≥± ψ(M + t) dt

0

T +M

=± ψ(r )dr

M

> 2µ.

Thus, (5.68) holds as well in this case by (5.60). Consequently, (5.68) holds for all

u ∈ ‚ω0 . Let ωT be the set of points σ (T, u) where u ∈ ω0 . Then ωT is a bounded,

open set in E with ‚ωT consisting of those points of the form σ (T )u, u ∈ ‚ω0 by

(5.57) and the continuous dependence of σ (T )u on u. Since 0 is in Q 2 and · ≡ 0

there, we see that σ (T )0 = 0 by the uniqueness of solutions of (5.56). Thus, 0 =

σ (T )0 ∈ ωT . Thus, ‚ωT ∈ K, and

G(u) > b + µ, u ∈ ‚ωT ,

(5.69)

by (5.68). But (5.69) contradicts (2.27). This completes the proof.

In proving Theorem 2.20, we merely replace G by ’G and follow the proof of

Theorem 2.18.

Chapter 6

Finding Linking Sets

6.1 Introduction

As we saw in Chapter 3, there are several suf¬cient conditions that imply that a set A

links a set B in the sense of De¬nition 2.9. Our goal is to ¬nd all subsets that link

according to this de¬nition. At the present, we are very close to achieving this goal.

In the present chapter we de¬ne two relationships that are close to each other. The

stronger one is suf¬cient for linking, while the weaker one is necessary. We use the

following maps.

De¬nition 6.1. We shall say that a map • : E ’ E is of class if it is a homeo-

morphism onto E and both •, • ’1 are bounded on bounded sets. If, furthermore,

•, • ’1 ∈ C 1 (E; E), we shall say that • ∈ U .

De¬nition 6.2. We shall say that a bounded set A is chained to a set B if A © B = φ

and

inf •(x) ¤ sup •(x) , •∈ U.

(6.1)

x∈B x∈A

De¬nition 6.3. We shall say that a bounded set A is strongly chained to a set B if

A © B = φ and (6.1) holds for every • ∈ .

De¬nition 6.4. For A ‚ E, we de¬ne

KU (A) = • ’1 (B R ) : • ∈ U, R > sup •(u) ,

u∈A

where

B R = {u ∈ E : u < R}.

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_6,

© Birkh¤user Boston, a part of Springer Science+Business Media, LLC 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 (C 2 ©CU )-functional on E, and let A, B be nonempty subsets

of E such that A is bounded and chained to B with

a0 := sup G < b0 := inf G.

(6.2)

B

A

Let

(6.3) a := inf sup G.

K ∈KU ( A) K

Then there is a sequence {u k } ‚ E such that

G(u k ) ’ a, G (u k ) ’ 0.

(6.4)

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

6.2 The strong case

De¬nition 6.8. For A ‚ E, we de¬ne

(A) = {• ∈ : •(u) = u, u ∈ A}

and

U (A) = {• ∈ : •(u) = u, u ∈ A}.

U

The following are easily proved.

Lemma 6.9.

•1 , •2 ∈ ’ •1 —¦ •2 ∈ .

•1 , •2 ∈ ’ •1 —¦ •2 ∈ U.

U

•1 , •2 ∈ (A) ’ •1 —¦ •2 ∈ (A).

•1 , •2 ∈ U (A) ’ •1 —¦ •2 ∈ U (A).

Lemma 6.10.

iff • ’1 ∈

•∈ .

iff • ’1 ∈

•∈ U.

U

(A) iff • ’1 ∈

•∈ (A).

iff • ’1 ∈

•∈ U (A) U (A).

De¬nition 6.11. For A ‚ E, we de¬ne

K(A) = • ’1 (B R ) : • ∈ , R > sup •(u) .

u∈A

6.2. The strong case 53

Lemma 6.12. If K ∈ K(A) and σ ∈ (A), then σ (K ) ∈ K(A).

such that K = • ’1 (B R ) with R > supu∈A •(u) . Let

Proof. There is a • ∈

• = • —¦ σ ’1 .

˜

Then • ∈

˜ (Lemma 6.9). Now,

• ’1 = σ —¦ • ’1 .

˜

Hence,

σ (K ) = σ [• ’1 (B R )] = • ’1 (B R ).

˜

Moreover,

•(u) = •(u),

˜ u ∈ A.

Hence,

sup •(x) = sup •(x) < R.

˜

x∈A x∈A

Therefore, σ (K ) ∈ K(A).

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