Recently, Qiu [18] obtained some versions of Ekeland's variational principle in locally convex spaces, which only need to assume local completeness of some related sets. Motivated by this paper we obtain some versions of EVP for vector-valued bifunctions in locally convex spaces. These results extend Qiu's results to vector-valued bifunctions.

Throughout this section is a locally convex space, is locally closed subset of , is a family of seminorms generating the locally convex topology on , is a Hausdorff locally convex space ordered by a closed convex cone with and . We consider a vector-valued bifunction , a family of positive real numbers and the following assumptions:

(A1) for all .

(A2) for any .

(A3) is C-bounded from below for all .

(A4) is -locally lower semicontinuous for any .

(A5)There exists such that the set is locally complete.

(A6)The set is locally complete.

Notice that if assumptions (A1) and (A2) hold, then is called half distance. The following result is a vectorial form of Ekeland-type variational principle.

Theorem 2.1.

Suppose that assumptions (A1)–(A4) are satisfied. If either assumption (A5) or assumption (A6) holds, then for any , there exists such that

(i), for any ;

(ii)For any , there exists such that .

Proof.

Without loss of generality, we may assume that and put with the product topology, then the topology can be generated by a family of seminorms, where , for all . If and , then since is -bounded from below for all we have . Take any fixed real number and put . Then is exactly the set , for all . If the set is locally complete, then is locally complete and if is locally complete, then is locally complete. Furthermore, is bounded closed convex subset of and . Hence, by Theorem 1.3, there exists

such that

According to (2.1), we have , so

and for each

Therefore, by (2.3) and (2.4), we have

Hence, the part (i) holds. We show that the point satisfies in the part (ii). Let . Since and is monotone, then . On the other hand we have . Hence,

But from (2.5) we have

Therefore, . Thus, Also, clearly . Hence, we have

Therefore, by (2.1), and so .

Supposing that and , we consider the following two cases.

Case 1.

If , then . Since is monotone, then for all we have

But is half distance and is sublinear, thus

Hence, by the part (ii) of Lemma 1.2;

Case 2.

Let , we will show that . If not, we assume that , that is,

Since and separates points in , we conclude that there exists such that thus . Put

Since is a cone,

that is,

By (2.1),

Since is a convex cone, by (2.15) and (2.16) we have

so

It is easy to verify that

Hence,

Therefore, and so , which it is a contradiction. This shows that . Thus, there exists such that

On the other hand by (A2) we have

Hence,

Therefore,

Remark 2.2.

In the above theorem, if assumption (A5) holds, then instead of assumption (A3), we can assume that is -bounded from below. Also, if assumption (A6) holds, assumption (A3) can be replaced by the following assumption: is -bounded from below for some .

As a consequence of the above theorem we can obtain the following result which is a vectorial version of Theorem 3.1 of [18].

Corollary 2.3.

Let be a function such that is -bounded from below and is -locally lower semicontinuous. Furthermore, let assumption (A6) holds or there exists such that the set is locally complete. Then there exists such that

(i), for any ;

(ii)for any , there exists such that .

Proof.

It is enough in Theorem 2.1 to consider for all .

In the following theorem we show that the previous results are equivalent to each other.

Theorem 2.4.

Corollary 2.3 implies Theorem 2.1.

Proof.

Let be defined as follows:

It is an easy task to derive the assumptions of Corollary 2.3 for the above function from the assumptions of Theorem 2.1. Therefore, there exists which satisfies the conditions (i) and (ii) of Corollary 2.3. Hence,

(i), for any ;

(ii)for any , there exists such that .

Also, by assumption (A2) we have . Thus,

Let be a convex subset of containing 0. The Minkowski functional of is defined as follows:

We extend by an additional element such that for all and for all .

By using Theorem 2.1 we obtain another version of vectorial form of Ekeland-type variational principle in which the perturbation function is the Minkowski functional of a bounded set.

Theorem 2.5.

Suppose that assumptions (A1)–(A4) are satisfied. Let be a locally closed, bounded convex set containing 0 and *α* be a positive real number. Let be locally complete or assumption (A5) holds. Then, for any , there exists such that:

(i);

(ii)For any , .

Proof.

Suppose that is the absolutely convex hull of the set , then is a normed space. Assume that

Since , then . Also, is -locally lower semicontinuous and is locally lower semicontinuous, then is closed in . Suppose that is restricted to . If is locally complete then is a Banach disk and is a Banach space. If the set is locally complete, then is a complete set in . Therefore, by Theorem 2.1 there exists such that:

(a).

(b)For any and ,

Since , then the part (a) holds. Now, we show that the part (b) holds. If and , then (2.29) becomes

Let and , then , so the part (b) holds. Let , , and . Since , then

Therefore, which is a contradiction. Hence,

Assuming that is a locally complete locally convex space, the condition on local completeness of some related subsets is automatically satisfied. However, we give the following examples of spaces which are not locally complete but the condition on local completeness of some related subsets is satisfied.

Example 2.6.

Let be the space of all continuous functions defined on . By Corollary 11-7-3, 11-7-4 of [29], with *weak*-topology is quasi barreled but it is not barreled. Therefore, by Proposition 11-2-5 of [29], with *weak**-topology is not locally complete.

Moreover,

But by Banach-Alaoglu theorem this set is *weak**-compact. Also is separable, so *weak**-topology on unit ball is metrizable. Hence, this set is locally complete.

Example 2.7.

Let be the space of all differentiable functions whose derivative is continuous. Then is not a complete space. Therefore, by Proposition [28] is not a locally complete space.

Also, the set is not locally complete. Suppose that is defined as follows:

where is ordered by the cone . If we choose , then the set is locally complete.