I (w0 + v 1 , a, b) = sup I (w0 + v, a, b).

(11.34)

v∈N ’1

From this and (11.32), it follows that

I (w0 + v 1 , a, b) ⊥ N

(11.35) ’1

11.2. The curves 133

and

I (w0 + v 1 , a, b) ≥ 0.

(11.36)

But (11.33) and (11.35) imply via Lemma 11.4 that v 1 = v 0 . Thus,

I (u 0 , a, b) ≥ 0,

and the lemma is proved.

Lemma 11.9. If (a, b) ∈ C ,2 , then

I (u, a, b) ¤ 0, u∈S ,2 ,

(11.37)

where

= {u ∈ E : I (u, a, b) ⊥ M }.

(11.38) S ,2

Proof. If (a, b) ∈ C then b = μ (a). Consequently, by (11.23),

,2 ,

m (a, b) = 0.

(11.39)

By (11.21), this implies

inf I (v + w, a, b) ¤ 0, v∈N.

(11.40)

w∈M

and write u 0 = v 0 + w0 with v 0 ∈ N , w ∈ M . Then

Let u 0 be any element in S ,2

I (v 0 + w0 , a, b) ⊥ M .

(11.41)

By Lemma 11.7, there is a w1 ∈ M such that

I (v 0 + w1 , a, b) = inf I (v 0 + w, a, b).

(11.42)

w∈M

Consequently,

I (v 0 + w1 , a, b) ⊥ M

(11.43)

and

I (v 0 + w1 , a, b) ¤ 0.

(11.44)

If we now apply Lemma 11.5, we see that w1 = w0 . Thus, I (u 0 , a, b) ¤ 0, and the

proof is complete.

© ‚ B R links S [hm] for each R > 0.

Lemma 11.10. N ’1 ,1

134 11. Fuˇ ´k Spectrum: Resonance

c±

Proof. We suppress the subscript ’ 1. Let P be the orthogonal projection of D onto

N, and de¬ne

F(u) = P I (u), u ∈ D.

(11.45)

I claim that the restriction F0 of F to N is a homeomorphism of N onto N. If this is so,

¯

then the image of N © B R under F0 is the closure of a bounded, open set. Moreover,

this open set contains the point 0 since F0 (0) = 0 and 0 is an interior point of N © B R .

Proposition 3.10 can then be used to show that N © ‚ B R links F ’1 (0) = S ,1 . Clearly,

F0 is a continuous map of N into itself. It is surjective. To see this, let h be any element

of N and take

G 0 (v) = I (v, a, b) ’ (h, v).

(11.46)

I claim that

G 0 (v) ’ ’∞ as v ’ ∞, v ∈ N.

(11.47)

Assuming this for the moment, we see that G 0 (v) has a maximum on N. At a point of

maximum we have G 0 (v) ⊥ N, producing a solution of

P I (v, a, b) = h.

(11.48)

Moreover, if

P I (v 0 , a, b) = h 0 , P I (v 1 , a, b) = h 1 ,

(11.49)

then

’ ’ + +

(h, v) = (Av, v) + a(v 1 ’ v 0 , v) ’ b(v 1 ’ v 0 , v)

(11.50)

where h = h 1 ’ h 0 , v = v 1 ’ v 0 . Thus,

+ + ’ ’

(b ’ » ’1 )(v 1 ’ v 0 , v) ’ (a ’ » ’1 )(v 1 ’ v 0 , v),

(11.51)

+ [» v ’ (Av, v)] = ’(h, v).

2

’1

Since a, b > » ’1 , the left-hand side of (11.51) is positive for all v ≡ 0. Consequently,

there is a δ > 0 such that

2

δv ¤h v

(11.52)

’1

showing not only that F0 is injective but that F0 is continuous. To prove (11.47), let

{v k } ‚ N be a sequence such that ρk = (Av k , v k ) ’ ∞, and let v k = v k /ρk . Then

˜

2

vk D

˜ 2 = (A v , v ) = 1 and there is a subsequence such that v ’ v in D. We have

˜k ˜k ˜k ˜

G 0 (v k )/ρk = I (v k , a, b) ’ (h, v k )/ρk

˜ ˜

2

(11.53)

’ I (v, a, b)

˜

= [v

˜ ’» v

˜

2 2

]

’1

D

’ a) v ’

+(» ˜ 2

’1

’ b) v + 2 .

+(» ˜

’1

11.3. Existence 135

Since v D = 1, the right-hand side of (11.53) is negative. This gives (11.47) and