A REMARK ON THE LOWER SEMICONTINUITY ASSUMPTION IN
THE EKELAND VARIATIONAL PRINCIPLE
Abstract. What happens to the conclusion of the Ekeland variational principle (briefly,
EVP) if a considered function f : X → R ∪ {+∞} is lower semicontinuous not on a whole
metric space X but only on its domain? We provide a straightforward proof showing that
it still holds but only for varying in some interval ]0, β − inf X f [, where β is a quantity
expressing quantitatively the violation of the lower semicontinuity of f outside its domain.
The obtained result extends EVP to a larger class of functions.
1. Introduction
Let X be a complete metric space, and f : X → R ∪ {+∞} a function. The domain,
the epigraph and a lower level set of f are denoted by domf := {x ∈ X | f (x) < +∞},
epif := {(x, t) ∈ X × R | f (x) ≤ t} and [f ≤ t] := {x ∈ X | f (x) ≤ t}, respectively. For any
nonempty set U ⊂ X, intU and bdU denote the interior and the boundary of U .
Recall that f is lower semicontinuous (briefly, lsc) at x ∈ X if the inequality
lim0 inf f (x0 ) ≥ f (x)
x →x
(1)
holds and f is lsc on X if it is lsc at every point x ∈ X [1, Definition 2, p. 11].
Let us recall the Ekeland variational principle (briefly, EVP), which has proved to be a
potent and flexible tool in analysis and in optimization theory, and a related result.
Theorem 1.1 ([3, Theorem 1.1]). Let X be a complete metric space, and f : X → R∪{+∞}
a function, 6≡ +∞, lsc on X and bounded from below. Let be a scalar satisfying
∈]0, +∞[.
Then for any point x0 ∈ X satisfying f (x0 ) ≤ inf X f + and any scalar λ > 0, there exists
x̄ ∈ domf such that:
(a) f (x̄) ≤ f (x0 );
(b) d(x0 ; x̄) ≤ λ;
(c) x̄ is a strict minimizer of the function x → f (x) + (/λ)d(x; x̄), i.e.
f (x̄) < f (x) + (/λ)d(x; x̄),
for all x 6= x̄.
(2)
Date: September 17, 2015.
1991 Mathematics Subject Classification. 49J40.
Key words and phrases. Ekeland variational principle, lower semicontinuity, Gâteaux differentiability.
1
Recall that f is said to be Gâteaux differentiable at a point x ∈int domf (X is assumed
to be a Banach space) if there exists f 0 (x) ∈ X ∗ such that for any u ∈ X
f (x + hu) − f (x)
= f 0 (x)(u).
h→0
h
lim
Proposition 1.1 ([1, Proposition 5, p. 258]). Assume that X is a Banach space and f is
finite and Gâteaux differentiable at x̄. Then condition (2) of Theorem 1.1 implies that
kf 0 (x̄)k∗ ≤ /λ,
where kf 0 (x̄)k∗ is the norm of f 0 (x̄) in the dual space X ∗ .
The main assumption in EVP is that f is “lsc on X” although Ekeland did not clarify
the meaning of the expression “lsc on X” in the paper [3]. His arguments, however, were
essentially based on the fact that the epigraph of f is closed. It is well-known that f is lsc
on X if and only if the epigraph and all lower level sets [f ≤ t] ( t ∈ R) of f are closed.
The assumption that f is lsc on the whole space X is in fact very important. As far
as it is known to the author, there is no attention paid to the question what happens to
the conclusion of EVP when f is only assumed to be lsc on its domain. Meanwhile, the
expression “f is lsc on X” sometimes is understood as “f is lsc on its domain” [2, p. 3]1.
In this note we study this question. Our main results are:
(i) We present examples illustrating that the conclusion of EVP may not hold at all or
it holds not for all positive when f is lsc only on its domain.
(i) We provide a straightforward proof showing that under a natural assumption, the
conclusion of EVP holds for varying in some interval ]0, β − inf X f [, where β is a
quantity expressing quantitatively the violation of Inequality (1) outside the domain
of f . This version of EVP collapses to the classical one when the function is lsc on
the whole space.
2. Main results
First, we consider some examples showing that the conclusion of EVP may fail when f is
lsc only on its domain.
Example 2.1. (i) Let f : R → R ∪ {+∞} be a function defined by f (x) = x if x > 0 and
f (x) = +∞ otherwise. This function is continuous on its domain but is not lsc at x = 0.
1This
has been recently corrected, see https://www.carma.newcastle.edu.au/jon/ToVA/errata.pdf
2
The conclusion of EVP does not hold for any > 0. Indeed, for λ = + 1 if (b) holds for
some x̄, then Proposition 1.1 yields |f 0 (x̄)| ≤ /( + 1) while we have f 0 (x̄) = 1.
(ii) Let f : R → R ∪ {+∞} be a function defined by


 −x + 1 if x < 0
f (x) =
5/x
if x > 0


+∞
if x = 0
This function is continuous on its domain but is not lsc at x = 0 and inf X f = 0. We show
that the conclusion of EVP does not hold for any > 1. Indeed, let x0 = − + δ + 1, where
δ is a scalar satisfying 0 < δ < min{ − 1, 1}. Then x0 = − + δ + 1 < 0 and we have
f (x0 ) = −x0 + 1 = − δ − 1 + 1 = − δ < = inf X f + . Let λ be a scalar satisfying
< λ < + 1 − δ. Suppose to the contrary that we can find a point x̄ 6= 0 satisfying the
assertions (a)-(b) of Theorem 1.1. Since x0 +λ = −+δ+1+λ < −+δ+1++1−δ = 2 and
x0 − λ < x0 < 0, the inequality d(x̄; x0 ) ≤ λ implies x̄ ∈] − ∞, 0[∪]0, 2[. If x̄ ∈] − ∞, 0[, then
f 0 (x̄) = −1 and if x̄ ∈]0, 2[, then f 0 (x̄) = −5/x2 < −1. In both cases, we have |f 0 (x̄)| ≥ 1
while Proposition 1.1 yields |f 0 (x̄)| ≤ λ < 1, a contradiction. The reader will see that the
conclusion of EVP holds for any ∈]0, 1].
To formulate a version of EVP for the case f is lsc on its domain, we need some characterization of how the lower semicontinuity is violated outside the domain of f and how close
f is to being “lsc on X”. For this end, let us establish an auxiliary result.
Lemma 2.1. Suppose that f is lsc on its domain. If [f ≤ t] is closed for some t > inf X f ,
then [f ≤ t0 ] is also closed for any t0 ∈] inf X f, t[.
Proof. Let {xi } be a sequence such that xi ∈ [f ≤ t0 ] for i = 1, 2, ... and xi → x. Since
[f ≤ t0 ] ⊂ [f ≤ t] and [f ≤ t] is closed, we get x ∈ [f ≤ t], which means that x ∈domf .
As f is lsc on its domain, we have f (x) ≤ lim inf x0 →x f (x0 ) ≤ lim inf i→+∞ f (xi ) ≤ t0 . Thus,
x ∈ [f ≤ t0 ] and the lower level set [f ≤ t0 ] is closed.
Now, let us define two quantities, which play an important role in the mentioned above
characterization. Set
β :=
inf
lim0 inf f (x0 )
x∈bd domf \domf
x →x
and (f is assumed to be bounded from below on X)
γ := sup{t > inf f | [f ≤ t] is closed}.
X
Here, we make a convention that sup ∅ = −∞ and inf ∅ = +∞.
Remark 2.1. (i) Assuming that f is lsc on its domain, we consider the question at which
points (outside the domain of f ) the lower semicontinuity of f is violated. It is clear that
3
f is lsc at any point x ∈ int(X \ domf ), because in such a case one has lim inf x0 →x f (x0 ) =
f (x) = +∞. So any point x at which f is not lsc must lie on the boundary of domf with
f (x) = +∞ and inequality (1) does not hold at x, i.e. we have
lim0 inf f (x0 ) < f (x) = +∞.
x →x
0
The value lim inf x0 →x f (x ) may be used as a tool measuring the violation of inequality (1)
at x and the quantity β expresses quantitatively the largest violation of inequality (1) of f
outside its domain. One can see that the larger the value β is, the smaller the violation of
inequality (1) of f outside its domain is.
(ii) It is known that f is lsc on X if and only if all its lower level sets are closed and
in such a case γ = +∞. Clearly, when f is lsc only on its domain (domf 6= X), we have
γ < +∞ and we could say that the larger the value γ is, the closer f is to being “lsc on X”.
The following proposition shows that γ and β are in fact equal.
Proposition 2.1. Let X be a complete metric space, and f : X → R ∪ {+∞} a function,
6≡ +∞, lsc on domf and bounded from below. If one of the two inequalities γ > inf X f and
β > inf X f holds, then the other also holds and γ = β.
Proof. First, suppose that γ > inf X f . We will prove that β ≥ γ. Suppose to the contrary
that β < γ. Then we can choose a scalar t̄ such that β < t̄ < γ and t̄ > inf X f . By the
definition of β, one can find u ∈ bd domf such that f (u) = +∞ and lim inf x0 →u f (x0 ) <
t̄. Then we can find a sequence {xi } such that xi → u, f (xi ) ≤ t̄ and limi→∞ f (xi ) =
lim inf x0 →u f (x0 ) < t̄. Thus, xi ∈ S̄ := [f ≤ t̄] for i = 1, 2, ..., xi → u but f (u) = +∞, i.e.
u 6∈ S̄ and S̄ is not closed. On the other hand, since inf X f < t̄ < γ, the definition of γ
implies that S̄ is closed. The obtained contradiction shows that the inequality β ≥ γ holds
and therefore, β ≥ γ > inf X f .
Next, suppose that β > inf X f . We show that γ ≥ β. It suffices to check that any lower
level set U := [f ≤ t] with t satisfying inf X f < t < β is closed. Observe that U is nonempty
because t > inf X f and that U ⊂ domf . Let {xi } be a sequence such that xi ∈ U for
i = 1, 2, ... and xi → u. We claim that u ∈ domf . Indeed, suppose to the contrary that
f (u) = +∞. Since xi ∈ U ⊂ domf and xi → u, we get u ∈ bd domf . Then we have
β=
inf
lim0 inf f (x0 ) ≤ lim0 inf f (x0 ) ≤ lim inf f (xi ) ≤ t < β,
x∈bd domf,f (x)=+∞ x →x
x →u
i→∞
which is a contradiction. Therefore, u ∈ domf . By the assumption, f is lsc at u. Hence,
f (u) ≤ lim inf x0 →u f (x0 ) ≤ lim inf i→∞ f (xi ) ≤ t, this means that u ∈ U and U is closed.
Thus, the inequality γ ≥ β holds and we obtain γ ≥ β > inf X f .
Finally, it is clear from the used arguments that the assertion is true.
A version of EVP for the case f is lsc on its domain reads as follows.
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Theorem 2.1. Let X be a complete metric space, and f : X → R ∪ {+∞} a function,
6≡ +∞, lsc on domf and bounded from below. Suppose that
β > inf f.
X
(3)
Then the conclusion of EVP stated in Theorem 1.1 holds for any scalar satisfying
∈]0, β − inf f [.
X
(4)
Proof. Let be a scalar satisfying relation (4) and x0 ∈ domf satisfying f (x0 ) ≤ inf X f + .
Set t := inf X f + . We will consider the lower level set S given by S := [f ≤ t]. Since
inf X f < t, the set S is nonempty. Further, since t < β and β = γ by Proposition 2.1, one
can find a scalar t0 such that t < t0 < γ and the set [f ≤ t0 ] is closed. Lemma 2.1 implies
that S is closed.
Define a function f¯ : X → R ∪ {+∞} by
(
f¯(x) =
f (x)
+∞
if x ∈ S
if x ∈
/ S.
We show that f¯ satisfies conditions of Theorem 1.1. First, we show that this function is
lsc on the whole space X. Let x ∈ X be given. If x ∈ domf¯ = S, the lower semicontinuity
of f at x and the equalities f¯(x) = f (x) and f¯(u) ≥ f (u) for all u ∈ X imply that
f¯ satisfies inequality (1) at x. If x ∈ X \ S, inequality (1) holds at x because the set
X \ S is open and, therefore, lim inf x0 →x f¯(x0 ) = f¯(x) = +∞. Thus, f¯ is lsc on X. Next,
since f (x0 ) ≤ inf X f + = t, we get x0 ∈ S = domf¯ and f¯(x0 ) = f (x0 ). We also have
inf X f¯ = inf X f > −∞. Therefore, f¯(x0 ) ≤ inf X f¯ + .
Applying Theorem 1.1 to the function f¯, we obtain the existence of x̄ ∈ domf¯ such
that f¯(x̄) ≤ f¯(x0 ), d(x0 ; x̄) ≤ λ, and x̄ is a strict minimizer of the function x → f¯(x) +
(/λ)d(x; x̄), i.e.
f¯(x̄) < f¯(x) + (/λ)d(x; x̄), ∀x ∈ domf¯, x 6= x̄.
Further, as x̄ ∈ domf¯ = S, we have f¯(x̄) = f (x̄). Recall that f¯(x0 ) = f (x0 ). Then it
follows from f¯(x̄) ≤ f¯(x0 ) that f (x̄) ≤ f (x0 ). It remains to check that the assertion (c)
stated in Theorem 1.1 holds. Let x ∈ domf, x 6= x̄ be given. Note that f¯(x̄) = f (x̄) because
x̄ ∈ S. If x ∈ S, then f¯(x) = f (x) and we have f (x̄) = f¯(x̄) < f¯(x) + (/λ)d(x; x̄) = f (x) +
(/λ)d(x; x̄). If x ∈
/ S, then f (x) > t ≥ f (x̄) and hence, f (x̄) < f (x) < f (x) + (/λ)d(x; x̄).
Thus, the assertion (b) holds.
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3. Some comments
(i) When f is lsc on X, we have
sup{t > inf f | [f ≤ t] is closed} =
X
inf
x∈bd domf \domf
lim0 inf f (x0 ) = +∞
x →x
and Theorem 2.1 reduces to Theorem 1.1.
(ii) Inequality (3) cannot be either dropped or replaced by the inequality β ≥ inf X f , see
Example 2.1 (i).
(iii) If γ = min{t > inf X f | [f ≤ t] is closed}, i.e. the set [f ≤ γ] is closed, the arguments
used the proof of Theorem 2.1 shows that relation (4) can be replaced by
∈]0, β − inf f ].
X
(iv) Estimate of given in relation (4) cannot be improved in the sense that the conclusion
of EVP may fail if we take > β − inf X f . Let us consider the function f in Example
2.1 (ii). We have γ = β = 1, inf X f = 0 and the lower level set [f ≤ 1] is closed. The
conclusion of EVP holds for any in ]0, 1] by Theorem 2.1 and comment (iii) but not
for > 1 as it has already been shown.
(v) Theorem 2.1 extends Theorem 1.1 to a larger class of functions. For instance, if
the domain of f is open (domf 6= X), and f is lsc and bounded from above on its
domain, then f cannot be lsc on the whole space X. Indeed, if x ∈bd domf , then
f (x) = +∞ while the boundedness from above of f implies lim inf x0 →x f (x0 ) < +∞
and therefore, f is not lsc at x. Then only Theorem 2.1 is applicable to the function
f while Theorem 1.1 is not. One can see that this also happens in Example 2.1(ii).
(vi) Theorem 2.1 may allow to get the same conclusion as in Theorem 1.1 but under the
weaker assumption that f is lsc on its domain. For instance, one can derive from
Theorem 2.1 results about the density of the range f 0 (X) similar to the ones stated
in [3, Corollaries 2.4 and 2.5] because only sufficiently small values of are needed for
the proof. But in general, if the value is a priori given, one should check whether
inequality (3) holds and satisfies relation (4) or not before using the conclusion of
EVP.
Acknowledgements: The author would like to express her gratitude to Prof. A.
Kruger for his useful comments and encouragement.
References
[1] J.-P. Aubin, I.Ekeland, Applied Nonlinear Analysis, John Wiley and Sons, 1984.
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[2] J.M. Borwein, Q.J. Zhu, Techniques of Variational Analysis, Springer, 2005.
[3] I. Ekeland, On the Variational Principle, J. Math. Anal. Appl. 47 (1974), 324-353.
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