I (w, a, b) = t w ’ w ’ w + (1 ’ t) w ≥ (1 ’ t) w D.

2 2 2 2 2

D D

t t

Lemma 16.7. If b > γl (a), then there is an > 0 such that

I (v, a, b) ¤ ’ v D, v ∈ Nl .

2

(16.18)

Proof. By the continuity of γl , there is a t > 1 such that b/t > γl (a/t). Hence,

a’ b+

I (v, a/t, b/t) = v ’ v ’ v ¤ 0, v ∈ Nl ,

2 2 2

D

t t

and

a’ b+

I (v, a, b) = t v ’ v ’ v + (1 ’ t) v ¤ (1 ’ t) v D.

2 2 2 2 2

D D

t t

16.4. Some lemmas 195

Lemma 16.8. If

t[ f (x, t1 ) ’ f (x, t0 )] ¤ a(t ’ )2 + b(t + )2 , t j ∈ R, t = t1 ’ t0 ,

(16.19)

then

(16.20) (G (v + w1 ) ’ G (v + w0 ), w) ≥ 2I (w, a, b), v, w j ∈ D, w = w1 ’ w0 .

Proof. We have

( f (x, v + w1 ) ’ f (x, v + w0 ), w) ¤ a w’ + b w+ 2 .

2

Hence,

(G (v + w1 ) ’ G (v+w0 ), w)/2

= w ’ ( f (x, v + w1 ) ’ f (x, v + w0 ), w)

2

D

≥ I (w, a, b).

Lemma 16.9. Suppose there are closed subspaces X,Y of E such that E = X • Y, and

(16.21) (G (x + y1 ) ’ G (x + y2 ), y1 ’ y2 ) ≥ m( y1 ’ y2 ), x ∈ X, y1 , y2 ∈ Y,

for some function m(t) satisfying

(a)

m(t) > 0, t > 0,

(b) there is a t0 > 0 such that

t0 m(t)

dt < ∞,

t

0

(c)

m(t)

’ 0 i mpli es t ’ 0,

t

and

(d)

M

1 m(t)

dt ’ ∞ M ’ ∞.

as

M t

0

Then there exists a continuous function • : X ’ Y satisfying

(i) G(x + •(x)) = min y∈Y G(x + y).

˜

(ii) The function G(x) = G(x + •(x)) : X ’ R is of class C 1 and satis¬es

˜

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

(16.22)

196 16. Multiple Solutions

Proof. For each x ∈ X, we de¬ne G x : Y ’ R by G x (y) = G(x + y). From (16.21),

we have

(G x (y1 ) ’ G x (y2 ), y1 ’ y2 ) ≥ m( y1 ’ y2 ).

In particular, G x (y) is strictly convex (Theorem 13.7). Thus, for each x ∈ X, G x (y)

has a unique critical point •(x). To see this, note that

(G x (y) ’ G x (0), y) ≥ m( y ), y ∈ Y.

Thus,

(G x (y), y) ≥ (G x (0), y) + m( y ).

Consequently,

1 d

G x (y) ’ G x (0) = G x (t y) dt

dt

0

1 1

= (G (t y), t y) dt

tx

0

1 1

≥ [(G x (0), t y) + m( t y )] dt

t

0

1 1

= (G x (0), y) + m( t y ) dt

t

0

y m(s)

= (G x (0), y) + ds

s

0

’ ∞ as y ’ ∞.

Hence, G x (y) has a unique minimum •(x) (Lemma 13.5). Moreover,

G x (•(x)) = min G x (y) = min G(x + y).

y∈Y y∈Y

Therefore, •(x) is the only element of Y such that

0 = (G x (•(x)), y) = (G (x + •(x)), y), y ∈ Y.

(16.23)

Note that • is continuous. Suppose, on the contrary, that {x n } is a sequence in X

such that x n ’ x ∈ X. Let tn = •(x n ) ’ •(x) . Since G is continuous, for each

µ > 0 and n suf¬ciently large, we have

|(G (x n + •(x)), y)| < µ y , y ∈ Y.

(16.24)

Because of (16.21), we obtain

(G (x n + •(x)) ’ G (x n + •(x n )), •(x) ’ •(x n )) ≥ m(tn ).

(16.25)

Since, by (16.23),

(G (x n + •(x n )), y) = 0, y ∈ Y,

16.4. Some lemmas 197