(5.35)

would hold for all u in the set

Q = {u ∈ E : b0 ’ 3δ ¤ G(u) ¤ a + 3δ}.

(5.36)

By reducing δ if necessary, we can ¬nd θ < 1, T > 0 such that

R+±

a0 ’ b0 + δ < θ T, T¤ ψ(s) ds.

(5.37)

δ+±

Thus, by Lemma 4.3, if u(t) is the solution of (4.6) with ρ(t) = 1/ψ(t), γ = 1,

t0 = 0, and u 0 = R, then

u(t) ≥ δ, t ∈ [0, T ].

Let

Q 0 = {u ∈ Q : b0 ’ 2δ ¤ G(u) ¤ a + 2δ},

(5.38)

Q 1 = {u ∈ Q : b0 ’ δ ¤ G(u) ¤ a + δ},

(5.39)

and

Q 2 = E\Q 0 , ·(u) = d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )].

(5.40)

46 5. The Method Using Flows

As before, we note that · satis¬es (5.4). There is a locally Lipschitz continuous map

ˆ

Y (u) of E = {u ∈ E : G (u) = 0} into itself such that

ˆ

Y (u) ¤ 1, θ G (u) ¤ (G (u), Y (u)), u∈E

(5.41)

(cf., e.g., [120]). Let σ (t) be the ¬‚ow generated by

W (u) = ’·(u)Y (u)ρ(d(u, B )),

(5.42)

˜

where ρ(„ ) = 1/ψ(„ ). Since W (u) ¤ ρ(d(u, B )) ¤ ρ( u ) = 1/ψ( u ) and is

˜

+ in view of Theorem 4.5. Since

locally Lipschitz continuous, σ (t) exists for all t ∈ R

t

σ (t)u ’ u = W (σ („ )u)d„,

(5.43)

0

we have

t

σ (t)u ’ σ (s)u ¤ ρ(d(σ (r )u, B )) dr.

s

If v ∈ B , we have

t

h(s) = d(σ (s)u, B ) ¤ σ (s)u ’ v ¤ σ (t)u ’ v + ρ(d(σ (r )u, B )) dr,

s

which implies

t

h(s) ¤ h(t) + ρ(h(r )) dr.

(5.44)

s

We also have

(5.45) d G(σ (t)u)/dt = (G (σ ), σ ) = ’·(σ )(G (σ ), Y (σ ))ρ(d(σ, B ))

¤ ’θ ·(σ ) G (σ ) ρ(d(σ, B ))

¤ ’θ ·(σ )ψ(d(σ, B ))ρ(d(σ, B ))

= ’θ ·(σ )

in view of (5.35) and (5.41). Now suppose u ∈ E a+δ is such that there is a t1 ∈ [0, T ]

for which σ (t1 )u ∈ Q 1 . Then

/

G(σ (t1 )u) < b0 ’ δ,

since we cannot have G(σ (t1 )u) > a + δ for such u by (5.45). But this implies

G(σ (T )u) < b0 ’ δ.

(5.46)

On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then

T

G(σ (T )u) ¤ G(u) ’ θ dt ¤ a ’ θ T < b0 ’ δ

0

5.5. Theorem 2.21 47

by (5.37). Thus, (5.46) holds for u ∈ E a+δ and, in particular, for any u ∈ A. By the

de¬nition (2.6) of a, there is a K ∈ K such that

sup G < a + (δ/2).

(5.47)

K

Note that σ (t)u = v ∈ B for u ∈ A and t ∈ [0, T ]. For v ∈ B , this follows from

(5.44) and the fact that

h(t) = d(σ (t)u, B ) ≥ u(t) ≥ δ, t ∈ [0, T ],

in view of Lemma 4.8. If v ∈ B\B , we have, by (5.45),

t

G(σ (t)u) ¤ a ’ θ ·(σ („ )u)d„ < a, t > 0,

0

¯

unless ·(σ („ )u) = 0 for 0 ¤ „ ¤ t. But this would mean that ·(σ („ )u) ∈ Q 2 in view

of (5.4). But then we would have G(u) ¤ b0 ’2δ since we cannot have G(u) ≥ a +2δ.

Thus,

G(σ (t)u) < a.

This shows that

B © σ (t)A = φ, 0 ¤ t ¤ T.

(5.48)

˜

Taking b = a + (δ/2), (2.14) tells us that there is a K ∈ K such that

˜

K‚ σ (t)A ∪ σ (T )[E b ∪ K ].

(5.49)

t ∈[0,T ]

˜

But (5.46), (5.47), and (5.48) imply that K © B = φ, contradicting the fact that A links

B [mm].

We also give the proof of Theorem 2.22.

Proof. Again, we may assume that a = a0 . We interchange A and B and consider the

˜

functional G(u) = ’G(u). Then

˜

a0 = sup G = ’ inf G = ’b0 < ∞

˜

B

B

and

˜ ˜

b0 = inf G = ’ sup G = ’a0 > ’∞.

A A

Moreover,

R+β

˜

a0 ’ b0 = a0 ’ b0 <

˜ ψ(t) dt,

β