Proof.
(a) Fixed any (ξ,η)∈Δ1×Δ2, we can easy see that all conditions of Theorem 2.3 of [15] hold, hence from Theorem 2.3 of [15], we know that Γ(ξ,η) is nonempty and compact.
(b) Let {(ξα,ηα)}⊂Δ1×Δ2 be a net such that (ξα,ηα)→(ξ,η) and {xα} be a net with xα∈Γ(ξα,ηα). Since xα∈K(ηα) and K:Δ2→2X are upper semicontinuous with nonempty compact values, there are an x∈K(η) and a subnet {xαι} of {xα} such that xαι→x. Since T:K→2Z is upper semicontinuous with nonempty compact values, T({xαι}∪{x}) is compact. Since sαι∈T(xαι), there is an s∈T(x) such that a subnet of {sαι} converges to s. Without loss of generality, we still denote the subnet by {sαι}, and hence sαι→s.
If x∉Γ(ξ,η), then there is a y∈K(η) such that
(3.2)f(ξ,s,x,y)∈-intC(x).
Since K(η) is compact, there is a net, say {yαι}, in K(η) converges to y. Since the mapping (ξ,s,x,y)→f(ξ,s,x,y) is continuous, and the mapping x→Y∖(-intC(x)) is closed, we have
(3.3)f(ξ,s,x,y)=limαι f(ξαι,sαι,xαι,yαι)∈Y∖(-intC(x)),
which contracts (3.2). Thus, x∈Γ(ξ,η).
In order to prove (c), we introduce Lemmas 3.2–3.4 as follows.
Let Y⋆ be the topological dual space of Y. For each x∈𝒦,
(3.4)C⋆(x)={g∈Y⋆:g(y)≥0 ∀y∈C(x)}.
Let C⋆=∩x∈𝒦C⋆(x), then C⋆ is nonempty and connected. If C⋆:𝒦→2Y⋆ is a constant mapping, then C⋆(x)=C⋆ for all x∈𝒦. In the sequel, we suppose that C⋆ is not a singleton. That is, C⋆∖{0}≠∅, and hence it is connected. For each g∈C⋆∖{0}, let us denote the set of g-efficient solutions to (PGVEP) by
(3.5)Sξ,η(g)={x∈K(η):sups∈T(x) g(f(ξ,s,x,y))≥0 for every y∈K(η)}.Lemma 3.2. Under the framework of Theorem 3.1,(3.6)Sξ,η(g)≠∅,for every g∈C⋆∖{0}.
Proof. From (a) of Theorem 3.1, we know that, for each (ξ,η)∈Δ1×Δ2, there is an x-∈K(η) with s-∈T(x-) such that
(3.7)g(f(ξ,s-,x-,y))≥0,
for all y∈K(η) and for all g∈C⋆∖{0}. Thus, x-∈Sξ,η(g) for every g∈C⋆∖{0}. Hence, Sξ,η(g)≠∅ for every g∈C⋆∖{0}.
Lemma 3.3. Suppose that for any (ξ,η)∈Δ1×Δ2 and y∈K(η), f(ξ,T(K(η)),K(η),y) are bounded. Then, the mapping Sξ,η:C⋆∖{0}→2K(η) is upper semicontinuous with compact values.
Proof. Fixed any (ξ,η)∈Δ1×Δ2. We first claim that the mapping Sξ,η:C⋆∖{0}→2K(η) is closed. Let xν∈Sξ,η(gν), xν→x and gν→g with respect to the strong topology σ(Y⋆,Y) in Y⋆.
Since xν∈Sξ,η(gν), there is an sν∈T(xν) such that g(f(ξ,sν,xν,y))≥0 for all y∈K(η). Since T is upper semicontinuous with nonempty compact values, by a similar argument in the proof of Theorem 3.1(b), there is an s∈T(x) such that a subnet of {sν} converges to s. Without loss of generality, we still denote the subnet by {sν}.
For each y∈K(η), we define Pf(ξ,T(K(η)),K(η),y)(g)=supz∈f(ξ,T(K(η)),K(η),y)|g(z)| for all g∈Y⋆. We note that the set f(ξ,T(K(η)),K(η),y) is bounded by assumption, hence Pf(ξ,T(K(η)),K(η),y)(g) is well defined and is a seminorm of Y⋆. For any ε>0, let 𝒰ε={g∈Y⋆:Pf(ξ,T(K(η)),K(η),y)(g)<ε} be a neighborhood of 0 with respect to σ(Y⋆,Y). Since gν→g, there is a α0∈Λ such that gν-g∈𝒰ε for every ν≥ν0. That is, Pf(ξ,T(K(η)),K(η),y)(gν-g)=supz∈f(ξ,T(K(η)),K(η),y)|(gν-g)(z)|<ε/2 for every ν≥ν0. This implies that
(3.8)|(gν-g)(f(ξ,sν,xν,y))|<ε2,
for all ν≥ν0. Since the mapping (s,x)→f(ξ,s,x,y) is continuous and (sν,xν)→(s,x), we have
(3.9)f(ξ,sν,xν,y)→f(ξ,s,x,y).
By the continuity of g, we have
(3.10)|g(f(ξ,sν,xν,y))-g(f(ξ,s,x,y))|<ε2,
for some ν1 and all ν≥ν1. Let us choose ν2=max{ν0,ν1}. Combining (3.8) and (3.10), we know that, for all ν≥ν2,
(3.11)|gν(f(ξ,sν,xν,y))-g(f(ξ,s,x,y))| ≤|gν(f(ξ,sν,xν,y))-g(f(ξ,sν,xν,y))| +|g(f(ξ,sν,xν,y))-g(f(ξ,s,x,y))| <ε2+ε2 =ε.
That is gν(f(ξ,sν,xν,y))→g(f(ξ,s,x,y)). Since gν(f(ξ,sν,xν,y))≥0, there is an s∈T(x) such that g(f(ξ,s,x,y))≥0, which proves that x∈Sξ,η(g). Therefore, the mapping Sξ,η:C⋆∖{0}→2K(η) is closed. By the compactness and Corollary 7 in [33, page 112], the mapping Sξ,η is upper semicontinuous with compact values.
Lemma 3.4. Suppose that for any (ξ,η)∈Δ1×Δ2, x∈K(η) and s∈T(K(η)), f(ξ,s,x,K(η))+C(x) is convex. Then (3.12)Γ(ξ,η)⊇⋃g∈C⋆∖{0}Sξ,η(g).
Furthermore, if
C
:
K
(
η
)
→
2
Y
is constant, then we have
(3.13)
Γ
(
ξ
,
η
)
=
⋃
g
∈
C
⋆
∖
{
0
}
S
ξ
,
η
(
g
)
.
Proof. We first claim that Γ(ξ,η)⊇∪g∈C⋆∖{0}Sξ,η(g).
If x∈∪g∈C⋆∖{0}Sξ,η(g), there is a g∈C⋆∖{0} such that x∈Sξ,η(g). Then, there is a g∈C⋆∖{0} such that
(3.14)g(f(ξ,s,x,y))≥0
for all y∈K(η). This implies that f(ξ,s,x,y)∉-intC(x) for all y∈K(η). Indeed, if there is a y-∈K(η) such that f(ξ,s,x,y-)∈-intC(x). Since g∈C⋆∖{0}, we have g(f(ξ,s,x,y))<0 which contracts (3.14). Thus, x∈Γ(ξ,η). This proves (3.12) holds.
Second, if C:K(η)→2Y is constant, we claim that Γ(ξ,η)⊆∪g∈C⋆∖{0}Sξ,η(g).
If x∈Γ(ξ,η), then x∈K(η) with s∈T(x) and f(ξ,s,x,y)∉-intC for all y∈K(η), that is, f(ξ,s,x,K(η))∩(-intC)=∅. Hence,
(3.15)(f(ξ,s,x,K(η))+C)∩(-intC)=∅.
Since f(ξ,s,x,K(η))+C is convex, by Eidelheit separation theorem, there is a g∈Y⋆∖{0} and ρ∈ℝ such that
(3.16)g(w′)<ρ≤g(f(ξ,s,x,y)+w),
for all y∈K(η), w∈C, w′∈-intC. Then,
(3.17)(g-ρ)(w′)<0≤(g-ρ)(f(ξ,s,x,y)+w),
for all y∈K(η), w∈C, w′∈-intC.
Without loss of generality, we denote g-ρ by g, then
(3.18)g(w′)<0≤g(f(ξ,s,x,y)+w),
for all y∈K(η), w∈C, w′∈-intC. By the left-hand side inequality of (3.18) and the linearity of g, we have g(m)>0 for all m∈intC. Since C is closed, for any m in the boundary of C, there is a net {mν}⊂intC such that mν→m. By the continuity of g, g(m)=g(limνmν)=limνg(mν)≥0. Hence, for all w∈C, g(w)≥0, that is g∈C⋆∖{0}.
By the right-hand side inequality of (3.18), for all w∈C, there is an s∈T(x) such that g(f(ξ,s,x,y)+w)≥0 for all y∈K(η). This implies that g(f(ξ,s,x,y))≥0 for all y∈K(η) if we choose w=0. Hence, sups∈T(x)g(f(ξ,s,x,y))≥0 for all y∈K(η). Thus, x∈Sξ,η(g). Therefore, x∈∪g∈C⋆∖{0}Sξ,η(g), and hence
(3.19)Γ(ξ,η)⊆⋃g∈C⋆∖{0}Sξ,η(g).
Combining this with (3.12), we have
(3.20)Γ(ξ,η)=⋃g∈C⋆∖{0}Sξ,η(g).