A primal-dual approach of weak vector equilibrium problems

Definition 2.1

Let K ⊆ X be convex. The function f : K → Z is called C-convex on K, iff for all x, y ∈ K and t ∈ [0, 1] one has

Definition 2.2

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.

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

   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, z ∈ K 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, y ∈ K such that F(x, y, (1 – t)x + ty) ∉ –int C for all t ∈]0, 1] one has

   F(x,y,x)∉−intC.

Definition 2.3 (Knaster-Kuratowski-Mazurkiewicz)

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

Lemma 2.4

(Fan [19]). Let X be a Hausdorff topological vector space, M ⊆ X and G : M ⇉ X be a KKM application. If G(x) is closed for every x ∈ M, and there exists x₀ ∈ M, such that G(x₀) is compact, then ⋂_x∈MG(x) ≠ ∅.

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.

1. 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).

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

1. Let x₀ ∈ K 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 y ∈ K.

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

Consequently, we have

for all t ∈ [0, 1].

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

or

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

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],

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

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, y ∈ K, x ≠ y 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

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,

wheref:[−1,1]⟶[0,1],f(x)=−2x−1,ifx∈−1,−122x+1,ifx∈−12,0−2x+1,ifx∈0,1,andg:[−1,1]⟶[−1,1],g(x)=−23x+13,ifx∈−1,12−2x+1,ifx∈12,1.

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 y ∈ K the closedness of the sets G(y) = {x ∈ K : F(x, y, x) ∉ –int C}.

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.

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

2. 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 = x₀. Assume that x₀ ∉ G(y). Then F(x₀, y, x₀) ∈ –int C. According to the assumption, the function x ⟶ F(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 x ∈ U. But then, for k = –F(x₀, y, x₀) ∈ int C, one obtains that there exists α₀ such that F(x_α, y, x_α) ∈ –int C, for α ≥ α₀, which contradicts the fact that (x_α) ⊆ G(y₀). Hence G(y) ⊆ K is closed.

For (b) consider the net (x_α) ⊆ G(y) and let lim x_α = x₀. Assume that x₀ ∉ G(y). Then F(x₀, y, x₀) ∈ –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(x₀, y, x₀) – 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 ≥ α₀. 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 ≥ α₀, contradiction.  □

Example 3.6

LetC={(x1,x2)∈R2:x12≤x22,x2≥0}.Obviously, C is a closed convex and pointed cone in ℝ²with nonempty interior. Consider the trifunction

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] ∖ { 12}. We show that x ⟶ F(x, y, x) is not C-upper semicontinuous at the point x = 12for 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 ∈ ] 12 – ϵ, 12 + ϵ[ such that (c₁, c₂) + F(12,y,12) – F(x, y, x) ∉ int C. Hence, for fixed ϵ > 0 (ϵ < 2) let (c₁, c₂) = (ϵ – 1, 1) ∈ int C. Consider x = x=12+ϵ2∈12−ϵ,12+ϵ.Then, (c₁, c₂) + F(12,y,12) – F(x, y, x) = −32,32 ∉ int C.

Next, we show that condition (b) in Lemma 3.5 holds for x = 12and every fixed y ∈ [0, 1]. Obviously, instead of nets one can consider sequences, hence let (x_n) ⊆ [0, 1], x_n ⟶ 12, n ⟶ ∞. We must show, that there exists a sequence (z_n) ⊆ ℝ², lim z_n = z such that F(x_n, y, x_n) – z_n ∈ –C and F(12,y,12) – 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 ≤ 12and F(x_n, y, x_n)–z_n = (x_n, –x_n) ∈ –C for every n ∈ ℕ, such that x_n > 12. Obviously lim z_n = z = (12+y,12), hence F(12,y,12)–z = (0, 0) ∈ C.

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

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

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

3. ∀x ∈ K, F(x, x, x) ∉ –int C,

4. There exists K₀ ⊆ X a nonempty and compact set and y₀ ∈ K, such that F(x, y₀, x) ∈ –int C, for all x ∈ K ∖ K₀.

Then, there exists an element x₀ ∈ K such that F(x₀, y, x₀) ∉ –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 y₁, y₂,…,y_n ∈ K and y ∈ co{y₁, y₂,…,y_n} such that y∉∪i=1nG(yi). In other words, there exists λ₁,λ₂,…, λ_n ≥ 0 with ∑i=1nλi=1 such that y=∑i=1nλiyi∉G(yi) for all i ∈ {1, 2,…,n}, that is F∑i=1nλiyi,yi,∑i=1nλiyi∈−intC, for all i ∈ {1, 2,…,n}. But then, since –int C is convex, one has

From assumption (ii), we have that

or equivalently, F∑i=1nλiyi,∑i=1nλiyi,∑i=1nλiyi∈∑i=1nλiF∑i=1nλiyi,yi,∑i=1nλiyi−C.

On the other hand, ∑i=1nλiF∑i=1nλiyi,yi,∑i=1nλiyi∈−intC and int C + C = int C, hence

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 z ∈ G(y₀) ∖ K₀. This implies that z ∈ K ∖ K₀, and according to (iv) F(z, y₀, z) ∈ –int C, which contradicts the fact that z ∈ G(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, ⋂_y∈KG(y) ≠ ∅. In other words, there exists x₀ ∈ K, such that F(x₀, y, x₀) ∉ –int C for all y ∈ K.  □

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 y₀ ∈ K ∩ K₀, hence K ∩ K₀ ≠ ∅, and since K ∩ K₀is compact one can assume directly that K₀ ⊆ K. Further, if K is compact condition (iv) is automatically satisfied with K₀ = K.

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

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

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

3. ∀x ∈ K, F(x, x, x) ∈ –C ∖ –int C,

4. There exists a nonempty compact convex subset K₀of K with the property that for every x ∈ K₀ ∖ core_K K₀, there exists an y₀ ∈ core_K K₀such that F(x, y₀, x) ∈ –C.

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

Proof

K₀ is compact, hence, according to Theorem 3.7 there exists x₀ ∈ K₀ such that F(x₀, y, x₀) ∉ –int C, ∀ y ∈ K₀. We show, that F(x₀, y, x₀) ∉ –int C, ∀ y ∈ K. First we show, that there exists z₀ ∈ core_KK₀ such that F(x₀, z₀, x₀) ∈ –C. Indeed, if x₀ ∈ core_KK₀ then let z₀ = x₀ and the conclusion follows from (iii). Assume now, that x₀ ∈ K₀ ∖ core_KK₀. Then, according to (iv), there exists z₀ ∈ core_KK₀ such that F(x₀, z₀, x₀) ∈ –C.

Let y ∈ K. Then, since z₀ ∈ core_KK₀, there exists λ ∈ [0, 1] such that λz₀ + (1 – λ)y ∈ K₀, consequently F(x₀, λz₀ + (1 – λ)y, x₀) ∉ –int C. From (ii) we have

or, equivalently

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

in other words

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

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).

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.

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

2. 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.

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

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

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

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

4. There exists a nonempty compact convex subset K₀of K with the property that for every x ∈ K₀ ∖ core_KK₀, there exists an y₀ ∈ core_KK₀such 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 x₀ ∈ K such that F(x₀, y, x₀) ∉ –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).  □

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, B_r : = {x ∈ X : ∥x∥ ≤ r}, is weakly compact. Therefore, if we endow the reflexive Banach space X with the weak topology, we can take K₀ = B_r ∩ K, hence, condition (iv) in Theorem 3.7 becomes : there exists r > 0 and y₀ ∈ K, such that for all x ∈ K satisfying ∥x∥ > 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 x ∈ K satisfying ∥x∥ > r, there exists some y₀ ∈ K, (which may depend by X), with ∥y₀∥ < ∥x∥ and for which the condition F(x, y₀, x) ∈ –C holds. However, in this case the diagonal condition (iii) becomes

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 x ∈ K satisfying ∥x∥ ≤ r, there exists y₀ ∈ K with ∥y₀∥ < r, and F(x, y₀, x) ∈ –C. Note that in this case one can work with the diagonal condition (iii)

Remark 3.14

Taking into account that for r > 0, K₀ = B_r ∩ K = {x ∈ K : ∥x∥ ≤ r}, core_KK₀ = {x ∈ K : ∥x∥ < r} and K ∖ core_KK₀ = {x ∈ K : ∥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 x ∈ K, ∥x∥ = r, there exists y₀ ∈ K with ∥y₀∥ < r and F(x, y₀, x) ∈ –C.

Theorem 4.1

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, convex and closed subset of X. Consider the mappings f, g : K × K ⟶ Z satisfying

1. ∀y ∈ K, it holds that 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) + g(x_α, y) – z_α ∈ –C and f(x, y) + g(x, y) – z ∈ C,

2. ∀x ∈ K, the mappings y ⟶ f(x, y) and y ⟶ g(x, y) are C-convex,

3. ∀x ∈ K, f(x, x) = 0 and g(x, x) ∉ –int C,

4. There exists K₀ ⊆ X a nonempty and compact set and y₀ ∈ K, such that f(x, y₀) + g(x, y₀) ∈ –int C, for all x ∈ K ∖ K₀.

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

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 y ∈ K, it holds that for every x ∈ K and for every net (x_α) ⊆ K, lim x_α = x there exist the netszα1,zα2⊆Z,limzα1=z1,limzα2=z2suchthatf(xα,y)−zα1∈−C,g(xα,y)−zα2∈−Cand f(x, y)−z¹ ∈ C, g(x, y)−z² ∈ C.

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’) ∀ x ∈ K, f(x, x) = 0 and g(x, x) ∈ −C ∖ −int C,

(iv’) There exists a nonempty compact convex subset K₀of K with the property that for every x ∈ K₀ ∖ core_K K₀, there exists an y₀ ∈ core_K K₀such 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 x ∈ K, but relative to the solution of (5).

Theorem 4.4

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 mappings f, g : K × K ⟶ Z. Let x₀be a solution of the problem (5), i.e. g(x₀, y) ∉ −int C for all y ∈ K. Assume that the following statements hold.

1. For all y ∈ K 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 y ∈ K 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 x ∈ K, f(x, x) = 0.

Then, x₀is a solution of (4), that is, f(x₀, y)+g(x₀, y) ∉ −int C for all y ∈ K.

Proof

Let y ∈ K. 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 y ∈ K. On the other hand, f(y, y)+g(x₀, y) = g(x₀, y) ∉ −int C, hence

From (ii) we obtain that f(x₀, y)+g(x₀, y) ∉ −int C. Since y ∈ K 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 × K ⟶ Z is said to be C-essentially quasimonotone relative to the second variable, iff for all y₁, y₂, …, y_n ∈ K and all λ₁, λ₂, …, λ_n ≥ 0 with∑i=1nλi=1one has

Lemma 4.6

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 F₁ : K × K × K ⟶ Z, F₁(x, y, z) = f(z, y)+g(x, y) and assume that the bifunctions f, g : K × K ⟶ Z satisfy

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

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

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

Proof

We show at first that for all y₁, y₂, …, y_n ∈ K and λ₁, λ₂, …, λ_n ≥ 0 with ∑i=1nλi=1, n ≥ 1 one has

Assume the contrary, that is, there exist y₁, y₂, …, y_n ∈ K and there exist λ₁, λ₂, …, λ_n ≥ 0, with ∑i=1nλi=1 such that

This assumption is equivalent to

From assumption (i) we have that ∑i=1nλif∑i=1nλiyi,yi∉−int C. But then, sinceint C + C =int C, we have

Now using the fact that y⟶g∑i=1nλiyi,y is C-convex and g(x, x) ∈ C for all x ∈ K we obtain

a contradiction.

Assume that G : K ⇉ K, G(y) = {x ∈ K : F(x, y, x) ∉ −int C} is not a KKM application. Then there exists y₁, y₂, …, y_n ∈ K and y ∈ co{y₁, y₂, …, y_n} such that y∉∪i=1nG(yi). In other words, there exists λ₁, λ₂, …, λ_n ≥ 0 with ∑i=1nλi=1 such that y=∑i=1nλiyi∉G(yi) for all i ∈ {1, 2, …, n}, that is

But then, since −int C is convex one has

which contradicts the fact that

□

An easy consequence is the following theorem.

Theorem 4.7

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 subset of X. Assume that the bifunctions f, g : K × K ⟶ Z satisfy

1. There exists a nonempty compact convex subset K₀of K with the property that for every x ∈ K₀ ∖ core_K K₀, there exists an y₀ ∈ core_K K₀such that f(x, y₀)+g(x, y₀) ∈ −C.

2. ∀ y ∈ K₀, it holds that for every x ∈ K₀and for every net (x_α) ⊆ K₀, lim x_α = x there exists a net z_α ⊆ Z, lim z_α = z such that f(x_α, y)+g(x_α, y) − z_α ∈ −C and f(x, y)+g(x, y)−z ∈ C,

3. f is C-essentially quasimonotone relative to the second variable on K₀, that is for all y₁, y₂, …, y_n ∈ K₀and all λ₁, λ₂, …, λ_n ≥ 0 with∑i=1nλi=1one has

   ∑i=1nλif∑i=1nλiyi,yi∉−int C,

4. y ⟶ g(x, y) and y ⟶ f(x, y) are C-convex on K for all x ∈ K₀,

5. ∀ x ∈ K₀, f(x, x) ∈ −C ∖−int C and g(x, x) = 0.

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

Proof

Consider the mapping F : K₀ × K₀ × K₀ ⟶ Z, F(x, y, z) = f(z, y)+g(x, y). Lemma 4.6 assures that

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

We show that the latter relation holds for every y ∈ K. First we show, that there exists z₀ ∈ core_KK₀ such that f(x₀, z₀)+g(x₀, z₀) ∈ −C. Indeed, if x₀ ∈ core_KK₀ then let z₀ = x₀ and the conclusion follows from (v). Assume now, that x₀ ∈ K₀ ∖ core_KK₀. Then, according to (i), there exists z₀ ∈ core_KK₀ such that f(x₀, z₀)+g(x₀, z₀) ∈ −C.

Let y ∈ K. Then, since z₀ ∈ core_KK₀, there exists λ ∈ [0, 1] such that λ z₀ + (1 − λ)y ∈ K₀, consequently f(x₀, λ z₀ + (1 − λ)y) + g(x₀, λ z₀ + (1 − λ)y) ∉ −int C. From (iv) we have

or, equivalently

Assume that f(x₀, y)+g(x₀, y) ∈ −int C. Then,

in other words

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

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 y ⟶ f(x, y) is C-convex for every x ∈ K 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.