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

52 6. Finding Linking Sets

We shall show

Theorem 6.5. If a bounded set A is strongly chained to a set B, then A links B.

Theorem 6.6. Let G be a (C2∩CU )-functional on E, and let A, B be nonempty subsets of E such that A is bounded and chained to B with

(6.2)	a		:= sup G < b0		inf G	.
		0			:= B
			A
Let
(6.3)			a := K	inf( A) sup G.
				KU	K
Then there is a sequence {uk } E such that
(6.4)	G(uk ) → a, G (uk ) → 0.

Theorem 6.7. If E is a Hilbert space and A links B, then it is chained to B.

6.2 The strong case

Definition 6.8. For A E, we define
( A) = {ϕ : ϕ(u) = u, u A}
and			U : ϕ(u) = u, u A}.
U ( A) = {ϕ
The following are easily proved.
Lemma 6.9.
ϕ1, ϕ2 ϕ1 ◦ ϕ2 .
ϕ1, ϕ2		U ϕ1 ◦ ϕ2					U .
ϕ1, ϕ2 ( A) ϕ1 ◦ ϕ2 ( A).
ϕ1, ϕ2	U ( A) ϕ1 ◦ ϕ2						U ( A).
Lemma 6.10.				ϕ−1	.
ϕ			iff
ϕ		U	iff	ϕ−1		U .
ϕ	( A)		iff	ϕ−1	( A).
ϕ	U ( A)		iff	ϕ−1		U ( A).

Definition 6.11. For A E, we define
K	( A)	=	ϕ−1	(BR ) : ϕ		, R > sup	ϕ(u)		.

u A

6.2. The strong case				53
Lemma 6.12. If K K( A) and σ	( A), then σ (K ) K( A).
Proof. There is a ϕ such that K = ϕ−1(BR) with R > supu A ϕ(u) . Let
ϕ		ϕ	◦	σ −1.
˜ =

Then ϕ˜ (Lemma 6.9). Now,

ϕ˜−1 = σ ◦ ϕ−1.

Hence,		σ [ϕ−1(BR)]	ϕ−1(BR).
σ (K )	=
			= ˜
Moreover,
ϕ(˜ u) = ϕ(u), u A.
Hence,
sup ϕ(˜ x) = sup ϕ(x) < R.
x A		x A
Therefore, σ (K ) K( A).

Corollary 6.13. K( A) is a minimax system for A.

Lemma 6.14. If A is strongly chained to B and K K( A), then K ∩ B = φ.

Proof. There is a ϕ such that K = ϕ−1(BR ) with R > supu A ϕ(u) . Since A is strongly chained to B, we have

x	B ϕ(x) ≤ sup ϕ(x) < R.
inf
	x A

Hence, there is a v B such that ϕ(v) BR. Thus, v = ϕ−1ϕ(v) ϕ−1(BR ).

Corollary 6.15. If A is strongly chained to B, then A links B relative to K( A).

Corollary 6.16. If A is strongly chained to B, then A links B strongly. Proof. Theorem 2.11.

We can now give the proof of Theorem 6.5.

Proof. Let G be a (C1 ∩ CU )-functional on E, and let A, B be nonempty subsets of E such that A is strongly chained to B and (6.2) holds. Then

(6.5)	a := K inf( A) sup G
	K	K

is finite. By Lemma 6.14, K ∩ B = φ for all K K( A). Hence, a ≥ b0. Since a0 < a, then the theorem follows from Corollary 6.16.

54 6. Finding Linking Sets

6.3 The remaining proofs

Lemma 6.17. If A is chained to B and K KU ( A), then K ∩ B = φ.

Proof. There is a ϕ U such that K = ϕ−1(BR ) with R > supu A ϕ(u) . Since A is chained to B, we have

x	B ϕ(x) ≤ sup ϕ(x) < R.
inf
	x A

Hence, there is a v B such that ϕ(v) BR . Hence, v = ϕ−1ϕ(v) ϕ−1(BR ).

Lemma 6.18. If K KU ( A) and σ U ( A), then σ (K ) KU ( A).

We can now give the proof of Theorem 6.6.

Proof. Let G be a (C2 ∩ CU )-functional on E, and let A, B be nonempty subsets of E such that A is chained to B and (6.2) holds. Then a given by (6.3) is finite. By Lemma 6.17, K ∩ B = φ for all K KU ( A). Hence, a ≥ b0. If (6.4) were false, there would exist a positive constant δ such that 3δ < a − b0 and

(6.6)		G (u) ≥ 3δ
whenever
(6.7)	u Q = {u E : |G(u) − a| ≤ 3δ}.
Let
	Q0	= {u E : |G(u) − a| ≤ 2δ},
	Q1	= {u E : |G(u) − a| ≤ δ},
	Q2	= E\Q0,

η(u) = d(u, Q2)/[d(u, Q1) + d(u, Q2)].

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies

η(

u

) =

1

,

u

Q

1; η(u) = 0, u

¯ 2;

0

< η(

u

) <

1

,

.

Q

otherwise

Consider the differential equation

(6.8)

σ (t) = W (σ (t)),

t R, σ (0) = u,

where

(6.9)

W (u)

= −

η(u)G

(u)/

G (u) .

Since G C2(E, R), the mapping W is locally Lipschitz continuous on the whole of E and is bounded in norm by 1. Hence, by Corollary 4.6, (6.8) has a unique solution

6.3.	The remaining proofs										55
1			R. Let us denote the solution of (6.8) by			σ (	)	u. The mapping	σ (	)	is in
for all t						t			t
CU (E × R, E) and is called the flow generated by (6.9). Note that
(6.10)			dG(σ (t)u)/dt =	(G (σ (t)u), σ (t)u)
			= −η(σ (t)u) G (σ (t)u)
			≤	−2δη(σ (t)u).
Thus,
			G(σ (t)u) ≤ G(u), t ≥ 0,
			G(σ (t)u) ≤ a0,		u A, t ≥ 0,
and			σ (t)u = u,		u A, t ≥ 0.
Again, this follows from the fact that
			G(σ (t) u) ≤ a0 ≤ b0 < a − 3δ, u A, t ≥ 0.
Hence, η(σ (t) u) = 0 for u A, t ≥ 0. This means that
			σ (t) u = 0,		u A, t ≥ 0,
and			σ (t) u = σ (0) u = u, u A, t ≥ 0.
Let
(6.11)			Eα = {u E : G(u) ≤ α}.
We note that there is a T > 0 such that
(6.12)			σ (T )Ea+δ Ea−δ .

(In fact, we can take T = 1.) Let u be any element in Ea+δ . If there is a t1 [0, T ] such that σ (t1)u / Q1, then

G(σ (T )u) ≤ G(σ (t1)u) < a − δ

by (6.10). Hence, σ (T )u Ea−δ . On the other hand, if σ (t)u Q1 for all t [0, T ],

then η(σ (t)u) = 1 for all t, and (6.10) yields

(6.13)

G(σ (T )u) ≤ G(u) − 2δT ≤ a − δ.

Hence, (6.12) holds. Now, by (6.3), there is a K KU ( A) such that

(6.14)

˜

K Ea+δ .

˜

Note that

σ (

T

)

U (

A

).

=

σ (T )(K ). Then

K

A

)

by Lemma 6.12.

U (

Let K

K

But

sup G = sup G(σ (T )u) < a − δ,

˜

u

K

K

which contradicts (6.3), proving the theorem.

56	6. Finding Linking Sets

Now we give the proof of Theorem 6.7.

Proof. Assume that ϕ		satisfy (6.1), and let G			(	u	) = ϕ(			u	2. Then, by
		U does not 1									)
the definition of the class		U , G CU (E, R), sup G( A) < inf G(B), and G has no
critical level a ≥ infx B G(x) > 0. To show the latter, assume that there is a sequence
uk satisfying
(6.15)		G(uk ) → a, G (uk ) → 0.
Then we have, for any bounded sequence vk E,
(6.16)		(ϕ (uk )vk , ϕ(uk )) → 0.	=					→
=	(ϕ (uk ))−1(ϕ(uk )). Then (ϕ (uk )vk , ϕ(uk ))
Let vk :				G(uk )					a. However, the

sequence vk is bounded: G(uk ) → a implies that ϕ(uk ) is bounded, which implies that uk , and consequently that ϕ (uk ))−1 and vk are also bounded. Hence, G (uk ) → 0 implies G(uk ) → 0, showing that a = 0. Thus, A does not link B.

6.4 Notes and remarks

The results of this chapter are from [138]. For related material cf. [137] and [156].

Chapter 7

Sandwich Pairs

7.1 Introduction

In this chapter we discuss the situation in which one cannot find linking sets that separate a functional G, i.e., satisfy

(7.1)	a0 :=	sup G	≤ b0 :=	inf G	.
		A		B

Are there weaker conditions that will imply the existence of a PS sequence

(7.2)

G(uk ) → a, G (uk ) → 0?

Our answer is yes, and we find pairs of subsets such that the absence of (7.1) produces a PS sequence. We have

Definition 7.1. We shall say that a pair of subsets A, B of a Banach space E forms a sandwich if, for any G C1(E, R), the inequality

(7.3)	−∞ <	b	0 :=	B	≤	a	0	=	A	< ∞
				inf G				:	sup G
implies that there is a sequence satisfying
(7.4)	G(uk ) → c,			b0 ≤ c ≤ a0,					G (uk ) → 0.

Unlike linking, the order of a sandwich pair is immaterial; i.e., if the pair A, B forms a sandwich, so does B, A. Moreover, we allow sets forming a sandwich pair to intersect. One sandwich pair has been studied in Chapter 3. In fact, Theorem 3.17 tells us that if M, N are closed subspaces of a Hilbert space E, and M = N , then M, N form a sandwich pair if one of them is finite-dimensional.

Until recently, only complementing subspaces have been considered. The purpose of the present chapter is to show that other sets can qualify as well. Infinite-dimensional sandwich pairs will be considered in Chapter 15.

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_7, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

58	7. Sandwich Pairs

7.2 Criteria

In this section we present sufficient conditions for sets to qualify as sandwich pairs. We have

Proposition 7.2. If A, B is a sandwich pair and J is a diffeomorphism on the entire space having a derivative J satisfying

(7.5)

J (u)−1 ≤ C, u E,

then JA, JB is a sandwich pair.

Proof. Suppose G C1 satisfies

(7.6)

−∞ <

b

0 := J B

≤ a0 := J A

< ∞.

inf G

sup G

Let

G1(u) = G( J u),

u E.

Then

−∞ < b0

:= J B

= J u J B

) =

(7.7)

(

J u

B

1

inf G

inf

G

inf G

≤ a0 := sup G = sup

G( J u) = sup G1 < ∞.

J A

J u J A

A

Since A, B form a sandwich pair, there is a sequence {hk } E such that

(7.8)

G1(hk )

→

c,

b0

≤

c

≤

a0

, G

(hk )

→

0.

1

If we set uk = J hk , this becomes

(7.9)

G(uk ) → c,

b0 ≤ c ≤ a0,

G (uk ) J (hk ) → 0.

In view of (7.5), this implies G (uk ) → 0. Thus, J A, J B is a sandwich pair.

Proposition 7.3. Let N be a closed subspace of a Hilbert space E with complement

M	=	M	{	v0	}	1
						, where v0 is an element in E having unit norm, and let δ be any positive
number. Let ϕ(t) C (R) be such that
							0 ≤ ϕ(t) ≤ 1, ϕ(0) = 1,
and							ϕ(t) = 0, |t| ≥ 1.
Let
(7.10)		F(v + w + sv0) = v + [s + δ − δϕ( w 2/δ2)]v0, v N, w M, s R.

Assume that one of the subspaces M, N is finite-dimensional. Then A = N = N {v0}, B = F−1(δv0) form a sandwich pair.

7.2. Criteria	59
Proof. Define
J (v + w + sv0) = v + w + [s + δ − δϕ( w 2/δ2)]v0,	v N, w M, s R.

Then J is a diffeomorphism on E with its inverse having a derivative satisfying (7.5). Moreover, J A = N and J B = M +δv0. Hence, J A, J B form a sandwich pair as long as one of them is finite-dimensional (Theorem 3.17). We now apply Proposition 7.2.

Theorem 7.4. Let N be a finite dimensional subspace of a Banach space E. Let F be a Lipschitz continuous map of E onto N such that F = I on N and

(7.11)

F(g) − F(h) ≤ K g − h , g, h E.

Let p be any point of N. Then A = N, B = F−1( p) form a sandwich pair.

Proof. Let G be a C1-functional on E satisfying (7.3), where A, B are the subsets of E specified in the theorem. If the theorem is not true, then there is a δ > 0 such that

(7.12)	G (u) ≥ 3δ
whenever
(7.13)	b0 − 3δ ≤ G(u) ≤ a0 + 3δ.

E

Since G C1(E, R), there is a locally Lipschitz continuous mapping Y (u) of

= {

u

E : G

(u)

=

0

}

into E such that

ˆ

(

) ≤

ˆ

u

1

,

u

,

Y

E

and

(G (u), Y (u)) ≥ 2δ

whenever u satisfies (7.13) (for the construction of such a map, cf., e.g., [112]). Let

Q0	= {u E : b0 − 2δ ≤ G(u) ≤ a0 + 2δ},
Q1	= {u E : b0 − δ ≤ G(u) ≤ a0 + δ},
Q2	=	E\Q0,
η(u) =		ρ(u, Q2)/[ρ(u, Q1) + ρ(u, Q2)].

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies

	η(u) = 1, u Q1; η(u) = 0, u	Q	2;	0	< η(u) < 1, otherwise.
Consider the differential equation
(7.14)	σ (t) = W (σ (t)), t R,			σ (0) = u N,
where	W (u) = −η(u)Y (u).

60 7. Sandwich Pairs

The mapping W is locally Lipschitz continuous on the whole of E and is bounded in norm by 1. Hence, by Theorem 4.5, (7.13) has a unique solution for all t R. Let us denote the solution of (7.13) by σ (t)u. The mapping σ (t) is in C(E × R, E) and is called the flow generated by W (u). Note that

(7.15)	dG(σ (t)u)/dt =	(G (σ (t)u), σ (t)u)
	=	−η(σ (t)u)(G (σ (t)u), Y (σ (t)u))
	≤	−2δη(σ (t)u).
Let	Eα = {u E : G(u) ≤ α}.
I claim that there is a T > 0 such that
(7.16)	σ (T )Ea0+δ Eb0−δ .

In fact, we can take T > (a0 − b0 + δ)/2δ. Let u be any element in Ea0+δ . If there is a t1 [0, T ] such that σ (t1)u / Q1, then

G(σ (T )u) ≤ G(σ (t1)u) < b0 − δ

by (7.15). Hence, σ (T )u Eb0−δ . On the other hand, if σ (t)u Q1 for all t [0, T ], then η(σ (t)u) = 1 for all t, and (7.15) yields

G(σ (T )u) ≤ G(u) − 2δT ≤ a0 − 2δT ≤ b0 − δ.

Hence, (7.16) holds. Let be a bounded open subset of N containing the point p such that

(7.17) ρ(∂ , p) > K T + δ,

where ρ is the distance in E and K is the constant in (7.11). If v ∂ , then

	v − p ≤ v − Fσ (t)v + Fσ (t)v − p .
Then
(7.18)	Fσ (t)v − p > K T + δ − t K > 0, v ∂ , 0 ≤ t ≤ T,
since			t
	Fσ (t)v − v ≤ K	0	σ (s)v ds ≤ K t.

Let

H (t) = Fσ (t).

Then H (t) is a continuous map of into N for 0 ≤ t ≤ T. Moreover, H (t)v = p for v ∂ by (7.18). Hence, the Brouwer degree d(H (t), , p) is defined. Consequently,

d(H (T ), , p) = d(H (0), , p) = d(I, , p) = 1.

7.3. Notes and remarks

61

This means that there is a v such that

Fσ (T )v = p.

But then

σ (T )v F−1( p) = B.

This is not consistent with (7.16). Hence, A, B form a sandwich pair.

7.3 Notes and remarks

The material of this chapter comes from [135].