Well-Posedness for Generalized Set Equilibrium Problems

Lemma 1. (i) Φ(s, x)∩int C = ∅ for all x ∈ K and for all s ∈ T(x).

(ii) $$ with $$ if and only if $$.

(iii) $$ with $$ if and only if $$.

Proof. (i) If not, there exists $$ for some $$ and $$. Then $$ and τ ∈ int C. Hence, $$ which contradicts the fact that 0 ∈ f(s, x, K) for all x ∈ K and for all s ∈ T(x).

(ii) $$ if and only if $$ if and only if $$.

(iii) By (i) and (ii), we have $$ with $$ if and only if $$ if and only if $$.

Definition 2. A sequence {(s_n, x_n) ∈ Z × K : s_n ∈ T(x_n)} is a minimizing sequence for Φ if for every neighborhood U_Y ∈ 𝔑_Y(0) of 0, there is n₀ ∈ ℕ, such that Φ(s_n, x_n)∩U_Y ≠ ∅ for all n ≥ n₀.

Definition 3. (GSEP) is M-well-posed if it satisfies the following conditions:

• (i)

  there exists at least one solution, that is, the set Ω ≠ ∅;

• (ii)

  for every minimizing sequence {(s_n, x_n)} and for every U_X ∈ 𝔑_X(0), there is n₀ ∈ ℕ such that x_n ∈ Ω + U_X for all n ≥ n₀.

Definition 4. For ϵ ∈ C, the ϵ-approximate solution set of (GSEP) is defined by Ω(ϵ) = {x ∈ K : Φ(s, x)∩(C − ϵ) ≠ ∅ for some s ∈ T(x)}.

Definition 5. (GSEP) is B-well-posed if it satisfies the following conditions:

• (i)

  there exists at least one solution, that is, the set Ω ≠ ∅;

• (ii)

  the mapping Ω : C⇉X is upper Hausdorff continuous at ϵ = 0; that is, for every U_X ∈ 𝔑_X(0), there exists U_Y ∈ 𝔑_Y(0) such that Ω(ϵ) ⊂ Ω + U_X for every ϵ ∈ U_Y∩C.

Definition 6 (see [16], [17].)A set-valued mapping T : X⇉Z is

• (i)

  upper semicontinuous if for every x ∈ X and every open set V in Y with T(x) ⊂ V, there exists a neighborhood W(x) of x such that T(W(x)) ⊂ V;

• (ii)

  lower semicontinuous if for every x ∈ X and every open neighborhood V(y) of every y ∈ T(x), there exists a neighborhood W(x) of x such that T(u)∩V(y) ≠ ∅ for all u ∈ W(x);

• (iii)

  continuous if it is both upper semicontinuous and lower semicontinuous.

Example 7. (i) There is an example that satisfies M-well-posed, but not B-well-posed. (ii) There is an example that satisfies both B-well-posed and M-well-posed.

Solution. (i) The first one is inspired by the example of [10]. Let X = Z = ℝ, Y = ℝ², $$, K = ℝ₊, T : K⇉Z satisfy T(x) = [x/2, x] for all x ∈ K. The set-valued mapping f : Z × K × K⇉Y is defined by

for all (s, x, y) ∈ Z × K × K with s ∈ T(x).

Then the set of all solutions for (GSEP) is Ω = {0}. For any x ∈ K and any s ∈ T(x), the minimizing mapping is

If we choose U_X = (−1/2,1/2), a neighborhood of 0, and x_n = n, ϵ_n = (0, 1/n²) for all n ∈ ℕ, we can easily see that ϵ_n → (0,0) as n → ∞, and n ∈ Ω(ϵ_n)∖(Ω + U_X) for all n ∈ ℕ. Thus, (GSEP) is not B-well-posed.

Nevertheless, for any minimizing sequence {(s_n, x_n)} for Φ with s_n ∈ T(x_n) for all n and for every U_Y ∈ 𝔑_Y(0), there exists n₀ ∈ ℕ such that Φ(s_n, x_n)∩U_Y ≠ ∅ for n ≥ n₀. This will force that x_n → 0 as n → ∞, and hence x_n ∈ Ω + U_X for all n ≥ n₀. Therefore, (GSEP) is M-well-posed.

(ii) We modify the above example as follows. Let X, Z, Y, C, K, T be given the same as in (i). The set-valued mapping f : Z × K × K⇉Y is defined by

for all (s, x, y) ∈ Z × K × K with s ∈ T(x).

Then the set of all solutions for (GSEP) is Ω = K. For any x ∈ K and any s ∈ T(x), the minimizing mapping is

Since for any minimizing sequence {(s_n, x_n)}, and for every U_X ∈ 𝔑_X(0), we always have x_n ∈ Ω  +  U_X. Thus, (GSEP) is M-well-posed. Furthermore, since Ω + U_X = K + U_X, for all $$, we always have Ω(ϵ) ⊂ Ω + U_X. Hence, (GSEP) is B-well-posed.

Proposition 8. (a) If (GSEP) is B-well-posed, then it is M-well-posed. (b) If (GSEP) is M-well-posed, and for every W_Y ∈ 𝔑_Y(0), there exists U_Y ∈ 𝔑_Y(0) such that

where h(x) = T(x)×{x} for all x ∈ K∖ cl (Ω), then (GSEP) is B-well-posed.

Proof. For the idea of the proof, we can use the similar direction of [10, Propositions 3 and 4]. For the sake of clarity, we give the proof of (b) as follows. Suppose that (GSEP) is not B-well-posed. Then there is a neighborhood $$ of 0, and sequences {ϵ_n} ⊂ C with ϵ_n → 0 and x_n ∈ Ω(ϵ_n) such that

This means Since x_n ∈ Ω(ϵ_n), there exists s_n ∈ T(x_n) such that Now, we separate into two cases.

Case 1. If the sequence {(s_n, x_n)} is a minimizing sequence, then by M-well-posedness, for this $$, there is a n₀ ∈ ℕ such that $$, for all n ≥ n₀, which contradicts (8).

Case 2. If the sequence {(s_n, x_n)} is not a minimizing sequence, then there is a W_Y ∈ 𝔑_Y(0) and a subsequence $$ of {x_n} with a corresponding subsequence $$ such that

By relation (10), we have $$, for all k ∈ ℕ. For this W_Y and condition (7), there is a symmetric neighborhood U_Y ∈ 𝔑_Y(0) such that $$, for all k ∈ ℕ. For k large enough, $$. Taking $$ for all k ∈ ℕ. This implies that, for k large enough, $$ which contradicts (11). This completes the proof.

Lemma 9. Let Y be a regular topological vector spaces, and let A be a nonempty compact subset of Y. Suppose that A∩(−C) = ∅; then there is a neighborhood W_Y of 0 such that (A + W_Y)∩(W_Y − C) = ∅. In particular, (A + W_Y)∩W_Y = ∅.

Proof. Suppose that A∩(−C) = ∅. For all η ∈ A, we have η ∉ −C. Since Y is regular, there is a neighborhood $$ of 0 such that

Since A is a nonempty compact subset, the set There exist η₁, η₂, …, η_n ∈ A, such that Let $$. Then Since $$, we have

Theorem 10. Let X, Y, Z be three Hausdorff topological vector spaces where X is a finite dimensional space and Y is regular, let K be a nonempty closed convex subset of X, and let C ⊂ Y be a closed convex and pointed cone with apex at the origin and int C ≠ ∅. The mapping f : Z × K × K⇉Y is upper semi-continuous with nonempty compact values and satisfies f(s, x, x) = {0} for all x ∈ K and for all s ∈ T(x), and the mapping T : K⇉Z is upper semi-continuous with nonempty compact values, such that

• (i)

  the solution set Ω of (GSEP) is nonempty and bounded;

• (ii)

  the minimizing mapping Φ is upper Hausdorff continuous on T(K) × K;

• (iii)

  f(s, x, y)∩(−C) = ∅ for all x ∈ Ω, for all s ∈ T(K) and for all y ∈ K∖Ω;

• (iv)

  the mapping (s, x) → f(s, x, y) is above-C-upper Hausdorff continuous on Z × K for every y ∈ K, and the mapping x → f(s, x, y) is above-C-concave [19] on K for every s ∈ T(K) and y ∈ K;

• (v)

  for every minimizing sequence {(s_n, x_n)} ⊂ T(K) × K, and for each (s, y) ∈ T(K) × K, there is a sequence {ζ_n} with ζ_n ∈ f(s, x_n, y) for each n ∈ ℕ is a bounded sequence in Y.

Then (GSEP) is M-well-posedness.

Proof. We prove it by contradiction. Suppose that (GSEP) is not M-well-posedness. Then there exists a minimizing sequence {(s_n, x_n)} ⊂ TK × K and ϵ > 0 such that

for infinitely many n, where B denotes the unit open ball in X. Let us choose a subsequence from {(s_n, x_n)} so that the relation (17) holds for all elements of the subsequence. Such a subsequence is still a minimizing sequence, and we still denote it by {(s_n, x_n)} if there is no any confusion. Now, we separate our discussion into two cases.

Case 1. If the sequence {x_n} is bounded, then it has a convergent subsequence $$ that converges to some point x^⋆ ∈ X with a corresponding subsequence $$ with $$ for every k ∈ ℕ. By the upper semi-continuity of T, there exists a convergent subsequence of $$ (without any confuse, we still denote it by $$) converges to some point s^⋆ ∈ T(x^⋆). From (17), $$ for every k ∈ ℕ. Hence, x^⋆ ∉ Ω, and by Lemma 1, we have 0 ∉ Φ(s^⋆, x^⋆). By Lemma 9, there is a neighborhood U_Y ∈ 𝔑_Y(0) such that

Since $$ is a minimizing sequence, for each k, we can choose $$ such that $$. Since Φ is upper Hausdorff continuity of Φ at (s^⋆, x^⋆), we have Hence, for k large enough, $$ which contradicts (18).

Case 2. If the sequence {x_n} is unbounded. Since Ω is bounded, so are Ω  +  ϵB and cl (Ω  +  ϵB). Then the set cl (Ω + ϵB) is compact. We denote that Ω∩∂(Ω + ϵB) = ∅, where ∂(Ω + ϵB) means the boundary of Ω + ϵB. Since the sequence {x_n} is unbounded, there is a subsequence $$ with $$ as k → ∞. Without loss of generality, we may assume that x_n ∉ ∂(Ω + ϵB) for all n ∈ ℕ. Fix any $$ and let $$ for any n ∈ ℕ, where $$. After a simple calculation, we can see that λ_n → 1 as n → ∞. Hence, {v_n} has a subsequence $$ that converges to some point v^⋆ ∈ ∂(Ω + ϵB). For the similar process in Case 1, we have a subsequence $$ of the corresponding sequence {t_n} with t_n ∈ T(v_n) converges to some point t^⋆ ∈ T(v^⋆). By the above C-concavity of f in x, we have

By condition (v), there is a bounded sequence $$ with $$ for each k ∈ ℕ in Y. Hence, by (20), we have Next, we claim that Indeed, if $$. By Lemma 9, there is a neighborhood U_Y ∈ 𝔑_Y(0) such that For this U_Y, by the above C-upper Hausdorff continuity of f and the fact that f(t^⋆, v^⋆, v^⋆) = {0}, we have for k large enough. Since the sequence $$ is bounded, the left-hand side of (21) will fell into $$ for k large enough. But in this situation, we can see that $$ which contradicts (23). Thus, the relation (22) holds. Since $$ and t^⋆ ∈ T(K), by condition (iii) we have v^⋆ ∈ Ω which contradicts the fact that v^⋆ ∈ ∂(Ω + ϵB).

From the discussions of above two cases, (GSEP) is M-well-posedness.

Remark 11. Theorem 10 generalize the Theorem 1 of [10] to (GSEP).

Lemma 12. Suppose that F : K⇉Y and f : K × K⇉Y satisfy f(x, y) = F(y) − F(x) for all x, y ∈ K, then

for all x ∈ K.

Proof. For any fixed x ∈ K. Choose any ξ ∈ Min _wf(x, K); we have ξ ∈ F(K) − F(x) and (F(K) − F(x))∩(ξ − int C) = ∅. There exist ξ₁ ∈ F(K) and ξ₂ ∈ F(x) such that ξ = ξ₁ − ξ₂ and (F(K) − F(x))∩(ξ₁ − ξ₂ − int C) = ∅. That is, (F(K) + ξ₂ − F(x))∩(ξ₁ − int C) = ∅. Since 0 ∈ ξ₂ − F(x) and F(K) ⊂ F(K) + ξ₂ − F(x), we know that F(K)∩(ξ₁ − int C) = ∅. Thus, ξ₁ ∈ Min _wF(K), and hence, ξ ∈ Min _wF(K) − F(x). Therefore,

Lemma 13. Let X, Y, C, K be given the same as in Theorem 10, let f be as given in Lemma 12, and let F : K⇉Y be upper semi-continuous with nonempty compact values such that the set Ω = w − Eff (F, K) is bounded, where w − Eff (F, K) = {x ∈ K : y ∈ F(x), F(K)∩(y − int C) = ∅}. Any sequence {x_n} satisfies for every neighborhood U_Y ∈ 𝔑_Y(0) of 0; there is n₀ ∈ ℕ, such that Min _wf(x_n, K)∩U_Y ≠ ∅ for all n ≥ n₀. Then there exists a bounded sequence {η_n} in Y with η_n ∈ F(x_n), for all n ∈ ℕ.

Proof. Since Ω is bounded, its closure cl (Ω) is compact. By the upper semi-continuity of F, F(cl (Ω)) is compact. Hence it is bounded, so is F(Ω). Fixed W_Y ∈ 𝔑_Y(0), a symmetric neighborhood of 0. Since the sequence {x_n} satisfies for every neighborhood U_Y ∈ 𝔑_Y(0) of 0, there is n₀ ∈ ℕ, such that Min _wf(x_n, K)∩U_Y ≠ ∅ for all n ≥ n₀. From Lemma 12, we have

for all n ≥ n₀. We can pick some points ξ_n ∈ Min _wF(K), and η_n ∈ F(x_n) such that for all n ≥ n₀. Since W_Y is symmetric, we have for all n ≥ n₀. Since F(Ω) is bounded, so is F(Ω) + W_Y. Therefore, the sequence {η_n} is bounded.

Theorem 14. Let X, Y, Z, C, K, f, T be given the same as in Theorem 10. Suppose that

• (i)

  the solution set Ω of (GSEP) is nonempty and compact;

• (ii)

  for every x, z ∈ K and s ∈ T(K), if f(s, x, z)∩C ≠ ∅, then f(s, z, x)∩(−C) ≠ ∅;

• (iii)

  f(s, x, z)∩(−C) = ∅ for all x ∈ Ω, for all s ∈ T(K) and for all z ∈ K∖Ω;

• (iv)

  the mapping (s, x) → f(s, x, y) is above C-upper Hausdorff continuous on Z × K for every y ∈ K, and the mapping x → f(s, x, y) is above C-concave on K for every s ∈ T(K) and y ∈ K;

• (v)

  for each minimizing sequence {(s_n, x_n)} ⊂ T(K) × K, and for each (y, s) ∈ K × Z, there is a sequence {ζ_n} with ζ_n ∈ f(s, x_n, y) for each n ∈ ℕ is a bounded sequence in Y;

• (vi)

  for every x ∈ K∖Ω and for every s ∈ T(x), Min _wf(s, x, K) ⊂ f(s, x, Ω).

Then (GSEP) is M-well-posedness.

Proof. We prove it by contradiction. Suppose that (GSEP) is not M-well-posedness. Hence there exists a minimizing sequence {(s_n, x_n)} ⊂ T(K) × K and ϵ > 0 such that the relation (17) holds. If the sequence {x_n} is unbounded, by a similar process in Case 2 of Theorem 10 we know that (GSEP) is M-well-posedness. If the sequence {x_n} is bounded, then it has a convergent subsequence that converges to some point x^⋆ ∈ X with a corresponding subsequence {s_n} with s_n ∈ T(x_n) for every n ∈ ℕ. We still denote it by {x_n} if there is no confusion. The relation (17) tells us that

Since {(s_n, x_n)} is a minimizing sequence, we can choose a sequence {τ_n} with τ_n → 0, where τ_n ∈ Φ(x_n) = Min _wf(s_n, x_n, K) for all n ∈ ℕ. For the same process as in Case 1 of Theorem 10, by the upper semi-continuity of T, there exists a convergent subsequence of {s_n} (without any confusion, we still denote it by {s_n}) that converges to some point s^⋆ ∈ T(x^⋆). Since x_n ∈ K∖Ω, by condition (vi), τ_n ∈ Min _wf(s_n, x_n, K) ⊂ f(s_n, x_n, Ω). Then, for each n ∈ ℕ, there is a z_n ∈ Ω such that τ_n ∈ f(s_n, x_n, z_n). Since Ω is compact, there is a subsequence $$ of {z_n} that converges to some point z^⋆ ∈ Ω. Now we claim that f(s^⋆, x^⋆, z^⋆)∩C ≠ ∅. Indeed, suppose that f(s^⋆, x^⋆, z^⋆)∩C = ∅. By Lemma 9, there is W_Y ∈ ℕ_Y(0) such that or By the above C-upper Hausdorff continuity of f, for this W_Y, there is k₀ ∈ ℕ such that for all k ≥ k₀. Thus, from (32), $$ which contradicts the fact that $$. Hence, f(s^⋆, x^⋆, z^⋆)∩C ≠ ∅. By condition (ii), f(s^⋆, z^⋆, x^⋆)∩(−C) ≠ ∅. Since z^⋆ ∈ Ω, by condition (iii), x^⋆ ∈ Ω which contradicts (30). Hence (GSEP) is M-well-posedness.

Theorem 15. Let X, Y, Z, C, K, T, f be given the same as in Theorem 10. Suppose that

• (i)

  the solution set Ω of (GSEP) is nonempty and bounded;

• (ii)

  f(s, x, z)∩(−C) = ∅ and f(s, z, x)∩(−C) = ∅ for all z ∈ cl (Ω), for all s ∈ T(K) and for all x ∈ K∖cl (Ω);

• (iii)

  the mapping x → f(s, x, y) is above C-concave on K for every (s, y) ∈ T(K) × K, and the mapping s → f(s, x, y) is lower semi-continuous on T(K), for every (x, y) ∈ K × K;

• (iv)

  for every x ∈ K∖cl (Ω) and for every s ∈ T(x), Min _wf(s, x, K) ⊂ f(s, x, cl (Ω)).

Then (GSEP) is M-well-posedness.

Proof. Suppose that (GSEP) is not M-well-posedness. Then we have a minimizing sequence {(s_n, x_n)} ⊂ T(K) × K and ϵ > 0 such that the relation (17) holds for all n ∈ ℕ. We can use the same process as in the proof of Theorem 10 under the situation when we replace Ω by cl (Ω). If the sequence {x_n} is unbounded, combining Lemmas 12 and 13, we have the sequences $$, {v_n}, $$, {λ_n}, {t_n}, $$, $$, and points $$, v^⋆, t^⋆ with the same properties as in the proof (Case 2) of Theorem 10. By condition (iii), we have the mapping x → f(s, x, y) which is above C-concave on K for every (s, y) ∈ T(K) × K and the relations (20) and (21) hold. Since the mapping s → f(s, x, y) is lower semi-continuous on T(K) for every (x, y) ∈ K × K and fix any $$, there exists $$ such that $$. For this $$, by (21), there exists $$ such that

Since the mapping (s, x) → f(s, x, y) is upper semi-continuous, hence it is upper Hausdorff continuous on Z × K for every y ∈ K, and $$, for any given W_Y ∈ 𝔑_Y(0), for k large enough. From (34) and (35) and the fact that $$, we have Next, we claim that Indeed, if not, $$. By Lemma 9, there exists N_Y ∈ 𝔑_Y(0) such that $$. But this contradicts (36), and hence (37) holds. Since $$, by condition (ii), v^⋆ ∈ cl (Ω) which contradicts the fact v^⋆ ∈ ∂(cl (Ω) + ϵB). Hence, (GSEP) is M-well-posed. On the other hand, if the sequence {x_n} is bounded, the sequences $$, {s_n}, $$, the points x^⋆, s^⋆, and the number ϵ are the same as in the Case 1 of Theorem 10, so that $$ for all k ∈ ℕ. Hence x^⋆ ∉ cl (Ω). Since {(s_n, x_n)} is minimizing, we can choose a sequence η_n ∈ Φ(s_n, x_n) = Min _wf(s_n, x_n, K) with η_n → 0. By condition (v), η_n ∈ Min _wf(s_n, x_n, K) ⊂ f(s_n, x_n, cl (Ω)). For each n ∈ ℕ, there is z_n ∈ cl (Ω) such that η_n ∈ f(s_n, x_n, z_n). Since cl (Ω) is compact, there is a subsequence $$ of {z_n} that converges to some point z^⋆ ∈ cl (Ω). Since f : Z × K × K⇉Y is upper semi-continuous with nonempty compact values, we have From condition (ii) and the fact that z^⋆ ∈ cl (Ω), we have x^⋆ ∈ Ω which contradicts the fact x^⋆ ∉ Ω. Hence, (GSEP) is M-well-posed.

Corollary 16. Under the framework of Theorem 14 (resp., Theorem 15) the following condition (A) holds:

• (A)

  for every W_Y ∈ 𝔑_Y(0), there is a U_Y ∈ 𝔑(0) such that

  $$ ()

where g(x) = h(x) × cl (Ω) and h is given in (7).

Then (GSEP) is B-well-posed.

Proof. From Theorem 14 (resp., Theorem 15), (GSEP) is M-well-posed. By condition (vi) of Theorem 14 (resp., condition (iv) of Theorem 15), we have

for every s ∈ T(x) and x ∈ K∖cl (Ω). Hence, Combining this with condition (A), the condition (7) holds. Hence, by Proposition 8, (GSEP) is B-well-posed.