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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# A primal-dual approach of weak vector equilibrium problems

Szilárd László
• Corresponding author
• Department of Mathematics, Technical University of Cluj-Napoca, Str. Memorandumului nr. 28, 400114, Cluj-Napoca, Romania
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Published Online: 2018-03-24 | DOI: https://doi.org/10.1515/math-2018-0028

## Abstract

In this paper we provide some new sufficient conditions that ensure the existence of the solution of a weak vector equilibrium problem in Hausdorff topological vector spaces ordered by a cone. Further, we introduce a dual problem and we provide conditions that assure the solution set of the original problem and its dual coincide. We show that many known problems from the literature can be treated in our primal-dual model. We provide several coercivity conditions in order to obtain the existence of the solution of the primal-dual problems without compactness assumption. We apply the obtained results to perturbed vector equilibrium problems.

MSC 2010: 47J20; 26B25; 90C33

## 1 Introduction

A considerable number of problems that arise in mathematics can be treated in the framework of equilibrium problems, just to mention optimization problems, fixed points, saddle points or variational inequalities, as well as many important problems in physics and mathematical economics, such as location problems or Nash equilibria in game theory.

The scalar equilibrium theory has been initiated by Ky Fan [1], his minimax inequality still being considered one of the most notable results in this field. Recall, that the classical scalar equilibrium problem [1, 2], defined by a bifunction φ : K × K ⟶ ℝ, consists in finding x0K such that

$φ(x0,y)≥0,∀y∈K.$

The pioneering work of Giannessi [3], led to several extensions of the scalar equilibrium problem to the vector case. These vector equilibrium problems, much like their scalar counterpart, offer a unified framework for treating vector optimization, vector variational inequalities or cone saddle point problems, to name just a few [4, 5, 6, 7, 8, 9, 10, 11].

Let X and Z be Hausdorff topological vector spaces, let KX be a nonempty set and let CZ be a convex and pointed cone. Assume that the interior of the cone C, denoted by int C, is nonempty and consider the mapping F : K × K × KZ. The weak vector equilibrium problem governed by the vector trifunction F consists in finding x0K, such that

$F(x0,y,x0)∉−intC,∀y∈K.$(1)

The dual vector equilibrium problem of (1) is defined as: Find x0K, such that

$F(x0,y,y)∉−intC,∀y∈K.$(2)

We underline that for Z = ℝ and C = ℝ+ = [0, ∞), the previous problems reduce to the scalar equilibrium problems studied by Inoan and Kolumbán in [12].

The study of the problems (1) and (2) is motivated by the following setting. Assume that the weak vector equilibrium problem, which consists in finding x0K such that f(x0, y) ∉ –int C for all yK, has no solution though the diagonal condition f(x, x) = 0, for all xK holds. Then, we may study instead a perturbed equilibrium problem (see also [13, 14]) and provide assumptions on the perturbation function g, such that the problem which consists in finding x0K such that f(x0, y) + g(x0, y) ∉ –int C for all yK, has a solution. But in this case the latter problem is the dual of the following problem: Find x0K, such that F(x0, y, x0) ∉ –int C, for all yK, with the trifunction F(x, y, z) = f(x, z) + g(x, y). Moreover, for an appropriate perturbation g the primal problem, that is, to find x0K such that g(x0, y) ∉ –int C for all yK has a solution. Hence, it is worthwhile to obtain conditions that assure that the solution sets of (1) and (2) coincide. This setting may have some important consequences. Indeed, by taking X a Banach space and g(x, y) = ϵxye, where ϵ > 0 and eC ∖ {0}, a solution of the perturbed vector equilibrium problem is called ϵ-equilibrium point, see [15, 16]. Further, special cases of the perturbed vector equilibrium problems lead to some deep results such as Deville-Godefroy-Zizler perturbed equilibrium principle or Ekeland vector variational principle, see [14].

Moreover, take F(x, y, z) = 〈Az, yx〉, where A : KL(x, z) is a given operator and L(x, z) denotes the set of all linear and continuous operators from X to Z. For x*L(x, z) and xX, we denote by 〈x*, x〉 the vector x*(x) ∈ Z. In this setting (1) becomes: find x0K, such that 〈Ax0, yx0〉 ∉ –int C for all yK, which is the weak vector variational inequality of Stampacchia, see [17]. On the other hand (2) becomes: find x0K, such that 〈Ay, yx0〉 ∉ –int C for all yK, which is the weak vector variational inequality of Minty, see [17].

In this paper, we obtain some existence results of the solution for the vector equilibrium problem (1) and (2). Some of our conditions are new in the literature. Several examples and counterexamples circumscribe our research and show that our conditions are essential. The paper is organized as follows. In the next section, we introduce some preliminary notions and the necessary apparatus that we need in order to obtain our results. In section 3 we state our results concerning on weak vector equilibrium problems. Our conditions, which ensure the existence of a solution of the above mentioned vector equilibrium problems are considerably weakening the existing conditions from the literature. We also emphasize the case when the set K is a closed subset of a reflexive Banach space. Finally, we apply our results to vector equilibrium problems given by the sum of two bifunctions which can be seen as perturbed equilibrium problems.

## 2 Preliminaries

Let X be a real Hausdorff topological vector space. For a non-empty set DX, we denote by int D its interior, by cl D its closure and by co D its convex hull. Recall that a set CX is a cone, iff tkC for all cC and t ≥ 0. The cone C is convex if C + C = C, and pointed if C ∩ (–C) = {0}. Note that a closed, convex and pointed cone C induce a partial ordering on Z, that is z1z2z2z1C. In the sequel when we use int C, we tacitly assume that the cone C has nonempty interior. Following the same logical approach, one can introduce the strict inequality z1 < z2z2z1 ∈ int C, or z1 < z2z2z1C ∖ {0}. These relations lead to saying, that z1z2z2z1 ∉ –int C, or z1z2z2z1 ∉ –C ∖ {0}. It is an easy exercise to show that int C + C = int C.

Let Z be another Hausdorff topological vector space, let KX be a nonempty set and let CZ be a convex and pointed cone.

A map f : KXZ is said to be C-upper semicontinuous at xK iff for any neighborhood V of f(x) there exists a neighborhood U of x such that f(u) ∈ VC for all uUK. Obviously, if f is continuous at xK, then it is also C-upper semicontinuous at xK. If the cone C has nonempty interior then according to [18], f is C-upper semicontinuous at xK, if and only if, for any k ∈ int C, there exists a neighborhood U of x such that

$f(u)∈f(x)+k−intC for all u∈U∩K.$

#### Definition 2.1

Let KX be convex. The function f : KZ is called C-convex on K, iff for all x, yK and t ∈ [0, 1] one has

$tf(x)+(1−t)f(y)−f(tx+(1−t)y)∈C.$

Note that the function f : KZ is C-convex, iff for all x1, x2,…, xnK, n ∈ ℕ and λi ≥ 0, i ∈ {1, 2,…,n}, with $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}=1,\end{array}$ one has

$∑i=1nλif(xi)−f∑i=1nλixi∈C.$

We will use the following notations for the open, respectively closed, line segments in X with the endpoints x and y

$]x,y[:={z∈X:z=x+t(y−x),t∈]0,1[},[x,y]:={z∈X:z=x+t(y−x),t∈[0,1]}.$

The line segments ]x, y], respectively [x, y[ are defined similarly. Further, we need the following notions, see [3].

#### Definition 2.2

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

1. We say that F is weakly C-pseudomonotone with respect to the third variable, if for all x, yK

$F(x,y,x)∉−intC⟹F(x,y,y)∉−intC.$

Assume now, that K is a nonempty convex subset of X.

2. We say that F is weakly explicitly C-quasiconvex with respect to the second variable, if for all x, y, zK and for all t ∈]0, 1[ one has

$F(x,(1−t)x+ty,z)−F(x,x,z)∈−intC,$

or

$F(x,(1−t)x+ty,z)−F(x,y,z)∈−intC.$

3. We say that F is weakly C-hemicontinuous with respect to the third variable, if for all x, yK such that F(x, y, (1 – t)x + ty) ∉ –int C for all t ∈]0, 1] one has

$F(x,y,x)∉−intC.$

In subsequent section, the notion of a KKM map and the well-known intersection Lemma due to Ky Fan [19] will be needed.

#### Definition 2.3 (Knaster-Kuratowski-Mazurkiewicz)

Let X be a Hausdorff topological vector space and let MX. The application G : MX is called a KKM application if for every finite number of elements x1, x2, …, xnM one has

$co{x1,x2,…,xn}⊆⋃i=1nG(xi).$

#### Lemma 2.4

(Fan [19]). Let X be a Hausdorff topological vector space, MX and G : MX be a KKM application. If G(x) is closed for every xM, and there exists x0M, such that G(x0) is compact, thenxMG(x) ≠ ∅.

## 3 The coincidence of solution sets and solution existence

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 CZ be a convex and pointed cone with nonempty interior and let K be a nonempty subset of X. Consider the mapping F : K × K × KZ. Then, the following statements hold.

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

2. Assume that K is convex and F(x, x, y) ∈ –C for all x, yK, xy. 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 xK which solves (2) is also a solution of (1).

#### Proof

• ”(i)”

Let x0K be a solution of (1). Then F(x0, y, x0) ∉ –int C for all y ∈ K. On the other hand F is weakly C-pseudomonotone with respect to the third variable, hence F(x0, y, x0) ∉ –int C implies F(x0, y, y) ∉ –int C for all yK.

• ”(ii)”

Let x0K be a solution of (2). Then F(x0, y, y) ∉ –int C for all yK. Let zK, zx0. Since K is convex we obtain that (1 – t)x0 + tzK for all t ∈ [0, 1].

Consequently, we have

$F(x0,(1−t)x0+tz,(1−t)x0+tz)∉−intC$

for all t ∈ [0, 1].

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

$F(x0,(1−t)x0+tz,(1−t)x0+tz)−F(x0,x0,(1−t)x0+tz)∈−int,C$

or

$F(x0,(1−t)x0+tz,(1−t)x0+tz)−F(x0,z,(1−t)x0+tz)∈−intC.$

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

$F(x0,(1−t)x0+tz,(1−t)x0+tz)−F(x0,z,(1−t)x0+tz)∈−intC.$

Since F(x0, (1 – t)x0 + tz, (1 – t)x0 + tz) ∉ –int C, for all t ∈ [0, 1], and by assumption F(x0, z, z) ∉ –int C, we have that for all t ∈]0, 1],

$F(x0,z,(1−t)x0+tz)∉−intC.$

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

$F(x0,z,x0)∉−intC.$

Since Z is arbitrary, it follows that x0 is a solution of (1).  □

#### Remark 3.2

Note that the assumptions F(x, x, y) ∈ –C for all x, yK, xy and F is weakly explicitly C-quasiconvex with respect to the second variable in the hypothesis (ii) of Theorem 3.1 can be replaced by the assumption

$F(x,(1−t)x+ty,z)−F(x,y,z)∈−intC,$

for all t ∈]0, 1[, as follows directly from the proof.

However, in what follows we show that the latter assumption in the hypothesis (ii) of Theorem 3.1 is essential. More precisely, we give an example of a trifunction F which is not weakly explicitly C-quasiconvex with respect to the second variable but all the other assumptions of Theorem 3.1 (ii) hold, meanwhile the problem (2) has a solution, but the problem (1) has no solution.

#### Example 3.3

[see also [21], Example 3.2). Let us consider the trifunction F : [–1, 1]×[–1, 1]×[–1, 1] ⟶ ℝ2,

$F(x,y,z)=(f(y)−f(x))g(z),(f(y)−f(x))g(z),$

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

Further, consider C = $\begin{array}{}{\mathbb{R}}_{+}^{2}\end{array}$ = {(x1, x2) ∈ ℝ2 : x1 ≥ 0, x2 ≥ 0} the nonnegative orthant of ℝ2, 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, yK 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 x0 = $\begin{array}{}-\frac{1}{2}\end{array}$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 yK, hence F ( $\begin{array}{}-\frac{1}{2}\end{array}$, y, y) ∉ –int C. In other words x0 = $\begin{array}{}-\frac{1}{2}\end{array}$ is a solution of (2).

On the other hand, for y = $\begin{array}{}\frac{3}{4}\end{array}$K we obtain $\begin{array}{}\left(f\left(y\right)-f\left(-\frac{1}{2}\right)\right)\cdot g\left(-\frac{1}{2}\right)=-\frac{1}{2}\cdot \frac{2}{3}<0\end{array}$ which shows that F( $\begin{array}{}-\frac{1}{2}\end{array}$, y, $\begin{array}{}-\frac{1}{2}\end{array}$) ∈-int C. Hence, x0 = $\begin{array}{}-\frac{1}{2}\end{array}$ is not a solution of (1).

#### Remark 3.4

In order to use Fan’s Lemma to obtain solution existence for the problem (1) we need conditions that assure for every yK the closedness of the sets G(y) = {xK : F(x, y, x) ∉ –int C}.

#### Lemma 3.5

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

1. The mapping xF(x, y, x) is C-upper semicontinuous on K.

2. For every xK and for every net (xα) ⊆ K, lim xα = x there exists a net zαZ, lim zα = z such that F(xα, y, xα) – zα ∈ –C and F(x, y, x) – zC.

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

#### Proof

Let us prove (a). Consider the net (xα) ⊆ G(y) and let lim xα = x0. Assume that x0G(y). Then F(x0, y, x0) ∈ –int C. According to the assumption, the function xF(x, y, x) is C-upper semicontinuous at x0, hence for every k ∈ int C there exists U, a neighborhood of x0, such that F(x, y, x) ∈ F(x0, y, x0) + k –int C for all xU. But then, for k = –F(x0, y, x0) ∈ int C, one obtains that there exists α0 such that F(xα, y, xα) ∈ –int C, for αα0, which contradicts the fact that (xα) ⊆ G(y0). Hence G(y) ⊆ K is closed.

For (b) consider the net (xα) ⊆ G(y) and let lim xα = x0. Assume that x0G(y). Then F(x0, y, x0) ∈ –int C. But by the assumption there exists a net zαK, lim zα = z such that F(xα, y, xα) – zα ∈ –C and F(x0, y, x0) – zC. From the latter relation we get z ∈ –int C, and since –int C is open we have zα ∈ -int C for every αα0. But then, F(xα, y, xα) ∈ zαC and int C + C = int C leads to F(xα, y, xα) ∈ –int C for αα0, 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 $\begin{array}{}C=\left\{\left({x}_{1},{x}_{2}\right)\in {\mathbb{R}}^{2}:{x}_{1}^{2}\le {x}_{2}^{2},\phantom{\rule{thinmathspace}{0ex}}{x}_{2}\ge 0\right\}.\end{array}$ Obviously, C is a closed convex and pointed cone in2 with nonempty interior. Consider the trifunction

$F:[0,1]×[0,1]×[0,1]⟶R2,F(x,y,z)=(x+y,2x−z)if0≤x≤12,(2x+y,z−x)if12

Then, for every fixed y ∈ [0, 1] the mapping xF(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 xF(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 (c1, c2) ∈ int C and x ∈ ] $\begin{array}{}\frac{1}{2}\end{array}$ϵ, $\begin{array}{}\frac{1}{2}\end{array}$ + ϵ[ such that (c1, c2) + F $\begin{array}{}\left(\frac{1}{2},y,\frac{1}{2}\right)\end{array}$F(x, y, x) ∉ int C. Hence, for fixed ϵ > 0 (ϵ < 2) let (c1, c2) = (ϵ – 1, 1) ∈ int C. Consider x = $\begin{array}{}x=\frac{1}{2}+\frac{ϵ}{2}\in \left]\frac{1}{2}-ϵ,\frac{1}{2}+ϵ\right[.\end{array}$ Then, (c1, c2) + F $\begin{array}{}\left(\frac{1}{2},y,\frac{1}{2}\right)\end{array}$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 (xn) ⊆ [0, 1], xn$\begin{array}{}\frac{1}{2}\end{array}$, n ⟶ ∞. We must show, that there exists a sequence (zn) ⊆ ℝ2, lim zn = z such that F(xn, y, xn) – zn ∈ –C and F $\begin{array}{}\left(\frac{1}{2},y,\frac{1}{2}\right)\end{array}$zC.

Let zn = (xn + y, xn). Then F(xn, y, xn) –zn = (0, 0) ∈ –C for every n ∈ ℕ, such that xn$\begin{array}{}\frac{1}{2}\end{array}$ and F(xn, y, xn)–zn = (xn, –xn) ∈ –C for every n ∈ ℕ, such that xn > $\begin{array}{}\frac{1}{2}\end{array}$. Obviously lim zn = z = $\begin{array}{}\left(\frac{1}{2}+y,\frac{1}{2}\right)\end{array}$, hence F $\begin{array}{}\left(\frac{1}{2},y,\frac{1}{2}\right)\end{array}$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 CZ 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 × KZ satisfying

1. yK, one of the conditions (a), (b) in Lemma 3.5 is satisfied,

2. xK, the mapping yF(x, y, x) is C-convex,

3. xK, F(x, x, x) ∉ –int C,

4. There exists K0X a nonempty and compact set and y0K, such that F(x, y0, x) ∈ –int C, for all xKK0.

Then, there exists an element x0K such that F(x0, y, x0) ∉ –int C, for all yK.

#### Proof

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

We show next that G is a KKM mapping. Assume the contrary. Then, there exists y1, y2,…,ynK and y ∈ co{y1, y2,…,yn} such that $\begin{array}{}y\notin {\cup }_{i=1}^{n}G\left({y}_{i}\right).\end{array}$ In other words, there exists λ12,…, λn ≥ 0 with $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}=1\end{array}$ such that $\begin{array}{}y=\sum _{i=1}^{n}{\lambda }_{i}{y}_{i}\notin G\left({y}_{i}\right)\end{array}$ 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

$∑i=1nλiF∑i=1nλiyi,yi,∑i=1nλiyi∈−intC.$

From assumption (ii), we have that

$∑i=1nλiF∑i=1nλiyi,yi,∑i=1nλiyi−F∑i=1nλiyi,∑i=1nλiyi,∑i=1nλiyi∈C,$

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

$F∑i=1nλiyi,∑i=1nλiyi,∑i=1nλiyi∈−intC,$

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

We show that G(y0) is compact. For this is enough to show that G(y0) ⊆ K0. Assume the contrary, that is G(y0) ⊈ K0. Then, there exits zG(y0) ∖ K0. This implies that zKK0, and according to (iv) F(z, y0, z) ∈ –int C, which contradicts the fact that zG(y0).

Hence, G(y0) is a closed subset of the compact set K0 which shows that G(y0) is compact.

Thus, according to Ky Fan’s Lemma, ⋂yK G(y) ≠ ∅. In other words, there exists x0K, such that F(x0, y, x0) ∉ –int C for all yK.  □

#### Remark 3.8

The approach, based on Ky Fan’s Lemma, in the proof of Theorem 3.7, is well known in the literature, see, for instance, [11, 21, 22, 23, 24, 25]. Note that condition (iv) combined with condition (iii) in Theorem 3.7 ensure that y0KK0, hence KK0 ≠ ∅, and since KK0 is compact one can assume directly that K0K. Further, if K is compact condition (iv) is automatically satisfied with K0 = K.

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

$coreVU={u∈U:U∩]u,v]≠∅,∀v∈V}.$

Note that coreV V = V. Our coercivity condition concerning the problem (1) becomes:

There exists a nonempty compact convex subset K0 of K such that for every xK0 ∖ coreK K0 there exists an y0 ∈ coreK K0 such that F(x, y0, 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 CZ 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 × KZ satisfying

1. yK, one of the conditions (a), (b) in Lemma 3.5 is satisfied,

2. xK, the mapping yF(x, y, x) is C-convex,

3. xK, F(x, x, x) ∈ –C ∖ –int C,

4. There exists a nonempty compact convex subset K0 of K with the property that for every xK0coreK K0, there exists an y0coreK K0 such that F(x, y0, x) ∈ –C.

Then, there exists an element x0K such that F(x0, y, x0) ∉ –int C, for all yK.

#### Proof

K0 is compact, hence, according to Theorem 3.7 there exists x0K0 such that F(x0, y, x0) ∉ –int C, ∀ yK0. We show, that F(x0, y, x0) ∉ –int C, ∀ yK. First we show, that there exists z0 ∈ coreKK0 such that F(x0, z0, x0) ∈ –C. Indeed, if x0 ∈ coreK K0 then let z0 = x0 and the conclusion follows from (iii). Assume now, that x0K0 ∖ coreK K0. Then, according to (iv), there exists z0 ∈ coreKK0 such that F(x0, z0, x0) ∈ –C.

Let yK. Then, since z0 ∈ coreKK0, there exists λ ∈ [0, 1] such that λz0 + (1 – λ)yK0, consequently F(x0, λz0 + (1 – λ)y, x0) ∉ –int C. From (ii) we have

$λF(x0,z0,x0)+(1−λ)F(x0,y,x0)−F(x0,λz0+(1−λ)y,x0)∈C$

or, equivalently

$(1−λ)F(x0,y,x0)−F(x0,λz0+(1−λ)y,x0)∈C−λF(x0,z0,x0)⊆C.$

Assume that F(x0, y, x0) ∈ –int C. Then,

$−F(x0,λz0+(1−λ)y,x0)∈−(1−λ)F(x0,y,x0)+C⊆intC,$

in other words

$F(x0,λz0+(1−λ)y,x0)∈−intC,$

a contradiction. Hence, F(x0, y, x0) ∉ –int C, for all yK.  □

#### Remark 3.10

According to Theorem 3.1, under the extra assumption that F is weakly C-pseudomonotone with respect to the third variable Theorem 3.7 and Theorem 3.9 provide the solution existence of (2).

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) = {xK : F(x, y, y) ∉ –int C.

#### Lemma 3.11

Let X and Z be Hausdorff topological vector spaces, let CZ be a convex and pointed cone with nonempty interior and let K be a nonempty, convex and closed subset of X. Let yK and consider the mapping F : K × K × KZ. Assume that one of the following conditions hold.

1. The mapping xF(x, y, y) is C-upper semicontinuous on K.

2. For every xK and for every net (xα) ⊆ K, lim xα = x, there exists a net zαZ, lim za = z, such that F(xα, y, y) – zα ∈ –C and F(x, y, y) – zC.

Then, the set G(y) = {xK : 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 K0 of K such that for every xK0 ∖ coreKK0 there exists an y0 ∈ coreK K0 such that F(x, y0, y0) ∈ –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 CZ 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 × KZ satisfying

1. yK, one of the conditions (a), (b) in Lemma 3.11 is satisfied,

2. xK, the mapping yF(x, y, y) is C-convex,

3. xK, F(x, x, x) ∈ –C ∖ –int C and F(x, x, y) ∈ –C for all x, yK, xy,

4. There exists a nonempty compact convex subset K0 of K with the property that for every xK0coreKK0, there exists an y0coreKK0 such that F(x, y0, y0) ∈ –C.

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

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

Then, there exists an element x0K such that F(x0, y, x0) ∉ –int C for all yK.

#### 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).  □

#### Remark 3.13

Note that Condition (iv) in the hypotheses of Theorem 3.7, Theorem 3.9 and Theorem 3.12 is usually hard to be verified. However, it is well known that in a reflexive Banach space X, the closed ball with radius r > 0, Br : = {xX : ∥x∥ ≤ r}, is weakly compact. Therefore, if we endow the reflexive Banach space X with the weak topology, we can take K0 = BrK, hence, condition (iv) in Theorem 3.7 becomes : there exists r > 0 and y0K, such that for all xK satisfyingx∥ > r one has that F(x, y0, x) ∈ –int C.

Furthermore, it can easily be checked, that in this setting condition (iv) in the hypothesis of Theorem 3.7 can be weakened by assuming that there exists r > 0, such that for all xK satisfyingx∥ > r, there exists some y0K, (which may depend by X), withy0∥ < ∥xand for which the condition F(x, y0, x) ∈ –C holds. However, in this case the diagonal condition (iii) becomes

$∀x∈K,F(x,x,x)∈−C∖−intC.$

Another coercivity condition (iv) which ensures the solution existence in a reflexive Banach space context is the following: assume that there exists r > 0, such that, for all xK satisfyingx∥ ≤ r, there exists y0K withy0∥ < r, and F(x, y0, x) ∈ –C. Note that in this case one can work with the diagonal condition (iii)

$∀x∈K,F(x,x,x)∉−intC.$

#### Remark 3.14

Taking into account that for r > 0, K0 = BrK = {xK : ∥x∥ ≤ r}, coreKK0 = {xK : ∥x∥ < r} and KcoreK K0 = {xK : ∥x∥ = r}, it is an easy verification that the condition (iv) in the hypothesis of Theorem 3.9, in a reflexive Banach space setting becomes: there exists r > 0 such that for all xK, ∥x∥ = r, there exists y0K withy0∥ < r and F(x, y0, x) ∈ –C.

## 4 On the perturbed weak vector equilibrium problems

In this section, we obtain the existence of a solution of a perturbed weak vector equilibrium problem. Let X and Z be the Hausdorff topological vector spaces and K be a nonempty, convex and closed subset of X. We consider further CZ a convex and pointed cone with nonempty interior.

Let f : K × KZ be a bifunction and assume that f is diagonal null, that is f(x, x) = 0 for all xK. Consider the weak vector equilibrium problem, which consists in finding x0K such that

$f(x0,y)∉−intC,∀y∈K.$(3)

Let g : K × KZ be another bifunction. We associate with (3) the following perturbed vector equilibrium problem. Find x0K such that

$f(x0,y)+g(x0,y)∉−intC,∀y∈K.$(4)

As it was emphasized before, (4) can be considered as a particular case of the primal problem (1) with the trifunction F1 : K × K × KZ, F1(x, y, z) = f(z, y) + g(x, y). Note that in this case the dual of (4) is the following problem. Find x0K such that

$g(x0,y)∉−intC,∀y∈K.$(5)

On the other hand, (4) can be considered as a particular instance of the dual problem (2) with the trifunction F2 : K × K × KZ, F2(x, y, z) = f(x, z) + g(x, y). In this case the primal problem is given by (5).

Hence, by using the results from the previous sections one can easily obtain the existence of a solution for (4). For instance, it is an easy exercise that the C-convexity of the mappings yf(x, y) and yg(x, y) for every xK assure the C-convexity of the mapping yF1(x, y, x) for every xK and the C-convexity of the mapping yF2 (x, y, y) for every xK, respectively. We will use condition (b) of Lemma 3.5, since this assumption is new in the literature. However in the forthcoming results this condition can always be replaced by condition (a) of Lemma 3.5, namely C-upper semicontinuity. An easy consequence of Theorem 3.7 is the following result.

#### Theorem 4.1

Let X and Z be Hausdorff topological vector spaces, let CZ be a convex and pointed cone with nonempty interior, and let K be a nonempty, convex and closed subset of X. Consider the mappings f, g : K × KZ satisfying

1. yK, it holds that for every xK and for every net (xα) ⊆ K, lim xα = x there exists a net zαZ, lim zα = z such that f(xα, y) + g(xα, y) – zα ∈ –C and f(x, y) + g(x, y) – zC,

2. xK, the mappings yf(x, y) and yg(x, y) are C-convex,

3. xK, f(x, x) = 0 and g(x, x) ∉ –int C,

4. There exists K0X a nonempty and compact set and y0K, such that f(x, y0) + g(x, y0) ∈ –int C, for all xKK0.

Then, there exists an element x0K such that F(x0, y) + g(x0, y) ∉ –int C, for all yK.

#### Proof

The conclusion follows by Theorem 3.7 by taking F1(x, y, z) = f(z, y) + g(x, y) in its hypothesis.  □

#### Remark 4.2

Note that condition (i) in Theorem 4.1 is satisfied if we assume separately for the bifunctions f and g the following: for all yK, it holds that for every xK and for every net (xα) ⊆ K, lim xα = x there exist the nets $\begin{array}{}{z}_{\alpha }^{1},\phantom{\rule{thinmathspace}{0ex}}{z}_{\alpha }^{2}\subseteq Z,\phantom{\rule{thinmathspace}{0ex}}lim{z}_{\alpha }^{1}={z}^{1},\phantom{\rule{thinmathspace}{0ex}}lim{z}_{\alpha }^{2}={z}^{2}such\phantom{\rule{thinmathspace}{0ex}}thatf\left({x}_{\alpha },y\right)-{z}_{\alpha }^{1}\in -C,\phantom{\rule{thinmathspace}{0ex}}g\left({x}_{\alpha },y\right)-{z}_{\alpha }^{2}\in -C\end{array}$ and f(x, y)−z1C, g(x, y)−z2C.

#### Remark 4.3

The existence of a solution of (4) also follows via Theorem 3.9., if we replace the conditions (iii) and (iv) in the hypothesis of Theorem 4.1 by the following.

(iii’)xK, f(x, x) = 0 and g(x, x) ∈ −C ∖ −int C,

(iv’) There exists a nonempty compact convex subset K0 of K with the property that for every xK0coreK K0, there exists an y0coreK K0 such that f(x, y0)+g(x, y0) ∈ −C.

Moreover, in this case we can drop the assumption that K is closed.

Next we obtain the existence of a solution of the perturbed problem (4) via duality. Note that in this case the conditions can be assumed not for all xK, but relative to the solution of (5).

#### Theorem 4.4

Let X and Z be Hausdorff topological vector spaces, let CZ be a convex and pointed cone with nonempty interior and let K be a nonempty convex subset of X. Consider the mappings f, g : K × KZ. Let x0 be a solution of the problem (5), i.e. g(x0, y) ∉ −int C for all yK. Assume that the following statements hold.

1. For all yK and t ∈]0, 1[ one has that g(x0, (1 − t)x0+ty) − f((1 − t)x0+ty, y) − g(x0, y) ∈ −int C,

2. For every yK the following implication holds. If f((1 − t)x0+ty, y)+g(x0, y) ∉ −int C for all t ∈]0, 1], then f(x0, y)+g(x0, y) ∉ −int C.

3. For all xK, f(x, x) = 0.

Then, x0 is a solution of (4), that is, f(x0, y)+g(x0, y) ∉ −int C for all yK.

#### Proof

Let yK. Since x0 is a solution of (5) one has, that g(x0, (1 − t)x0+ty) ∉ −int C for all t ∈ [0, 1]. Hence, by using the fact that C+int C =int C, from (i) we have that f((1 − t)x0+ty, y)+g(x0, y) ∉ − C, for all t ∈]0, 1[ and yK. On the other hand, f(y, y)+g(x0, y) = g(x0, y) ∉ −int C, hence $f((1−t)x0+ty,y)+g(x0,y)∉−int C, for all t∈]0,1].$

From (ii) we obtain that f(x0, y)+g(x0, y) ∉ −int C. Since yK was arbitrary chosen the conclusion follows. □

In what follows, we obtain the existence of a solution of (4) by assuming different conditions for the bifunctions f and g. We need the following notion.

#### Definition 4.5

A bifunction f: K × KZ is said to be C-essentially quasimonotone relative to the second variable, iff for all y1, y2, …, ynK and all λ1, λ2, …, λn ≥ 0 with $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}=1\end{array}$ one has $∑i=1nλif∑i=1nλiyi,yi∉−int C.$

#### Lemma 4.6

Let X and Z be Hausdorff topological vector spaces, let CZ be a convex and pointed cone with nonempty interior and let K be a nonempty, convex subset of X. Consider the mapping F1 : K × K × KZ, F1(x, y, z) = f(z, y)+g(x, y) and assume that the bifunctions f, g : K × KZ satisfy

1. f is C-essentially quasimonotone relative to the second variable,

2. yg(x, y) is C-convex for all xK and g(x, x) ∈ C for all xK.

Then, the map G : KK, G(y) = {xK : F1(x, y, x) ∉ −int C} is a KKM application.

#### Proof

We show at first that for all y1, y2, …, ynK and λ1, λ2, …, λn ≥ 0 with $\begin{array}{}{\sum }_{i=1}^{n}{\lambda }_{i}=1\end{array}$, n ≥ 1 one has $∑i=1nλiF1∑i=1nλiyi,yi,∑i=1nλiyi∉−int C.$

Assume the contrary, that is, there exist y1, y2, …, ynK and there exist λ1, λ2, …, λn ≥ 0, with $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}=1\end{array}$ such that $∑i=1nλiF1∑i=1nλiyi,yi,∑i=1nλiyi∈−int C.$

This assumption is equivalent to $∑i=1nλif∑i=1nλiyi,yi+g∑i=1nλiyi,yi∈−int C.$

From assumption (i) we have that $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}f\left(\sum _{i=1}^{n}{\lambda }_{i}{y}_{i},{y}_{i}\right)\notin -\text{int\hspace{0.17em}}C.\end{array}$ But then, sinceint C + C =int C, we have $∑i=1nλig∑i=1nλiyi,yi∉C.$

Now using the fact that $\begin{array}{}y⟶g\left({\sum }_{i=1}^{n}{\lambda }_{i}{y}_{i},y\right)\end{array}$ is C-convex and g(x, x) ∈ C for all xK we obtain $∑i=1nλig∑i=1nλiyi,yi−g∑i=1nλiyi,∑i=1nλiyi∈C,$

Assume that G : KK, G(y) = {xK : F(x, y, x) ∉ −int C} is not a KKM application. Then there exists y1, y2, …, ynK and y ∈ co{y1, y2, …, yn} such that $\begin{array}{}y\notin {\cup }_{i=1}^{n}G\left({y}_{i}\right).\end{array}$ In other words, there exists λ1, λ2, …, λn ≥ 0 with $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}=1\end{array}$ such that $\begin{array}{}y=\sum _{i=1}^{n}{\lambda }_{i}{y}_{i}\notin G\left({y}_{i}\right)\end{array}$ for all i ∈ {1, 2, …, n}, that is $F∑i=1nλiyi,yi,∑i=1nλiyi∈−int C,∀i∈{1,2,...,n}.$

But then, since −int C is convex one has $∑i=1nλiF∑i=1nλiyi,yi,∑i=1nλiyi∈−int C,$

which contradicts the fact that $∑i=1nλiF1∑i=1nλiyi,yi,∑i=1nλiyi∉−int C.$

An easy consequence is the following theorem.

#### Theorem 4.7

Let X and Z be Hausdorff topological vector spaces, let CZ be a convex and pointed cone with nonempty interior and let K be a nonempty convex subset of X. Assume that the bifunctions f, g : K × KZ satisfy

1. There exists a nonempty compact convex subset K0 of K with the property that for every xK0coreK K0, there exists an y0coreK K0 such that f(x, y0)+g(x, y0) ∈ −C.

2. yK0, it holds that for every xK0 and for every net (xα) ⊆ K0, lim xα = x there exists a net zαZ, lim zα = z such that f(xα, y)+g(xα, y) − zα ∈ −C and f(x, y)+g(x, y)−zC,

3. f is C-essentially quasimonotone relative to the second variable on K0, that is for all y1, y2, …, ynK0 and all λ1, λ2, …, λn ≥ 0 with $\begin{array}{}\sum _{i=1}^{n}{\lambda }_{i}=1\end{array}$ one has $∑i=1nλif∑i=1nλiyi,yi∉−int C,$

4. yg(x, y) and yf(x, y) are C-convex on K for all xK0,

5. xK0, f(x, x) ∈ −C ∖−int C and g(x, x) = 0.

Then, there exists x0K, such that f(x0, y)+g(x0, y) ∉ −int C for all yK.

#### Proof

Consider the mapping F : K0 × K0 × K0Z, F(x, y, z) = f(z, y)+g(x, y). Lemma 4.6 assures that $G:K0⇉K0,G(y)={x∈K0:F(x,y,x)∉−int C}$

is a KKM mapping. On the other hand, (ii) assures that G(y) is closed for every yK0. Since K0 is compact we have that G(y) is compact for every yK0, hence according to Lemma 2.4, ∩yK0G(y) ≠ ∅. In other words, there exists x0K0 such that f(x0, y)+g(x0, y) ∉ −int C for all yK0.

We show that the latter relation holds for every yK. First we show, that there exists z0 ∈ coreK K0 such that f(x0, z0)+g(x0, z0) ∈ −C. Indeed, if x0 ∈ coreK K0 then let z0 = x0 and the conclusion follows from (v). Assume now, that x0K0 ∖ coreK K0. Then, according to (i), there exists z0 ∈ coreK K0 such that f(x0, z0)+g(x0, z0) ∈ −C.

Let yK. Then, since z0 ∈ coreK K0, there exists λ ∈ [0, 1] such that λ z0 + (1 − λ)yK0, consequently f(x0, λ z0 + (1 − λ)y) + g(x0, λ z0 + (1 − λ)y) ∉ −int C. From (iv) we have $λ(f(x0,z0)+g(x0,z0))+(1−λ)(f(x0,y)+g(x0,y))−(f(x0,λz0+(1−λ)y)+g(x0,λz0+(1−λ)y))∈C$

or, equivalently $(1−λ)(f(x0,y)+g(x0,y))−(f(x0,λz0+(1−λ)y)+g(x0,λz0+(1−λ)y))∈C−λ(f(x0,z0)+g(x0,z0)⊆C.$

Assume that f(x0, y)+g(x0, y) ∈ −int C. Then, $−(f(x0,λz0+(1−λ)y)+g(x0,λz0+(1−λ)y))∈−(1−λ)(f(x0,y)+g(x0,y))+C⊆int C,$

in other words $f(x0,λz0+(1−λ)y)+g(x0,λz0+(1−λ)y)∈−int C,$

a contradiction. Hence, f(x0, y)+g(x0, y) ∉ −int C, for all yK. □

#### Remark 4.8

If K is also compact, then one can take K0 = K, thus, one can drop the assumption (i) and the assumption that the map yf(x, y) is C-convex for every xK in the hypothesis of Theorem 4.7. Moreover, the assumptions imposed on the bifunctions f and g can be permuted, which might become useful in order to chose the right perturbation bifunction, when we perturb a concrete problem.

## Acknowledgement

The author expresses his sincere thanks to an anonymous referee for his/her helpful comments and suggestions, which led to improvement of the originally submitted version of this work.

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Accepted: 2018-02-28

Published Online: 2018-03-24

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 276–288, ISSN (Online) 2391-5455,

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