In this section, we discuss the existence and closedness of the solution sets of symmetric generalized quasivariational inclusion problems by using the KakutaniFanGlicksberg fixed point theorem.
Theorem 8 For each \{i=1,2\}, assume for the problem (SQVIP_{
α
}) that

(i)
{K}_{i} is usc in A\times A with nonempty convex closed values and {P}_{i} is lsc in A\times A with nonempty closed values;

(ii)
{T}_{i} is usc in A\times A with nonempty convex compact values if \alpha =w (or \alpha =m) and {T}_{i} is lsc in A\times A with nonempty convex values if \alpha =s;

(iii)
for all (x,z,u)\in A\times B\times A, 0\in {F}_{i}(x,z,{P}_{i}(x,u));

(iv)
for all (x,z,u)\in A\times B\times A, the set \{a\in {K}_{i}(x,u):0\in {F}_{i}(a,z,y),\mathrm{\forall}y\in {P}_{i}(x,u)\} is convex;

(v)
the set \{(x,z,y)\in A\times B\times A:0\in {F}_{i}(x,z,y)\} is closed.
Then the (SQVIP_{
α
}) has a solution, i.e., there exist (\overline{x},\overline{u})\in A\times A such that \overline{x}\in {K}_{1}(\overline{x},\overline{u}), \overline{u}\in {K}_{2}(\overline{x},\overline{u}) and
\begin{array}{c}(y,\overline{z})\alpha {P}_{1}(\overline{x},\overline{u})\times {T}_{1}(\overline{x},\overline{u})\phantom{\rule{1em}{0ex}}\text{satisfying}0\in {F}_{1}(\overline{x},\overline{z},y),\hfill \\ (y,\overline{v})\alpha {P}_{2}(\overline{x},\overline{u})\times {T}_{2}(\overline{x},\overline{u})\phantom{\rule{1em}{0ex}}\text{satisfying}0\in {F}_{2}(\overline{u},\overline{v},y).\hfill \end{array}
Moreover, the solution set of the (SQVIP_{
α
}) is closed.
Proof Similar arguments can be applied to three cases. We present only the proof for the case where \alpha =m.
Indeed, for all (x,z,u,v)\in A\times B\times A\times B, define mappings: {\mathrm{\Phi}}_{m},{\mathrm{\Pi}}_{m}:A\times B\times A\to {2}^{A} by
{\mathrm{\Phi}}_{m}(x,z,u)=\{a\in {K}_{1}(x,u):0\in {F}_{1}(a,z,y),\mathrm{\forall}y\in {P}_{1}(x,u)\},
and
{\mathrm{\Pi}}_{m}(x,v,u)=\{b\in {K}_{2}(x,u):0\in {F}_{2}(b,v,y),\mathrm{\forall}y\in {P}_{2}(x,u)\}.

(a)
Show that {\mathrm{\Phi}}_{m}(x,z,u) and {\mathrm{\Pi}}_{m}(x,v,u) are nonempty convex sets.
Indeed, for all (x,z,u)\in A\times B\times A and (x,v,u)\in A\times B\times A, for each \{i=1,2\}, {K}_{i}(x,u), {P}_{i}(x,u) are nonempty. Thus, by assumptions (i), (ii) and (iii), we have {\mathrm{\Phi}}_{m}(x,z,u) and {\mathrm{\Pi}}_{m}(x,v,u) are nonempty. On the other hand, by the condition (iv), we also have {\mathrm{\Phi}}_{m}(x,z,u), {\mathrm{\Pi}}_{m}(x,v,u) are convex.

(b)
We will prove {\mathrm{\Phi}}_{m} and {\mathrm{\Pi}}_{m} are upper semicontinuous in A\times B\times A with nonempty closed values.
First, we show that {\mathrm{\Phi}}_{m} is upper semicontinuous in A\times B\times A with nonempty closed values. Since A is a compact set, by Lemma 6(ii), we need only to show that {\mathrm{\Phi}}_{m} is a closed mapping. Indeed, let a net \{({x}_{n},{z}_{n},{u}_{n}):n\in I\}\subset A\times B\times A such that ({x}_{n},{z}_{n},{u}_{n})\to (x,z,u)\in A\times B\times A, and let {a}_{n}\in {\mathrm{\Phi}}_{m}({x}_{n},{z}_{n},{u}_{n}) such that {a}_{n}\to {a}_{0}. Now we need to show that {a}_{0}\in {\mathrm{\Phi}}_{m}(x,z,u). Since {a}_{n}\in {K}_{1}({x}_{n},{u}_{n}) and {K}_{1} is upper semicontinuous with nonempty closed values by Lemma 6(i), hence {K}_{1} is closed, thus we have {a}_{0}\in {K}_{1}(x,u). Suppose the contrary {a}_{0}\notin {\mathrm{\Phi}}_{m}(x,z,u). Then \mathrm{\exists}{y}_{0}\in {P}_{1}(x,u) such that
0\notin {F}_{1}({a}_{0},z,{y}_{0}).
(1)
By the lower semicontinuity of {P}_{1}, there is a net \{{y}_{n}\} such that {y}_{n}\in {P}_{1}({x}_{n},{u}_{n}), {y}_{n}\to {y}_{0}. Since {a}_{n}\in {\mathrm{\Phi}}_{m}({x}_{n},{z}_{n},{u}_{n}), we have
0\in {F}_{1}({a}_{n},{z}_{n},{y}_{n}).
(2)
By the condition (v) and (2), we have
0\in {F}_{1}({a}_{0},z,{y}_{0}).
(3)
This is a contradiction between (3) and (1). Thus, {a}_{0}\in {\mathrm{\Phi}}_{m}(x,z,u). Hence, {\mathrm{\Phi}}_{m} is upper semicontinuous in A\times B\times A with nonempty closed values. Similarly, we also have {\mathrm{\Pi}}_{m}(x,v,u) is upper semicontinuous in A\times B\times A with nonempty closed values.

(c)
Now we need to prove the solution set {\mathrm{\Xi}}_{m}(F)\ne \mathrm{\varnothing}.
Define the setvalued mappings {\mathrm{\Theta}}_{m},{\mathrm{\Omega}}_{m}:A\times B\times A:\to {2}^{A\times B} by
{\mathrm{\Theta}}_{m}(x,z,u)=({\mathrm{\Phi}}_{m}(x,z,u),{T}_{1}(x,u)),\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,z,u)\in A\times B\times A
and
{\mathrm{\Omega}}_{m}(x,v,u)=({\mathrm{\Pi}}_{m}(x,v,u),{T}_{2}(x,u)),\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,v,u)\in A\times B\times A.
Then {\mathrm{\Theta}}_{m}, {\mathrm{\Omega}}_{m} are upper semicontinuous and \mathrm{\forall}(x,z,u)\in A\times B\times A, \mathrm{\forall}(x,v,u)\in A\times B\times A, {\mathrm{\Theta}}_{m}(x,z,u) and {\mathrm{\Theta}}_{m}(x,v,u) are nonempty closed convex subsets of A\times B\times A.
Define the setvalued mapping H:(A\times B)\times (A\times B)\to {2}^{(A\times B)\times (A\times B)} by
H((x,z),(u,v))=({\mathrm{\Theta}}_{m}(x,z,u),{\mathrm{\Omega}}_{m}(x,v,u)),\phantom{\rule{1em}{0ex}}\mathrm{\forall}((x,z),(u,v))\in (A\times B)\times (A\times B).
Then H is also upper semicontinuous and \mathrm{\forall}((x,z),(u,v))\in (A\times B)\times (A\times B), H((x,z),(u,v)) is a nonempty closed convex subset of (A\times B)\times (A\times B).
By Lemma 7, there exists a point (({x}^{\ast},{z}^{\ast}),({v}^{\ast},{u}^{\ast}))\in (A\times B)\times (A\times B) such that (({x}^{\ast},{z}^{\ast}),({u}^{\ast},{v}^{\ast}))\in H(({x}^{\ast},{z}^{\ast}),({u}^{\ast},{v}^{\ast})), that is,
({x}^{\ast},{z}^{\ast})\in {\mathrm{\Theta}}_{m}({x}^{\ast},{z}^{\ast},{u}^{\ast}),\phantom{\rule{2em}{0ex}}({u}^{\ast},{v}^{\ast})\in {\mathrm{\Omega}}_{m}({x}^{\ast},{v}^{\ast},{u}^{\ast}),
which implies that {x}^{\ast}\in {\mathrm{\Phi}}_{m}({x}^{\ast},{z}^{\ast},{u}^{\ast}), {z}^{\ast}\in {T}_{1}({x}^{\ast},{u}^{\ast}), {u}^{\ast}\in {\mathrm{\Pi}}_{m}({x}^{\ast},{v}^{\ast},{u}^{\ast}) and {v}^{\ast}\in {T}_{2}({x}^{\ast},{u}^{\ast}). Hence, there exists ({x}^{\ast},{u}^{\ast})\in A\times A, {z}^{\ast}\in {T}_{1}({x}^{\ast},{u}^{\ast}), {v}^{\ast}\in {T}_{2}({x}^{\ast},{u}^{\ast}) such that {x}^{\ast}\in {K}_{1}({x}^{\ast},{u}^{\ast}), {u}^{\ast}\in {K}_{2}({x}^{\ast},{u}^{\ast}), satisfying
0\in {F}_{1}({x}^{\ast},{z}^{\ast},y),\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in {P}_{1}({x}^{\ast},{u}^{\ast}),
and
0\in {F}_{2}({u}^{\ast},{v}^{\ast},y),\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in {P}_{2}({x}^{\ast},{u}^{\ast}),
i.e., (SQVIP_{
α
}) has a solution.

(d)
Now we prove that {\mathrm{\Xi}}_{m}(F) is closed. Indeed, let a net \{({x}_{n},{u}_{n}),n\in I\}\in {\mathrm{\Xi}}_{m}(F): ({x}_{n},{u}_{n})\to ({x}_{0},{u}_{0}). We need to prove that ({x}_{0},{u}_{0})\in {\mathrm{\Xi}}_{m}(F). Indeed, by the lower semicontinuity of {P}_{i},i=1,2, for any {y}_{0}\in {P}_{i}({x}_{0},{u}_{0}), there exists {y}_{n}\in {P}_{i}({x}_{n},{u}_{n}) such that {y}_{n}\to {y}_{0}. Since ({x}_{n},{u}_{n})\in {\mathrm{\Xi}}_{m}(F), there exists {z}_{n}\in {T}_{1}({x}_{n},{u}_{n}), {v}_{n}\in {T}_{2}({x}_{n},{u}_{n}), {x}_{n}\in {K}_{1}({x}_{n},{u}_{n}), {u}_{n}\in {K}_{2}({x}_{n},{u}_{n}) such that
0\in {F}_{1}({x}_{n},{z}_{n},{y}_{n}),
and
0\in {F}_{2}({u}_{n},{v}_{n},{y}_{n}).
Since {K}_{1}, {K}_{2} are upper semicontinuous in A\times A with nonempty closed values, by Lemma 6(i), we have {K}_{1}, {K}_{2} are closed. Thus, {x}_{0}\in {K}_{1}({x}_{0},{u}_{0}), {u}_{0}\in {K}_{2}({x}_{0},{u}_{0}). Since {T}_{1}, {T}_{2} are upper semicontinuous in A\times A with nonempty compact values, there exists {z}_{0}\in {T}_{1}({x}_{0},{u}_{0}) and {v}_{0}\in {T}_{2}({x}_{0},{u}_{0}) such that {z}_{n}\to {z}_{0}, {v}_{n}\to {v}_{0} (taking subnets if necessary). By the condition (v) and ({x}_{n},{z}_{n},{u}_{n},{v}_{n})\to ({x}_{0},{z}_{0},{u}_{0},{v}_{0}), we have
0\in {F}_{1}({x}_{0},{z}_{0},{y}_{0}),
and
0\in {F}_{2}({u}_{0},{v}_{0},{y}_{0}).
This means that ({x}_{0},{u}_{0})\in {\mathrm{\Xi}}_{m}(F). Thus, {\mathrm{\Xi}}_{m}(F) is a closed set.
□
If {K}_{1}(x,u)={P}_{1}(x,u)={S}_{1}(x,u), {K}_{2}(x,u)={P}_{2}(x,u)={S}_{2}(x,u), \alpha =m, and {F}_{1}(x,z,y)={G}_{1}(x,z,y)C, {F}_{2}(u,z,y)={G}_{2}(u,z,y)C, with {S}_{1},{S}_{2}:A\times A\to {2}^{A}, {G}_{1},{G}_{2}:A\times B\times A\to {2}^{Z} are setvalued mappings, and C\subset Z is a nonempty closed convex cone. Then (SQVIP_{
α
}) becomes (SSVQEP) studied in [6].
In this special case, we have the following corollary.
Corollary 9 For each \{i=1,2\}, assume for the problem (SSVQEP) that

(i)
{S}_{i} is continuous in A\times A with nonempty convex closed values;

(ii)
{T}_{i} is usc in A\times A with nonempty convex compact values;

(iii)
for all (x,z,u)\in A\times B\times A, {G}_{i}(x,z,{S}_{i}(x,u))\subset C;

(iv)
for all (x,z,u)\in A\times B\times A, the set \{a\in {S}_{i}(x,u):{G}_{i}(a,z,y)\subset C,\mathrm{\forall}y\in {S}_{i}(x,u)\} is convex;

(v)
the set \{(x,z,y)\in A\times B\times A:{G}_{i}(x,z,y)\subset C\} is closed.
Then the (SSVQEP) has a solution, i.e., there exist (\overline{x},\overline{u})\in A\times A and \overline{z}\in {T}_{1}(\overline{x},\overline{u}), \overline{v}\in {T}_{2}(\overline{x},\overline{u}) such that \overline{x}\in {S}_{1}(\overline{x},\overline{u}), \overline{u}\in {S}_{2}(\overline{x},\overline{u}) and
{G}_{1}(\overline{x},\overline{z},y)\subset C,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in {S}_{1}(\overline{x},\overline{u}),
and
{G}_{2}(\overline{u},\overline{v},y)\subset C,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in {S}_{2}(\overline{x},\overline{u}).
Moreover, the solution set of the (SSVQEP) is closed.
Remark 10 Chen et al. [6] obtained an existence result of (SSVQEP). However, the assumptions in Theorem 3.1 in [6] are different from the assumptions in Corollary 9. The following example shows that all assumptions of Corollary 9 are satisfied. But Theorem 3.1 in [6] is not fulfilled.
Example 11 Let X=Y=Z=\mathbb{R}, A=B=[0,1], C=[0,+\mathrm{\infty}) and let {S}_{1}(x)={S}_{2}(x)=[0,1], {G}_{1},{G}_{2},F:[0,1]\times [0,1]\times [0,1]\to {2}^{\mathbb{R}} and
{T}_{1}(x,u)={T}_{2}(x,u)=\{\begin{array}{cc}[0,2]\hfill & \text{if}{x}_{0}={u}_{0}=\frac{1}{2},\hfill \\ [0,1]\hfill & \text{otherwise}.\hfill \end{array}
and
{G}_{1}(x,z,y)={G}_{2}(u,z,y)=F(x,z,y)=\{\begin{array}{cc}[\frac{1}{2},1]\hfill & \text{if}{x}_{0}={z}_{0}={y}_{0}=\frac{1}{2},\hfill \\ [1,2]\hfill & \text{otherwise}.\hfill \end{array}
We show that assumptions of Corollary 9 are easily seen to be fulfilled. Hence, by Corollary 9, (SSVQEP) has a solution. But F is neither type II Clower semicontinuous nor Cconcave at {x}_{0}=\frac{1}{2}. Thus, Theorem 3.1 in [6] does not work.
If {K}_{1}(x,u)={P}_{1}(x,u)={S}_{1}(x), {K}_{2}(x,u)={P}_{2}(x,u)={S}_{2}(u), {T}_{1}(x,u)={T}_{1}(x), {T}_{2}(x,u)={T}_{2}(u), \alpha =m and {F}_{1}(x,z,y)={G}_{1}(x,z,y)C, {F}_{2}(u,z,y)={G}_{2}(u,z,y)C, with {S}_{1},{S}_{2}:A\to {2}^{A}, {G}_{1},{G}_{2}:A\times B\times A\to {2}^{Z} are setvalued mappings, and C\subset Z is a nonempty closed convex cone. Then (SQVIP_{
α
}) becomes (SGSVQEP) studied in [5].
In this special case, we have the following corollary.
Corollary 12 For each \{i=1,2\}, assume for the problem (SGSVQEP) that

(i)
{S}_{i} is continuous in A with nonempty convex closed values;

(ii)
{T}_{i} is usc in A with nonempty convex compact values;

(iii)
for all (x,z)\in A\times B, {G}_{i}(x,z,{S}_{i}(x))\subset C;

(iv)
for all (x,z)\in A\times B, the set \{a\in {S}_{i}(x):{G}_{i}(a,z,y)\subset C,\mathrm{\forall}y\in {S}_{i}(x)\} is convex;

(v)
the set \{(x,z,y)\in A\times B\times A:{G}_{i}(x,z,y)\subset C\} is closed.
Then the (SGSVQEP) has a solution, i.e., there exist (\overline{x},\overline{u})\in A\times A and \overline{z}\in {T}_{1}(\overline{x}), \overline{v}\in {T}_{2}(\overline{u}) such that \overline{x}\in {S}_{1}(\overline{x}), \overline{u}\in {S}_{2}(\overline{u}) and
{G}_{1}(\overline{x},\overline{z},y)\subset C,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in {S}_{1}(\overline{x})
and
{G}_{2}(\overline{u},\overline{v},y)\subset C,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in {S}_{2}(\overline{u}).
Moreover, the solution set of the (SGSVQEP) is closed.
Remark 13 In [5], PlubtiengSitthithakerngkiet also obtained an existence result of (SGSVQEP). However, the assumptions in Theorem 3.1 in [5] are different from the assumptions in Corollary 12. The following example shows that in this special case, all assumptions of Corollary 12 are satisfied. But Theorem 3.1 in [5] is not fulfilled.
Example 14 Let X=Y=Z=\mathbb{R}, A=B=[0,1], C=[0,+\mathrm{\infty}) and let {S}_{1}(x)={S}_{2}(x)=[0,1], F:[0,1]\times [0,1]\times [0,1]\to {2}^{\mathbb{R}} and
{T}_{1}(x)={T}_{2}(x)=\{\begin{array}{cc}[0,2]\hfill & \text{if}{x}_{0}=\frac{1}{2},\hfill \\ [0,1]\hfill & \text{otherwise}.\hfill \end{array}
and
{G}_{1}(x,z,y)={G}_{2}(u,z,y)=F(x,z,y)=\{\begin{array}{cc}[\frac{1}{2},1]\hfill & \text{if}{x}_{0}={z}_{0}={y}_{0}=\frac{1}{2},\hfill \\ [1,2]\hfill & \text{otherwise}.\hfill \end{array}
We show that all assumptions of Corollary 12 are satisfied. So, by this corollary, the considered problem has solutions. However, F is not lower (C)continuous at {x}_{0}=\frac{1}{2}. Also, Theorem 3.1 in [5] does not work.
If {K}_{1}(\overline{x},\overline{u})={P}_{1}(\overline{x},\overline{u})={K}_{2}(\overline{x},\overline{u})={P}_{2}(\overline{x},\overline{u})=S(\overline{x}), {T}_{1}(\overline{x},\overline{u})={T}_{2}(\overline{x},\overline{u})=\{z\} and {F}_{1}(x,z,y)={F}_{2}(u,z,y)=G(x,y)C, for each \overline{x},\overline{u}\in A and S:A\to {2}^{A}, G:A\times A\to {2}^{Z} are setvalued mappings, and C\subset Z is a nonempty closed convex cone. Then (SQVIP_{
α
}) becomes (SVQEP) studied in [21].
In this special case, we also have the following corollary.
Corollary 15 Assume for the problem (SVQEP) that

(i)
S is continuous in A with nonempty convex closed values;

(ii)
for all x\in A, G(x,S(x))\subset C;

(iii)
for all x\in A, the set \{a\in S(x):G(a,y)\subset C,\mathrm{\forall}y\in S(x)\} is convex;

(iv)
the set \{(x,y)\in A\times A:G(x,y)\subset C\} is closed.
Then the (SVQEP) has a solution, i.e., there exists \overline{x}\in S(\overline{x}) such that
G(\overline{x},y)\subset C,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in S(\overline{x}).
Moreover, the solution set of the (SVQEP) is closed.
The following example shows that in this special case, all assumptions of Corollary 15 are satisfied. But Theorem 3.3 in [21] is not fulfilled.
Example 16 Let X, Y, Z, A, B, C as in Example 14, and let S(x)=[0,1], G:[0,1]\times [0,1]\to {2}^{\mathbb{R}} and
G(x,y)=\{\begin{array}{cc}[1,\frac{5}{2}]\hfill & \text{if}{x}_{0}={z}_{0}={y}_{0}=\frac{1}{2},\hfill \\ [\frac{1}{5},\frac{3}{4}]\hfill & \text{otherwise}.\hfill \end{array}
We show that all assumptions of Corollary 15 are satisfied. So, (SVQEP) has a solution. However, G is not upper Ccontinuous at {x}_{0}=\frac{1}{2}. Also, Theorem 3.3 in [21] does not work.
The following example shows that all assumptions of Corollary 9, Corollary 12 and Corollary 15 are satisfied. However, Theorem 3.1 in [6], Theorem 3.1 in [5] and Theorem 3.3 in [21] are not fulfilled. The reason is that G is not properly Cquasiconvex.
Example 17 Let A, B, X, Y, Z, C as in Example 14, and let S:[0,1]\to {2}^{\mathbb{R}}, G:[0,1]\times [0,1]\to {2}^{\mathbb{R}}, {S}_{1}(x,u)={S}_{2}(x,u)=S(x)=[0,1], {T}_{1}(x,u)={T}_{2}(x,u)=T(x,u)=\{z\} and
{G}_{1}(x,z,y)={G}_{2}(u,z,y)=G(x,y)=\{\begin{array}{cc}[1,2]\hfill & \text{if}{x}_{0}={y}_{0}=\frac{1}{2},\hfill \\ [\frac{1}{2},1]\hfill & \text{otherwise}.\hfill \end{array}
We show that all assumptions of Corollary 9, Corollary 12 and Corollary 15 are satisfied. However, G is not properly Cquasiconvex at {x}_{0}=\frac{1}{2}. Thus, it gives the case where Corollary 9, Corollary 12 and Corollary 15 can be applied but Theorem 3.1 in [6], Theorem 3.1 in [5] and Theorem 3.3 in [21] do not work.