In this section, we introduce AGVQEP in Hadamard manifolds and present a sufficient condition for the existence of solutions to AGVQEP. As applications, we obtain results to solve AVQEP, GSEP, SEP, and the perturbed saddle point problem in noncompact Hadamard manifolds.
Let K be a nonempty subset of E, W be a nonempty set, and let \(H:E\rightrightarrows E\), \(C:E\rightrightarrows W\), \(\psi:E \times E\rightrightarrows W\) be setvalue mappings. We consider the following AGVQEP as follows: find \(\widehat{x}\in K\) such that
$$\widehat{x}\in H(\widehat{x}) \quad \mbox{and}\quad \psi(\widehat{x},y) \nsubseteq C( \widehat{x}),\quad \forall y\in H(\widehat{x}). $$
It is worthwhile noting that AGVQEP is motivated by the generalized vector quasiequilibrium problem introduced by Ansari and FloresBazán [7]. In particular, let \(E=\mathbb{R}^{n}\), W be a Hausdorff topological vector space, and let \(P:\mathbb{R} ^{n}\rightrightarrows W\) be a setvalued mapping such that, for each \(x\in\mathbb{R}^{n}\), \(P(x)\) is a closed and convex cone with \(\operatorname{int}P(x)\neq\emptyset\). Moreover, let \(C:\mathbb{R}^{n} \rightrightarrows W\) be a setvalued mapping defined by \(C(x)= \operatorname{int}P(x)\) for every \(x\in\mathbb{R}^{n}\). Then AGVQEP retrieves a particular instance of the equilibrium problem in [7]. Here we would like to point out that the feasible set of AGVQEP is controlled by a setvalued mapping. In the real world, there are important problems which can be regarded as AGVQEPs in which the condition that the feasible set of AGVQEP is controlled by a setvalued mapping must be satisfied; for example, the equilibrium problems of the generalized games in Dasgupta and Maskin [38], Smeers et al. [39], Krawczyk [40], and Ansink and Houba [41].
Remark 3.1
If \(H(x)\equiv E\) for every \(x\in E\), W is a Hausdorff topological vector space, and each \(C(x)\) is replaced by \(\operatorname{int}C(x)\), where \(C(x)\) is a closed and convex cone with \(\operatorname{int}C(x)\neq\emptyset\), then AGVQEP reduces to the generalized vector equilibrium problem investigated by Batista et al. [25]. By the arguments in [25], we can see that AGVQEP also includes the equilibrium problems in [21, 23, 26] as its special cases.
Remark 3.2
Some other special cases of AGVQEP are given as follows.

(I)
Let \(H(x)\equiv E\) for every \(x\in E\). Then AGVQEP reduces to the abstract vector quasiequilibrium problem (for short, AVQEP), which consists in finding \(\widehat{x}\in E\) such that \(\psi(\widehat{x},y) \nsubseteq C(\widehat{x})\) for every \(y\in E\).

(II)
If \(W=\mathbb{R}\), \(C(x)\equiv(\infty,0)\) for every \(x\in E\), and \(F=f\), where \(f:E\times E\rightarrow\mathbb{R}\) is a bifunction, then GVQEP reduces to the generalized scalar equilibrium problem (for short, GSEP), which is to find \(\widehat{x}\in E\) such that \(\widehat{x}\in H(\widehat{x})\) and \(f(\widehat{x},y)\geq0\) for every \(y\in H(\widehat{x})\). Furthermore, if \(H(x)\equiv E\) for every \(x\in E\), then GSEP reduces to SEP.
Now, we are ready, by using Lemma 2.5, to present the following existence theorem of solutions to AGVQEP in noncompact Hadamard manifolds.
Theorem 3.1
Let
\(K\subseteq E\)
be a nonempty compact set and
W
be a nonempty set. Let
\(\varsigma, \psi:E\times E\rightrightarrows W\), \(C:E\rightrightarrows W\)
be three setvalued mappings and
\(H:E\rightrightarrows E\)
be a FanBrowder mapping. Assume that
 (i):

the set
\(E^{*}=\{x\in E:x\notin H(x)\}\)
is open in
E;
 (ii):

for each
\((x,y)\in E\times E\), \(\varsigma(x,y)\subseteq C(x)\)
implies
\(\psi(x,y)\subseteq C(x)\);
 (iii):

for each
\(x\in E\), \(\psi(x,x)\nsubseteq C(x)\);
 (iv):

for each
\(x\in E\), the set
\(\{y\in E:\psi(x,y)\subseteq C(x)\}\)
is convex;
 (v):

\(\bigcup_{y\in E}\{x\in H^{1}(y):\psi(x,y)\subseteq C(x)\}= \bigcup_{y\in E}\operatorname{int}\{x\in H^{1}(y):\psi(x,y)\subseteq C(x)\}\);
 (vi):

one of the following conditions holds:
 (vi)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset
\(E_{N}\)
of
E
containing
N
such that
$$E_{N}\setminus K\subseteq\bigcup_{y\in E_{N}} \operatorname{int}\bigl( \bigl(E^{*}\cap H^{1}(y) \bigr) \cup \bigl(H^{1}(y)\cap \bigl\{ x\in E:\psi(x,y) \subseteq C(x) \bigr\} \bigr) \bigr); $$
 (vi)_{2}
:

there exists a point
\(y_{0}\in E\)
such that
$$E\setminus K\subseteq \operatorname{int}\bigl( \bigl(E^{*}\cap H^{1}(y_{0}) \bigr)\cup \bigl(H^{1}(y_{0}) \cap \bigl\{ x\in E:\psi(x,y_{0})\subseteq C(x) \bigr\} \bigr) \bigr). $$
Then AGVQEP has at least a solution in
K.
Proof
Define two setvalued mappings \(L, Q:E\rightrightarrows E\) by
$$L(x)= \bigl\{ y\in E:\varsigma(x,y)\subseteq C(x) \bigr\} \quad \mbox{and}\quad Q(x)= \bigl\{ y \in E:\psi(x,y)\subseteq C(x) \bigr\} ,\quad \forall x\in E. $$
Moreover, let us define two setvalued mappings \(F, G:E\rightrightarrows E\) by
$$\begin{aligned} F(x)= \textstyle\begin{cases} H(x)\cap L(x), & \mbox{if } x\notin E^{*}, \\ H(x), & \mbox{if } x\in E^{*}, \end{cases}\displaystyle \end{aligned}$$
and
$$\begin{aligned} G(x)= \textstyle\begin{cases} H(x)\cap Q(x), & \mbox{if } x\notin E^{*}, \\ H(x), & \mbox{if } x\in E^{*}. \end{cases}\displaystyle \end{aligned}$$
By (ii), we have \(F(x)\subseteq G(x)\) for every \(x\in E\). Since each \(Q(x)\) is convex by (iv) and each \(H(x)\) is convex by the definition of a FanBrowder mapping, it follows that \(G(x)\) is convex for every \(x\in E\).
For each \(y\in E\), we have
$$\begin{aligned} G^{1}(y) =& \bigl\{ x\in E^{*}:y\in H(x) \bigr\} \cup \bigl\{ x\notin E^{*}:y\in H(x) \cap Q(x) \bigr\} \\ =& \bigl(E^{*}\cap H^{1}(y) \bigr)\cup \bigl( \bigl(E \setminus E^{*} \bigr)\cap \bigl(H^{1}(y) \cap Q^{1}(y) \bigr) \bigr) \\ =& \bigl(E^{*}\cap H^{1}(y) \bigr)\cup \bigl(H^{1}(y)\cap Q^{1}(y) \bigr) \\ =& \bigl(E^{*}\cap H^{1}(y) \bigr)\cup \bigl(H^{1}(y)\cap \bigl\{ x\in E:\psi(x,y) \subseteq C(x) \bigr\} \bigr). \end{aligned}$$
We claim that \(\bigcup_{y\in E}G^{1}(y)=\bigcup_{y\in E}\operatorname{int}G ^{1}(y)\). In fact, it is clear that \(\bigcup_{y\in E}\operatorname{int}G^{1}(y) \subseteq\bigcup_{y\in E}G^{1}(y)\). In order to prove that \(\bigcup_{y\in E}G^{1}(y)\subseteq\bigcup_{y\in E}\operatorname{int}G^{1}(y)\), we take \(x_{0}\in \bigcup_{y\in E}G^{1}(y)\). Then there exists \(y^{*}\in E\) such that
$$x_{0}\in G^{1} \bigl(y^{*} \bigr)= \bigl(E^{*}\cap H^{1} \bigl(y^{*} \bigr) \bigr)\cup \bigl(H^{1} \bigl(y ^{*} \bigr)\cap \bigl\{ x\in E:\psi \bigl(x,y^{*} \bigr)\subseteq C(x) \bigr\} \bigr). $$
If \(x_{0}\in E^{*}\cap H^{1}(y^{*})\), then by (i) and the definition of a FanBrowder mapping, we have
$$x_{0}\in E^{*}\cap H^{1} \bigl(y^{*} \bigr)=\operatorname{int}\bigl(E^{*}\cap H^{1} \bigl(y ^{*} \bigr) \bigr)\subseteq \operatorname{int}G^{1} \bigl(y^{*} \bigr). $$
We notice that, according to (v) and the definition of Q, one has
$$\bigcup_{y\in E} \bigl(H^{1}(y)\cap Q^{1}(y) \bigr)=\bigcup_{y\in E} \operatorname{int}\bigl(H^{1}(y)\cap Q^{1}(y) \bigr). $$
So, if \(x_{0}\in H^{1}(y^{*})\cap\{x\in E:\psi(x,y^{*})\subseteq C(x)\}=H^{1}(y^{*})\cap Q^{1}(y^{*})\), then there exists \(\widetilde{y}\in E\) such that
$$x_{0}\in \operatorname{int}\bigl(H^{1}(\widetilde{y})\cap Q^{1}(\widetilde{y}) \bigr) \subseteq \operatorname{int}G^{1}( \widetilde{y}). $$
Combining these two cases, we can conclude that \(\bigcup_{y\in E}G ^{1}(y)\subseteq\bigcup_{y\in E}\operatorname{int}G^{1}(y)\). Therefore, we have \(\bigcup_{y\in E}G^{1}(y)=\bigcup_{y\in E}\operatorname{int}G^{1}(y)\).
Now, we show that \(x\notin G(x)\) for every \(x\in E\). Indeed, if \(x\in E^{*}\), then by the definition of \(E^{*}\), we have \(x\notin H(x)=G(x)\); if \(x\notin E^{*}\), then by (iii), \(x\notin Q(x)\) and so, \(x\notin H(x)\cap Q(x)=G(x)\).
By (vi) and the above fact, we can see that one of the following conditions holds:

for each \(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset \(E_{N}\) of E containing N such that
$$\begin{aligned} E_{N}\setminus K \subseteq&\bigcup_{y\in E_{N}} \operatorname{int}\bigl( \bigl(E^{*} \cap H^{1}(y) \bigr) \cup \bigl(H^{1}(y)\cap \bigl\{ x\in E:\psi(x,y) \subseteq C(x) \bigr\} \bigr) \bigr) \\ =&\bigcup_{y\in E_{N}}\operatorname{int}G^{1}(y); \end{aligned}$$

there exists a point \(y_{0}\in E\) such that
$$\begin{aligned} K \supseteq&E\setminus \operatorname{int}\bigl( \bigl(E^{*}\cap H^{1}(y_{0}) \bigr)\cup \bigl(H^{1}(y_{0}) \cap \bigl\{ x\in E:\psi(x,y_{0})\subseteq C(x) \bigr\} \bigr) \bigr) \\ =&\operatorname{cl}\bigl(E\setminus G^{1}(y_{0}) \bigr). \end{aligned}$$
Therefore, by Lemma 2.5 and Remark 2.3, there exists \(\widehat{x} \in K\) such that \(G(\widehat{x})=\emptyset\). Since \(H(x)\neq\emptyset \) for every \(x\in E\), it follows that \(\widehat{x}\notin E^{*}\) and \(G(\widehat{x})=H(\widehat{x})\cap Q(\widehat{x})=\emptyset\); i.e., \(\widehat{x}\in H(\widehat{x})\) and \(\psi(\widehat{x},y)\nsubseteq C(\widehat{x})\) for every \(y\in H(\widehat{x})\). Thus, the conclusion of Theorem 3.1 holds and the proof is complete. □
Corollary 3.1
Let
\(K\subseteq E\)
be a nonempty compact set and
W
be a nonempty set. Let
\(\psi:E\times E\rightrightarrows W\), \(C:E\rightrightarrows W\)
be two setvalued mappings and
\(H:E\rightrightarrows E\)
be a FanBrowder mapping. Assume that
 (i):

the set
\(E^{*}=\{x\in E:x\notin H(x)\}\)
is open in
E;
 (ii):

for each
\(x\in E\), \(\psi(x,x)\nsubseteq C(x)\);
 (iii):

for each
\(x\in E\), the set
\(\{y\in E:\psi(x,y)\subseteq C(x)\}\)
is convex;
 (iv):

\(\bigcup_{y\in E}\{x\in H^{1}(y):\psi(x,y)\subseteq C(x)\}= \bigcup_{y\in E}\operatorname{int}\{x\in H^{1}(y):\psi(x,y)\subseteq C(x)\}\);
 (v):

one of the following conditions holds:
 (v)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset
\(E_{N}\)
of
E
containing
N
such that
$$E_{N}\setminus K\subseteq\bigcup_{y\in E_{N}} \operatorname{int}\bigl( \bigl(E^{*}\cap H^{1}(y) \bigr) \cup \bigl(H^{1}(y)\cap \bigl\{ x\in E:\psi(x,y) \subseteq C(x) \bigr\} \bigr) \bigr); $$
 (v)_{2}
:

there exists a point
\(y_{0}\in E\)
such that
$$E\setminus K\subseteq \operatorname{int}\bigl( \bigl(E^{*}\cap H^{1}(y_{0}) \bigr)\cup \bigl(H^{1}(y_{0}) \cap \bigl\{ x\in E:\psi(x,y_{0})\subseteq C(x) \bigr\} \bigr) \bigr). $$
Then AGVQEP has at least a solution in
K.
Proof
Let \(\varsigma=\psi\). It is easy to see that all the conditions of Theorem 3.1 are satisfied. Therefore, it follows from Theorem 3.1 that AGVQEP has at least a solution in K. Thus, the result holds and the proof of Corollary 3.1 is complete. □
Corollary 3.2
Let
\(K\subseteq E\)
be a nonempty compact set and
W
be a nonempty set. Let
\(\psi:E\times E\rightrightarrows W\), \(C:E\rightrightarrows W\)
be two setvalued mappings and
\(H:E\rightrightarrows E\)
be a FanBrowder mapping. Assume that
 (i):

the set
\(E^{*}=\{x\in E:x\notin H(x)\}\)
is open in
E;
 (ii):

for each
\(x\in E\), \(\psi(x,x)\nsubseteq C(x)\);
 (iii):

for each
\(x\in E\), the set
\(\{y\in E:\psi(x,y)\subseteq C(x)\}\)
is convex;
 (iv):

for each
\(y\in E\), the set
\(\{x\in E:\psi(x,y)\subseteq C(x)\}\)
is open in
E;
 (v):

one of the following conditions holds:
 (v)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset
\(E_{N}\)
of
E
containing
N
such that
$$E_{N}\setminus K\subseteq\bigcup_{y\in E_{N}} \bigl( \bigl(E^{*}\cap H^{1}(y) \bigr) \cup \bigl(H^{1}(y)\cap \bigl\{ x\in E:\psi(x,y)\subseteq C(x) \bigr\} \bigr) \bigr); $$
 (v)_{2}
:

there exists a point
\(y_{0}\in E\)
such that
$$E\setminus K\subseteq \bigl( \bigl(E^{*}\cap H^{1}(y_{0}) \bigr)\cup \bigl(H^{1}(y _{0})\cap \bigl\{ x\in E: \psi(x,y_{0})\subseteq C(x) \bigr\} \bigr) \bigr). $$
Then AGVQEP has at least a solution in
K.
Proof
By (iv), (v), and the fact that \(H^{1}(y)\) is open in E for every \(y\in E\), we can see that (iv) and (v) of Corollary 3.1 hold. Therefore, by Corollary 3.1, AGVQEP has at least a solution in K. This completes the proof. □
Remark 3.3
Corollary 3.2 extends Theorem 3.1 of Batista et al. [25] in the following aspects: (a) concerns the more general abstract generalized vector quasiequilibrium problems instead of the generalized vector equilibrium problems; (b) from one coercivity condition to two alternative coercivity conditions; (c) since W in Corollary 3.2 does not need to be a real Hausdorff topological vector space, it is not required for each \(C(x)\) to be a closed and convex cone; (d) (iii) is weaker than h3 of Theorem 3.1 of Batista et al. [25]; (e) by the fact that W in Corollary 3.2 may be any nonempty set without topological structure, we adopt the assumption that the set \(\{x\in E:\psi(x,y) \subseteq C(x)\}\) is open in E for every \(y\in E\), which is weaker than h2 of Theorem 3.1 of Batista et al. [25]. In addition, the proof of Corollary 3.2 originates from the existence of maximal elements in noncompact Hadamard manifolds, while the authors of [25] used the KKM property to prove their result. Therefore, the proof technique of Corollary 3.2 is different from that of Theorem 3.1 of Batista et al. [25].
By using Corollary 3.2, we can prove the existence of an equilibrium for the generalized water market game model under the condition that there are a river structure, water balances, and heterogeneous water users via a water delivery infrastructure. We would like to point out that our convex and continuous conditions are weaker than the corresponding conditions in Proposition 2.1 due to Ansink and Houba [41].
Corollary 3.3
Let
\(K\subseteq E\)
be a nonempty compact set, W
be a nonempty set, and let
\(C:E\rightrightarrows W\), \(\psi:E\times E\rightrightarrows W\)
be two setvalued mappings. Assume that
 (i):

for each
\(x\in E\), \(\psi(x,x)\nsubseteq C(x)\);
 (ii):

for each
\(x\in E\), the set
\(\{y\in E:\psi(x,y)\subseteq C(x)\}\)
is convex;
 (iii):

for each
\(y\in E\), the set
\(\{x\in E:\psi(x,y)\subseteq C(x)\}\)
is open in
E;
 (iv):

one of the following conditions holds:
 (iv)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset
\(E_{N}\)
of
E
containing
N
such that
\(E_{N}\setminus K\subseteq\bigcup_{y\in E_{N}}\{x\in E:\psi(x,y) \subseteq C(x)\}\);
 (iv)_{2}
:

there exists a point
\(y_{0}\in E\)
such that
\(E\setminus K\subseteq\{x\in E:\psi(x,y_{0})\subseteq C(x)\}\).
Then AVQEP has at least a solution in
K.
Proof
The conclusion of Corollary 3.3 follows from Corollary 3.2 by letting \(H(x)\equiv E\) for every \(x\in E\). This completes the proof. □
Remark 3.4
Let us give the following items:
(1) If W is a real Hausdorff topological vector space and \(\{C(x):y\in E\}\) is a family of nonempty convex cones, then (iv) of Theorem 3.1, (iii) of Corollaries 3.13.2, and (ii) of Corollary 3.3 can be replaced by one of the following conditions:

for each \(x\in E\), \(\psi(x,\cdot)\) is convex with respect to \(C(x)\);

for each \(x\in E\), \(\psi(x,\cdot)\) is quasiconvexlike with respect to \(C(x)\).
In fact, we consider the first assumption. Let \(x\in E\) be any given. In order to prove that the set \(\{y\in E:\psi(x,y)\subseteq C(x) \}\) is convex, we assume that \(y_{1}, y_{2}\in\{y\in E:\psi(x,y) \subseteq C(x)\}\) and \(t\in[0,1]\). Applying this assumption and the fact that each \(C(x)\) is a convex cone, we have
$$\begin{aligned} \psi \bigl(x,\operatorname{exp}_{y_{1}}t\operatorname{exp}_{y_{1}}^{1}y_{2} \bigr) \subseteq&t \psi(x,y_{1})+(1t)\psi(x,y_{2})+C(x) \\ \subseteq&tC(x)+(1t)C(x)+C(x) \\ \subseteq&C(x), \end{aligned}$$
which implies that \(\operatorname{exp}_{y_{1}}t\operatorname{exp}_{y_{1}}^{1}y_{2} \in\{y\in E:\psi(x,y)\subseteq C(x)\}\) for every \(t\in[0,1]\) and so, the set \(\{y\in E:\psi(x,y)\subseteq C(x)\}\) is convex for every \(x\in E\). Now, we prove that the desired conclusion holds under the condition that the second assumption is satisfied. Indeed, let \(x\in E\) be fixed and then let \(y_{1}, y_{2}\in\{y\in E:\psi(x,y) \subseteq C(x)\}\) and \(t\in[0,1]\) be any given. By the definition of a quasiconvexlike mapping, we have
$$\psi \bigl(x,\operatorname{exp}_{y_{1}}t\operatorname{exp}_{y_{1}}^{1}y_{2} \bigr)\subseteq \psi(x,y_{1})+C(x), $$
or
$$\psi \bigl(x,\operatorname{exp}_{y_{1}}t\operatorname{exp}_{y_{1}}^{1}y_{2} \bigr)\subseteq \psi(x,y_{2})+C(x). $$
Since \(y_{1}, y_{2}\in\{y\in E:\psi(x,y)\subseteq C(x)\}\), we have \(\psi(x,y_{1})\subseteq C(x)\) and \(\psi(x,y_{2})\subseteq C(x)\). Therefore, given \(C(x)+C(x)\subseteq C(x)\), which is obtained by using the property of a convex cone, it follows from the above formulas that \(\psi(x,\operatorname{exp}_{y_{1}}t\operatorname{exp}_{y_{1}}^{1}y_{2})\subseteq C(x)\), and this shows that the set \(\{y\in E:\psi(x,y)\subseteq C(x)\}\) is convex for every \(x\in E\).
(2) If W is a Hausdorff topological space, then (iv) of Corollary 3.2 and (iii) of Corollary 3.3 can be replaced by the following conditions:

for each \(y\in E\), \(\psi(\cdot,y)\) is upper semicontinuous on E with compact values;

the graph \(G_{r}(C)\) of C; i.e., \(\{(x,w)\in E\times W:w\in C(x)\}\) is an open set in \(E\times W\).
Indeed, it suffices to prove that the set \(\{x\in E:\psi(x,y)\nsubseteq C(x)\}\) is closed in E for every \(y\in E\). Let \(\{x_{ \alpha}\}\) be a net in \(\{x\in E:\psi(x,y)\nsubseteq C(x)\}\) such that \(x_{\alpha}\rightarrow x_{0}\). Since \(\psi(x_{\alpha},y) \nsubseteq C(x_{\alpha})\), there exists \(z_{\alpha}\in\psi(x _{\alpha},y)\) such that \(z_{\alpha}\notin C(x_{\alpha})\). Hence, we have \(z_{\alpha}\in W\setminus C(x_{\alpha})\). By the upper semicontinuity and compact values of ψ on E, it follows from Proposition 1 in [42] that there exists a subnet of \(\{z_{\alpha}\}\) with limit \(z_{0}\) and \(z_{0}\in\psi(x_{0},y)\). Without loss of generality, let us assume that \(z_{\alpha}\rightarrow z_{0}\in\psi(x_{0},y)\). On the other hand, since the graph \(G_{r}(C)\) of C is an open set in \(E\times W\), the setvalued mapping \(x\rightrightarrows W\setminus C(x)\) has a closed graph in \(E\times W\). It follows that \(z_{0}\in W\setminus C(x_{0})\) and so, \(z_{0}\notin C(x_{0})\). Thus, \(x_{0}\in\{x\in E: \psi(x,y)\nsubseteq C(x)\}\), which implies that the set \(\{x\in E:\psi(x,y)\nsubseteq C(x)\}\) is closed in E for every \(y\in E\). Therefore, the set \(\{x\in E:\psi(x,y)\subseteq C(x)\}\) is open in E for every \(y\in E\).
(3) If (iv)_{2} of Corollary 3.3 holds, then the solution set of AVQEP is a nonempty compact set, which can be written as follows:
$$\bigcap_{y\in E} \bigl\{ x\in E:\psi(x,y)\nsubseteq C(x) \bigr\} . $$
In fact, by the conclusion of Corollary 3.3, we can see that the above set is nonempty. Furthermore, it follows from (iv)_{2} of Corollary 3.3 that
$$\begin{aligned} \emptyset\neq\bigcap_{y\in E} \bigl\{ x\in E:\psi(x,y) \nsubseteq C(x) \bigr\} \subseteq& \bigl\{ x\in E:\psi(x,y_{0}) \subseteq C(x) \bigr\} \subseteq K. \end{aligned}$$
Together with (iii) of Corollary 3.3, we can see that the solution set of AVQEP is a nonempty closed subset of K. Therefore, the solution set of AVQEP is a nonempty compact set.
If \(W=\mathbb{R}\), \(C(x)\equiv(\infty,0)\) for every \(x\in E\) and \(F=f\), where \(f:E\times E\rightarrow\mathbb{R}\) is a bifunction, then Corollaries 3.2 and 3.3 reduce to the following existence results of solutions to GSEP and SEP, respectively.
Corollary 3.4
Let
\(K\subseteq E\)
be a nonempty compact set, \(H:E\rightrightarrows E\)
be a FanBrowder mapping, and
\(f:E\times E\rightarrow\mathbb{R}\)
be a bifunction. Assume that
 (i):

the set
\(E^{*}=\{x\in E:x\notin H(x)\}\)
is open in
E;
 (ii):

for each
\(x\in E\), \(f(x,x)\geq0\);
 (iii):

for each
\(x\in E\), the set
\(\{y\in E:f(x,y)< 0\}\)
is convex;
 (iv):

for each
\(y\in E\), the set
\(\{x\in E:f(x,y)\geq0\}\)
is closed in
E;
 (v):

one of the following conditions holds:
 (v)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset
\(E_{N}\)
of
E
containing
N
such that
$$E_{N}\setminus K\subseteq\bigcup_{y\in E_{N}} \bigl( \bigl(E^{*}\cap H^{1}(y) \bigr) \cup \bigl(H^{1}(y)\cap \bigl\{ x\in E:f(x,y)< 0 \bigr\} \bigr) \bigr); $$
 (v)_{2}
:

there exists a point
\(y_{0}\in E\)
such that
$$E\setminus K\subseteq \bigl( \bigl(E^{*}\cap H^{1}(y_{0}) \bigr)\cup \bigl(H^{1}(y _{0})\cap \bigl\{ x\in E:f(x,y_{0})< 0 \bigr\} \bigr) \bigr). $$
Then GSEP has at least a solution in
K.
Corollary 3.5
Let
\(K\subseteq E\)
be a nonempty compact set and
\(f:E\times E\rightarrow\mathbb{R}\)
be a bifunction. Assume that
 (i):

for each
\(x\in E\), \(f(x,x)\geq0\);
 (ii):

for each
\(x\in E\), \(\{y\in E:f(x,y)<0\}\)
is convex;
 (iii):

for each
\(y\in E\), the set
\(\{x\in E:f(x,y)\geq0\}\)
is closed in
E;
 (iv):

one of the following conditions holds:
 (iv)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exists a nonempty compact convex subset
\(E_{N}\)
of
E
containing
N
such that
\(E_{N}\setminus K\subseteq\bigcup_{y\in E_{N}}\{x\in E:f(x,y)<0\}\);
 (iv)_{2}
:

there exists a point
\(y_{0}\in E\)
such that
\(E\setminus K\subseteq\{x\in E:f(x,y_{0})<0\}\).
Then SEP has at least a solution in
K.
Remark 3.5
Corollary 3.5 improves Theorem 3.2 of Colao et al. [21] because there are two alternative coercivity conditions in Corollary 3.5, while there is only one coercivity condition in Theorem 3.2 of Colao et al. [21]. Furthermore, the coercivity condition (iv)_{2} of Corollary 3.5 is weaker than the coercivity condition (iv) of Theorem 3.2 of Colao et al. [21]. To see this, we can consider K in Theorem 3.2 of Colao et al. [21] as a Hadamard submanifold, and then let \(L'=L\cap K\), which is a nonempty compact subset of K. Then it follows from \(K\setminus L=K\setminus L'\) and \(y_{0}\in L\cap K\subseteq K\) that (iv) of Theorem 3.2 of Colao et al. [21] implies (iv)_{2} of Corollary 3.5.
As an application of Corollary 3.5, we have the following perturbed saddle point theorem in noncompact Hadamard manifolds.
Theorem 3.2
Let
\(K_{1}, K_{2}\subseteq E\)
be two nonempty compact sets and
\(f, g:E\times E\rightarrow\mathbb{R}\)
be two bifunctions. Assume that
 (i):

for each
\(x\in E\), \(f(x,x)g(x,x)=0\);
 (ii):

for each
\(x\in E\), \(\{y\in E:f(x,y)< g(x,y)\}\)
is convex;
 (iii):

for each
\(y\in E\), \(\{x\in E:f(x,y)>g(x,y)\}\)
is convex;
 (iv):

the bifunction
\(E\times E\ni(x,y)\mapsto f(x,y)g(x,y)\)
is continuous;
 (v):

one of the following conditions holds:
 (v)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exist two nonempty compact convex subsets
\(E_{N}\), \(\widetilde{E}_{N}\)
of
E
containing
N
such that
\(E_{N}\setminus K_{1}\subseteq \bigcup_{y\in E_{N}}\{x\in E:f(x,y)< g(x,y)\}\)
and
\(\widetilde{E}_{N} \setminus K_{2}\subseteq\bigcup_{x\in\widetilde{E}_{N}}\{y\in E:f(x,y)>g(x,y) \}\);
 (v)_{2}
:

there exist two points
\(y_{0}, x_{0}\in E\)
such that
\(E\setminus K_{1}\subseteq\{x\in E:f(x,y_{0})< g(x,y_{0})\}\)
and
\(E\setminus K_{2}\subseteq\{y\in E:f(x_{0},y)>g(x_{0},y)\}\).
Then
f
has a perturbed saddle point
\((\widehat{x},\widehat{y}) \in K_{1}\times K_{2}\); i.e.,
$$f(x,\widehat{y})+ \bigl[g(\widehat{x},\widehat{y})g(x,\widehat{y}) \bigr]\leq f( \widehat{x},\widehat{y})\leq f(\widehat{x},y)+ \bigl[g(\widehat{x}, \widehat{y})g( \widehat{x},y) \bigr],\quad \forall (x,y)\in E\times E. $$
In particular, \(\inf_{y\in E}\sup_{x\in E}[f(x,y)g(x,y)]=\sup_{x \in E}\inf_{y\in E}[f(x,y)g(x,y)]\).
Proof
Define a bifunction \(h_{1}:E\times E\rightarrow \mathbb{R}\) by \(h_{1}(x,y)=f(x,y)g(x,y)\) for every \((x,y)\in E\times E\). By (i), (ii), (iv) and the first parts of (v)_{1} and (v)_{2}, we can see that all the conditions of Corollary 3.5 are satisfied. Thus, by Corollary 3.5, there exists \(\widehat{x}\in K_{1}\) such that \(h_{1}(\widehat{x},y)\geq0\) for every \(y\in E\). Define a bifunction \(h_{2}:E\times E\rightarrow\mathbb{R}\) by \(h_{2}(y,x)=g(x,y)f(x,y)\) for every \((y,x)\in E\times E\). Then it follows from (i), (iii), (iv), and the second parts of (v)_{1} and (v)_{2} that all the hypotheses of Corollary 3.5 are fulfilled. Thus, by Corollary 3.5 again, there exists \(\widehat{y}\in K_{2}\) such that \(h_{2}(\widehat{y},x)\geq0\) for every \(x\in E\). Therefore, we have \(f(\widehat{x},\widehat{y})g( \widehat{x},\widehat{y})=0\) and
$$f(x,\widehat{y})g(x,\widehat{y})\leq0=f(\widehat{x},\widehat{y})g( \widehat{x},\widehat{y})\leq f(\widehat{x},y)g(\widehat{x},y),\quad \forall (x,y) \in E \times E. $$
It follows from the above inequality that
$$f(x,\widehat{y})+ \bigl[g(\widehat{x},\widehat{y})g(x,\widehat{y}) \bigr]\leq f( \widehat{x},\widehat{y})\leq f(\widehat{x},y)+ \bigl[g(\widehat{x}, \widehat{y})g( \widehat{x},y) \bigr], \quad \forall (x,y)\in E\times E, $$
and
$$\inf_{y\in E}\sup_{x\in E} \bigl[f(x,y)g(x,y) \bigr]\leq\sup_{x\in E}\inf_{y \in E} \bigl[f(x,y)g(x,y) \bigr]. $$
Since \(\inf_{y\in E}\sup_{x\in E}[f(x,y)g(x,y)]\geq\sup_{x\in E} \inf_{y\in E}[f(x,y)g(x,y)]\) is always true, we get
$$\inf_{y\in E}\sup_{x\in E} \bigl[f(x,y)g(x,y) \bigr]=\sup_{x\in E}\inf_{y\in E} \bigl[f(x,y)g(x,y) \bigr]. $$
This completes the proof. □
By setting \(g(x,y)\equiv0\) for every \((x,y)\in E\times E\), we obtain the following saddle point theorem from Theorem 3.2.
Theorem 3.3
Let
\(K_{1}, K_{2}\subseteq E\)
be two nonempty compact sets and
\(f:E\times E\rightarrow\mathbb{R}\)
be a bifunction. Assume that
 (i):

for each
\(x\in E\), \(f(x,x)=0\);
 (ii):

for each
\(x\in E\), \(\{y\in E:f(x,y)<0\}\)
is convex;
 (iii):

for each
\(y\in E\), \(\{x\in E:f(x,y)>0\}\)
is convex;
 (iv):

the bifunction
\(E\times E\ni(x,y)\mapsto f(x,y)\)
is continuous;
 (v):

one of the following conditions holds:
 (v)_{1}
:

for each
\(N\in{ \mathcal {F}}(E)\), there exist two nonempty compact convex subsets
\(E_{N}\), \(\widetilde{E}_{N}\)
of
E
containing
N
such that
\(E_{N}\setminus K_{1}\subseteq \bigcup_{y\in E_{N}}\{x\in E:f(x,y)<0\}\)
and
\(\widetilde{E}_{N}\setminus K_{2}\subseteq\bigcup_{x\in\widetilde{E}_{N}}\{y\in E:f(x,y)>0\}\);
 (v)_{2}
:

there exist two points
\(y_{0}, x_{0}\in E\)
such that
\(E\setminus K_{1}\subseteq\{x\in E:f(x,y_{0})<0\}\)
and
\(E\setminus K _{2}\subseteq\{y\in E:f(x_{0},y)>0\}\).
Then
f
has a saddle point
\((\widehat{x},\widehat{y})\in K_{1} \times K_{2}\); i.e.,
$$f(x,\widehat{y})\leq f(\widehat{x},\widehat{y})\leq f(\widehat{x},y),x\quad \forall (x,y)\in E\times E. $$
In particular, \(\inf_{y\in E}\sup_{x\in E}f(x,y)=\sup_{x\in E} \inf_{y\in E}f(x,y)\).