Schechter Minimax Systems and Critical Point Theory (Springer, 2009)

104					9. Superlinear Problems
1		1	∩	B	=	φ for all s		[0, 1]. This
Then		. Moreover, by (9.53) and (9.60), (s) A

gives the contradiction that completes the proof.

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

For this case, the map λ → a(λ) is nonincreasing and a (λ) = da(λ)/dλ exists for almost all λ > 0. Therefore, we consider those λ where a (λ) exists. We choose

λn (0, λ) ∩	and λn → λ as n → ∞. Then (9.31) is still true for n large enough.
We also prove that there exists a n , k0 = k0(λ) > 0 such that
(9.64)	n (s)u ≤ k0 if	Gλ( n (s, u)) ≥ a(λ) − (λ − λn ).
In fact, by the definition of a(λn ), there exists a n such that
(9.65)	sup Gλ( n (s)u)	≤	sup Gλn ( n (s)u)
s [0,1],u A			s [0,1],u A
		≤ a(λn ) + (λ − λn )
		≤	a(λ) + (a (λ) − 2)(λn − λ).

If Gλ( n (s)u) ≥ a(λ) − (λ − λn ), then by (9.31), (9.64), and (9.65),

(9.66)	I (	(s, u))	=		Gλn ( n (s)u) − Gλ( n (s)u)
					λn − λ
	n
			≥		a(λn ) − a(λ) + 2(λ − λn )
					λn	−	λ
			≥	a (λ) − 3.

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

(9.67) J ( n (s)u) = λn I ( n (s)u) − Gλn ( n (s)u) ≥ −λ|a (λ) − 3| − λ − a(λn )

≥ −λ|a (λ) − 3| − λ − λ|a (λ) − 1| − a(λ)

and

(9.68) J ( n (s)u) ≤ −Gλn ( n (s)u) ≤ −Gλ( n (s)u) ≤ −a(λ) + λ.

Hence, (9.66)–(9.68) imply that n (s)u ≤ k0 = k0(λ), a constant depending only on λ. 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 Gλ by −Gλ, 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

(9.69)

0 < μF(x, t) ≤ t f (x, t), |t| ≥ r,

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 condition. 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 assumption. 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(t2 + 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 first introduced by Struwe [148] for minimization 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 C1(E, R) satisfying

(10.1)	a	0 :=	A	≤ b0	:= B	,
			sup G		inf G

there are a sequence {uk } E and a constant c such that

(10.2)	b0 ≤ c < ∞
and
(10.3)	G(uk ) → c, G (uk ) → 0.

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 finite-dimensional manifold. It is not clear if it is possible for A to link B if neither is contained in such a manifold. For instance, if E = M N, where M, N are closed, infinite-dimensional subspaces of E and BR is the ball centered at the origin of radius R in E, it is unknown if the set A = M ∩ ∂ BR links B = N. (If either M or N is finite-dimensional, then A does link B; cf. Example 2 of Section 3.4.) Very little is known for the infinitedimensional case. Unfortunately, this situation arises in some important applications, including Hamiltonian systems, the wave equation, and elliptic systems, to name a few.

The purpose of the present chapter is to study linking when both M and N are infinite-dimensional and G has some additional continuity property. The property we have chosen is that of weak-to-weak continuity:

Definition 10.1. Let E be a Banach space. We shall call a functional G C1(E, R) weak-to-weak continuously differentiable if, for each sequence

(10.4) uk → u weakly in E,

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_10, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

108	10. Weak Linking
there exists a renamed subsequence such that
(10.5)	G (uk ) → G (u) weakly.

We restrict our attention to Hilbert space and prove the following.

Theorem 10.2. Let E be a separable Hilbert space, and let G be a weak-to-weak continuously differentiable functional on E. Let N be a closed subspace of E, and let Q be a bounded, convex, open subset of N containing the point p. Let F be a continuous map of E onto N such that

(a)F|Q = I, and

(b)for each finite-dimensional subspace S of E containing p such that F S = {0}, there is a finite-dimensional subspace S0 = {0} of N containing p such that

(10.6)	v	¯	∩	0, w	S		F	(v + w) 0.
		Q		S				S

(The restriction F S = {0} is made in case p = 0.) Set A = ∂ Q, B = F−1( p). If

(10.7) a1 = sup G < ∞

¯

Q

and (10.1) holds, then there is a sequence {uk } E such that (10.2), (10.3) hold and c ≤ a1.

	If F does not satisfy (a), but the restriction F		of			Q is a homeomorphism
of	¯	0		F to ¯
				N	,	then this can replace (a).
	Q onto the closure of a convex, open subset

Moreover, most, if not all, sets A, B known to link when one of the subspaces M, N is finite-dimensional will link now as well.

This leads to the following.

Definition 10.3. A subset A of a Banach space E links a subset B weakly if, for every G C1(E, R) that is weak-to-weak continuously differentiable and satisfies (10.1), there are a sequence {uk } E and a constant c such that (10.2) and (10.3) hold.

Thus we have

Corollary 10.4. Under the hypotheses of Theorem 10.2, the set A = ∂ Q links the set B = F−1( p) weakly.

Theorem 10.2 will be proved in the next section.

10.2 Another norm

In this section we prove Theorem 10.2 by introducing a different norm that is equivalent to the weak topology on bounded sets.

10.2. Another norm

109

Proof. Assume that there is no such sequence. Then there is a positive number δ > 0 such that

(10.8)					G (u) ≥ 2δ
whenever u belongs to the set
(10.9)	ˆ	= {	u		E : b	0 − 2δ ≤ G(u) ≤ a1 + 2δ}.
	E

Since E is separable, we can norm it with a norm |u|w satisfying

(10.10) |u|w ≤ u , u E,

and such that the topology induced by this norm is equivalent to the weak topology of E on bounded subsets of E.

This can be done as follows. Let {ek } be an orthonormal basis for E. We then set

|	u 2	∞	|(u, ek )|2	.
	|w =		k2
		k=1

˜

We denote E equipped with this norm by E. Let

E = {u E : G (u) = 0}.

For u E , let h(u) = G (u)/ G (u) . Then, by (10.8),

(10.11)

(G (u), h(u))

≥

2δ,

u

ˆ

E.

Let

T = (a1 − b0 + 4δ)/δ,

(10.12)

BR = {u E : u < R},

R = sup u + T,

ˆ =

Q

ˆ .

¯ R ∩

B

B

E

B

there is an E neighborhood W

(

u

)

of u such that

For each u ˆ ,

˜

(10.13)

(G (v), h(u)) > δ,

v

W (u)

∩

ˆ

B.

Otherwise, there would be a sequence

{vk }

ˆ

B such that

(10.14)

|vk − u|w → 0

and

(G (vk ), h(u)) ≤ δ.

B is bounded in

E, vk

→

u weakly in E and (10.5) implies that

Since ˆ

(10.15)

(G (vk ), h(u)) → (G (u), h(u)) ≤ δ

110										10. Weak Linking
									B be the set B with the inherited
in view of (10.14). This contradicts (10.11). Let ˜										ˆ
topology of		E. It is a metric space, and W				(	u	) ∩	B is an open set in this space. Thus,
			˜		˜				˜	˜
{W (u) ∩	˜ },		u
	B				B, is an open covering of the paracompact space B (cf., e.g., [112]).

Consequently, there is a locally finite refinement {Wτ } of this cover. For each τ, there is an element uτ such that Wτ W (uτ ). Let {ψτ } be a partition of unity subordinate to this covering. Each ψτ is locally Lipschitz continuous with respect to the norm |u|w and consequently with respect to the norm of E. Let

(10.16)

Y

(

u

) =

ψτ (

)

(

u

τ ), u

˜ .

u

h

B

Then Y (u) is locally Lipschitz continuous with respect to both norms. Moreover,

(10.17)

Y (u) ≤

ψτ (u) h(uτ ) ≤ 1

and

=

≥

ˆ

(G (u), Y (u))

τ

τ

(10.18)

ψ

(u)(G (u), h(u

))

δ,

u

B.

For u

Q

E

,

let

σ (t)u be the solution of

¯

∩ ˆ

(10.19)

σ (t) = −Y (σ (t)),

t ≥ 0, σ (0) = u.

Note that

σ ( )

u will exist as long as

σ (t)u is in

ˆ

(

u

,

t

)

t

B. Moreover, it is continuous in

with respect to both topologies.

ˆ,

σ (

)

ˆ

(σ ( )

) >

0 − δ

Next we note that if u

¯ ∩

we cannot have

u

u

b

Q

E

t

B and G

t

for 0 ≤ t ≤ T : for by (10.18), (10.19),

(10.20)

dG(σ (t)u)/dt = (G (σ ), σ ) = −(G (σ ), Y (σ )) ≤ −δ

as long as σ ( )

u

ˆ

σ ( )

u

ˆ

≤

t

≤

T , we would have

t

B. Hence, if

t

B for 0

(10.21)

G(σ (T )u) − G(u) ≤ −δT = −(a1 − b0 + 4δ).

Thus, we would have G(σ (T )u) < b0 − 4δ. On the other hand, if σ (s)u exists for
0 ≤ s < T, then σ (t)u		ˆ .	To see this, note that
		B
				t
(10.22)	u − σ (t)u = zt (u) :=			0	Y (σ (s)u)ds.
By (10.17),
(10.23)			zt (u) ≤ t.
Consequently,
(10.24)			σ (t)u ≤ u + t < R.

σ ( ) ˆ

Thus, t u B. We can now conclude that for that σ (s)u exists for 0 ≤ s ≤ t and G(σ (t)u) ≤

Q	E	there is a t	≥	0 such
each u ¯ ∩	ˆ,

b0 − δ. Let

(10.25)	u := inf{t ≥ 0 : G(σ (t)u) ≤ b0 − δ},	u		¯	∩ ˆ .
	T			Q	E

10.2. Another norm

111

Then σ (t)u exists for 0 ≤ t ≤ Tu and Tu < T . Moreover, Tu is continuous in u. Define

σˆ (t)u =

σ (t)u,

0 ≤ t ≤ Tu,

σ (Tu )u,

Tu ≤ t ≤ T,

¯

ˆ

¯

ˆ

∩

\

σ ( )

=

,

≤

≤

σ ( )

for u

Q

E. For u

Q

E, define

ˆ

t u

u

0

t

T . Then

ˆ

t u is

continuous in (u, t), and

(10.26)

G

ˆ

T

)

u

)

≤

b

0

−

δ,

u

¯ .

(σ (

Q

Let

=

ˆ

≤

≤

(10.27)

ϕ(v,

)

v

¯

,

0

t

T.

t

F

σ (t)v,

Q

Then

ϕ is a continuous map of

Q

× [0, T ] to

N. Let

¯

¯

}

= {

(

,

)

: u

= ˆ

, t

[0,

T ]

.

K

u

t

Q

Then K

is a compact subset of

E

×

R

. To see this, let (uk , tk ) be any sequence in K .

k

˜

Then u

= σ ( k )vk ,

where

vk

¯ .

Since Q is bounded, there is a subsequence such

t

Q

that

vk

→ v0

weakly in E and t

→ t0 in [0, T ]. Since

Q is convex and bounded,

v0

is

−

|

→

k

¯

in ¯

|

vk

v0

w

0

.

Since

ˆ

)

˜

×

,

we have

σ (

R

Q and

t

is continuous in E

uk = σˆ (tk )vk σˆ (t0)v0 = u0 K .

Each u0

B has a neighborhood W

(

u

0)

in E and a finite-dimensional subspace S

u

0)

ˆ

∩

˜

ˆ

(

such that Y (u)

S(u0) for u

W (u0)

ˆ .

Since

t

)

u is continuous in

(

u

,

t

),

for

σ (

(

0

B

each

u

, t0)

K there are a neighborhood

W (u0

, t0)

˜ ×

R and a finite-dimensional

E

subspace S(u0, t0) E such that zˆt (u) S(u0, t0) for (u, t) W (u0, t0), where

t

ˆ

E

u

ˆ ,

(10.28)

zˆt (u) := u − σˆ (t)u =

0 Y (σ (s)u)ds,

0,

u

ˆ .

E

Since K is compact, there are a finite number of points (u j , t j ) K such that K

W = W (u j , t j ). Let S be a finite-dimensional subspace of E containing p and all
the S(u j , t j ) and such that F S = {0}. Then, for each v	¯	,	ˆt (v)	S	.	Then,
	Q		we have z

by assumption (b) of Theorem 10.2, there is a finite dimensional-subspace S0 = {0} of

N containing p such that F

z

(v)) S0

Q

S

.

We note that

ϕ(

u

,

t

)

(v − ˆt

for all v ¯ ∩

0

Q

S

×

[0

,

T ] into S

.

For t in [0

,

T ]

,

let

ϕt (v) = ϕ(v,

t

)

. Then

maps ¯ ∩

0

0

(10.29)

ϕt (v) = p,

v ∂ (Q ∩ S0) = ∂ Q ∩ S0, 0 ≤ t ≤ T,

b0 by

for, if ϕ(v, t)

=

p, then σ (t)v

F−1( p)

=

B. This implies G(σ (t)v)

≥

ˆ

ˆ

(10.1). But (10.20) and (10.1) imply that G(σˆ (t)v) < b0 for t > 0. Since p ∂ Q by hypothesis, we obtain a contradiction. Thus, (10.29) holds. Consequently, the Brouwer degree d(ϕt , Q ∩ S0, p) is defined. Since ϕt is continuous, we have

(10.30) d(ϕT , Q ∩ S0, p) = d(ϕ0, Q ∩ S0, p) = d(I, Q ∩ S0, p) = 1.

Hence, there is a v Q such that Fσˆ (T )v = p. Consequently, σˆ (T )v F−1( p) = B. In view of (10.1), this implies

G(σˆ (T )v) ≥ b0,

contradicting (10.26). This completes the proof.

112	10. Weak Linking

10.3 Some examples

The following are examples of sets that link weakly.

Example 1. Let M, N be closed subspaces such that E = M N (both can be infinite-

dimensional). Let

BR = {u E : u < R}

and take A = ∂ BR ∩ N,							B = M. Then A links B weakly. To see this, take =
B	R ∩	N	,	Q	= ¯		E, we write
					. For u
(10.31)							u = v + w, v N, w M,

and take F to be the projection

Fu = Pu = v.

For any finite-dimensional subspace S of E such that F S = {0}, take S0 = P S. Since F|Q = I and M = F−1(0), we see by Theorem 10.2 that A links B weakly.

Example 2. We take M, N as in Example 1. Let w0 = 0 be an element of M, and take

A= {v N : v ≤ R} {sw0 + v : v N, s ≥ 0, sw0 + v = R},

B= ∂ Bδ ∩ M, 0 < δ < R.

Then A links B weakly. Again, we may assume that w0 = 1. Let

Q = {sw0 + v : v N, s ≥ 0, sw0 + v ≤ R}.

Then A

= ∂ Q in

N

=

N

{w0

}. If u is given by (10.31), we define

˜

Fu = v + w w0.

=

˜

|

=

I, and B

F

−1(δw0). For any

Then F is a continuous map of E onto N , F

Q

finite-dimensional subspace S of E, take S0 = P S {w0}. We can now apply Theorem 10.2 to conclude that A links B weakly.

Example 3. Take M, N as before and let v0 = 0 be an element of N. We write N = {v0} N . We take

A= {v N : v ≤ R} {sv0 + v : v N , s ≥ 0, sv0 + v = R},

B= {w M : w ≥ δ} {sv0 + w : w M, s ≥ 0, sv0 + w = δ},

where 0 < δ < R. Then A links B weakly. To see this, we let

Q = {sv0 + v : v N , s ≥ 0, sv0 + v ≤ R}

and reason as before. For simplicity, we assume that v0 = 1, that E is a Hilbert space, and that the splitting E = N {v0} M is orthogonal. If

(10.32) u = v + w + sv0, v N , w M, s R,

10.3. Some examples		113
we define
		v0, w ≤ δ
F(u) = v + s + δ − δ2 − w 2
= v + (s + δ)v0,		w > δ.

Note that F|Q = I while F−1(δv0) is precisely the set B. For any finite-dimensional subspace S of E containing (δv0), take S0 = F S {δv0}. Hence, we can conclude via Theorem 10.2 that A links B weakly.

Example 4. This is the same as Example 3 with A replaced by A = ∂ BR ∩ N. The

= ¯ ∩ proof is the same with Q replaced by Q BR N.

Example 5. Let M, N be as in Example 1. Take A = ∂ Bδ ∩ N, and let v0 be any element in ∂ B1 ∩ N. Take B to be the set of all u of the form

u = w + sv0, w M,

satisfying any of the following:

(a)w ≤ R, s = 0;

(b)w ≤ R, s = 2R0;

(c)w = R, 0 ≤ s ≤ 2R0,

where 0 < δ < min( R, R0). Then A links B weakly. To see this, take N = {v0} N . Then any u E can be written in the form (10.32). Define

F(u)

=

R0

−

max

R0

w ,

s

−

R0

|

v0

+

v

R

|

and Q

=

¯

∩

N. Then F

|

Q

=

I . Moreover, A

=

F

−1(0). For any

Bδ

C(E, N) and F

finite-dimensional subspace S of E such that F S = {0}, take S0 = P S {v0}. Hence, A links B by Theorem 10.2.

Example 6.

Let M, N be as in Example 1. Let v0 be in ∂ B1 ∩ N and write N =

{

v0

}

N

. Let A

=

∂ Bδ

∩

N, Q

=

¯

∩

N, and

Bδ

B = {w M : w ≤ R} {w + sv0 : w M, s ≥ 0, w + sv0 = R},

where 0 < δ < R. Then A links B weakly. To see this, write u = w + v + sv0, w M, v N , s R and take

F(u) = (c R − max{c w + sv0 , |c R − s|})v0 + v ,

where c = δ/( R − δ). Then F is the identity operator on Q, and F−1(0) = B. For any finite-dimensional subspace S of E such that F S = {0}, take S0 = P S {v0}. Apply Theorem 10.2.

114	10. Weak Linking

10.4 Some applications

In this section we apply Theorem 10.2 to semilinear boundary-value problems. Let be a domain in Rn and let A be a self-adjoint operator in L2( ) having 0 in its resolvent set.Thus, there is an interval (a, b) in its resolvent set satisfying a < 0 < b. Let f (x, t) be a Carath´eodory function on × R such that

(10.33)				| f (x, t)| ≤ V (x)2|t| + W (x)V (x), x , t R,
and
(10.34)			1/2			f (x, t)/t → α±(x)		as	t → ±∞,
where V	,				2
		W		L2	( ), and multiplication by V (x) > 0 is a compact operator from
D = D(| A|				) to L ( ). Let
						∞		N =	a
						M =	d E(λ)D,		d E(λ)D,
						b			−∞

where {E(λ)} is the spectral measure of A. Then M, N are invariant subspaces for A and D = M N. If

(10.35)	α(u, v) =	(α+u+ − α−u−)vdx, α(u) = α(u, u),
then we assume that
(10.36)		α(v) ≥ ( Av, v),	v N,
(10.37)		( Aw, w) ≥ α(w),	w M.
We also assume that the only solution of
(10.38)		Au = α+u+ − α−u−

is u ≡ 0, where u± = max{±u, 0}. We have

Theorem 10.5. Under the above hypotheses, there is at least one solution of

(10.39)	Au = f (x, u), u D( A).
Proof.	Let
(10.40)	a(u, v) = ( Au, v),	a(u) = ( Au, u),	u, v D,
and
(10.41)	G(u) = a(u) − 2	F(x, u) dx,	u D,