and consequently with respect to the norm of E. Let

˜

Y (u) = ψ„ (u)h(u „ ), u ∈ B.

(10.16)

Then Y (u) is locally Lipschitz continuous with respect to both norms. Moreover,

Y (u) ¤ ψ„ (u) h(u „ ) ¤ 1

(10.17)

and

ˆ

(G (u), Y (u)) = ψ„ (u)(G (u), h(u „ )) ≥ δ, u ∈ B.

(10.18)

¯ ˆ

For u ∈ Q © E, let σ (t)u be the solution of

σ (t) = ’Y (σ (t)), t ≥ 0, σ (0) = u.

(10.19)

ˆ

Note that σ (t)u will exist as long as σ (t)u is in B. Moreover, it is continuous in (u, t)

with respect to both topologies.

¯ˆ ˆ

Next we note that if u ∈ Q © E, we cannot have σ (t)u ∈ B and G(σ (t)u) > b0 ’ δ

for 0 ¤ t ¤ T : for by (10.18), (10.19),

d G(σ (t)u)/dt = (G (σ ), σ ) = ’(G (σ ), Y (σ )) ¤ ’δ

(10.20)

ˆ ˆ

as long as σ (t)u ∈ B. Hence, if σ (t)u ∈ B for 0 ¤ t ¤ T , we would have

G(σ (T )u) ’ G(u) ¤ ’δT = ’(a1 ’ b0 + 4δ).

(10.21)

Thus, we would have G(σ (T )u) < b0 ’ 4δ. On the other hand, if σ (s)u exists for

ˆ

0 ¤ s < T, then σ (t)u ∈ B. To see this, note that

t

u ’ σ (t)u = z t (u) := Y (σ (s)u)ds.

(10.22)

0

By (10.17),

z t (u) ¤ t.

(10.23)

Consequently,

σ (t)u ¤ u + t < R.

(10.24)

ˆ ¯ ˆ

Thus, σ (t)u ∈ B. We can now conclude that for each u ∈ Q © E, there is a t ≥ 0 such

that σ (s)u exists for 0 ¤ s ¤ t and G(σ (t)u) ¤ b0 ’ δ. Let

¯ ˆ

Tu := inf{t ≥ 0 : G(σ (t)u) ¤ b0 ’ δ}, u ∈ Q © E.

(10.25)

10.2. Another norm 111

Then σ (t)u exists for 0 ¤ t ¤ Tu and Tu < T . Moreover, Tu is continuous in u. De¬ne

σ (t)u, 0 ¤ t ¤ Tu ,

σ (t)u =

ˆ

σ (Tu )u, Tu ¤ t ¤ T,

¯ ˆ ¯ ˆ

for u ∈ Q © E. For u ∈ Q \ E, de¬ne σ (t)u = u, 0 ¤ t ¤ T . Then σ (t)u is

ˆ ˆ

continuous in (u, t), and

¯

G(σ (T )u) ¤ b0 ’ δ, u ∈ Q.

ˆ

(10.26)

Let

¯

•(v, t) = F σ (t)v, v ∈ Q, 0 ¤ t ¤ T.

ˆ

(10.27)

¯

Then • is a continuous map of Q — [0, T ] to N. Let

¯

K = {(u, t) : u = σ (t)v, v ∈ Q, t ∈ [0, T ]}.

ˆ

˜

Then K is a compact subset of E — R. To see this, let (u k , tk ) be any sequence in K .

¯

Then u k = σ (tk )v k , where v k ∈ Q. Since Q is bounded, there is a subsequence such

¯

that v k ’ v 0 weakly in E and tk ’ t0 in [0, T ]. Since Q is convex and bounded, v 0 is

¯ ˜

in Q and |v k ’ v 0 |w ’ 0. Since σ (t) is continuous in E — R, we have

ˆ

u k = σ (tk )v k

ˆ σ (t0 )v 0 = u 0 ∈ K .

ˆ

ˆ ˜

Each u 0 ∈ B has a neighborhood W (u 0 ) in E and a ¬nite-dimensional subspace S(u 0 )

ˆ

such that Y (u) ‚ S(u 0 ) for u ∈ W (u 0 ) © B. Since σ (t)u is continuous in (u, t), for

ˆ

˜

each (u 0 , t0 ) ∈ K there are a neighborhood W (u 0 , t0 ) ‚ E—R and a ¬nite-dimensional

subspace S(u 0 , t0 ) ‚ E such that z t (u) ‚ S(u 0 , t0 ) for (u, t) ∈ W (u 0 , t0 ), where

ˆ

ˆ

t

Y (σ (s)u)ds,

ˆ u ∈ E,

0

z t (u) := u ’ σ (t)u =

ˆ ˆ

(10.28)

ˆ

u ∈ E.

0,

Since K is compact, there are a ¬nite number of points (u j , t j ) ∈ K such that K ‚

W = ∪W (u j , t j ). Let S be a ¬nite-dimensional subspace of E containing p and all

¯

the S(u j , t j ) and such that F S = {0}. Then, for each v ∈ Q, we have z t (v) ∈ S. Then,

ˆ

by assumption (b) of Theorem 10.2, there is a ¬nite dimensional-subspace S0 = {0} of

¯

N containing p such that F(v ’ z t (v)) ∈ S0 for all v ∈ Q © S0 . We note that •(u, t)

ˆ

¯

maps Q © S0 — [0, T ] into S0 . For t in [0, T ], let •t (v) = •(v, t). Then

•t (v) = p, v ∈ ‚(Q © S0 ) = ‚ Q © S0 , 0 ¤ t ¤ T,

(10.29)

for, if •(v, t) = p, then σ (t)v ∈ F ’1 ( p) = B. This implies G(σ (t)v) ≥ b0 by

ˆ ˆ

(10.1). But (10.20) and (10.1) imply that G(σ (t)v) < b0 for t > 0. Since p ∈ ‚ Q by

ˆ

hypothesis, we obtain a contradiction. Thus, (10.29) holds. Consequently, the Brouwer

degree d(•t , Q © S0 , p) is de¬ned. Since •t is continuous, we have

d(•T , Q © S0 , p) = d(•0 , Q © S0 , p) = d(I, Q © S0 , p) = 1.

(10.30)

Hence, there is a v ∈ Q such that F σ (T )v = p. Consequently, σ (T )v ∈ F ’1 ( p) = B.

ˆ ˆ

In view of (10.1), this implies

G(σ (T )v) ≥ b0 ,

ˆ

contradicting (10.26). This completes the proof.

112 10. Weak Linking

10.3 Some examples

The following are examples of sets that link weakly.

Example 1. Let M, N be closed subspaces such that E = M • N (both can be in¬nite-

dimensional). Let