Minimax Results with Respect to Different Altitudes in the Situation of Linking

Definition 1. Let f : X → ℝ be a continuous function defined on a metric space X. An element u ∈ X is called a critical point of f if |df | (u) = 0, where by |df | (u) we denote the supremum of the real numbers σ ≥ 0 such that there exist δ > 0 and a continuous map ℋ : B_δ(u)×[0, δ] → X satisfying

Remark 2. (a) Note that |df | (u)∈[0, +∞] (take σ = 0, any δ > 0, and ℋ(v, t) ≡ v in (1)) and the map X → [0, +∞], u↦|df | (u) is lower semicontinuous. The quantity |df | (u) is usually called the weak slope of f at u.

(b) If (X, ∥·∥) is a real Banach space, d is the distance induced by the norm ∥·∥, and f ∈ C¹(X), then |df | (u) coincides with the norm of the differential f^′(u). Thus, Definition 1 reduces to the usual notion of critical point in this context.

Definition 3. A compact pair (D, S) in X and a closed subset A of X are linking if the following property holds:

where Γ : = {γ ∈ C(D, X) : γ_|S = id_S}.

Proposition 4. Assume that a > −∞ (see (4)) and that there exist γ ∈ Γ and a number ρ > 0 such that

Then K_a∩A ≠ ∅.

Theorem 5. Assume that c = a > −∞. Then

In particular, there exists a sequence (u_n) ⊂ X such that

Theorem 6. Assume that c > a > −∞. Then

• (a)

  for all δ > 0 and ρ ≥ 2d(S, A), one has

  $$ ()

• (b)

  for all δ > 0 and ρ ≥ 2d(S∩[f > a], A), one has

  $$ ()

Remark 7. The estimate in part (b) of Theorem 6 is better than the one in part (a) when ρ ∈ [2d(S∩[f > a], A), +∞). Nevertheless, Theorem 6(a) provides an estimate for all ρ ∈ [2d(S, A), +∞).

Theorem 8. (a) Let (D, S) be a compact pair in X. Assume that c > b. Then

In particular, there exists a sequence (u_n) ⊂ X such that

(b) Let (D, S) be a compact pair in X and let A be a closed subset of X satisfying (3). Assume that c = b > a > −∞. Then

Corollary 9. (a) Assume that −∞ < b ≤ a. Then (10) holds. If c = a, then (6) holds.

(b) Assume that b > a > −∞. Then (12) holds. If c > b, then (10) holds.

Definition 10. We say that a continuous function f : X → ℝ satisfies the Palais-Smale condition at level ℓ (condition (PS) _ℓ for short), with ℓ ∈ ℝ, if every sequence (u_n) ⊂ X such that f(u_n) → ℓ and |df | (u_n) → 0 as n → ∞ has a convergent subsequence in X.

Corollary 11. If conditions (3) and (PS) _c hold, then one has the following.

• (a)

  In the case c = a > −∞, then K_c∩A ≠ ∅.

• (b)

  In the case c > a, if one assumes that c > b (for instance, if b ≤ a), then K_c ≠ ∅.

Remark 12. (a) Corollary 11 actually holds under a weaker condition than (PS) _c, namely, $$: every sequence (u_n) ⊂ X such that f(u_n) ↓ c and |df | (u_n) → 0 as n → ∞ has a convergent subsequence in X.

(b) Corollary 11(a) holds under a weaker condition than $$, namely, $$: every sequence (u_n) ⊂ X such that f(u_n) ↓ c, d(u_n, A) → 0, and |df | (u_n) → 0 as n → ∞ has a convergent subsequence in X. The conclusion in Corollary 11(a) is also more precise since it establishes the location of the critical point in the set A.

Corollary 13. Let f : X → ℝ be a continuous function on the metric space X, and let (A_n), (D_n), and (S_n) be sequences of subsets of X such that for all n ≥ 1, A_n is closed, (D_n, S_n) is a compact pair in X, (D_n, S_n) and A_n satisfy (3) with Γ replaced by $$. Assume that

Then, denoting $$ and $$, there exists a sequence (u_n) ⊂ X satisfying In particular, if f satisfies condition (PS) _ℓ for all $$, then there exists u ∈ X such that $$ and |df | (u) = 0.

Remark 14 (assume that b ≤ a). Considering particular situations of linking as in (3) in a Banach space (X, ∥·∥), we obtain through Corollary 11 generalizations of classical minimax results in smooth critical point theory.

• (a)

  If D = {te : t ∈ [0,1]}, S = {0, e}, and A = {u ∈ X : ∥u∥ = ρ} (with ρ > 0 and e ∈ X such that ∥e∥ > ρ), then we get the mountain pass theorem (see [1]).

• (b)

  If X = X₁ ⊕ X₂, with dim X₁ < +∞, D = {u ∈ X₁ : ∥u∥ ≤ 1}, S = {u ∈ X₁ : ∥u∥ = 1}, and A = X₂, then we get the saddle point theorem (see [2]).

• (c)

  If X = X₁ ⊕ X₂, with dim X₁ < +∞, D = {u ∈ X₁ : ∥u∥ ≤ r}  +  {te : t ∈ [0, r]}, S = ∂D, and A = {u ∈ X₂ : ∥u∥ = ρ} (with r > ρ > 0 and e ∈ X₂ such that ∥e∥ = 1), we get the generalized mountain pass theorem (see [2]).

It is worth noting that generalizations of the minimax results cited in (a)–(c), relative to critical point theories for certain classes of nondifferentiable functionals, can already be found in the literature and have numerous applications to partial differential equations, differential inclusions, variational inequalities, hemivariational inequalities, and variational-hemivariational inequalities. In this respect, motivated by the study of existence of solutions of the so-called variational-hemivariational inequalities, introduced by Motreanu-Panagiotopoulos [6], we mention the critical point theory developed by these authors in [6] for functionals f = φ + ψ : X → ℝ ∪ {+∞}, with φ : X → ℝ locally Lipschitz and ψ : X → ℝ ∪ {+∞} lower semicontinuous, convex, and not identically +∞ (see also Chang [7] for the case when ψ = 0, and see Szulkin [8] for the case when φ : X → ℝ is of class C¹ and ψ : X → ℝ ∪ {+∞} is lower semicontinuous, convex, and not identically +∞).

Theorem 15. Let X be a metric space endowed with the metric d, and let f : X → ℝ be a continuous function. For all ρ > 0, there exists a deformation η : X × [0,1] → X (i.e., η is continuous and satisfies η(u, 0) = u for all u ∈ X) such that for all u ∈ X and t ∈ [0,1], one has

• (a)

  t ∈ (0,1]⇒d(η(u, t), u) < ρt;

• (b)

  |df | (u) = 0⇒η(u, t) = u;

• (c)

  |df | (u) > 0, t ∈ (0,1]⇒f(η(u, t)) < f(u).

Proof. First, assume that |df | ≢+∞. We consider the continuous function σ : X → [0, +∞) given by

We have σ(u) = 0 for all u ∈ X with |df | (u) = 0. Using that u↦|df | (u) is lower semicontinuous (see Remark 2(a)), we see that Applying [3, Theorem (2.8)], we find a deformation $$ and a continuous map $$ with the properties (for all u ∈ X and t ∈ [0,1]): (i) $$; (ii) $$; (iii) $$. Then the deformation η : X × [0,1] → X defined by has properties (a)–(c) of the statement. If |df | ≡+∞, we make the same reasoning taking an arbitrary positive constant in place of σ.

Proof of Proposition 4. Let $$, and let η : X × [0,1] → X be a deformation satisfying properties (a)–(c) of Theorem 15 (with $$ in place of ρ). Note that $$. Let ϑ : X → [0,1] be a continuous function such that

Defining $$ by with γ as in the statement of Proposition 4, we have $$. Fix u₀ ∈ D with $$ (see (3)). Then, using property (a) in Theorem 15, we infer that Hence, using properties (b), (c) of Theorem 15, relation (5), and the assumption that a ∈ ℝ, we obtain that and |df | (γ(u₀)) = 0. This combined with property (b) of Theorem 15 yields $$, which completes the proof.

Theorem 16. Let X be a metric space endowed with the metric d, let f : X → ℝ be a continuous function, let C be a subset of X, and let ℓ ∈ ℝ, σ > 0, and ρ > 0 be real numbers. Assume that the metric space f⁻¹([ℓ, s]) (endowed with the induced metric) is complete for all s ∈ (ℓ, ℓ + σρ) and that

Then there exist a continuous function τ : C∩[f < ℓ + σρ]→[0, ρ) and a continuous map η : (C∩[f < ℓ + σρ])  ×  [0,1]  →  [f < ℓ + σρ] such that for all u ∈ C∩[f < ℓ + σρ] and t ∈ [0,1], one has

• (a)

  τ(u) ≤ (1/σ)max {f(u) − ℓ, 0};

• (b)

  d(η(u, t), u) ≤ τ(u)t;

• (c)

  f(η(u, t)) ≤ f(u) − στ(u)t;

• (d)

  f(u) ≥ ℓ⇒f(η(u, 1)) = ℓ.

Proof of Theorem 5. For a fixed δ > 0, it suffices to prove that

To this end, let σ be a number such that Let ρ > 0 such that In particular, these choices combined with (24) guarantee that Hereafter, we denote $$. Thus, applying Theorem 16 with the numbers ρ and σ, $$, and ℓ : = c, we find a deformation $$ such that

Using that 3ρ ≤ d(S, A), we see that S ⊂ (X∖B_3ρ(A))∩[f < c + σρ]. Let ϑ : [f < c + σρ]→[0,1] be a continuous function such that ϑ(u) = 0 if u ∈ X∖B_3ρ(A) and ϑ(u) = 1 if $$. Define the deformation η : [f < c + σρ]  ×  [0,1]  →  [f < c + σρ] by

To prove the continuity of η, we note that the restriction of η to the sets ([f < c + σρ]∖B_3ρ(A))×[0,1] and $$, which are closed in [f < c + σρ], is continuous. Indeed, the restriction of η to ([f < c + σρ]∖B_3ρ(A))×[0,1] coincides with the projection (u, t) ↦ u, whereas the restriction of η to $$ coincides with the map $$ (since ϑ ≡ 0 on $$) which is continuous (by the continuity of ϑ).

Let γ₀ ∈ Γ with max _D(f∘γ₀) < c + σρ. Then the map γ : D → X defined by

is well defined and γ ∈ Γ. Finally, let u ∈ D such that γ(u) ∈ B_ρ(A). Since (by the first part of (27)), we have that γ₀(u) ∈ B_2ρ(A), so ϑ(γ₀(u)) = 1; thus (by the second part of (27)). Consequently, we can apply Proposition 4. Therefore, K_a∩A ≠ ∅. This concludes the proof.

Proof of Theorem 6. (a) By (3), we see that d(S, A) > 0. For a fixed δ > 0 and $$, it suffices to prove that

To this end, let σ be a number such that Let ρ, ε be positive numbers such that Setting θ : = a + σρ − c ∈ (0, δ), since $$, we have This allows us to apply Theorem 16 with the numbers ρ and σ, $$, and ℓ : = a. Hence we find a deformation $$  ×  [0,1]  →  [f < c + θ] such that

The fact that θ > 0 in conjunction with the inequality ρ + 2ε < d(S, A) yields S ⊂ (X∖B_ρ+2ε(A))∩[f < c + θ]. Let ϑ : [f < c + θ]→[0,1] be a continuous function such that ϑ(u) = 0 if u ∈ X∖B_ρ+2ε(A) and ϑ(u) = 1 if $$. Define the deformation η : [f < c + θ]  ×  [0,1]  →[f < c + θ] by

The continuity of η can be shown as in the proof of Theorem 5. Let γ₀ ∈ Γ with max _D(f ∘ γ₀) < c + θ. Then the map γ : D → X given by is well defined, and we have γ ∈ Γ. Let u ∈ D such that γ(u) ∈ B_ε(A). Since it follows that γ₀(u) ∈ B_ρ+ε(A), which yields Consequently, Proposition 4 (with ρ replaced by ε) can be applied, and we conclude that K_a∩A ≠ ∅.

(b) We see that d(S∩[f > a], A) ≥ d(S, A) > 0. For a fixed δ > 0 and $$, it suffices to show that

To this end, let σ be a number such that Let ρ, ε be positive numbers such that Setting θ : = a + σρ − c ∈ (0, δ), we have Thus, we can apply Theorem 16 with the numbers ρ and σ, $$, and ℓ : = a. So we find a deformation $$ such that

Let ϑ : [f < c + θ]→[0,1] be a continuous function such that ϑ(u) = 0 if u ∈ X∖B_ρ+2ε(A) and ϑ(u) = 1 if $$. Define the deformation η : [f < c + θ]  ×  [0,1]  →  [f < c + θ] by

The continuity of η can be shown as in the proof of Theorem 5.

Let γ₀ ∈ Γ with max _D(f∘γ₀) < c + θ. Define γ : D → X by

We show that γ ∈ Γ. To prove that γ is continuous, it suffices to note that the restrictions of γ to the closed sets {u ∈ D : f(γ₀(u)) ≥ a} and {u ∈ D : f(γ₀(u)) ≤ a} are continuous. This is immediate for the latter closed set. Concerning the former, this follows from the fact that the equality γ(u) = η(γ₀(u), 1) holds whenever f(γ₀(u)) = a (which is a consequence of the definition of η and relation (45)). Whence γ is continuous. Using that $$, we infer that γ ∈ Γ.

Let an arbitrary point u ∈ D with d(γ(u), A) < ε. If f(γ₀(u)) > a, then

(by the first part of (46)), so γ₀(u) ∈ B_ρ+ε(A), and thus $$ (by the second part of (46)). If f(γ₀(u)) ≤ a, then it is clear that f(γ(u)) ≤ a. Consequently, in both cases, we have f(γ(u)) ≤ a. This allows us to apply Proposition 4 (with ρ replaced by ε) and thus to obtain that K_a∩A ≠ ∅, which completes the proof.

Proof of Theorem 8. (a) The hypotheses allow us to apply Theorem 5 with A : = {u ∈ X : f(u) ≥ c}. Indeed, since c > b, we have that S∩A = ∅. Moreover, since D is compact, for each γ ∈ Γ, there exists u_γ ∈ D with max _D(f∘γ) = f(γ(u_γ)) ≥ c. It follows that condition (3) is satisfied and

so inf _Af = c > −∞. The conclusion follows now from Theorem 5.

(b) This follows from Theorem 6(b).

Proof of Corollary 9. (a) If c > a, then we have c > b. Thus, we can apply Theorem 8(a). In the case c = a, we apply Theorem 5.

(b) If c > b, then we obtain from Theorem 8(a) that (10) holds (so that a fortiori (12) holds since b − a ≥ 0). If c = b, then we apply Theorem 8(b).

Proof of Corollary 11. Parts (a) and (b) follow from Theorems 5 and 8(a), respectively, taking into account the lower semicontinuity of the map u↦|df | (u) (see Remark 2(a)).

Proof of Corollary 13. Fix an arbitrary n ≥ 1. If b_n ≤ a_n, then applying Corollary 9(a), we have

If b_n > a_n, then using Corollary 9(b), we have Let Then we have σ_n → 0 as n → ∞ and This completes the proof.