R ¤ d = d(A , B).

48 5. The Method Using Flows

Since

˜

A = {u ∈ A : G(u) < a0 },

˜

we can apply Theorem 2.21 to conclude that for each δ > 0, there is a u ∈ E such that

˜ ˜ ˜

b0 ’ δ ¤ G(u) ¤ a0 + δ,

˜ G (u) < ψ(d(u, A )).

(5.50)

This implies (2.42).

Theorems 2.24 and 2.19 follow from the same arguments used in the proof of Theo-

rem 2.4 (in the case of Theorem 2.24 we substitute ’G for G). Now we give the proof

of Theorem 2.18.

Proof. If the conclusion of the theorem were not true, then there would exist µ > 0 and

ψ(r ) ∈ such that

G (u) ≥ ψ( u )

(5.51)

whenever

|G(u) ’ b| ¤ 3µ.

(5.52)

Let

Q = {u ∈ E : |G(u) ’ b| ¤ 2µ},

Q 1 = {u ∈ E : |G(u) ’ b| ¤ µ},

Q 2 = B\Q,

and

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

Note that ·(u) is locally Lipschitz continuous and satis¬es

§

⎪·(u) = 1, u ∈ Q1,

⎪

⎨

¯

·(u) = 0, u ∈ Q2,

(5.53)

⎪

⎪

©·(u) ∈ (0, 1), otherwise.

Let Y (u) be a locally Lipschitz continuous pseudo-gradient for G satisfying

G (u)Y (u) ≥ ± G (u) , Y (u) ¤ 1,

(5.54)

for some ± > 0. Let W (u) = ·(u)Y (u). Then W (u) is a locally Lipschitz continuous

mapping of E into itself such that

G (u)W (u) ≥ 0, W (u) ¤ 1.

(5.55)

Let σ (t)u be the solution of the initial-value problem

dσ (t)u/dt = W (σ (t)u), σ (0)u = u.

(5.56)

5.5. Theorem 2.21 49

The solution of (5.56) exists for every u ∈ E and t ≥ 0 by Theorem 4.5. By (5.55),

σ (t)u ’ u ¤ t,

(5.57)

and by (5.54) and (5.56),

d G(σ (t)u)/dt = G (σ (t)u)W (σ (t)u) ≥ ±·(σ (t)u) G (σ (t)u) .

(5.58)

This implies that

G(σ (t1 )u) ¤ G(σ (t2 )u), 0 ¤ t1 < t2 .

(5.59)

By hypothesis,

G(u) ≥ b, u ∈ ‚ω0 .

(5.60)

Let

M = sup

(5.61) u

u∈‚ω0

and let T be given by

T = d(0, ‚ω0 )/2.

(5.62)

Take µ > 0 so small that

T +M

2µ < ± ψ(t) dt.

(5.63)

M

By (5.57) and (5.62),

σ (t)u = 0, u ∈ ‚ω0 , 0 ¤ t ¤ T.

(5.64)

If u ∈ ‚ω0 , but u ∈ Q 1 , we must have

G(u) > b + µ

(5.65)

since we cannot have

G(u) < b ’ µ

(5.66)

by (5.60). Thus, by (5.59),

G(σ (t)u) ≥ G(u) > b + µ, u ∈ ‚ω0 , u ∈ Q 1 , 0 ¤ t ¤ T.

(5.67)

On the other hand, if u ∈ ‚ω0 © Q 1 , let t1 be the largest number not greater than T

such that σ (t)u ∈ Q 1 for 0 ¤ t ¤ t1 . If t1 < T, then for δ > 0 suf¬ciently small,

G(σ (t1 + δ)u) ≥ G(σ (t1 )u) ≥ b

50 5. The Method Using Flows

and since σ (t1 + δ)u ∈ Q 1 , we must have

G(σ (t1 + δ)u) > b + µ.

Consequently,

G(σ (T )u) > b + µ.

(5.68)

If t1 = T , then by (5.51), (5.53), (5.57), (5.58), (5.61),and (5.63),

T

G(σ (T )u) ’ G(u) ≥ G (σ (t)u) dt

0

T

≥± ψ( σ (t)u ) dt

0

T

≥± ψ( u + t) dt

0