function on [0, ∞) such that

∞ d„

=∞

(5.5)

ρ(„ )

0

by (2.8). Let

W (u) = ’·(u)Y (u)ρ( u ).

Then

W (u) ¤ ρ( u ), u ∈ E.

By (5.5), for each u ∈ E, there is a unique solution of

t ∈ R+ , σ (0) = u

σ (t) = W (σ (t)),

(5.6)

(cf. Theorem 4.5). We have

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

(5.7)

¤ ’θ ·(σ ) G (σ ) ρ( σ )

¤ ’θ ·(σ ).

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.

(5.8)

This follows from the fact that

G(σ (t) u) ¤ a0 < 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.

5.3. Theorem 2.12 41

Let

E ± = {u ∈ E : G(u) ¤ ±}.

(5.9)

There is a T > 0 such that

σ (T )E a+δ ‚ E a’δ .

(5.10)

In fact, we can take T > 2δ/θ. To see this, 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 (5.7). 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 such t, and (5.7) yields

G(σ (T )u) ¤ G(u) ’ θ T < a ’ δ.

(5.11)

Hence, (5.10) holds. Now, by (2.6), there is a K ∈ K such that

K ‚ E a+δ .

(5.12)

˜ ˜

As we saw, σ (T ) ∈ (A). Let K = σ (T )(K ). Then K ∈ K by de¬nition. But

sup G = sup G(σ (T )u) < a ’ δ,

˜ u∈K

K

which contradicts (2.6), proving the theorem.

Next, we prove Theorem 2.8.

Proof. By (2.6) and (2.17), we see that b0 ¤ a. This implies (2.7) in view of (2.18).

The result now follows from Theorem 2.4.

Theorem 2.11 follows obviously from Theorem 2.8.

5.3 Theorem 2.12

Now we give the proof of Theorem 2.12.

Proof. If a0 < a, the conclusion follows from Theorem 2.4. Assume that a0 = a.

If there did not exist a sequence satisfying (2.9), then there would be positive numbers

δ, θ, T such that 2δ < θ T and (5.1) holds whenever u ∈ Q, where Q is given by (5.2).

Since a = a0 , we see by (2.6), (2.17), and (2.20) that b0 = a. De¬ne Q 0 , Q 1 , Q 2 , and

·(u) as before and let σ (t) be the ¬‚ow generated by the mapping (5.6). Let u be any

element in E a+δ . If there is a t1 ¤ T such that σ (t1 )u ∈ Q 1 , then

/

G(σ (t1 )u) < a ’ δ.

(5.13)

42 5. The Method Using Flows

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

G(σ (t)u) ¤ G(u) ’ θ t ¤ a + δ ’ θ t, t ∈ [0, T ].

(5.14)

Thus we have

G(σ (T )u) < a ’ δ.

(5.15)

Since b0 := inf B G = a, this shows that

σ (T )E a+δ © B = φ.

(5.16)

Moreover,

G(σ (t)u) ¤ a ’ θ t, u ∈ A, t ∈ [0, T ].

(5.17)

It therefore follows that (2.12), (2.13), and (2.21) hold for b = a +δ. Let K ∈ K satisfy

˜

(5.12). Then, by hypothesis, there is a K ∈ K satisfying (2.14) with b = a + δ. Now

˜

(2.21), (5.14), (5.16), and (5.17) imply that K © B = φ, contradicting (2.17).

To prove the last statement, we take ψ(t) = ρ(t) ≡ 1. Still assuming a0 = a, we

note that if there did not exist a sequence satisfying both (1.4) and (2.22), then there

would be positive numbers , δ, T such that δ < T and (5.1) holds whenever

u ∈ Q = {u ∈ E : d(u, B) ¤ 4T, |G(u) ’ a| ¤ 3δ}.

Let

= {u ∈ E : d(u, B) ¤ 3T, |G(u) ’ a| ¤ 2δ},

Q0

= {u ∈ E : d(u, B) ¤ 2T, |G(u) ’ a| ¤ δ}.

Q1

Since a = b0 , we see that Q 1 = φ. De¬ne Q 0 , Q 1 , Q 2 , and ·(u) as before and let

σ (t) be the ¬‚ow generated by the mapping

W (u) = ’·(u)Y (u),

with everything now with respect to the new sets Q j . In this case