Motivated by Takahashi and Zembayashi [22], and Ceng and Yao [23], we next prove the following crucial lemma concerning the GEP in a strictly convex, reflexive, and smooth Banach space.
Theorem 3.1. Let C be a nonempty, bounded, closed, and convex subset of a smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C × C to
satisfying (A1)-(A4), where
(A1) f(x, x) = 0 for all x ∈ C;
(A2) f is monotone, i.e. f(x, y) + f(y, x) ≤ 0 for all x, y ∈ C;
(A3) for all y ∈ C, f(., y) is weakly upper semicontinuous;
(A4) for all x ∈ C, f(x,.) is convex.
Let A be α-inverse strongly monotone of C into E*. For all r > 0 and × ∈ E, define the mapping S
r
: E → 2C as follows:
Then, the following statements hold:
-
(1)
for each x ∈ E, S
r
(x) ≠∅;
-
(2)
S
r
is single-valued;
-
(3)
〈S
r
(x) - S
r
(y), J(S
r
x - x)〉 ≤ 〈S
r
(x) - S
r
(y), J(S
r
y - y)〉 for all x, y ∈ E;
-
(4)
F (S
r
) = GEP (f);
-
(5)
GEP(f) is nonempty, closed, and convex.
Proof. (1) Let x0 be any given point in E. For each y ∈ C, we define the mapping G : C → 2E by
It is easily seen that y ∈ G(y), and hence G(y). ≠∅
-
(a)
First, we will show that G is a KKM mapping. Suppose that there exists a finite subset {y1, y2,..., y
m
} of C and α
i
> 0 with
such that
for all i = 1, 2,..., m. It follows that
By (A1) and (A4), we have
which is a contradiction. Thus, G is a KKM mapping on C.
-
(b)
Next, we show that G(y) is closed for all y ∈ C. Let {z
n
} be a sequence in G(y) such that z
n
→ z as n → ∞. It then follows from z
n
∈ G(y) that,
By (A3), the continuity of J, and the lower semicontinuity of || · ||2, we obtain from (3.2) that
This shows that z ∈ G(y), and hence G(y) is closed for all y ∈ C.
-
(c)
We prove that G(y) is weakly compact. We now equip E with the weak topology. Then, C, as closed, bounded convex subset in a reflexive space, is weakly compact. Hence, G(y) is also weakly compact.
Using (a), (b), and (c) and Lemma 2.5, we have ⋂x∈CG(y) ≠∅. It is easily seen that
Hence, s
r
(x0) ≠∅. Since x0 is arbitrary, we can conclude that s
r
(x) ≠∅ for all x ∈ E.
(2) We prove that S
r
is single-valued. In fact, for x ∈ C and r > 0, let z1, z2 ∈ S
r
(x). Then,
and
Adding the two inequalities and from the condition (A2) and monotonicity of A, we have
and hence,
Hence,
Since J is monotone and E is strictly convex, we obtain that z1 - x = z2 - x and hence z1 = z2.
Therefore S
r
is single-valued.
(3) For x, y ∈ C, we have
and
Again, adding the two inequalities, we also have
It follows from monotonicity of A that
(4) It is easy to see that
Hence, F (S
r
) = GEP (f).
(5) Finally, we claim that GEP (f) is nonempty, closed, and convex. For each y ∈ C, we define the mapping Θ : C → 2E by
Since y ∈ Θ (y), we have Θ(y) ≠∅ We prove that Θ is a KKM mapping on C. Suppose that there exists a finite subset {z1, z2,..., z
m
} of C and α
i
> 0 with
such that
for all i = 1, 2,..., m. Then,
From (A1) and (A4), we have
which is a contradiction. Thus, Θ is a KKM mapping on C.
Next, we prove that Θ (y) is closed for each y ∈ C. For any y ∈ C, let {x
n
} be any sequence in Θ (y) such that x
n
→ x0. We claim that x0 ∈ Θ (y). Then, for each y ∈ C, we have
By (A3), we see that
This shows that x0 ∈ Θ (y) and Θ(y) is closed for each y ∈ C. Thus,
is also closed.
We observe that Θ (y) is weakly compact. In fact, since C is bounded, closed, and convex, we also have Θ(y) is weakly compact in the weak topology. By Lemma 2.5, we can conclude that
.
Finally, we prove that GEP (f) is convex. In fact, let u, v ∈ F (S
r
) and z
t
= tu+(1 - t)v for t ∈ (0, 1). From (3), we know that
This yields that
Similarly, we also have
It follows from (3.4) and (3.5) that
Hence, z
t
∈ F (S
r
) = GEP (f) and hence GEP (f) is convex. This completes the proof.