued mapping S,T:K\to {2}^{K} by
\begin{array}{c}S(x)=\{y\in K\mid F(y,x)\u2288C\},\hfill \\ T(x)=\{y\in K\mid F(x,y)\subseteq C\setminus \{0\}\},\hfill \end{array}
and for any y\in K, we denoted the complement of {S}^{1}(y) by {({S}^{1}(y))}^{C}=\{x\in K\mid F(y,x)\subseteq C\}. For each x\in K, we define multivalued maps G,H:K\to {2}^{K} by
G(x)=\{\begin{array}{ll}S(x)\cap A(x)& \text{if}x\in P,\\ A(x)& \text{if}x\in K\setminus P\end{array}
and
H(x)=\{\begin{array}{ll}T(x)\cap A(x)& \text{if}x\in P,\\ A(x)& \text{if}x\in K\setminus P.\end{array}
Clearly, G(x) and H(x) are nonempty sets for all x\in K, and by the Cstrong pseudomonotonicity of F, we have G(x)\subseteq H(x) for all x\in K. We claim that H(x) is convex. Let {y}_{1},{y}_{2}\in T(x) and \theta \in (0,1). Since F is Cconvex in the second argument, we have
\begin{array}{rcl}F(x,\theta {y}_{1}+(1\theta ){y}_{2})& \subseteq & \theta F(x,{y}_{1})+(1\theta )F(x,{y}_{2})C\\ \subseteq & (C\setminus \{0\})C\\ \subseteq & C\setminus \{0\}.\end{array}
Then we have T(x) is convex and so H(x) is convex by the convexity of A(x). Next, we will show that {G}^{1}(y) is open in K for each y\in K. Since F is l.s.c. in the second argument and by the definition of {({S}^{1}(y))}^{C}, we have {({S}^{1}(y))}^{C} closed and so {S}^{1}(y) is open in K. By assumption, we obtain that
{G}^{1}(y)=({S}^{1}(y)\cap {A}^{1}(y))\cup ({A}^{1}(y)\cap K\setminus P)
is open in K. It is easy to see that the mapping H has no fixed point because 0\in F(x,x), \mathrm{\forall}x\in K. From the contrapositive of the generalization of the FanBrowder fixed point theorem and Lemma 2.4, we have
K\u2288\bigcup _{y\in K}{G}^{1}(y).
Hence, there exists \overline{x}\in K such that G(\overline{x})=\mathrm{\varnothing}. If \overline{x}\in K\setminus P, we have A(\overline{x})=\mathrm{\varnothing}, which contradicts the assumptions. Then \overline{x}\in P and hence S(\overline{x})\cap A(\overline{x})=\mathrm{\varnothing}. This means that \overline{x}\in A(\overline{x}) and F(y,\overline{x})\subseteq C for all y\in A(\overline{x}). This completes the proof by Lemma 3.1. □
The following example shows that GSVQEP has a solution under the condition of Theorem 3.2.
Example 3.3 Let Y=\mathbb{R}, C=[0,\mathrm{\infty}) and K=[1,1]. Define the mapping A:K\to {2}^{K} and F:K\times K\to {2}^{Y} by
A(x)=\{\begin{array}{ll}[0.5,x+0.5)& \text{if}1\le x0,\\ (0.5,0.5)& \text{if}x=0,\\ (x0.5,0.5]& \text{if}0x\le 1\end{array}
and
F(x,y)=\{\begin{array}{ll}[0,yx]& \text{if}xy,\\ [yx,0]& \text{if}x\ge y,\end{array}
respectively. By the definition of A, see Figure 1, we have the set P=\{x\in X\mid x\in A(x)\}=[0.5,0.5] which is closed and for each y\in K, {A}^{1}(y) is open in K.
We see that F is Cstrongly pseudomonotone. Indeed, if F(x,y)\u2288C\setminus \{0\}, then we only consider the case x<y, so F(x,y)=[0,yx]. That is,
F(y,x)=[xy,0]\subseteq C\phantom{\rule{1em}{0ex}}\text{for all}xy.
Let x,y,z\in K and \lambda \in [0,1]. If x<\lambda y+(1\lambda )z, then
\begin{array}{rcl}F(x,\lambda y+(1\lambda )z)& =& [0,\lambda y+(1\lambda )zx]\\ =& [0,\lambda (yx)+(1\lambda )(zx)]\\ \subseteq & [0,\lambda (yx)+(1\lambda )(zx)]C\\ =& \lambda [0,yx]+(1\lambda )[0,zx]C\\ =& \lambda F(x,y)+(1\lambda )F(x,y)C.\end{array}
Similarly, in another case, we have F is Cconvex in the second argument. Clearly, F is g.h.c. in the first argument and l.s.c. in the second argument.
Moreover, this example asserts that −0.5 is one of the solutions because if x=0.5, then A(x)=[0.5,0). Note that for all y\in A(x), y>x. Therefore F(0.5,y)[0,y+0.5]\u2288C\setminus \{0\} for all y\in [0.5,0).
Now, we present an existence theorem for GSVQEP when F is not necessarily monotone.
Theorem 3.4 Let K be a nonempty compact convex subset of X, let A:K\to {2}^{K} be a setvalued mapping such that for each x\in K, A(x) is a nonempty convex subset of K, and let the set P:=\{x\in X\mid x\in A(x)\} be closed. Assume that F:K\times K\to {2}^{Y} is Cconvex in the second argument and for each y\in K, the set \{x\in K\mid F(x,y)\subseteq C\setminus \{0\}\} is open. Then GSVQEP has a solution.
Proof We proceed with the contrary statements, that is, for each x\in X, x\notin A(x) or there exists y\in A(x) such that
F(x,y)\subseteq C\setminus \{0\}.
(3.1)
For every y\in K, we define the sets {N}_{y} and {M}_{y} as follows:
{N}_{y}:=\{x\in K:F(x,y)\subseteq C\setminus \{0\}\}
and
{M}_{y}:={N}_{y}\cup {P}^{C}.
By the assumption, we have the set {M}_{y} is open in K and we see that {\{{M}_{y}\}}_{y\in K} is an open cover of K. Since K is compact, there exists a finite subcover {\{{M}_{{y}_{i}}\}}_{i=1}^{n} such that K={\bigcup}_{i=1}^{n}{M}_{{y}_{i}}. By a partition of unity, there exists a family {\{{\beta}_{i}\}}_{i=1}^{n} of realvalued continuous functions subordinate to {\{{M}_{{y}_{i}}\}}_{i=1}^{n} such that for all x\in K, 0\le {\beta}_{i}(x)\le 1 and {\sum}_{i=1}^{n}{\beta}_{i}(x)=1 and for each x\notin {M}_{{y}_{i}}, {\beta}_{i}(x)=0. Let C:=co\{{y}_{1},{y}_{2},\dots ,{y}_{n}\}\subseteq K. Then C is a simplex of a finite dimensional space. Define a mapping S:C\to C by
S(x)=\sum _{i=1}^{n}{\beta}_{i}(x){y}_{i},\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in C.
(3.2)
Hence, we have S is continuous since {\beta}_{i} is continuous for each i. From Brouwer’s fixed point theorem, there exists {x}_{0}\in C such that {x}_{0}=S({x}_{0}). We define a setvalued mapping T:K\to {2}^{Y} by
T(x)=F(x,S(x))\phantom{\rule{1em}{0ex}}\text{for all}x\in K.
(3.3)
Now, we note that for any x\in K, \{{y}_{i}\mid x\in {M}_{{y}_{i}}\}\ne \mathrm{\varnothing}. Since F is Cconvex in the second argument, it follows from (3.1), (3.2) and (3.3) that we have
\begin{array}{rcl}T(x)& =& F(x,\sum _{i=1}^{n}{\beta}_{i}(x){y}_{i})\\ \subseteq & \sum _{i=1}^{n}{\beta}_{i}(x)F(x,{y}_{i})C\\ \subseteq & C\setminus \{0\}C\\ =& C\setminus \{0\}\end{array}
for all x\in K. Since {x}_{0}\in K and it is a fixed point of S, 0=F(x,x)=F(x,S(x))=T(x)\subseteq C\setminus \{0\}, which is a contradiction. This completes the proof. □
If we set A\equiv I, then Theorem 3.2 and Theorem 3.4 are reduced to Theorem 1 and Theorem 3 in Kum and Wong [27], respectively. Moreover, Theorem 3.2 is a multivalued version of Theorem 2.3 in Kazmi and Khan [20].
Let F(x,y)=\u3008Tx,\eta (y,x)\u3009 for all x,y\in K, where \eta :K\times K\to X and T:K\to {2}^{L(X,Y)}. As a consequence of Theorem 3.2 and using the same argument as in Kum and Wang ([27], Theorem 2), we have the following existence result for GQVLIP.
Corollary 3.5 Let K be a nonempty compact convex subset of X, let A:K\to {2}^{K} be a setvalued mapping such that for any x\in K, A(x) is a nonempty convex subset of K and for each y\in K, {A}^{1}(y) is open in K. Let the set P:=\{x\in X\mid x\in A(x)\} be closed, let \eta :K\times K\to X be affine and continuous in the first argument and hemicontinuous in the second argument, and let T:K\to {2}^{L(X,Y)} be a Cstrongly pseudomonotone and g.h.c. with nonempty compact values where L(X,Y) is equipped with topology of bounded convergence. Then GQVLIP has a solution.
As a consequence of Theorem 3.4, we obtain the following existence result for GQVLIP.
Corollary 3.6 Let K be a nonempty compact convex subset of X. Let A:K\to {2}^{K} be a setvalued mapping such that for each x\in X, A(x) is a nonempty convex subset of K and let the set P:=\{x\in X\mid x\in A(x)\} be closed. Assume that \eta :K\times K\to X is affine in the first argument and T:K\to {2}^{L(X,Y)} is a nonlinear mapping such that, for every y\in K, the set \{x\in K\mid \u3008T(x),\eta (y,x)\u3009\subseteq C\setminus \{0\}\} is open. Then GQVLIP has a solution.