Existence Results for Nonsmooth Vector Quasi-Variational-Like Inequalities

Definition 1. Let K⊆ℝⁿ be a nonempty set and η : K × K → X a mapping. The set K is said to be invex with respect to η if, for all x, y ∈ K and all t ∈ [0,1], we have x + tη(y, x) ∈ K.

Condition C. Let K⊆ℝⁿ be an invex set with respect to η : K × K → ℝⁿ. Then, for all x, y ∈ K, t ∈ [0,1], we have

Lemma 2 (see [17].)Let K be a nonempty convex subset of a vector space X and η : K × K → X a mapping. If η is affine in the first argument and skew, then it is also affine in the second argument.

Definition 3 (see [18], [19].)Let K⊆ℝⁿ be a nonempty set. A vector-valued function g : K → ℝ^ℓ is said to be C-lower semicontinuous (resp., C-upper semicontinuous) at x ∈ K if for any neighborhood V of g(x), there exists a neighborhood U of x such that g(y) ∈ V + C for all y ∈ U∩K (resp., g(y) ∈ V − C for all y ∈ U∩K). g is said to be C-lower semicontinuous (resp., C-upper semicontinuous) on K if it is C-lower semicontinuous (resp., C-upper semicontinuous) at every point x ∈ K.

Definition 4. Let K⊆ℝⁿ be a nonempty convex set. A vector-valued function g : K → ℝ^ℓ is said to be C-convex if, for all x, y ∈ K and all t ∈ [0,1],

Definition 5 (see [18], [19].)Let K⊆ℝⁿ be a nonempty convex set. A vector-valued function g : K → ℝ^ℓ is said to be C-quasiconvex if, for all α ∈ ℝⁿ, the set {x ∈ K : g(x) ≤_C α} is convex.

Definition 6. Let U be a nonempty subset of a topological vector space E. A set-valued map T : U → 2^U is said to be a KKM map provided and co  (M)⊆T(M) = ⋃_x∈M T(x) for each finite subset M of U, where co  (M) denotes the convex hull of M.

Theorem 7 (see [20].)Let U be a nonempty subset of a Hausdorff topological vector space E. Assume that T : U → 2^U∖{∅} is a KKM map satisfying the following conditions:

• (i)

  for each x ∈ U, T(x) is closed;

• (ii)

  for at least one x ∈ U, T(x) is compact.

Then, ⋂_x∈U T(x) ≠ ∅.

Theorem 8 (see [21], Corollary 3.2.)Let K be a nonempty convex subset of a Hausdorff topological vector space X and S, T : K → 2^K two set-valued maps. Assume that the following conditions hold:

• (i)

  for all x ∈ K, co  (S(x))⊆T(x);

• (ii)

  for all x ∈ K, x ∉ T(x) and S⁻¹(y) = {x ∈ K : y ∈ S(x)} is open in K;

• (iii)

  there exist a nonempty compact convex subset D⊆K and a nonempty compact subset B of K such that for each x ∈ K∖B, there exists $$ such that $$.

Then, there exists $$ such that $$.

Definition 9 (see [8].)Let K⊆ℝⁿ be a nonempty set and η : K × K → ℝⁿ a mapping. A vector-valued bifunction h = (h₁, …, h_ℓ) : K × ℝⁿ → ℝ^ℓ is said to be

• (a)

  C-pseudomonotone with respect to η on K if, for all x, y ∈ K,

  $$ ()

• (b)

  C-properly subodd if

  $$ ()
  for every d_i ∈ ℝⁿ, i = 1,2, …, m with $$ and x ∈ K.

Definition 10. Let K be a nonempty convex subset of ℝⁿ. A function g : K → ℝ^ℓ is said to be hemicontinuous if, for all x, y ∈ K, the mapping t ↦ g(y + t(x − y)) is continuous. The upper and lower hemicontinuity can be defined analogously.

Definition 11. Let K⊆ℝⁿ be an invex set with respect to η : K × K → ℝⁿ. A function g : K → ℝ^ℓ is said to be η-hemicontinuous if, for all x, y ∈ K, the mapping t ↦ g(y + tη(x, y)) is continuous. The upper and lower η-hemicontinuity can be defined analogously.

Definition 12 (see [8].)Let K⊆ℝⁿ be a nonempty invex set with respect to η : K × K → ℝⁿ. A vector-valued bifunction h = (h₁, …, h_ℓ) : K × ℝⁿ → ℝ^ℓ is said to be η-upper sign continuous if, for all x, y ∈ K and t ∈ (0,1),

Remark 13. It can be easily seen that if η is skew and h is η-upper hemicontinuous in the first argument, then it is η-upper sign continuous, but the converse is not true in general.

Proposition 14. Let K⊆ℝⁿ be a nonempty invex set with respect to  η : K × K → ℝⁿ such that Condition C holds. Let A : K → 2^K be a set-valued map such that, for each x ∈ K, A(x) is a nonempty and invex set with respect to η. Let the vector-valued bifunction h : K × ℝⁿ → ℝ^ℓ be C-properly subodd, C-pseudomonotone with respect to η, and η-upper sign continuous such that, for each fixed x ∈ K, h(x; ·) is positively homogeneous. Then, $$ is a solution of NSVQVLIP if and only if it is a solution of NMVQVLIP.

Proof. It is similar to the proof of Proposition 7.7 in [8]. However, we include it for the sake of completeness of the paper.

The C-pseudomonotonicity of h with respect to η implies that every solution of NSVQVLIP is a solution of NMVQVLIP.

Conversely, let $$ be a solution of NMVQVLIP. Then, $$, and

Since $$ is invex, we have $$ for all t ∈ (0,1), and therefore, (13) becomes

By Condition C, $$, and thus,

By positive homogeneity and C-proper suboddness of h, we have

Thus, the η-upper sign continuity of h yields $$ is a solution of NSVQVLIP.

Proposition 15. Let K⊆ℝⁿ be a nonempty convex set, and let η : K × K → ℝⁿ be affine in the first argument and skew. Let A : K → 2^K be a set-valued map such that, for each x ∈ K, A(x) is a nonempty and convex set. Let the vector-valued bifunction h : K × ℝⁿ → ℝ^ℓ be C-properly subodd, C-pseudomonotone with respect to η, and η-upper sign continuous such that, for each fixed x ∈ K, h(x; ·) is positively homogeneous. Then, $$ is a solution of NSVQVLIP if and only if it is a solution of NMVQVLIP.

Proof. It is similar to the proof of Proposition 7.8 in [8]. However, we include it for the sake of completeness of the paper.

The C-pseudomonotonicity of h with respect to η implies that every solution of NSVQVLIP is a solution of NMVQVLIP.

Conversely, let $$ be a solution of NMVQVLIP. Then, $$, and

Since $$ is convex, we have $$ for all t ∈ (0,1), and therefore, (17) becomes

Since η is affine in the first argument and skew, by Lemma 2, η is also affine in the second argument. Since η(x, x) = 0 by skewness of η, we obtain

By positive homogeneity of h in the second argument, we have

Since $$ by skewness of η, the C-proper suboddness of h implies that

The η-upper sign continuity of h yields $$ is a solution of NSVQVLIP.

Theorem 16. Let K⊆ℝⁿ be a nonempty convex set, and η : K × K → ℝⁿ be skew, affine, and lower semicontinuous in the first argument. Let h = (h₁, …, h_ℓ) : K × ℝⁿ → ℝ^ℓ be C-properly subodd, positively homogeneous in the second argument, and C-pseudomonotone with respect to η such that x ↦ h(y; η(x, y)) is continuous. Assume that there exist a nonempty compact convex subset D of K and $$ such that, for all x ∈ K∖D, $$ and $$ Then, there exists a solution $$ of MVQVLIP.

Furthermore, if  h is η-upper sign continuous, then $$ is a solution of SVQVLIP.

Proof. For all x ∈ K, we define two set-valued maps S₁, S₂ : K → 2^K by

For all x, y ∈ K and for each i = 1,2, we also define other two set-valued maps P_i : K → 2^K and Q_i : K → 2^K by

For each i = 1,2 and for all y ∈ K, we have (see, e.g., [22])

and therefore,

The rest of the proof is divided into the following four steps.

(a) We claim that Q₁ is a KKM map on K.

Assume the contrary that Q₁ is not a KKM map. Then, there exist a finite set {y₁, …, y_m} in K and t₁, …, t_m ≥ 0 with $$ such that $$ for all i = 1, …, m; that is,

If $$, then $$, and therefore,

Hence,

Since $$ is a convex cone and t_i ≥ 0 with $$, we have

Since η is skew, we have η(x, x) = 0. By the affineness of η in the first argument, we have

Since h is C-proper subodd, we have

By positive homogeneity of h, we obtain

a contradiction of (29).

If $$, then $$. By the definition of Q₁, we have $$, and therefore, $$ for all i = 1, …, m. Since $$ is convex, we obtain $$, again a contradiction. Hence, Q₁ is a KKM map.

(b) We show that $$, where $$ and D are the same as in the hypothesis.

Indeed, if $$, then $$; that is, either $$ or $$.

If $$, then x ∈ ℱ and $$; that is, x ∈ A(x) and $$, a contradiction to our assumption that $$.

If $$, then $$ if and only if $$, again a contradiction to our assumption that $$. Hence, $$.

(c) We show that $$.

Since D is compact, $$ is also compact. Moreover, since Q₁ is a KKM map,

for each finite subset {y₁, …, y_m} of K. Then by Theorem 7, we get $$.

(d) Next, we claim that $$.

Let $$; then $$ for each y ∈ K. For an arbitrary element y ∈ K, we have to show that z ∈ Q₂(y).

Since $$, there exists a sequence {z_m}⊆Q₁(y) such that {z_m} converges to z. Since {z_m}⊆Q₁(y), we have

Then, either $$ or {z_m}⊆K∖A⁻¹(y).

If $$, then {z_m}⊆ℱ and $$. It follows that {z_m}⊆ℱ and $$. Since ℱ is closed and z_m → z, we have z ∈ ℱ; that is, z ∈ A(z). By C-pseudomonotonicity of h, we obtain

By the continuity of x ↦ h(y; η(x, y))h, we get $$. This implies that z ∈ A(z) and y ∉ S₂(z); that is, $$, and hence, $$. Therefore, z ∈ A(z), and $$.

Let {z_m}⊆K∖A⁻¹(y). Since A⁻¹(y) is open in K, for all y ∈ K, K∖A⁻¹(y) is closed in K. Since z_m → z, we have z ∈ K∖A⁻¹(y). Hence, z ∉ A⁻¹(y)⇔y ∉ A(z), which implies that

From (c), we get ⋂_y∈K Q₂(y) ≠ ∅. Hence, there exists $$ such that

This implies that $$.

If $$, then $$, a contradiction. Otherwise, $$; then $$. Therefore, $$ such that $$ for all $$. From Proposition 15, $$ is a solution of SVQVLIP.

Remark 17. (a) Theorem 16 extends and generalizes [8, Theorem 7.34], [9, Theorem 5.1], and [11, Theorem 2.2].

(b) If A is a closed map, then the set ℱ = {x ∈ K : x ∈ A(x)} is closed.

Theorem 18. Let K⊆ℝⁿ be a nonempty convex set, and let η : K × K → ℝ be skew. Let h : K × ℝⁿ → ℝ^ℓ be C-pseudomonotone with respect to η such that h(x; 0) = 0 for all x ∈ K; the set $$ is convex, and the set $$ is closed in K. Assume that there exist a nonempty compact convex subset D⊆K and a nonempty compact subset B of K such that, for each x ∈ K∖B, there exists $$ such that $$ and $$. Then, there exists a solution $$ of NMVQVLIP.

Furthermore, if h is C-properly subodd, η-upper sign continuous, and for each fixed x ∈ K, h(x; ·) is positively homogeneous, then $$ is a solution of NSVQVLIP.

Proof. For each x ∈ K, define two set-valued maps P, Q : K → 2^K by

Then, x ∉ Q(x) for all x ∈ K. Indeed, by skewness of η, η(x, x) = 0 for all x ∈ K. By assumption, $$. Thus, x ∉ Q(x).

By hypothesis, the complement of P⁻¹(y) in K,

is closed in K for each y ∈ K. Therefore, P⁻¹(y) is open in K for all y ∈ K.

Define other two set-valued maps S, T : K → 2^K by

Since, for all x ∈ K, x ∉ Q(x), we have x ∉ T(x).

By C-pseudomonotonicity of h, we have P(x)⊆Q(x) for all x ∈ K. Since A(x) and Q(x) are convex, for all x ∈ K, we have

Since, for each y ∈ K, A⁻¹(y) and P⁻¹(y) are open in K, (A∩P) ⁻¹(y) = A⁻¹(y)∩P⁻¹(y) is open in K. Also, since, for each y ∈ K,

(see, e.g., the proof of [23, Lemma 2.3]) and K∖ℱ is open in K, we have that S⁻¹(y) is open in K. Therefore, by Theorem 8, there exists $$ such that $$. If$$, then $$, a contradiction to our assumption. So, $$, and thus, $$. Therefore,

Thus, $$ is a solution of NMVQVLIP.

Remark 19. If, for each fixed x ∈ K, the vector-valued function y ↦ h(x, η(y, x)) is C-quasiconvex, then the set $$ is convex.

Remark 20. For all x ∈ K, the set $$ is convex, if η is affine in the first argument and h is C-convex in the second argument.

Remark 21. The set $$ is closed in K if the vector-valued function x ↦ h(y, η(x, y)) is C-lower semicontinuous for each fixed y ∈ K.