2.2. The strong Ekeland variational principle
Let X be a Banach space and a function. A point is called
a minimum point for f if for all ;
a strict minimum point for f if for all ;
a strong minimum point for if and every sequence in X such that is norm-convergent to .
A sequence satisfying is called a minimizing sequence for f.
Remark 2.3. A strong minimum point is a strict minimum point, but the converse is not true.
Indeed, if there exist such that , where then the sequence satisfies but it is not convergent. Also, the function has a strict minimum at 0, but the sequence does not converge to 0.
Condition (b′) in Theorem 2.2 asserts that, in fact,
z is strict minimum point for the perturbed function
. Georgiev [
16,
17] proved a stronger variant of Ekeland variational principle, guaranteeing the existence of a strong minimum point
z for
.
Theorem 2.4 (Strong Ekeland Variational Principle).
Let be a complete metric space and a lsc function bounded from below on X. Then for every and there exists such that
Geogiev,
loc. cit., also showed the equivalence of this strong form of EkVP with stronger forms of Danes’ drop theorem, flower petal theorem, Phelps lemma, and others, extending so the results obtained by Penot [
24]. He gave a direct proof to the strong drop theorem, the strong EkVP being a consequence of the equivalence mentioned above. Later Turinici [
35] has shown that this strong form can be deduced from Theorem 2.2.
Observe that there is a discrepancy between the conditions (a′) in Theorem 2.2 and condition (a) in Theorem 2.4, condition (a′) being stronger than (a). As was remarked by Suzuki [
33,
34], a strong version of the Ekeland variational principle with condition (a′) instead of (a) can be proved by imposing supplementary conditions on the underlying metric (or Banach) space
X, which are, in some sense, also necessary.
Let
be a proper function defined on a metric space
. For
and
consider an element
satisfying the following conditions:
If is a normed space, then is replaced by
A metric space is called boundedly compact if every bounded closed subset of X is compact, or equivalently, if every bounded sequence in X contains a convergent subsequence.
Remark 2.5. It is obvious that a boundedly compact metric space is complete, and that a normed space is boundedly compact if and only if it is finite dimensional.
Theorem 2.6 ([
33]).
Let be a boundedly compact metric space, a lsc bounded from below function, and
Then there exists a point satisfying the conditions (2.5).
Remark 2.7. 1. Let
X be a vector space. A function
is called
quasi-convex if
for all
and
This is equivalent to the fact that the sublevel sets
are convex for all
(see [
23]).
2. One says that a Banach space X is a dual Banach space if there exists a Banach space Y such that Obviously, a reflexive Banach space is a dual Banach space with and, in this case, the weak (i.e. ) and the weak* (i.e. ) topologies on X agree.
In the Banach space case the following results can be proved.
Theorem 2.8 ([
33]).
Let X be a Banach space, a bounded from below function, and
- 1.
If X is a dual Banach space and f is -lsc, then there exists a point satisfying (2.5) with in the condition (iii).
- 2.
Suppose that the Banach space X is reflexive. If f is weakly lsc, then there exists a point satisfying the conditions (2.5). The same is true if f is quasi-convex and norm-lsc.
As it was shown by Suzuki [
34], in some sense, the results from Theorems 2.6 and 2.8 are the best that can be expected.
Theorem 2.9. For a metric space the following are equivalent.
- 1.
The metric space X is boundedly compact.
- 2.
For every proper lsc bounded from below function , and there exists a point satisfying the conditions (2.5).
- 3.
For every Lipschitz function , and there exists a point satisfying the conditions (2.5).
A similar result holds in the case of normed spaces.
Theorem 2.10. For a normed space the following are equivalent.
- 1.
X is a reflexive Banach space.
- 2.
For every proper lsc bounded from below quasi-convex function , and there exists a point satisfying the conditions (2.5).
- 3.
For every Lipschitz convex function , and there exists a point satisfying the conditions (2.5).