by hypothesis (D ). Hence, for R suf¬ciently large, one of the inequalities

G » (±R•0 )/R 2 ¤ K •0 ’2 {F(x, ±R•0 )/R 2 ¤ 0

2

D

holds. Thus,

a0 (») ¤ 0, »∈ .

Moreover, it follows from Theorem 9.16 that (9.24) holds. Hence,

b0 (») ≥ (» ’ σ )ρ 2 , »∈ .

This shows that hypothesis (H3 ) holds. We can now apply Theorem 9.15 to con-

clude that for almost all » ∈ , there exists u k (») ∈ D such that supk u k (») <

∞, G » (u k (»)) ’ 0, and

G » (u k (»)) ’ a(») ≥ b0 (»).

Once it is known that the sequence {u k } is bounded, we can apply the usual theory to

conclude that there is a solution of

G » (u) = 0, G » (u) = a(»)

(cf., e.g., [122, p. 64]). Moreover, from the de¬nition, we see that a(») ≥ (» ’ σ )ρ 2 .

Hence, the equation G » (u) = 0 has a nontrivial solution for almost every » ∈ . This

is equivalent to (9.10) having a nontrivial solution for almost every β ∈ (K ’1 , σ ’1 ).

Since K was arbitrary, the result follows.

˜

To prove Theorem 9.6, it suf¬ces to take » = »0 and show that hypothesis (D)

implies hypothesis (D ). To see this, we note that

F(x, ±R•0 )

F(x, ±R•0 ) d x/R 2 = 2

•0 d x ’ ∞

R 2 •0

2

by hypothesis (D) and the fact that •0 (x) = 0 a.e.

To prove Corollary 9.8, we let µ be any positive number. By hypothesis (C ), there

is a δ > 0 such that

F(x, t)/t 2 ¤ µ, |t| ¤ δ, x ∈ .

By Theorem 9.7, (9.10) has a nontrivial solution for a.e. β ∈ (0, »0 /µ). Since µ was

arbitrary, the result follows.

We now give the proof of Theorem 9.4.

96 9. Superlinear Problems

Proof. By Theorem 2.4, for each arbitrary K > 1, and a.e. » ∈ (1, K ), there exists u »

such that G » (u » ) = 0, G » (u » ) = a(») ≥ (» ’ 1)ρ 2 . Choose »n ’ 1, »n > 1. Then

there exists u n such that

G »n (u n ) = 0, G »n (u n ) = a(»n ) ≥ a(1) ≥ b0 (1).

By Theorem 3.4, we may assume that b0 (1) ≥ µ > 0. Therefore,

2F(x, u n )

d x ¤ c.

2

un D

Now we prove that {u n } is bounded. If u n D ’ ∞, let wn = u n / u n D; then

wn ’ w weakly in D, strongly in L 2 ( ), and a.e. in .

We now have two cases:

Case 1: w = 0 in D. We get a contradiction as follows:

2F(x, u n ) 2F(x, u n )

c≥ dx = |wn |2 d x

un 2 2

un

D

2F(x, u n )

≥ |wn |2 d x ’ W1 (x) d x ’ ∞.

2

un

w=0 w=0

Case 2: w = 0 in D. We de¬ne tn ∈ [0, 1] by

G »n (tn u n ) = max G »n (tu n ).

t ∈[0,1]

For any c > 0 and wn = cwn , we have

¯

F(x, wn ) d x ’ 0

¯

(cf., e.g., [122, p. 64]). Thus,

G »n (tn u n ) ≥ G »n (cwn ) = c2 »n ’ 2 F(x, wn ) d x ≥ c2 /2

¯

for n large enough. That is, limn’∞ G »n (tn u n ) = ∞ and (G »n (tn u n ), u n ) = 0. There-

fore,

G »n (tn u n ) = f (x, tn u n )tn u n ’ 2F(x, tn u n ) d x

= H (x, tn u n ) d x ’ ∞.

By hypothesis (E ),

G »n (u n ) = H (x, u n ) d x ≥ H (x, tn u n ) d x ’ ∞.

9.6. The monotonicity trick 97

But

G »n (u n ) = a(»n ) ¤ G »n ((1 ’ s)u)

sup

s∈[0,1] ,u∈A

¤ G K ((1 ’ s)u)

sup

s∈[0,1] ,u∈A

< c,

¤ C. It now follows that

a contradiction. Thus, u n D

G (u n ) ’ 0, G(u n ) ’ a(1) ≥ b0 (1).

We can now apply Theorem 3.4.1 in [120, p. 64] to obtain the desired solution.

9.6 The monotonicity trick

We now give the proof of Theorem 9.15.

Proof. First, we prove conclusion (1) with the ¬rst alternative (H1 ).

Evidently, a(») ≥ b0 (») since A links B. By (H1 ), the map » ’ a(») is non-

decreasing. Hence, a (») := da(»)/d» exists for almost every » ∈ . From this point

on, we consider those » where a (») exists. For ¬xed » ∈ , let »n ∈ (», 2») ©

, »n ’ » as n ’ ∞. Then there exists n(») such that

¯

a(»n ) ’ a(»)

a (») ’ 1 ¤ ¤ a (») + 1 forn ≥ n(»).

¯

(9.31)

»n ’ »

∈ , k0 := k0 (») > 0 such that

Next, we note that there exist n

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

(9.32) whenever

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

n

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

(9.33) sup sup

s∈[0,1],u∈A s∈[0,1],u∈A