Schechter Minimax Systems and Critical Point Theory (Springer, 2009)

52 6. Finding Linking Sets

We shall show

Theorem 6.5. If a bounded set A is strongly chained to a set B, then A links B.

Theorem 6.6. Let G be a (C2∩CU )-functional on E, and let A, B be nonempty subsets of E such that A is bounded and chained to B with

Theorem 6.7. If E is a Hilbert space and A links B, then it is chained to B.

6.2 The strong case

u A

Then ϕ˜ (Lemma 6.9). Now,

ϕ˜−1 = σ ◦ ϕ−1.

Corollary 6.13. K( A) is a minimax system for A.

Lemma 6.14. If A is strongly chained to B and K K( A), then K ∩ B = φ.

Proof. There is a ϕ such that K = ϕ−1(BR ) with R > supu A ϕ(u) . Since A is strongly chained to B, we have

Hence, there is a v B such that ϕ(v) BR. Thus, v = ϕ−1ϕ(v) ϕ−1(BR ).

Corollary 6.15. If A is strongly chained to B, then A links B relative to K( A).

Corollary 6.16. If A is strongly chained to B, then A links B strongly. Proof. Theorem 2.11.

We can now give the proof of Theorem 6.5.

Proof. Let G be a (C1 ∩ CU )-functional on E, and let A, B be nonempty subsets of E such that A is strongly chained to B and (6.2) holds. Then

is finite. By Lemma 6.14, K ∩ B = φ for all K K( A). Hence, a ≥ b0. Since a0 < a, then the theorem follows from Corollary 6.16.

54 6. Finding Linking Sets

6.3 The remaining proofs

Lemma 6.17. If A is chained to B and K KU ( A), then K ∩ B = φ.

Proof. There is a ϕ U such that K = ϕ−1(BR ) with R > supu A ϕ(u) . Since A is chained to B, we have

Hence, there is a v B such that ϕ(v) BR . Hence, v = ϕ−1ϕ(v) ϕ−1(BR ).

Lemma 6.18. If K KU ( A) and σ U ( A), then σ (K ) KU ( A).

We can now give the proof of Theorem 6.6.

Proof. Let G be a (C2 ∩ CU )-functional on E, and let A, B be nonempty subsets of E such that A is chained to B and (6.2) holds. Then a given by (6.3) is finite. By Lemma 6.17, K ∩ B = φ for all K KU ( A). Hence, a ≥ b0. If (6.4) were false, there would exist a positive constant δ such that 3δ < a − b0 and

η(u) = d(u, Q2)/[d(u, Q1) + d(u, Q2)].

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies

Since G C2(E, R), the mapping W is locally Lipschitz continuous on the whole of E and is bounded in norm by 1. Hence, by Corollary 4.6, (6.8) has a unique solution

Now we give the proof of Theorem 6.7.

sequence vk is bounded: G(uk ) → a implies that ϕ(uk ) is bounded, which implies that uk , and consequently that ϕ (uk ))−1 and vk are also bounded. Hence, G (uk ) → 0 implies G(uk ) → 0, showing that a = 0. Thus, A does not link B.

6.4 Notes and remarks

The results of this chapter are from [138]. For related material cf. [137] and [156].

Chapter 7

Sandwich Pairs

7.1 Introduction

In this chapter we discuss the situation in which one cannot find linking sets that separate a functional G, i.e., satisfy

Are there weaker conditions that will imply the existence of a PS sequence

Our answer is yes, and we find pairs of subsets such that the absence of (7.1) produces a PS sequence. We have

Definition 7.1. We shall say that a pair of subsets A, B of a Banach space E forms a sandwich if, for any G C1(E, R), the inequality

Unlike linking, the order of a sandwich pair is immaterial; i.e., if the pair A, B forms a sandwich, so does B, A. Moreover, we allow sets forming a sandwich pair to intersect. One sandwich pair has been studied in Chapter 3. In fact, Theorem 3.17 tells us that if M, N are closed subspaces of a Hilbert space E, and M = N , then M, N form a sandwich pair if one of them is finite-dimensional.

Until recently, only complementing subspaces have been considered. The purpose of the present chapter is to show that other sets can qualify as well. Infinite-dimensional sandwich pairs will be considered in Chapter 15.

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_7, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

7.2 Criteria

In this section we present sufficient conditions for sets to qualify as sandwich pairs. We have

Proposition 7.2. If A, B is a sandwich pair and J is a diffeomorphism on the entire space having a derivative J satisfying

then JA, JB is a sandwich pair.

Proof. Suppose G C1 satisfies

Since A, B form a sandwich pair, there is a sequence {hk } E such that

In view of (7.5), this implies G (uk ) → 0. Thus, J A, J B is a sandwich pair.

Proposition 7.3. Let N be a closed subspace of a Hilbert space E with complement

Assume that one of the subspaces M, N is finite-dimensional. Then A = N = N {v0}, B = F−1(δv0) form a sandwich pair.

Then J is a diffeomorphism on E with its inverse having a derivative satisfying (7.5). Moreover, J A = N and J B = M +δv0. Hence, J A, J B form a sandwich pair as long as one of them is finite-dimensional (Theorem 3.17). We now apply Proposition 7.2.

Theorem 7.4. Let N be a finite dimensional subspace of a Banach space E. Let F be a Lipschitz continuous map of E onto N such that F = I on N and

Let p be any point of N. Then A = N, B = F−1( p) form a sandwich pair.

Proof. Let G be a C1-functional on E satisfying (7.3), where A, B are the subsets of E specified in the theorem. If the theorem is not true, then there is a δ > 0 such that

and

(G (u), Y (u)) ≥ 2δ

whenever u satisfies (7.13) (for the construction of such a map, cf., e.g., [112]). Let

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies

60 7. Sandwich Pairs

The mapping W is locally Lipschitz continuous on the whole of E and is bounded in norm by 1. Hence, by Theorem 4.5, (7.13) has a unique solution for all t R. Let us denote the solution of (7.13) by σ (t)u. The mapping σ (t) is in C(E × R, E) and is called the flow generated by W (u). Note that

In fact, we can take T > (a0 − b0 + δ)/2δ. Let u be any element in Ea0+δ . If there is a t1 [0, T ] such that σ (t1)u / Q1, then

G(σ (T )u) ≤ G(σ (t1)u) < b0 − δ

by (7.15). Hence, σ (T )u Eb0−δ . On the other hand, if σ (t)u Q1 for all t [0, T ], then η(σ (t)u) = 1 for all t, and (7.15) yields

G(σ (T )u) ≤ G(u) − 2δT ≤ a0 − 2δT ≤ b0 − δ.

Hence, (7.16) holds. Let be a bounded open subset of N containing the point p such that

(7.17) ρ(∂ , p) > K T + δ,

where ρ is the distance in E and K is the constant in (7.11). If v ∂ , then

Let

H (t) = Fσ (t).

Then H (t) is a continuous map of into N for 0 ≤ t ≤ T. Moreover, H (t)v = p for v ∂ by (7.18). Hence, the Brouwer degree d(H (t), , p) is defined. Consequently,

d(H (T ), , p) = d(H (0), , p) = d(I, , p) = 1.

This means that there is a v such that

Fσ (T )v = p.

But then

σ (T )v F−1( p) = B.

This is not consistent with (7.16). Hence, A, B form a sandwich pair.

7.3 Notes and remarks

The material of this chapter comes from [135].