Variational inequality theory has played a fundamental and important role in the study of a wide range of problems arising in physics, mechanics, differential equations, contact problems in elasticity, optimization, economics and engineering sciences, etc. As a useful and important branch of variational inequality theory, vector variational inequalities were initially introduced and considered by Giannessi [1] in a finite-dimensional space in 1980. Ever since then, vector variational inequalities have been extensively studied and generalized in infinite-dimensional spaces.

Very recently, Fang and Huang [2] studied some existence results for strong vector variational inequalities in Banach spaces. Long, Huang and Teo [3] established an existence theorem of solutions for a generalized strong vector quasi-equlilbrium problem by using KakutaniFan-Glicksberg fixed point theorem. Li and He [4] studied the existence of solutions for VVI with a single-valued function and a continuous selection theorem and obtained the existence theorem for the GVVI under the assumption of -pseudomonotonicity.

Motivated and inspired by research works mentioned above, in this paper, we consider a generalized system, which includes as special cases the strong vector variational inequalities [1,2] and problems so called equilibrium problems [5,6]. In order to derive the existence of solutions for the generalized system, Using Brouwer’s fixed point theorem, we obtain some existence results for the generalized system without monotonicity. Furthermore, by the concepts of -C-continuous, C-strong pseudomonotonicity and hemicontinuity, we extend Minty’s lemma, moreover, with the help of Minty’s lemma and KKM-Fan lemma, we establish an existence theorem for the generalized system with monotonicity in real reflexive Banach spaces. The results presented in this paper extend the corresponding results of [2,6], and Theorems 3.1-3.3 can be considered as a generalization of Theorems 2.1-2.3 in [6].

Throughout the paper, otherwise special statement, let X and Y be two real Banach spaces, let be a nonempty, closed and convex set, let be a solid, pointed, closed and convex cone with apex at the origin. Let, be the space of all continuous linear mappings from X to Y. Let be a trifunction such that for all and. Then we consider the following generalized systems (for short, GS): Find, for each there exists such that  

And find such that

We know that a solution is called a weak solution and a strong solution to GS(1) and GS(2), respectively. It is easy to see that a solution to GS(2) is a solution to GS(1), but in general the converse is not true. Noting, , where is a single-valued mapping, is a nonlinear mapping, then GS(1)and GS(2) both reduces to the following strong vector variational-like inequality problem (for short, SVVLIP): Find such that

Now, we recall the following definitions.

Definition 2.1 [7]. Let K be a subset of a topological vector space X. A set-valued mapping is called a KKM mapping if, for each nonempty finite subset, we have where CoA denotes the convex hull of the set A.

Definition 2.2. Let X and Y be topological vector spaces, let K be a nonempty, convex subset of X and let C be a convex cone in Y. A mapping is said to be C-convex function if, for all and, ,

Remark 2.1. 1) If C contains (or is equal to,or is contained in) the non-negative orthant, then the C-convex function is called C-convex (or convex, or strictly Cconvex);

2) If C contains (or is equal to, or is contained in) the nonpositive orthant, then the C-convex function is called C-concave (or concave, or strictly C-concave).

Theorem 2.1. (Brouwer’s fixed point theorem [8]) Let K be a nonempty, compact and convex subset of a finite-dimensional space X and let be a continuous mapping. Then, there exists such that.

Theorem 2.2. (KKM-Fan Lemma [7]) Let K be a subset of a topological vector space X and let be a KKM mapping. If, for each, is closed and for at least one, is compact, then.

In this section, we shall obtain the following existence theorems for GS(1).

Theorem 3.1. Let K be a nonempty, compact and convex subset of X and let the mapping be a C-convex function in the second argument. Assume that, for every, the set is open in K. Then, GS(1) has a solution.

Proof. We proceed by contradiction. Assume that GS(1) admits no solution, then for each, there exists some and for all such that

For every, define the set as follows:

By given assumption, the set is open in K and hence from (4), it follows that is an open cover of K. Since K is compact, there exists a finite set such that. So there exists a continuous partition of unity subordinate to such that for all, with and whenever, whenever.

Define a mapping by

Since is continuous for each j, it follows from (6) that h is also continuous. Let. Then S is a simplex of a finite dimensional space and h maps S into S. By Theorem 1.1, there exists some such that.

Next define a mapping by . Since f is a C-convex function in the second argument, by (6), we have

For any given, let. Clearly, . since is a fixed point of h, it follows from (4) and (7) that for all,

That is, , a contradiction. Hence, GS(1) has a solution. This completes the proof.

Example 3.1. Let, , , ,

, ,.

For all and,

i.e.,

Then it is easy to see that f is a C-convex function in the second argument, and the

is open in K. Now we could say that all conditions of Theorem 2.1 hold. And also, the solutions set for GS(1)

in Theorem 2.1 is,.

Theorem 3.2. Let K be a nonempty, closed and convex subset of X and let the mapping be a C-convex function in the second argument. Assume that:

1) For every, the set

is open.

2) K is locally compact and there is an and, , such that, for all, , there exists an such that.

Then, GS(1) has a solution.

Proof. Let. Since K is locally compact, is compact; hence, it follows from Theorem 2.1 that there exists an, for all there exists such that

We claim that is the desired solution of GS(1). Indeed:

1) If, by the assumption 2), there exists such that

For any, choose such that . Then, from (7), it follows that. Since f is a C-convex function in the second argument, we have

Implying that

2) If, for any, choose such that. Then, it follows from (7) that. Since f is a C-convex function in the second argument, we have

This completes the proof.

The following example shows that the assumption that the set is open in K, for every, is not trivial. For two similar examples for a vector-valued function, see [2,6].

Example 3.2. Let, , , ,

, , ,

is continuous and monotone in the sense that for all. Also, it is easy to see that, for each, the set and hence is open in K.

Next, we define following concepts which will be used in the sequel.

Definition 3.1. A mapping is said to be C-strong pseudomonotone respect to if, for any, there exists, implies,.

Remark 3.1. This definition generalizes corresponding definitions of [9,10].

Example 3.3. Let, , ,

, ,

, ,.

That implies that, it follows that

Hence, f is C-strong pseudomonotone. For a similar example for a vector-valued function, see [6].

Definition 3.2. A mapping, is said to be −C-continuous at if, for each neighbourhood of nullelement in Y, there exists a neighbourhood of X, such that

If f is −C-continuous at each point of K, then f is −C-continuous on K.

Definition 3.3. Let W, E be two topological spaces, a set-valued mapping is upper semicontinuous at if, for each neighbourhood V of Tx, there exists a neighbourhood U of x, such that. If T is upper semicontinuous at each point of W, then T is upper semicontinuous on W.

Lemma 3.1 [11]. Let W, E be two topological spaces, a set-valued mapping is compact valued, then T is said to be upper semicontinuous at if and only if: for any net, , and, , there exists a setnet of, such that.

Now we prove the existence result for GS(2) with monotonicity. First, we prove the following Minty’s type lemma for GS(2).

Lemma 3.2. Let K be a nonempty closed and convex subset of X, Y be a real Banach space ordered by a nonempty closed convex pointed cone C with apex at the origin and. Let be a trifunction, be a nonempty compact set-valued mapping. Suppose the following conditions hold:

1), ,;

2) is C-convex function in the second argument;

3) T is upper semicontinous;

4) is -C-continuous with respect to, ,.

Then there exists and, such that

if and only if there exists such that

Proof. (9) (10): It follows from the C-strong pseudomonotonicity of f respect to T.

(10) (9): Suppose that there exists such that

If is not the solution of (9), that is to say, , , we have

For any given, we set, , for each, it follows that

Since f is C-convex function in the second argument, we have

It follows from inclusions (13), (14) and condition 1), we obtain

Implying that

From T is upper semicontinous and, , as, we obtain that has a setnet (for simplicity, still denoted) and there exists, such that.

If we could prove

we will obtain a contradiction between (16) and (12), as C is convex pointed cone. That will complete the proof.

In fact, suppose on the contrary that

Since C is closed, we know that there exists a origin neighbourhood V in Y, such that

Since C is convex cone, it follows that

From 4), there exists neighbourhood of such that

Since, , as, , such that

Thus from (18) we get

it follows (17) and (19) that

which leads a contradiction to (15). This completes the proof.

Theorem 3.3. Let K be a nonempty, closed, bounded and convex subset of a real reflexive Banach space X and let the mapping be a C-convex function in the second argument and C-strong pseudomonotone. be a nonempty compact setvalued mapping. Then, GS(2) has a solution.

Proof. Define two set-valued mappings, by

and

Clearly, and both are nonempty since for all.

We claim that A is KKM-mapping. If assertion were false, then there would exists and, , with such that

. Hence for any, we have, for each i.

Implying that,. Thus,

. Since f is a C-convex function in the second argument, we have

which is a contradiction. Hence A is a KKM-mapping.

Since f is C-strong pseudomonotone, it follows that for all, and hence B is also a KKM-mapping. By Lemma 2.1, we see that

.

Next we claim that for each fixed, is bounded, convex and closed in K. Indeed, is bounded as. Let, then we have

Since f is C-convex function in second argument, we have that, for,

This implies that is convex for each fixed. The continuity of f in the second argument and closedness of -C give the closedness of. We now equip X with that weak topology. Since is closed, bounded and convex subset of the reflexive Banach space X, then it turns out to be weakly compact for all. Hence by Theorem 1.2, we have. This implies that there exists, such that

Therefore by Minty’s type Lemma 3.2, we conclude that there exists and such that

That is to say GS(2) has a solution. This completes the proof.

Remark 3.2. Let, , where is a single-valued mapping, is a nonlinear mapping, then GS(1) and GS(2) both reduces to the SVVLIP(3). As applications, we have the following existence result for SVVLIP(3).

Corollary 3.1. Let K, X, T, be as in Theorem 3.1 and Remark 3.2, let be affine in the first argument. For each, the set is open in K. Then the SVVLIP(3) has a solution.

Corollary 3.2. Let K, X, T, be as in Corollary 3.1. Assume that:

1) For every, the set

is open.

2) K is locally compact and there is an and, , such that for all, ,

Then, the SVVLIP(3) has a solution.