m(tn )/tn < µ.

Thus,

m(tn )/tn ’ 0,

and consequently, tn ’ 0. Hence, • is continuous. This proves part (i).

For t > 0 and h ∈ X, we have

˜ ˜

G(x + th) ’ G(x) G(x + th + •(x + th)) ’ G(x + •(x))

=

(16.26)

t t

G(x + th + •(x)) ’ G(x + •(x))

¤

t

1

= (G (x + •(x) + sth), h) ds.

0

In a similar manner, we see that

˜ ˜

G(x + th) ’ G(x) G(x + th + •(x + th)) ’ G(x + •(x))

=

t t

G(x + th + •(x + th)) ’ G(x + •(x + th))

≥

t

1

≥ (G (x + •(x + th) + sth), h) ds.

0

Therefore, since G and • are continuous, we have

˜ ˜

G(x + th) ’ G(x)

= (G (x + •(x)), h).

lim

t ’0+ t

˜

This shows that G has a continuous Gateaux derivative and hence is of class C 1 .

From the above, we see that

˜

(G (x), h) = (G (x + •(x)), h), h ∈ X.

(16.27)

Lemma 16.10. Let M, N be closed subspaces of a Banach space E such that

dim N < ∞ and E = M • N. Let w0 = 0 be an element of M, and take

= {v ∈ N : v ¤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

K0

= ‚ Bδ © M, 0 < δ < R.

B

Let

Q = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v ¤ R},

198 16. Multiple Solutions

Then K 0 = ‚ Q. Let ξ(u) be a continuous map from Q to E such that

ξ(v) = v, v ∈ N © BR

and

ξ(u) ≥ r > δ > 0, u ∈ K0 © ‚ BR .

Then

ξ(Q) © B = φ.

Proof. Let P be the projection of E onto N via M. Then the conclusion of the lemma

is true iff there is a u ∈ Q such that

S(u) := Pξ(u) + ξ(u) w0 = δw0 .

Let

ξt (u) = (1 ’ t)ξ(u) + tu

and

St (u) = Pξt (u) + [(1 ’ t) ξ(u) + t u ]w0 .

Then S0 = S and

S1 (u) = Pu + u w0 .

Clearly, there is a unique u ∈ Q such that S1 (u) = δw0 . The St map Q into E, but there

is no point on ‚ Q that gets mapped into δw0 . If v ∈ N © B R , then St (v) = v + v w0 ,

which cannot equal δw0 . If u ∈ ‚ Q\N, then

[(1 ’ t) ξ(u) + t u ] ≥ r > δ.

Thus, no such point can be mapped into δw0 . Hence, the Brouwer degree d(St , Q, δw0 )

is de¬ned for t ∈ [0, 1], and d(S0 , Q, δw0 ) = d(S1 , Q, δw0 ) = 1. This completes the

proof.

Corollary 16.11. A = ξ(‚ Q) links B [mm].

Proof. Let K be the collection of all sets

K = {·(Q) : · ∈ C(Q, E), ·(u) = ξ(u), u ∈ ‚ Q}.

It is easily checked that K is a minimax system. If • ∈ (A), then •(K ) ∈ K for

K ∈ K. Since A © B = φ and K © B = φ for K ∈ K, we see that A links B [mm].

Lemma 16.12. If f (x, t) satis¬es (16.19) and b < m (a), then there is a continuous

map • from Nm ’ Mm such that

J (v) ≡ G(v + •(v)) = min G(v + w) ∈ C 1 (Nm , R), v ∈ Nm ,

(16.28)

w∈Mm

and

J (v) = G (v + •(v)), v ∈ Nm .

(16.29)

16.4. Some lemmas 199

Proof. In view of Lemmas 16.6 and 16.8, we have

(G (v + w1 ) ’ G (v + w0 ), w) ≥ w D, w ∈ Mm .

2

We can now apply Lemma 16.9 to arrive at the conclusion.

Lemma 16.13. If, in addition,

a0 (t ’ )2 + b0 (t + )2 ¤ 2F(x, t), |t| < δ,

(16.30)

for some δ > 0, with a0 , b0 < »l+1 , b0 > μl (a0 ), l ¤ m, then there are µ > 0, r > 0

such that

2

J (v) ¤ ’µ v D, v ∈ Nl © Br ,

(16.31)

where

Br = {u ∈ D : u ¤ r }.

D

Proof. Let q be any number satisfying

2 < q ¤ 2n/(n ’ 2T ), 2T < n

(16.32)

2 < q < ∞, n ¤ 2T.

(16.33)

It follows from Theorem 13.6 that there is a continuous map „ : Nl ’ Ml such that

„ (s v) = s „ (v), s ≥ 0,

(16.34)

I (v + „ (v), a0 , b0 ) = inf I (v + w, a0 , b0 ), v ∈ Nl ,

(16.35)

w∈Ml

„ (v) ¤C v D, v ∈ Nl .

(16.36) D