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.