In proving Lemma 13.1, we shall make use of

u in E, then there is a renamed subsequence such that u k ’ u,

¯

Lemma 13.2. If u k

where

u k = (u 1 + · · · + u k )/k.

¯

(13.5)

Proof. We may assume that u = 0. Take n 1 = 1, and inductively pick n 2 , n 3 , . . . , so

that

1 1

|(u nk , u n1 )| ¤ , . . . , |(u nk , u nk’1 )| ¤ .

k k

This can be done since

(u n , u n j ) ’ 0 as n ’ ∞, 1 ¤ j ¤ k.

Since

uk ¤ C

for some C, we have

⎡ ¤

j ’1

k k

⎣ (u i , u j )¦ /k 2

2 2

¯ = +2

uk uj

j =1 j =1 i=1

⎡ ¤

j

k

1

⎣kC 2 + 2 ¦ /k 2

¤

j

j =1 i=1

(C 2 + 2)/k ’ 0.

¤

Lemma 13.3. If G(u) is convex and l.s.c. on E, and u k u, then

G(u) ¤ lim inf G(u k ).

Proof. Let

L = lim inf G(u k ).

Then there is a renamed subsequence such that G(u k ) ’ L. Let µ > 0 be given. Then

L ’ µ < G(u k ) < L + µ

(13.6)

13.2. Convexity and lower semi-continuity 151

for all but a ¬nite number of k. Remove a ¬nite number and rename the subsequence so

that (13.6) holds for all k. Moreover, there is a renamed subsequence such that u k ’ u

¯

¯

by Lemma 13.2, where u k is given by (13.5). Thus,

⎛ ⎞

k

1

G(u) ¤ lim inf G(u k ) = lim inf G ⎝ u j⎠

¯

k

j =1

k

1 1

¤ G(u j ) ¤ lim inf · k(L + µ) = L + µ.

lim inf

k k

j =1

Since µ was arbitrary, we see that G(u) ¤ L, and the proof is complete.

A subset M ‚ E is called weakly closed if u ∈ M whenever there is a sequence

{u k } ‚ M converging weakly to u in E. A weakly closed set is closed in a stronger

sense than an ordinary closed set. It follows from Lemma 13.2 that

Lemma 13.4. If M is a closed, convex subset of E, then it is weakly closed in E.

Proof. Suppose {u k } ‚ M and u k u in E. Then, by Lemma 13.2, there is a renamed

subsequence such that u k ’ u, where u k is given by (13.5). Since M is convex, each

¯ ¯

u k is in M. Since M is closed, we see that u ∈ M.

¯

We can now give the proof of Lemma 13.1.

Proof. Let

± = inf G.

M

(At this point we do not know if ± = ’∞.) Let {wk } ‚ M be a sequence such that

G(wk ) ’ ±. Since M is bounded, we see that there is a renamed subsequence such

that wk w0 . Since M is closed and convex, it is weakly closed (Lemma 13.4).

Hence, w0 ∈ M. By Lemma 13.3, G(w0 ) ¤ lim inf G(wk ) = ±. Since G(w0 ) ≥ ±, we

see that (13.4) holds. So far, we have only used the convexity of G. We use the strict

convexity to show that w0 is unique. If there were another element w1 ∈ M such that

G(w1 ) = ±, then we would have

1 1 1

w0 + w1 < [G(w0 ) + G(w1 )] = ±,

G

2 2 2

which is impossible from the de¬nition of ±. This completes the proof.

We also have

Lemma 13.5. If M is closed and convex, G is convex, is l.s.c., and satis¬es

G(u) ’ ∞ as u ’ ∞, u∈M

(13.7)

(if M is unbounded), then G is bounded from below on M and has a minimum there.

If G is strictly convex, this minimum is unique.

152 13. Semilinear Wave Equations

Proof. If M is bounded, then Lemma 13.5 follows from Lemma 13.1. Otherwise, let

u 0 be any element in M. By (13.7), there is an R ≥ u 0 such that

G(u) ≥ G(u 0 ), u ∈ M, u ≥ R.

By Lemma 13.1, G is bounded from below on the set

M R = {w ∈ M : w ¤ R}

and has a minimum there. A minimum of G on M R is a minimum of G on M. Hence,

G is bounded from below on M and has a minimum there.

13.3 Existence of saddle points

We say that (v 0 , w0 ) is a saddle point of G if

G(v, w0 ) ¤ G(v 0 , w0 ) ¤ G(v 0 , w), v ∈ N, w ∈ M.

(13.8)

We now present some suf¬cient conditions for the existence of saddle points. Let

M, N be closed, convex subsets of a Hilbert space, and let G(v, w) : M — N ’ R be

a functional such that G(v, w) is convex and l.s.c. in w for each v ∈ N, and concave

and u.s.c. in v for each w ∈ M. If M is unbounded, assume also that there is a v 0 ∈ N

such that

G(v 0 , w) ’ ∞ as w ’ ∞, w ∈ M.

(13.9)