In this section we provide several conditions, some of them new in the literature, that assure the existence of solution of problem (1) and (2), respectively. Further, we give conditions that assure the coincidence of the solution sets of these problems. Hence, we can deduce the existence of a solution of the dual problem from the nonemptyness of the solution set of the primal problem and vice versa.

In what follows we provide a Minty type result (see [17, 20) for the problems (1) and (2). More precisely, we provide conditions that assure the coincidence of the solutions set of problems (1) and (2), respectively.

#### Theorem 3.1

*Let X and Z be Hausdorff topological vector spaces*, *let C* ⊆ *Z be a convex and pointed cone with nonempty interior and let K be a nonempty subset of X*. *Consider the mapping F* : *K* × *K* × *K* ⟶ *Z*. *Then*, *the following statements hold*.

*If F is weakly C*-*pseudomonotone with respect to the third variable*, *then every x* ∈ *K which solves* (1) *is also a solution of* (2).

*Assume that K is convex and F*(*x*, *x*, *y*) ∈ –*C for all x*, *y* ∈ *K*, *x* ≠ *y*. *If F is weakly explicitly C*-*quasiconvex with respect to the second variable and is weakly C*-*hemicontinuous with respect to the third variable*, *then every x* ∈ *K which solves* (2) *is also a solution of* (1).

#### Proof

”(i)”

Let *x*_{0} ∈ *K* be a solution of (1). Then *F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C* for all y ∈ *K*. On the other hand *F* is weakly C-pseudomonotone with respect to the third variable, hence *F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C* implies *F*(*x*_{0}, *y*, *y*) ∉ –int *C* for all *y* ∈ *K*.

”(ii)”

Let *x*_{0} ∈ *K* be a solution of (2). Then *F*(*x*_{0}, *y*, *y*) ∉ –int *C* for all *y* ∈ *K*. Let *z* ∈ *K*, *z* ≠ *x*_{0}. Since *K* is convex we obtain that (1 – *t*)*x*_{0} + *tz* ∈ *K* for all *t* ∈ [0, 1].

Consequently, we have

$$\begin{array}{}F({x}_{0},(1-t){x}_{0}+tz,(1-t){x}_{0}+tz)\notin -\text{int}\phantom{\rule{thinmathspace}{0ex}}C\end{array}$$

for all *t* ∈ [0, 1].

But *F* is weakly explicitly quasiconvex relative the second variable, hence for all *t* ∈]0, 1[ one has

$$\begin{array}{}F({x}_{0},(1-t){x}_{0}+tz,(1-t){x}_{0}+tz)-F({x}_{0},{x}_{0},(1-t){x}_{0}+tz)\in -\text{int},\phantom{\rule{thinmathspace}{0ex}}C\end{array}$$

or

$$\begin{array}{}F({x}_{0},(1-t){x}_{0}+tz,(1-t){x}_{0}+tz)-F({x}_{0},z,(1-t){x}_{0}+tz)\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C.\end{array}$$

Since *F*(*x*, *x*, *y*) ∈ –*C* for all *x*, *y* ∈ *K*, *x* ≠ *y* and *C* + int *C* = int *C* the first relation cannot hold. Hence, for all *t* ∈]0, 1[ one has

$$\begin{array}{}F({x}_{0},(1-t){x}_{0}+tz,(1-t){x}_{0}+tz)-F({x}_{0},z,(1-t){x}_{0}+tz)\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C.\end{array}$$

Since *F*(*x*_{0}, (1 – *t*)*x*_{0} + *tz*, (1 – *t*)*x*_{0} + *tz*) ∉ –int *C*, for all *t* ∈ [0, 1], and by assumption *F*(*x*_{0}, *z*, *z*) ∉ –int *C*, we have that for all *t* ∈]0, 1],

$$\begin{array}{}F({x}_{0},z,(1-t){x}_{0}+tz)\notin -\text{int}\phantom{\rule{thinmathspace}{0ex}}C.\end{array}$$

Taking into account the fact that *F* is weakly C-hemicontinuous with respect to the third variable, we obtain

$$\begin{array}{}F({x}_{0},z,{x}_{0})\notin -\text{int}\phantom{\rule{thinmathspace}{0ex}}C.\end{array}$$

Since *Z* is arbitrary, it follows that *x*_{0} is a solution of (1). □

#### Example 3.3

[see also [21], Example 3.2). *Let us consider the trifunction F* : [–1, 1]×[–1, 1]×[–1, 1] ⟶ ℝ^{2},

$$\begin{array}{}F(x,y,z)=\left((f(y)-f(x))g(z),(f(y)-f(x))g(z)\right),\end{array}$$

*where*
$\begin{array}{}f:[-1,1]\u27f6[0,1],f(x)=\left\{\begin{array}{l}-2x-1,\phantom{\rule{thinmathspace}{0ex}}if\phantom{\rule{thinmathspace}{0ex}}x\in \left[-1,-\frac{1}{2}\right]\\ 2x+1,\phantom{\rule{thinmathspace}{0ex}}if\phantom{\rule{thinmathspace}{0ex}}x\in \left(-\frac{1}{2},0\right]\\ -2x+1,\phantom{\rule{thinmathspace}{0ex}}if\phantom{\rule{thinmathspace}{0ex}}x\in \left(0,1\right],\end{array}\right.and\phantom{\rule{thinmathspace}{0ex}}g:[-1,1]\u27f6[-1,1],g(x)=\left\{\begin{array}{l}-\frac{2}{3}x+\frac{1}{3},\phantom{\rule{thinmathspace}{0ex}}if\phantom{\rule{thinmathspace}{0ex}}x\in \left[-1,\frac{1}{2}\right]\\ -2x+1,\phantom{\rule{thinmathspace}{0ex}}if\phantom{\rule{thinmathspace}{0ex}}x\in \left(\frac{1}{2},1\right].\end{array}\right.\end{array}$

Further, consider *C* =
$\begin{array}{}{\mathbb{R}}_{+}^{2}\end{array}$ = {(*x*_{1}, *x*_{2}) ∈ ℝ^{2} : *x*_{1} ≥ 0, *x*_{2} ≥ 0} the nonnegative orthant of ℝ^{2}, which is obviously a convex and pointed cone, with nonempty interior. We consider the problems (1) and (2) defined by the trifunction *F* and by the cone *C*. Obviously the set *K* = [–1, 1] is convex. Further *F*(*x*, *x*, *y*) = (0, 0) ∈ –*C* for all *x*, *y* ∈ *K* and since the functions *f* and *g* are continuous, from *F*(*x*, *y*, (1 – *t*)*x* + *ty*) ∉ –int *C* for all *t* ∈]0, 1] one has *F*(*x*, *y*, *x*) ∉ –int *C*, by taking the limit *t* ⟶ 0. Hence, *F* is weakly C-hemicontinuous with respect to the third variable. We show that *F* is not weakly explicitly C-quasiconvex with respect to the second variable. Indeed, for *x* = –1, *y* =
$\begin{array}{}-\frac{1}{2}\end{array}$, *z* =
$\begin{array}{}\frac{1}{2}\end{array}$ and all *t* ∈]0, 1[ one has *F*(*x*,(1 – *t*)*x* + *ty*, *z*) – *F*(*x*, *x*, *z*) = (0, 0) ∉ –int *C* and *F*(*x*,(1 – *t*)*x* + *ty*, *z*) – *F*(*x*, *y*, *z*) = (0, 0) ∉ –int *C*.

We show that *x*_{0} =
$\begin{array}{}-\frac{1}{2}\end{array}$ ∈ *K* is a solution of the (2), but is not a solution of (1). Indeed, it can easily be verified that (*f*(*y*) – *f*(
$\begin{array}{}-\frac{1}{2}\end{array}$))· *g*(*y*) ≥ 0 for all *y* ∈ *K*, hence *F* (
$\begin{array}{}-\frac{1}{2}\end{array}$, *y*, *y*) ∉ –int *C*. In other words *x*_{0} =
$\begin{array}{}-\frac{1}{2}\end{array}$ is a solution of (2).

On the other hand, for *y* =
$\begin{array}{}\frac{3}{4}\end{array}$ ∈ *K* we obtain
$\begin{array}{}\left(f(y)-f\left(-\frac{1}{2}\right)\right)\cdot g\left(-\frac{1}{2}\right)=-\frac{1}{2}\cdot \frac{2}{3}<0\end{array}$ which shows that *F*(
$\begin{array}{}-\frac{1}{2}\end{array}$, *y*,
$\begin{array}{}-\frac{1}{2}\end{array}$) ∈-int *C*. Hence, *x*_{0} =
$\begin{array}{}-\frac{1}{2}\end{array}$ is not a solution of (1).

#### Lemma 3.5

*Let X and Z be Hausdorff topological vector spaces*, *let C* ⊆ *Z be a convex and pointed cone with nonempty interior and let K be a nonempty*, *convex and closed subset of X*. *Let y* ∈ *K and consider the mapping F* : *K* × *K* × *K* ⟶ *Z*}. *Assume that one of the following conditions hold*.

*The mapping x* ⟶ *F*(*x*, *y*, *x*) *is C*-*upper semicontinuous on K*.

*For every x* ∈ *K and for every net* (*x*_{α}) ⊆ *K*, lim *x*_{α} = *x there exists a net z*_{α} ⊆ *Z*, lim *z*_{α} = *z such that F*(*x*_{α}, *y*, *x*_{α}) – *z*_{α} ∈ –*C and F*(*x*, *y*, *x*) – *z* ∈ *C*.

*Then*, *the set G*(*y*) = {*x* ∈ *K* : *F*(*x*, *y*, *x*) ∉ –int *C*} *is closed*.

#### Proof

Let us prove (a). Consider the net (*x*_{α}) ⊆ *G*(*y*) and let lim *x*_{α} = *x*_{0}. Assume that *x*_{0} ∉ *G*(*y*). Then *F*(*x*_{0}, *y*, *x*_{0}) ∈ –int *C*. According to the assumption, the function *x* ⟶ *F*(*x*, *y*, *x*) is C-upper semicontinuous at *x*_{0}, hence for every *k* ∈ int *C* there exists *U*, a neighborhood of *x*_{0}, such that *F*(*x*, *y*, *x*) ∈ *F*(*x*_{0}, *y*, *x*_{0}) + *k* –int *C* for all *x* ∈ *U*. But then, for *k* = –*F*(*x*_{0}, *y*, *x*_{0}) ∈ int *C*, one obtains that there exists *α*_{0} such that *F*(*x*_{α}, *y*, *x*_{α}) ∈ –int *C*, for *α* ≥ *α*_{0}, which contradicts the fact that (*x*_{α}) ⊆ *G*(*y*_{0}). Hence *G*(*y*) ⊆ *K* is closed.

For (b) consider the net (*x*_{α}) ⊆ *G*(*y*) and let lim *x*_{α} = *x*_{0}. Assume that *x*_{0} ∉ *G*(*y*). Then *F*(*x*_{0}, *y*, *x*_{0}) ∈ –int *C*. But by the assumption there exists a net *z*_{α} ⊆ *K*, lim *z*_{α} = *z* such that *F*(*x*_{α}, *y*, *x*_{α}) – *z*_{α} ∈ –*C* and *F*(*x*_{0}, *y*, *x*_{0}) – *z* ∈ *C*. From the latter relation we get *z* ∈ –int *C*, and since –int *C* is open we have *z*_{α} ∈ -int *C* for every *α* ≥ *α*_{0}. But then, *F*(*x*_{α}, *y*, *x*_{α}) ∈ *z*_{α} –*C* and int *C* + *C* = int *C* leads to *F*(*x*_{α}, *y*, *x*_{α}) ∈ –int *C* for *α* ≥ *α*_{0}, contradiction. □

In what follows we provide an example to emphasize that the condition (b) in Lemma 3.5 is in general weaker than condition (a).

#### Example 3.6

*Let*
$\begin{array}{}C=\{({x}_{1},{x}_{2})\in {\mathbb{R}}^{2}:{x}_{1}^{2}\le {x}_{2}^{2},\phantom{\rule{thinmathspace}{0ex}}{x}_{2}\ge 0\}.\end{array}$ *Obviously*, *C is a closed convex and pointed cone in* ℝ^{2} *with nonempty interior*. *Consider the trifunction*

$$\begin{array}{}F:[0,1]\times [0,1]\times [0,1]\u27f6{\mathbb{R}}^{2},F(x,y,z)=\left\{\begin{array}{l}(x+y,2x-z)if\phantom{\rule{thinmathspace}{0ex}}0\le x\le \frac{1}{2},\\ (2x+y,z-x)if\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}<x\le 1.\end{array}\right.\end{array}$$

*Then*, *for every fixed y* ∈ [0, 1] *the mapping x* ⟶ *F*(*x*, *y*, *x*) *is continuous*, *hence it is also C*-*upper semicontinuous*, *at every x* ∈ [0, 1] ∖ {
$\begin{array}{}\frac{1}{2}\end{array}$}. *We show that x* ⟶ *F*(*x*, *y*, *x*) *is not C*-*upper semicontinuous at the point x* =
$\begin{array}{}\frac{1}{2}\end{array}$ *for every fixed y* ∈ [0, 1]. *For this it is enough to show that for all ϵ* > 0 *there exists* (*c*_{1}, *c*_{2}) ∈ int *C and x* ∈ ]
$\begin{array}{}\frac{1}{2}\end{array}$ – *ϵ*,
$\begin{array}{}\frac{1}{2}\end{array}$ + *ϵ*[ *such that* (*c*_{1}, *c*_{2}) + *F*
$\begin{array}{}(\frac{1}{2},y,\frac{1}{2})\end{array}$ – *F*(*x*, *y*, *x*) ∉ int *C*. *Hence*, *for fixed ϵ* > 0 (*ϵ* < 2) *let* (*c*_{1}, *c*_{2}) = (*ϵ* – 1, 1) ∈ int *C*. *Consider x* =
$\begin{array}{}x=\frac{1}{2}+\frac{\u03f5}{2}\in \left]\frac{1}{2}-\u03f5,\frac{1}{2}+\u03f5\right[.\end{array}$ *Then*, (*c*_{1}, *c*_{2}) + *F*
$\begin{array}{}(\frac{1}{2},y,\frac{1}{2})\end{array}$ – *F*(*x*, *y*, *x*) =
$\begin{array}{}\left(-\frac{3}{2},\frac{3}{2}\right)\end{array}$ ∉ int *C*.

*Next*, *we show that condition (b) in Lemma 3.5 holds for x* =
$\begin{array}{}\frac{1}{2}\end{array}$ *and every fixed y* ∈ [0, 1]. *Obviously*, *instead of nets one can consider sequences*, *hence let* (*x*_{n}) ⊆ [0, 1], *x*_{n} ⟶
$\begin{array}{}\frac{1}{2}\end{array}$, *n* ⟶ ∞. *We must show*, *that there exists a sequence* (*z*_{n}) ⊆ ℝ^{2}, lim *z*_{n} = *z such that F*(*x*_{n}, *y*, *x*_{n}) – *z*_{n} ∈ –*C and F*
$\begin{array}{}(\frac{1}{2},y,\frac{1}{2})\end{array}$ – *z* ∈ *C*.

*Let z*_{n} = (*x*_{n} + *y*, *x*_{n}). *Then F*(*x*_{n}, *y*, *x*_{n}) –*z*_{n} = (0, 0) ∈ –*C for every n* ∈ ℕ, *such that x*_{n} ≤
$\begin{array}{}\frac{1}{2}\end{array}$ *and F*(*x*_{n}, *y*, *x*_{n})–*z*_{n} = (*x*_{n}, –*x*_{n}) ∈ –*C for every n* ∈ ℕ, *such that x*_{n} >
$\begin{array}{}\frac{1}{2}\end{array}$. *Obviously* lim *z*_{n} = *z* =
$\begin{array}{}(\frac{1}{2}+y,\frac{1}{2})\end{array}$, *hence F*
$\begin{array}{}(\frac{1}{2},y,\frac{1}{2})\end{array}$–*z* = (0, 0) ∈ *C*.

Now we are able to prove the following existence result concerning on the solution of the problem (1).

#### Theorem 3.7

*Let X and Z be Hausdorff topological vector spaces*, *let C* ⊆ *Z be a convex and pointed cone with nonempty interior*, *and let K be a nonempty*, *convex and closed subset of X*. *Consider the mapping F* : *K* × *K* × *K* ⟶ *Z satisfying*

∀*y* ∈ *K*, *one of the conditions (a)*, *(b) in Lemma 3.5 is satisfied*,

∀*x* ∈ *K*, *the mapping y* ⟶ *F*(*x*, *y*, *x*) *is C*-*convex*,

∀*x* ∈ *K*, *F*(*x*, *x*, *x*) ∉ –int *C*,

*There exists K*_{0} ⊆ *X a nonempty and compact set and y*_{0} ∈ *K*, *such that F*(*x*, *y*_{0}, *x*) ∈ –int *C*, *for all x* ∈ *K* ∖ *K*_{0}.

*Then*, *there exists an element x*_{0} ∈ *K such that F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C*, *for all y* ∈ *K*.

#### Proof

We consider the set-valued map *G* : *K* ⇉ *K*, *G*(*y*) = {*x* ∈ *K* : *F*(*x*, *y*, *x*) ∉ –int *C*}. From (i) via Lemma 3.5 we obtain that *G*(*y*) is closed for all *y* ∈ *K*. Moreover, (iii) assures that *G*(*y*) ≠ ∅, since *y* ∈ *G*(*y*).

We show next that *G* is a KKM mapping. Assume the contrary. Then, there exists *y*_{1}, *y*_{2},…,*y*_{n} ∈ *K* and *y* ∈ co{*y*_{1}, *y*_{2},…,*y*_{n}} such that
$\begin{array}{}y\notin {\cup}_{i=1}^{n}G({y}_{i}).\end{array}$ In other words, there exists λ_{1},λ_{2},…, λ_{n} ≥ 0 with
$\begin{array}{}\sum _{i=1}^{n}{\lambda}_{i}=1\end{array}$ such that
$\begin{array}{}y=\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\notin G({y}_{i})\end{array}$ for all *i* ∈ {1, 2,…,*n*}, that is
$\begin{array}{}F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C,\end{array}$ for all *i* ∈ {1, 2,…,*n*}. But then, since –int *C* is convex, one has

$$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{\lambda}_{i}F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C.}\end{array}$$

From assumption (ii), we have that

$$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{\lambda}_{i}F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)-F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)\in C,}\end{array}$$

or equivalently,
$\begin{array}{}F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)\in \sum _{i=1}^{n}{\lambda}_{i}F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)-C.\end{array}$

On the other hand,
$\begin{array}{}\sum _{i=1}^{n}{\lambda}_{i}F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C\end{array}$ and int *C* + *C* = int *C*, hence

$$\begin{array}{}{\displaystyle F\left(\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i},\sum _{i=1}^{n}{\lambda}_{i}{y}_{i}\right)\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C,}\end{array}$$

which contradicts (iii). Consequently, *G* is a KKM application.

We show that *G*(*y*_{0}) is compact. For this is enough to show that *G*(*y*_{0}) ⊆ *K*_{0}. Assume the contrary, that is *G*(*y*_{0}) ⊈ *K*_{0}. Then, there exits *z* ∈ *G*(*y*_{0}) ∖ *K*_{0}. This implies that *z* ∈ *K* ∖ *K*_{0}, and according to (iv) *F*(*z*, *y*_{0}, *z*) ∈ –int *C*, which contradicts the fact that *z* ∈ *G*(*y*_{0}).

Hence, *G*(*y*_{0}) is a closed subset of the compact set *K*_{0} which shows that *G*(*y*_{0}) is compact.

Thus, according to Ky Fan’s Lemma, ⋂_{y∈K} *G*(*y*) ≠ ∅. In other words, there exists *x*_{0} ∈ *K*, such that *F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C* for all *y* ∈ *K*. □

In what follows, inspired from [26], we provide another coercivity condition concerning a compact set and its algebraic interior. Let *U*, *V* ⊆ *X* be convex sets and assume that *U* ⊆ *V*. We recall that the algebraic interior of *U* relative to *V* is defined as

$$\begin{array}{}{\text{core}}_{V}U=\{u\in U:U\cap ]u,v]\ne \mathrm{\varnothing},\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall}v\in V\}.\end{array}$$

Note that core_{V} *V* = *V*. Our coercivity condition concerning the problem (1) becomes:

There exists a nonempty compact convex subset *K*_{0} of *K* such that for every *x* ∈ *K*_{0} ∖ core_{K} *K*_{0} there exists an *y*_{0} ∈ core_{K} *K*_{0} such that *F*(*x*, *y*_{0}, *x*) ∈ –*C*.

In the following results we use the coercivity conditions emphasized above and we drop the closedness condition on *K*. However, condition (iii) also changes.

#### Theorem 3.9

*Let X and Z be Hausdorff topological vector spaces*, *let C* ⊆ *Z be a convex and pointed cone with nonempty interior*, *and let K be a nonempty*, *convex subset of X*. *Consider the mapping F* : *K* × *K* × *K* ⟶ *Z satisfying*

∀*y* ∈ *K*, *one of the conditions (a)*, *(b) in Lemma 3.5 is satisfied*,

∀*x* ∈ *K*, *the mapping y* ⟶ *F*(*x*, *y*, *x*) *is C*-*convex*,

∀*x* ∈ *K*, *F*(*x*, *x*, *x*) ∈ –*C* ∖ –int *C*,

*There exists a nonempty compact convex subset K*_{0} *of K with the property that for every x* ∈ *K*_{0} ∖ *core*_{K} K_{0}, *there exists an y*_{0} ∈ *core*_{K} K_{0} *such that F*(*x*, *y*_{0}, *x*) ∈ –*C*.

*Then*, *there exists an element x*_{0} ∈ *K such that F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C*, *for all y* ∈ *K*.

#### Proof

*K*_{0} is compact, hence, according to Theorem 3.7 there exists *x*_{0} ∈ *K*_{0} such that *F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C*, ∀ *y* ∈ *K*_{0}. We show, that *F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C*, ∀ *y* ∈ *K*. First we show, that there exists *z*_{0} ∈ core_{K}*K*_{0} such that *F*(*x*_{0}, *z*_{0}, *x*_{0}) ∈ –*C*. Indeed, if *x*_{0} ∈ core_{K} *K*_{0} then let *z*_{0} = *x*_{0} and the conclusion follows from (iii). Assume now, that *x*_{0} ∈ *K*_{0} ∖ core_{K} *K*_{0}. Then, according to (iv), there exists *z*_{0} ∈ core_{K}*K*_{0} such that *F*(*x*_{0}, *z*_{0}, *x*_{0}) ∈ –*C*.

Let *y* ∈ *K*. Then, since *z*_{0} ∈ core_{K}*K*_{0}, there exists λ ∈ [0, 1] such that λ*z*_{0} + (1 – λ)*y* ∈ *K*_{0}, consequently *F*(*x*_{0}, λ*z*_{0} + (1 – λ)*y*, *x*_{0}) ∉ –int *C*. From (ii) we have

$$\begin{array}{}\lambda F({x}_{0},{z}_{0},{x}_{0})+(1-\lambda )F({x}_{0},y,{x}_{0})-F({x}_{0},\lambda {z}_{0}+(1-\lambda )y,{x}_{0})\in C\end{array}$$

or, equivalently

$$\begin{array}{}(1-\lambda )F({x}_{0},y,{x}_{0})-F({x}_{0},\lambda {z}_{0}+(1-\lambda )y,{x}_{0})\in C-\lambda F({x}_{0},{z}_{0},{x}_{0})\subseteq C.\end{array}$$

Assume that *F*(*x*_{0}, *y*, *x*_{0}) ∈ –int *C*. Then,

$$\begin{array}{}-F({x}_{0},\lambda {z}_{0}+(1-\lambda )y,{x}_{0})\in -(1-\lambda )F({x}_{0},y,{x}_{0})+C\subseteq \text{int}\phantom{\rule{thinmathspace}{0ex}}C,\end{array}$$

in other words

$$\begin{array}{}F({x}_{0},\lambda {z}_{0}+(1-\lambda )y,{x}_{0})\in -\text{int}\phantom{\rule{thinmathspace}{0ex}}C,\end{array}$$

a contradiction. Hence, *F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C*, for all *y* ∈ *K*. □

Using the same technique as in the proof of Theorem 3.7, based on Fan’s Lemma, on can easily obtain solution existence of (2). However, note that depending on the structure of the trifunction *F*, the conditions may significantly differ to those assumed in the hypothesis of Theorem 3.7 or Theorem 3.9. In what follows we state a result concerning the closedness of the set *G*(*y*) = {*x* ∈ *K* : *F*(*x*, *y*, *y*) ∉ –int *C*.

#### Lemma 3.11

*Let X and Z be Hausdorff topological vector spaces*, *let C* ⊆ *Z be a convex and pointed cone with nonempty interior and let K be a nonempty*, *convex and closed subset of X*. *Let y* ∈ *K and consider the mapping F* : *K* × *K* × *K* ⟶ *Z*. *Assume that one of the following conditions hold*.

*The mapping x* ⟶ *F*(*x*, *y*, *y*)* is C*-*upper semicontinuous on K*.

*For every x* ∈ *K and for every net* (*x*_{α}) ⊆ *K*, lim *x*_{α} = *x*, *there exists a net z*_{α} ⊆ *Z*, lim *z*_{a} = *z*, *such that F*(*x*_{α}, *y*, *y*) – *z*_{α} ∈ –*C and F*(*x*, *y*, *y*) – *z* ∈ *C*.

*Then*, *the set G*(*y*) = {*x* ∈ *K* : *F*(*x*, *y*, *y*) ∉ –int *C*} *is closed*.

The proof is similar to the proof of Lemma 3.5 therefore we omit it. Our coercivity condition concerning the problem (2) is the following:

There exists a nonempty compact convex subset *K*_{0} of *K* such that for every *x* ∈ *K*_{0} ∖ core_{K}*K*_{0} there exists an *y*_{0} ∈ core_{K} *K*_{0} such that *F*(*x*, *y*_{0}, *y*_{0}) ∈ –*C*. As we have mentioned before, it is an easy exercise to provide solution existence of (2) under similar conditions to those in the hypotheses of Theorem 3.7 and Theorem 3.9.

However, by using Theorem 3.1 we obtain the following existence result concerning the existence of the solution of (1).

#### Theorem 3.12

*Let X and Z be Hausdorff topological vector spaces*, *let C* ⊆ *Z be a convex and pointed cone with nonempty interior and let K be a nonempty*, *convex subset of X*. *Consider the mapping F* : *K* × *K* × *K* ⟶ *Z satisfying*

∀*y* ∈ *K*, *one of the conditions (a)*, *(b) in Lemma 3.11 is satisfied*,

∀*x* ∈ *K*, *the mapping y* ⟶ *F*(*x*, *y*, *y*) *is C*-*convex*,

∀*x* ∈ *K*, *F*(*x*, *x*, *x*) ∈ –*C* ∖ –int *C and F*(*x*, *x*, *y*) ∈ –*C for all x*, *y* ∈ *K*, *x* ≠ *y*,

*There exists a nonempty compact convex subset K*_{0} *of K with the property that for every x* ∈ *K*_{0} ∖ *core*_{K}*K*_{0}, *there exists an y*_{0} ∈ *core*_{K}*K*_{0} *such that F*(*x*, *y*_{0}, *y*_{0}) ∈ –*C*.

*F is weakly explicitly C*-*quasiconvex with respect to the second variable*,

*F is weakly C*-*hemicontinuous with respect to the third variable*.

*Then*, *there exists an element x*_{0} ∈ *K such that F*(*x*_{0}, *y*, *x*_{0}) ∉ –int *C for all y* ∈ *K*.

#### Proof

Similarly to the proof of Theorem 3.9 one can prove that (i)-(iv) assure the nonemptyness of the solution set of (2). On the other hand, (iii), (v) and (vi) via Theorem 3.1 assure the nonemptyness of the solution set of (1). □

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