Ekeland-type variational principle with applications to nonconvex minimization and equilibrium problems

Abstract. The aim of the present paper is to establish a variational principle in metric spaces without assumption of completeness when the involved function is not lower semicontinuous. As consequences, we derive many fixed point results, nonconvex minimization theorem, a nonconvex minimax theorem, a nonconvex equilibrium theorem in noncomplete metric spaces. Examples are also given to illustrate and to show that obtained results are proper generalizations.


Introduction and preliminaries
Ekeland [17,18] formulated a variational principle, which is considered the basis of modern calculus of variations.Since its discovery, there are many generalizations and equivalent formulations of it (see [19,20,28,37,40,45] and references therein).This principle says that when a function is not guaranteed to have a minimum, there is a "good" approximate substitute.Under the conditions of lower semicontinuity and boundedness below, the best we can get is an approximate minimum.These results are essential in areas of nonlinear analysis and optimization theory.A visualization of Ekeland's variational principle is shown in Fig. 1.
In Fig. 1, by taking λ = 1 we draw a line with slope equal to − = − tan θ.Then the theorem guarantees that for any given , there is a point (v, F (v)) such that if we create an open downwards cone with that point as its vertex and having angle 2θ, the function values for all other inputs will stay above the cone.Here F : X → R ∪ {∞} is a lower semicontinuous function, which is bounded below.
Ekeland's variational principle has been widely used to prove the existence of approximate solutions of minimization problems for lower semicontinuous functions on c Vilnius University, 2019 complete metric spaces (see, for instance, [5]).Since minimization problems are particular cases of equilibrium problems, many authors deal to derive existence results for solutions of equilibrium problems using Ekeland-type variational principles.There are many improvements and generalizations of Takahashi's nonconvex minimization theorem, Caristi's fixed point theorem and Ekeland's variational principle in complete metric spaces by using generalized distances: for example, w-distances, τ -distances, τ -functions, weak τ -functions and Q-functions (see [2,22,30,42,44]).
Meanwhile, the lower semicontinuous condition plays a key role in finding the solution of min x∈X f (x), but it is not essential for solving some minimization problems.A function, which may not be necessarily lower semicontinuous, can still obtain its infimum.By getting motivation from above mentioned work, in present paper, we prove some generalizations of Ekeland's variational principle in the setting of T -orbitally complete metric spaces for functions, which are not necessarily lower semicontinuous, by introducing T -orbitally lower semicontinuity.As an application, we deduce some fixed point results, Takahashi's nonconvex minimization theorem, a nonconvex minimax theorem, a nonconvex equilibrium theorem in the setting of T -orbitally complete metric spaces.Our result generalizes the results of [9, 10, 12-14, 17, 23, 32, 34, 37].
In the sequel, let (X, d) be a metric space and T : X → X be a self mapping.For A ⊆ X, the diameter of A is diam(A) = sup{d(a, b): a, b ∈ A}, and for each x ∈ X, orbit of T is O(x; n) = {x, T x, T 2 x, T 3 x, . . ., T n x}, n = 1, 2, 3, . . ., and O(x) = O(x; ∞) = {x, T x, T 2 x, . . .}.A metric space X is said to be T -orbitally complete if every Cauchy sequence, which is contained in O(x) for some x ∈ X, converges in X [13].Note that every complete metric space is T -orbitally complete space, but converse is not true in general.For example, X = (0, 1) with usual metric is not a complete metric space but a T -orbitally complete, where T : X → X is defined by T x = 1/2 for all x ∈ X. Definition 1. (See [7].)Let (X, d) be a metric space and T : X → X be a self mapping.T is said to be orbitally continuous at a point z in X if for any sequence The Banach contraction principle [8] is an elementary result in metric fixed point theory.Many interesting generalizations of this golden principle have been obtained by https://www.mii.vu.lt/NA considering several conditions on metric spaces (see [1,6,12,15,21,23,24,31,32,[34][35][36]43]).An interesting generalization is given by Berinde [10] by introducing a comparison function.
Definition 2. A function φ : R + → R + is said to be comparison function if it satisfies the following: (φ1) φ is monotone increasing; (φ2) {φ n (t)} converges to 0 as n → ∞ for all t 0.
Denote the set all comparison functions φ : R + → R + by Φ.

Auxiliary results
We begin with the following definitions.Definition 3. Let (X, d) be a metric space and T : X → X be a self mapping.A function f : X → R is said to be T -orbitally continuous at a point z in X if for any sequence Proof.Suppose that T is orbitally continuous function at z ∈ X, then there exists a sequence {x n } contained in O(x) for some x ∈ X such that Definition 4. Let (X, d) be a metric space and T be a self mapping on X.A function f : X → R is said to be T -orbitally lower semicontinuous mapping if where {x n } is a sequence contained in O(x) for some x ∈ X. Definition 5. Let (X, d) be a metric space and T be a self mapping on X.A function f : X → R is said to be T -orbitally upper semicontinuous mapping if where {x n } is a sequence contained in O(x) for some x ∈ X.
Remark 1.Every lower semicontinuous (upper semicontinuous) mapping is T -orbitally lower semicontinuous (upper semicontinuous) mapping, but converse needs not to be true as shown in example below.
Lemma 3. Let T be a self mapping on a metric space (X, d) and f : X → R be a function.Then f is T -orbitally continuous if and only if f is T -orbitally lower semicontinuous and T -orbitally upper semicontinuous.
Proof.Suppose that f is T -orbitally lower semicontinuous and T -orbitally upper semicontinuous, then for a sequence {x n } ⊆ O(x) for some x ∈ X such that x n → z ∈ X, we have and Combining ( 1) and (2), we get f x n → f z as n → ∞.Hence f is T -orbitally continuous.Conversely, suppose that f is T -orbitally continuous, then there exists a sequence Hence f is T -orbitally lower semicontinuous and T -orbitally upper semicontinuous.Definition 6.Let (X, d) be a metric space and T be a self mapping on X.A subset Y of X is said to be T -orbitally complete if every Cauchy sequence, which is contained in O(y) for some y ∈ Y , converges in Y .https://www.mii.vu.lt/NALemma 4. Let (X, d) be a metric space and T be a self mapping on X. Assume that X is T -orbitally complete and Y is closed subset of X.Then Y is T -orbitally complete.
Proof.Suppose {y n } is a Cauchy sequence contained in O(y) for some y ∈ Y ⊆ X.Since X is T -orbitally complete, so y n → z ∈ X.But Y is closed, so z ∈ Y .This completes the proof.

Variational principle
Let (X, d) be a metric space and T be a self mapping on X such that O(x) = {x, T x, T 2 x, T 3 x, . . .}. Assume the following: (A) There are two functions α : and (O1) If x n is a sequence contained in orbit of T such that x n → z, then z belongs to some orbit of T .
In addition, if f : X → R ∪ {+∞}, δ 0 > 0, δ n 0, n ∈ N, is a sequence of nonnegative integers and u 0 ∈ O(x 0 ) for some x 0 ∈ X, then for all w ∈ X, denote Theorem 1.Let (X, d) be a metric space, T be a self mapping on X and f : X → R ∪ {+∞}.Assume that ρ ∈ Ω, (α1) is satisfied and for u 0 ∈ O(x 0 ) for some x 0 ∈ X and > 0, following assumptions hold: Then there exists a sequence u n in X and u ∈ O(x) such that when for infinitely many n, δ n > 0, and when δ k > 0 and δ j = 0 for all j > k 0, conclusion (iv) is replaced by Proof.There arises two cases for δ n : Case 1: Infinitely many δ n > 0.
In this case, without loss of generality, we assume that δ n > 0 for all n.Then for u 0 ∈ O(x 0 ) for some x 0 ∈ X, consider Since u 0 ∈ G(u 0 ), so G(u 0 ) is nonempty.For every y ∈ G(u 0 ), by using assumption (A1), we have Choose similarly as above, G(u 1 ) is nonempty.Continuing in this process, we can choose u n−1 ∈ G(u n−2 ) and consider Let us choose and define which is nonempty.From ( 8) and ( 9), for each y ∈ G(u n ), we get https://www.mii.vu.lt/NAThis implies that for all y ∈ G(u n ), we have Now we have sequence of nonempty sets {G(u n )} such that Thus, the intersection contains only one point.From ( 5)-( 11) we obtain that u satisfies (ii) and u n → u as n → ∞.Further, for all w = u , we have From ( 4), ( 7) and ( 8), for every q m, we obtain Combining ( 12) and (13) gives Letting q, m → ∞ gives (iii) and (iv).
Assume that δ k > 0 and δ j = 0 for all j > k 0. Without loss of generality, we suppose that δ i > 0 for all i k.Thus, when n k, we choose the same u n and G(u n ) as in Case 1.When n > k, we take and put Then by following the same steps as in Case 1, hypotheses (ii)-(iv) hold.When w = u , there exists an m > k such that This completes the proof.
There exists Take δ 0 = 1 and δ n = 0 for all n ∈ N. Since S m = {0} for each m ∈ N ∪ {0}, therefore S 0 is nonempty and T -orbitally complete, also for each m ∈ N, S m is closed.Hence hypothesis (A), (A1), (A2) and (A3) of Theorem 1 hold true.Further, there exists a sequence g n in X such that g n (t) = 1/n for all t ∈ [0, 1] and This implies that Hence conclusions (i), (ii) and (iii) of Theorem 1 hold.Since δ 0 = 1 and δ j = 0 for all j > 0, so we have to satisfy conclusion (v).For this, there arise two cases.Case 1.For 0 = g ∈ X such that g(t) > 0, there exists m 0 such that https://www.mii.vu.lt/NA(a) If u m (t) > 0 and u m (t) > g(t) for all t, then (16) gives (b) If u m (t) > 0 and u m (t) < g(t) for all t, then (16) gives (c) If u m (t) < 0, then (16) gives Case 2. For 0 = g ∈ X such that g(t) < 0, there exists m 0 such that (a) If u m (t) > 0 for all t, then (17) gives (b) If u m (t) < 0 and u m (t) < g(t) for all t, then (17) gives (c) If u m (t) < 0 and u m (t) > g(t) for all t, then (17) gives Thus, conclusion (v) of Theorem 1 holds for all g = 0 = u .
Remark 2. Note that, in Example 2, metric space (X, d) is not complete.Indeed, consider the sequence {g n } in Fig. 2. Since d(g n , g m ) < for n, m > 1/ , therefore {g n } is a Cauchy sequence such that for every g ∈ X which is not continuous.Hence {g n } is not convergent.Therefore, Theorem 3.1 of [3], Theorem 1.1 of [17], Theorem 4.2 of [16] and Theorem 1 of [37] cannot be applied for this example.Proof.Let u 0 ∈ O(x 0 ) for some x 0 ∈ X and Since u 0 ∈ S 0 , so G(u 0 ) is nonempty.Now, if z n is a sequence in S 0 that converges to z ∈ X, then z belongs to orbit of T and ∆(z n , 0) f (u 0 ).From T -orbitally lower semicontinuity of f , T -orbitally lower semicontinuity of ρ(., u 0 ) and (α1) we get This shows that z ∈ S 0 and S 0 is closed subset of X.Since X is T -orbitally complete, so from Lemma 4, S 0 is T -orbitally complete.Next, suppose that for m ∈ N, y ∈ S m−1 , w n is a sequence contained in the set S m = {w ∈ S m−1 | ∆(w, m) ∆(y, m − 1)} such that w n → w * ∈ X.Then w * belongs to some orbit of T , and ∆(w n , m) ∆(y, m−1).
From T -orbitally lower semicontinuity of f , T -orbitally lower semicontinuity of ρ(., u 0 ) and (α1) we get that is, ∆(w * , m) ∆(y, m − 1).Consequently, S m is closed.https://www.mii.vu.lt/NAFrom Lemma 5 and Theorem 1 we obtain the following: Theorem 2. Let (X, d) be a metric space, T be a self mapping on X and f : X → R ∪ {+∞} be a T -orbitally lower semicontinuous function, bounded from below.Assume that X is T -orbitally complete and (A) holds.If for u 0 ∈ O(x 0 ) for some x 0 ∈ X and > 0, Then there exists u ∈ O(x) such that all conclusions (i)-(v) of Theorem 1 hold.
Taking α(x, y) = 1 for all x, y ∈ X in Theorem 2, we get the following: Corollary 1.Let (X, d) be a metric space, T be a self mapping on X and f : X → R ∪ {+∞} be a T -orbitally lower semicontinuous function, bounded from below.Assume that X is T -orbitally complete and (A) holds.If for u 0 ∈ O(x 0 ) for some x 0 ∈ X and > 0, Then there exists u ∈ O(x) such that when for infinitely many n, δ n > 0, and when δ k > 0 and δ j = 0 for all j > k 0, conclusion (iv) is replaced by Since there is always a some point x with f (x) inf F + and also x ∈ O(x), therefore form Corollary 1 we have the following: Corollary 2. Let (X, d) be a metric space, T be a self mapping on X and f : X → R ∪ {+∞} be a T -orbitally lower semicontinuous function, bounded from below.Assume that X is T -orbitally complete and (A) holds.Then there exists u ∈ O(x) such that conclusions (i)-(v) of Corollary 1 hold.
If we consider ρ(y, z) = ( /λ)d(y, z), δ 0 = 1 and δ n = 0 for all n > 0 in Corollary 1, then we get the following corollary, which is the Ekeland's -variational principle in the context of T -orbitally complete metric spaces.
Corollary 3 [Ekeland's -variational principle].Let (X, d) be a metric space, T be a self mapping on X and f : X → R ∪ {+∞} be a T -orbitally lower semicontinuous function, bounded from below.Assume that X is T -orbitally complete and (O1) holds.If for u 0 ∈ O(x 0 ) for some x 0 ∈ X and , λ > 0, Taking ρ(y, z) = √ d(y, z), δ 0 = 1 and δ n = 0 for all n > 0 in Corollary 1, we get the following: Corollary 4. Let (X, d) be a metric space, T be a self mapping on X and f : X → R ∪ {+∞} be a T -orbitally lower semicontinuous function bounded from below.Assume that X is T -orbitally complete and (O1) holds.If for u 0 ∈ O(x 0 ) for some x 0 ∈ X and > 0, then there exists u ∈ O(x) such that ), for every w = u .
Remark 3. Since every metric space is T -orbitally complete metric space, therefore, in the light of Remark 1, Corollary 1 is the generalization of Theorem 1 of [37], and Corollary 3 is the generalization of Theorem 1.1 of [17].
Theorem 3. Let (X, d) be a metric space and T be a self mapping on X. Assume that f : for all x ∈ X.Then for every > 0, there exists u 0 ∈ O(x 0 ) for some x 0 ∈ X such that Proof.For n ∈ N and x ∈ X, from (19) and Lemma 1 we get https://www.mii.vu.lt/NAThis shows that {f (T n x)} is a decreasing sequence of nonnegative integers and bounded below by 0. Also, Letting limit as n → ∞ in (20) and using (φ2), we obtain It follows that there exists > 0 such that This completes the proof.
Proposition 1.Let (X, d) be a metric space and T be a self mapping on X. Assume that X is T -orbitally complete and, for φ ∈ Φ, T satisfy (19) for all x ∈ X.If {T n x} is any sequence contained in O(x) for some x ∈ X and converges to z ∈ X, then z is a fixed point of T , provided that f : Proof.For n ∈ N and x ∈ X, as in Theorem 2, we get Since f is T -orbitally semicontinuous, so from above inequality we obtain This implies that f (z) = 0, consequently, z is a fixed point of T .
Taking α(x, y) = 1 for all x, y ∈ X in Corollary 5, we get the following: Corollary 6.Let (X, d) be a metric space and T be a self mapping on X. Assume that is T -orbitally lower semicontinuous.If there exists φ ∈ Φ such that for all x ∈ X, then there exists u ∈ O(x) such that conclusions (i)-(iv) of Corollary 1 hold.
If we consider ρ(y, z) = ( /λ)d(y, z), δ 0 = 1 and δ n = 0 for all n > 0 in Corollary 6, then we get the following: Corollary 7. Let (X, d) be a metric space and T be a self mapping on X. Assume that X is T -orbitally complete and f : for all x ∈ X, then for , λ > 0, there exists a sequence {u n } and u ∈ O(x) such that

Fixed point results
In this section, we derive some fixed point results from Section 3.
Theorem 5 [ Ćirić-type fixed point theorem].Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and there exists q ∈ (0, 1) such that the following holds: Then T has a unique fixed point in X, provided that f : Proof.Put y = T x in (24), we get where which implies Now, if which leads to contradiction.Therefore, Here arises two cases: This implies that where r = q/(2 − q) ∈ (0, 1).
For uniqueness, suppose that x = u is another fixed point of T , then from (24) we obtain d(u , x) qd(u , x), a contradiction.Thus, x = u .https://www.mii.vu.lt/NATheorem 6.Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and T satisfy the following: d(T x, T y) αd(x, y) + βd(x, T x) + γd(y, T y) + δd(x, T y) + Ld(y, T x), where Proof.From (29) and by using Lemma of [34], there exists 0 < k < 1 such that Relation (30) implies that ( 19) holds true for φ(t) = kt, k ∈ (0, 1).Therefore, by using Corollary 7, there exists a sequence {u n } and , λ > 0 such that u n → u ∈ O(x) and for every w ∈ X with w = u .Now we claim that f (u ) = 0.If not, then choose and λ such that /λ ∈ (0, 1 − k).Considering w = T u in (31), we obtain Hence f (u ) = 0, and thus, u is a fixed point of T in X.
In similar fashion, we can deduce the following: Theorem 7. Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and T satisfy the following: Theorem 8. Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and T satisfy the following: Theorem 9. Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and T satisfy the following: Theorem 10.Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and there exists δ ∈ (0, 1) and L 0 such that following holds: d(T x, T y) δ(x, y) + Ld(y, T x).
Then T has a unique fixed point in X, provided that f : Theorem 11.Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and there exists q ∈ [0, 1) such that the following holds: for all x, y ∈ X, where τ (x, y) is a nonnegative real function.Then T has a fixed point in X, provided that f : In addition, if τ (x, y) (d(x, y)) −1 , then T has a unique fixed point.
Theorem 12. Let (X, d) be a metric space and T : X → X be a mapping.Assume that (X, d) is T -orbitally complete and there exists L 0 such that following holds: + Ld(y, T x).
Then T has a fixed point in X, provided that f : X → R ∪ {+∞} such that f (x) = d(x, T x) is T -orbitally lower semicontinuous.
Proof.From Corollary 5 we have that for each > 0, there exists a sequence δ j of positive real numbers and a sequence u n such that u n → u ∈ O(x) as n → ∞, and for every w ∈ X, w = u , we have ∆(w, ∞) > ∆(u , ∞) https://www.mii.vu.lt/NA or f (w) + α(u 0 , w) We suppose that w = T w for all w ∈ X, then there exists T u such that T u = u , so (32) implies Relation (33) with hypothesis (i) and (ii) gives This leads to contradiction.Thus, there exists w ∈ X such that w = T w.

Nonconvex minimax theorems and equilibrium problem
In many existing general topological minimax theorems, the convexity assumptions on the sets or on the functions are essential.Some applications of Ekeland's variational principles to minimax problems in Banach spaces are obtained by McLinden [26].Ansari et al. [38] and Lin [25] studied minimax theorems for a family of multivalued mappings in locally convex topological vector spaces.For detail in this direction, see [4].Recall that equilibrium problem is to find x ∈ X such that F (x, y) 0 for all y ∈ X, where X is a metric space and F : X × X → R. The equilibrium problem is a unified model of optimization problems, saddle point problems, Nash equilibrium problems, variational inequality problems, nonlinear complementarity problems and fixed point problems.Blum [27], Oettli and Thera [41] first gave the existence of a solution of an equilibrium problem in the setting of complete metric spaces.
In this section, we obtain minimax theorems in incomplete metric spaces without assumption of convexity and also obtain the existence of a solution of equilibrium problem in incomplete metric spaces.The obtained results also present the importance of the T -orbitally lower semicontinuous mappings by showing that a function, which is T -orbitally lower semicontinuous but may not be necessarily lower semicontinuous, can still obtain its infimum and play an important role to solve minimization problems and equilibrium problems.
Theorem 14 [Takahashi's-type nonconvex minimization theorem].Let (X, d) be a metric space, T be a self mapping on X and f : X → R∪{+∞} be a T -orbitally lower semicontinuous function bounded from below.Assume that X is T -orbitally complete, (A) hold and for any u ∈ O(x) for x ∈ X with f (u) > inf z∈X f (z), there exists y ∈ X such that χ(y, ∞) χ(u, ∞).Then there exists v ∈ O(x) for x ∈ X such that f (v) = inf z∈X f (z).
Proof.From Corollary 2 there exists u ∈ O(x) for x ∈ X such that u n → u and for all We claim that for u By assumption, we get there exists y ∈ X such that χ(y, ∞) χ(u , ∞), then from (35) we get χ(y, ∞) χ(u , ∞) < χ(y, ∞), which leads to contradiction.
Theorem 15 [Nonconvex minimax theorem].Let (X, d) be a metric space, T be a self mapping on X and F : X ×X → R∪{+∞} be a T -orbitally lower semicontinuous function and bounded from below in the first argument.Assume that X is T -orbitally complete, (A) hold and for any u where Proof.It follows from Theorem 14 that for all for v ∈ X, there exists u Taking the supremum over v on both sides gives Replacing u(v) by an arbitrary u ∈ X and taking infimum, we obtain The following result is a nonconvex equilibrium theorem in T -orbitally complete metric spaces.https://www.mii.vu.lt/NATheorem 16 [Nonconvex equilibrium theorem].Let (X, d) be a metric space, T be a self mapping on X and F : X ×X → R∪{+∞} be a T -orbitally lower semicontinuous function and bounded from below in the first argument.Assume that X is T -orbitally complete, (A) hold and for each u ∈ O(x) for x ∈ X with {b ∈ X: F (u, b) < 0} = ∅, there exists w = w(u) ∈ X with w = u such that (36) holds for all r ∈ X.Then there exists v ∈ O(x) such that F (v, y) 0 for all y ∈ X.
Proof.By using Corollary 2, for each z ∈ X, there exists u (z) ∈ O(x) for x ∈ X such that u n → u and for all a = u (z), We claim that there exists v ∈ O(x) such that F (v, y) 0 for all y ∈ X.If not, then for each u ∈ O(x), there exists y ∈ X such that F (u, y) < 0. This implies that for each u ∈ O(x), the set {b ∈ X: F (u, b) < 0} is nonempty.From our assumption there exists Combining ( 37) and (38) gives which leads to contradiction.
Hence (36) holds true.All the hypothesis of Theorem 16 are satisfied, and there exists 0 ∈ O(g) such that F (0, h) 0 for all h ∈ X. Hence in all cases, (36) holds true.All the hypothesis of Theorem 16 are satisfied, and there exists 2 ∈ O(x) such that F (2, y) 0 for all y ∈ X. Remark 6.In Example 4, (X, d) is not a complete metric space because x n = 1 + 1/n is a Cauchy sequence in X that converges to 1 as n → ∞, but 1 / ∈ X.Also, the function F is not lower semicontinuous in the first argument.Indeed, x n = 1.5 + 1/n is a sequence in X, which converges to 1.5 ∈ X, but for y ∈ {0, −1}, Therefore, equilibrium formulations of Ekeland's variational principles given in [11,29,33,41,42,44] cannot be applied for this example.

Conclusion
A variational principle is obtained in metric spaces, which are not necessarily complete, by introducing the notion of T -orbitally lower semicontinuous functions.We also obtain a nonconvex minimization theorem, a nonconvex minimax theorem and a nonconvex equilibrium theorem in such metric spaces.The existence of a solution of an equilibrium problem is also proved from obtained results.As consequences, the fixed points for many contractions in the existing literature are explored.