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

Proof. There is a • ∈ such that K = • ’1 (B R ) with R > supu∈A •(u) . Since A

is strongly chained to B, we have

inf •(x) ¤ sup •(x) < R.

x∈B x∈A

Hence, there is a v ∈ B such that •(v) ∈ B R . Thus, v = • ’1 •(v) ∈ • ’1 (B R ).

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 (C 1 © 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

(6.5) a := inf sup G

K ∈K( A) K

is ¬nite. 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 • ∈ U such that K = • ’1 (B R ) with R > supu∈A •(u) . Since A

is chained to B, we have

inf •(x) ¤ sup •(x) < R.

x∈B x∈A

Hence, there is a v ∈ B such that •(v) ∈ B R . Hence, v = • ’1 •(v) ∈ • ’1 (B R ).

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 (C 2 © 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 ¬nite. 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

G (u) ≥ 3δ

(6.6)

whenever

u ∈ Q = {u ∈ E : |G(u) ’ a| ¤ 3δ}.

(6.7)

Let

= {u ∈ E : |G(u) ’ a| ¤ 2δ},

Q0

= {u ∈ E : |G(u) ’ a| ¤ δ},

Q1

= E\Q 0 ,

Q2

·(u) = d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )].

It is easily checked that ·(u) is locally Lipschitz continuous on E and satis¬es

¯

·(u) = 1, u ∈ Q 1 ; ·(u) = 0, u ∈ Q2; 0 < ·(u) < 1, otherwise.

Consider the differential equation

σ (t) = W (σ (t)), t ∈ R, σ (0) = u,

(6.8)

where

W (u) = ’·(u)G (u)/ G (u) .

(6.9)

Since G ∈ C 2 (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

6.3. The remaining proofs 55

for all t ∈ R. Let us denote the solution of (6.8) by σ (t) u. The mapping σ (t) is in

CU (E — R, E) and is called the ¬‚ow generated by (6.9). Note that

1

d G(σ (t)u)/dt = (G (σ (t)u), σ (t)u)

(6.10)

= ’·(σ (t)u) G (σ (t)u)

¤ ’2δ·(σ (t)u).

Thus,

G(σ (t)u) ¤ G(u), t ≥ 0,

G(σ (t)u) ¤ a0 , u ∈ A, t ≥ 0,

and

σ (t)u = u, u ∈ A, t ≥ 0.

Again, this follows from the fact that

G(σ (t) u) ¤ a0 ¤ b0 < a ’ 3δ, u ∈ A, t ≥ 0.

Hence, ·(σ (t) u) = 0 for u ∈ A, t ≥ 0. This means that

σ (t) u = 0, u ∈ A, t ≥ 0,

and

σ (t) u = σ (0) u = u, u ∈ A, t ≥ 0.

Let

E ± = {u ∈ E : G(u) ¤ ±}.

(6.11)

We note that there is a T > 0 such that

σ (T )E a+δ ‚ E a’δ .

(6.12)

(In fact, we can take T = 1.) Let u be any element in E a+δ . If there is a t1 ∈ [0, T ]

such that σ (t1 )u ∈ Q 1 , then

/

G(σ (T )u) ¤ G(σ (t1 )u) < a ’ δ

by (6.10). Hence, σ (T )u ∈ E a’δ . On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ],

then ·(σ (t)u) = 1 for all t, and (6.10) yields

G(σ (T )u) ¤ G(u) ’ 2δT ¤ a ’ δ.

(6.13)

Hence, (6.12) holds. Now, by (6.3), there is a K ∈ KU (A) such that

K ‚ E a+δ .

(6.14)

˜ ˜

Note that σ (T ) ∈ U (A). Let K = σ (T )(K ). Then K ∈ KU (A) by Lemma 6.12.

But

sup G = sup G(σ (T )u) < a ’ δ,

˜ u∈K

K

which contradicts (6.3), proving the theorem.

56 6. Finding Linking Sets

Now we give the proof of Theorem 6.7.

Proof. Assume that • ∈ U does not satisfy (6.1), and let G(u) = •(u) 2 . Then, by

the de¬nition of the class U , G ∈ CU (E, R), sup G(A) < inf G(B), and G has no

1

critical level a ≥ infx∈B G(x) > 0. To show the latter, assume that there is a sequence

u k satisfying

G(u k ) ’ a, G (u k ) ’ 0.

(6.15)

Then we have, for any bounded sequence v k ∈ E,

(• (u k )v k , •(u k )) ’ 0.

(6.16)

Let v k := (• (u k ))’1 (•(u k )). Then (• (u k )v k , •(u k )) = G(u k ) ’ a. However, the

sequence v k is bounded: G(u k ) ’ a implies that •(u k ) is bounded, which implies