the following alternatives holds:

either

σ1 (t0 ) n (2s ’ 1)u > k0 + 2 + d A ;

(9.61)

or

|G » (σ1 (t0 ) n (2s ’ 1)u) ’ a(»)| > 2µ1 ;

(9.62)

or

dist(σ1 (t0 ) n (2s ’ 1)u, B) > 3T1 .

(9.63)

Assume that (9.61) holds. If σ1 (T1 ) n (2s ’ 1)u ∈ B, then

b0 (») = a(») ¤ G » (σ1 (T1 ) n (2s ’ 1)u) ¤ G » ( n (2s ’ 1)u).

n (2s ’ 1)u ¤ k0 . Furthermore, since

By (9.32),

σ1 (t0 ) n (2s ’ 1)u ’ σ1 (0) n (2s ’ 1)u ¤ t0 ,

it follows that

σ1 (t0 ) n (2s ’ 1)u ¤ k0 + t0 ¤ k0 + 1,

which contradicts (9.61). Hence, σ1 (T1 ) n (2s ’ 1)u ∈ B.

Assume that (9.62) holds. Note, by (9.32), that

G » (σ1 (t0 ) n (2s ’ 1)u) ¤ G » (σ1 (0) n (2s ’ 1)u)

= G »( n (2s ’ 1)u)

¤ a(») + µ1 .

Therefore, (9.62) implies that

G » (σ1 (T1 ) n (2s ’ 1)u) ¤ G » (σ1 (t0 ) n (2s ’ 1)u) ¤ a(») ’ 2µ1.

It follows that σ1 (T1 ) n (2s ’ 1)u ∈ B since a(») = b0 (»).

Assume (9.63) holds. Note that σ1 (t)u ’ σ1 (t )u ¤ |t ’ t |. It therefore follows

that

σ1 (t) n (2s ’ 1)u ’ w ≥ σ1 (t0 ) n (2s ’ 1)u ’ w ’ |t ’ t0 |

for all w ∈ B, t ∈ [0, T1 ]. Hence, dist(σ1 (t) n (2s ’ 1)u, B) ≥ 2T1 for all t ∈ [0, T1 ].

In particular,

σ1 (T1 ) n (2s ’ 1)u ∈ B.

Hence, (9.60) holds in this case as well.

De¬ne

σ1 (2sT1 )u, 0 ¤ s ¤ 1/2,

—

1 (s)u :=

σ1 (T1 ) n (2s ’ 1)u, 1/2 ¤ s ¤ 1.

104 9. Superlinear Problems

— —

Then 1 ∈ . Moreover, by (9.53) and (9.60), 1 (s)A © B = φ for all s ∈ [0, 1]. This

gives the contradiction that completes the proof.

Now, we consider conclusion (1) with the second alternative (H2 ).

For this case, the map » ’ a(») is nonincreasing and a (») = da(»)/d» exists

for almost all » > 0. Therefore, we consider those » where a (») exists. We choose

»n ∈ (0, ») © and »n ’ » as n ’ ∞. Then (9.31) is still true for n large enough.

We also prove that there exists a n ∈ , k0 = k0 (») > 0 such that

n (s)u ¤ k0 if G » ( n (s, u)) ≥ a(») ’ (» ’ »n ).

(9.64)

In fact, by the de¬nition of a(»n ), there exists a ∈ such that

n

G »( n (s)u) ¤ G »n ( n (s)u)

(9.65) sup sup

s∈[0,1],u∈A s∈[0,1],u∈A

¤ a(»n ) + (» ’ »n )

¤ a(») + (a (») ’ 2)(»n ’ »).

If G » ( n (s)u) ≥ a(») ’ (» ’ »n ), then by (9.31), (9.64), and (9.65),

G »n ( n (s)u) ’

G » ( n (s)u)

I( n (s, u)) =

(9.66)

»n ’ »

a(»n ) ’ a(») + 2(» ’ »n )

≥

»n ’ »

≥ a (») ’ 3.

On the other hand, by (H2 ), (9.31), (9.65), (9.66),

J( n (s)u) = »n I ( n (s)u) ’ G »n ( n (s)u)

(9.67)

≥ ’»|a (») ’ 3| ’ » ’ a(»n )

≥ ’»|a (») ’ 3| ’ » ’ »|a (») ’ 1| ’ a(»)

and

J( n (s)u) ¤ ’G »n ( n (s)u) ¤ ’G » ( n (s)u) ¤ ’a(») + ».

(9.68)

Hence, (9.66)“(9.68) imply that n (s)u ¤ k0 = k0 (»), a constant depending only

on ». The rest is similar to the proof under the assumption (H1 ); we omit the details.

Finally, conclusion (2) can be proved immediately by interchanging A and B,

replacing G » by ’G » , and using conclusion (1).

9.7 Notes and remarks

Problem (9.6) has been studied by many people. The vast majority of results obtained

concern sublinear problems. Much less has been proved for the superlinear case. In [7]

the basic assumption was

0 < μF(x, t) ¤ t f (x, t), |t| ≥ r,

(9.69)

9.7. Notes and remarks 105

for some μ > 2 and r ≥ 0. This is a very convenient hypothesis since it readily

achieves mountain pass geometry as well as satisfaction of the Palais“Smale condi-

tion. However, it is a severe restriction; it strictly controls the growth of f (x, t) as

|t| ’ ∞. Almost every author discussing superlinear problems has made this assump-

tion. We have been able to weaken assumption (9.69) considerably, but not to our

complete satisfaction. We assume either that

μF(x, t) ’ t f (x, t) ¤ C(t 2 + 1), |t| ≥ r,

for some μ > 2 and r ≥ 0 or that (9.7) is convex in t. These allow much more freedom

for the function f (x, t). But they do not allow as much freedom as we would like.

We were able to weaken it much further and assume only hypothesis (D). However, we

paid a heavy price for this generalization: We were only able to solve (9.10) for almost

all positive values of β.

The method (called the monotonicity trick), which allowed us to solve (9.10) for

almost all values of β in some interval, was ¬rst introduced by Struwe [148] for mini-

mization problems. It was applied by Jeanjean [76] and others for various types of

problems.

The material of this chapter comes from [141] and [139]. See also [49].

Chapter 10

Weak Linking

10.1 Introduction

As we noted, a subset A of a Banach space E links a subset B of E if, for every

G ∈ C 1 (E, R) satisfying

a0 := sup G ¤ b0 := inf G,

(10.1)

B

A

there are a sequence {u k } ‚ E and a constant c such that

b0 ¤ c < ∞

(10.2)

and

G(u k ) ’ c, G (u k ) ’ 0.

(10.3)

We also saw that there are several criteria that imply that a set A links a set B. However,

all of them require that at least one of the sets A, B be contained in a ¬nite-dimensional