W (u) ¤ 1.

(5.18)

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

/

G(σ (t1 )u) < a ’ δ

(5.19)

or

d(σ (t1 )u, B) > 2T.

(5.20)

Since

σ (t)u ’ σ (t )u ¤ |t ’ t |

5.3. Theorem 2.12 43

by (5.18), (5.20) implies

d(σ (t)u, B) > T, 0 ¤ t ¤ T.

(5.21)

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

G(σ (T )u) ¤ G(u) ’ 2 T ¤ a + δ ’ 2δ = a ’ δ.

(5.22)

Thus, either we have

G(σ (T )u) < a ’ δ

(5.23)

or (5.21) holds. Since b0 = a, this shows that

σ (T )E a+δ © B = φ.

(5.24)

We also note that

σ (t)A © B = φ, 0 ¤ t ¤ T.

(5.25)

To see this, we observe, by (5.7), that

t

G(σ (t)u) ¤ a0 ’ 2 ·(σ („ )u)d„, u ∈ A.

0

If σ (t)u ∈ B, we must have G(σ (t)u) ≥ b0 ≥ a0 . The only way this can happen is if

·(σ („ )u) ≡ 0, 0 ¤ „ ¤ t.

¯

But this implies σ („ )u ∈ Q 2 for such „ , and this in turn implies either

G(σ („ )u) < a ’ δ, 0 ¤ „ ¤ t,

or

d(σ („ )u, B) > 2T, 0 ¤ „ ¤ t.

In either case we cannot have σ (t)u ∈ B. Thus, (5.25) holds. Now, by (2.6), there is a

K ∈ K such that

K ‚ E a+δ ,

(5.26)

˜ ˜

and there is a K ∈ K such that (2.14) holds. By (5.25), K © B = φ, contradicting the

fact that A links B [mm]. This completes the proof of the theorem.

Next, we give the proof of Theorem 2.6.

Proof. De¬ne

B= v∈ K \A : G(v) ≥ a0 .

K ∈K

By (2.23), A links B relative to K, and (2.20) holds. Apply Theorem 2.12.

44 5. The Method Using Flows

5.4 Theorem 2.14

Now we give the proof of Theorem 2.14.

Proof. We assume that a = a0 and follow the proof of Theorem 2.4. In this case we

cannot take 3δ < a ’ a0 and we cannot conclude that (5.8) holds. Because of this, it

does not follow that σ (T )K ∈ K for all K ∈ K. Let K ∈ K be such that

a ¤ sup G < a + δ,

(5.27)

K \A

and let K ρ be given by (2.25). Note that

a0 ¤ sup G, K ∈ K,

(5.28)

K \A

by (2.20). Since G(u) is Lipschitz continuous on K ρ and a = a0 , there is a constant C

such that

G(u) ¤ a + Cd(u), u ∈ Kρ ,

(5.29)

where d(u) = d(u, A). Pick T > 2δ/θ and T > ρC/θ. As in the proof of Theorem 2.4,

we have

σ (T )K ‚ E a’δ .

(5.30)

Let ζ (t) be a nondecreasing, continuous function on R+ satisfying

t = 0,

0,

ζ(t) =

(5.31)

t ≥ ρ,

T,

and

T t/ρ < ζ(t) < T, 0 < t < ρ.

(5.32)

Note that the ¬‚ow σ (t) = σ (ζ(t)) satis¬es (2.12) and (2.13). Consider the map

˜

Su = σ (ζ(d(u)))u, u ∈ K.

(5.33)

Consequently, by hypothesis, S(K ) ∈ K. Since d(u) = 0 for u ∈ A, and

σ (0)u = u, we have

Su = u, u ∈ A.

If u ∈ K \K ρ , then Su = σ (T )u ∈ E a’δ . If u ∈ K ρ \A and Su ∈ E a’δ , then

·(σ (t)u) = 1, 0 ¤ t ¤ ζ (d(u)).

Consequently,

G(Su) ¤ G(u) ’ θ ζ (d(u))

(5.34)

¤ a + Cd(u) ’ θ T d(u)/ρ

< a.

5.5. Theorem 2.21 45

Thus,

G(Su) < a, u ∈ K \A,

contradicting (5.28). This completes the proof of the theorem.

Now, we prove Theorem 2.13.

Proof. Let B be given by (2.26). Then A links B relative to the system K. Now apply

Theorem 2.14.

5.5 Theorem 2.21

We can now prove Theorem 2.21.

Proof. We may assume that a = a0 . Otherwise, by Theorem 2.4, a sequence (2.9)

˜ ˜

exists with ψ replaced by ψ(t) = ψ(t + ±). Since ψ satis¬es the hypotheses of Theo-

rem 2.4 and

d(u, B ) ¤ u + ±,

for each δ > 0, we can ¬nd a u ∈ E such that

˜

a ’ δ ¤ G(u) ¤ a + δ, G (u) < ψ( u ) ¤ ψ(d(u, B )),

which certainly implies (2.33).

If the conclusion of the theorem were not true, there would be a δ > 0 such that