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

42 5. The Method Using Flows

On the other hand, if σ (t)u Q1 for all t [0, T ], then η(σ (t)u) ≡ 1 in [0, T ] and

(5.14)	G(σ (t)u) ≤ G(u) − θ t ≤ a + δ − θ t, t [0, T ].
Thus we have
(5.15)	G(σ (T )u) < a − δ.
Since b0 := infB G = a, this shows that
(5.16)	σ (T )Ea+δ ∩ B = φ.
Moreover,
(5.17)	G(σ (t)u) ≤ a − θ t, u A, t [0, T ].

It therefore follows that (2.12), (2.13), and (2.21) hold for b = a + δ. Let K K satisfy

K		satisfying (2.14) with b				=	a	+ δ.	Now
(5.12). Then, by hypothesis, there is a ˜ K			B
K	∩			= φ,	contradicting (2.17).
(2.21), (5.14), (5.16), and (5.17) imply that ˜

To prove the last statement, we take ψ(t) = ρ(t) ≡ 1. Still assuming a0 = a, we note that if there did not exist a sequence satisfying both (1.4) and (2.22), then there would be positive numbers , δ, T such that δ < T and (5.1) holds whenever

u Q = {u E : d(u, B) ≤ 4T, |G(u) − a| ≤ 3δ}.

Let
Q0	=	{u E : d(u, B) ≤ 3T, |G(u) − a| ≤ 2δ},
Q1	=	{u E : d(u, B) ≤ 2T, |G(u) − a| ≤ δ}.

Since a = b0, we see that Q1	= φ. Define Q0, Q1, Q2, and η(u) as before and let
σ (t) be the flow generated by the mapping
	W (u) = −η(u)Y (u),
with everything now with respect to the new sets Q j . In this case
(5.18)	W (u) ≤ 1.

Let u be any element in Ea+δ . If there is a t1 ≤ T such that σ (t1)u / Q1, then either

(5.19)	G(σ (t1)u) < a − δ
or
(5.20)	d(σ (t1)u, B) > 2T.
Since	σ (t)u − σ (t )u ≤ |t − t |

5.3. Theorem 2.12					43
by (5.18), (5.20) implies
(5.21)		d(σ (t)u, B) > T,			0 ≤ t ≤ T.
On the other hand, if σ (t)u Q1 for all t [0, T ], then
(5.22)	G(σ (T )u) ≤ G(u) − 2 T ≤ a + δ − 2δ = a − δ.
Thus, either we have
(5.23)		G(σ (T )u) < a − δ
or (5.21) holds. Since b0 = a, this shows that
(5.24)		σ (T )Ea+δ ∩ B = φ.
We also note that
(5.25)		σ (t) A ∩ B = φ,			0 ≤ t ≤ T.
To see this, we observe, by (5.7), that
				t
	G(σ (t)u) ≤ a0 − 2		0	η(σ (τ )u)dτ, u A.

If σ (t)u B, we must have G(σ (t)u) ≥ b0 ≥ a0. The only way this can happen is if

	σ (τ )u	¯		η(σ (τ )u) ≡ 0, 0 ≤ τ ≤ t.
But this implies			2	for such	τ	, and this in turn implies either
		Q

G(σ (τ )u) < a − δ, 0 ≤ τ ≤ t,

or

d(σ (τ )u, B) > 2T, 0 ≤ τ ≤ t.

In either case we cannot have σ (t)u B. Thus, (5.25) holds. Now, by (2.6), there is a

K K such that
(5.26)	K Ea+δ ,
K	such that (2.14) holds. By (5.25), K	∩	B	= φ,	contradicting the
and there is a ˜ K	˜

fact that A links B [mm]. This completes the proof of the theorem.

Next, we give the proof of Theorem 2.6.

Proof. Define

B = v	K \ A : G(v) ≥ a0 .
	K K

By (2.23), A links B relative to K, and (2.20) holds. Apply Theorem 2.12.

44	5. The Method Using Flows

5.4 Theorem 2.14

Now we give the proof of Theorem 2.14.

Proof. We assume that a = a0 and follow the proof of Theorem 2.4. In this case we cannot take 3δ < a − a0 and we cannot conclude that (5.8) holds. Because of this, it does not follow that σ (T )K K for all K K. Let K K be such that

(5.27)	a ≤ sup G < a + δ,
	K \ A
and let Kρ be given by (2.25). Note that
(5.28)	a0 ≤ sup G, K K,
	K \ A

by (2.20). Since G(u) is Lipschitz continuous on Kρ and a = a0, there is a constant C such that

(5.29) G(u) ≤ a + Cd(u), u Kρ ,

where d(u) = d(u, A). Pick T > 2δ/θ and T > ρC/θ . As in the proof of Theorem 2.4, we have

(5.30)	σ (T )K Ea−δ .
Let ζ (t) be a nondecreasing, continuous function on R+ satisfying
(5.31)	ζ (t) = 0,		t	= 0,
	T,		t ≥ ρ,
and
(5.32)	T t/ρ < ζ (t) < T,		0 < t < ρ.
Note that the flow σ˜ (t) = σ (ζ (t)) satisfies (2.12) and (2.13). Consider the map
(5.33)	Su = σ (ζ (d(u)))u,			u K .
Consequently,	by hypothesis, S(K )	K.	Since d(u) = 0 for u A, and
σ (0)u = u, we have		u A.
	Su = u,
If u K \Kρ , then Su = σ (T )u Ea−δ . If u Kρ \ A and Su Ea−δ , then
	η(σ (t)u) = 1,	0 ≤ t ≤ ζ (d(u)).
Consequently,
(5.34)	G(Su) ≤ G(u) − θ ζ (d(u))
	≤ a + Cd(u) − θ T d(u)/ρ
	< a.

5.5. Theorem 2.21	45
Thus,	G(Su) < a, u K \ A,

contradicting (5.28). This completes the proof of the theorem.

Now, we prove Theorem 2.13.

Proof. Let B be given by (2.26). Then A links B relative to the system K. Now apply Theorem 2.14.

5.5 Theorem 2.21

We can now prove Theorem 2.21.

Proof. We may assume that a

= a0.

Otherwise, by Theorem 2.4, a sequence (2.9)

˜

=

ψ(t

+ α).

˜

exists with ψ replaced by ψ (t)

Since ψ satisfies the hypotheses of Theo-

rem 2.4 and

d(u, B ) ≤ u + α,

for each δ > 0, we can find a u E such that

˜

≤

−

≤

≤

+

G (u)

ψ(d(u, B )),

a

δ

G(u)

a

δ,

< ψ (

u )

which certainly implies (2.33).

If the conclusion of the theorem were not true, there would be a δ > 0 such that

(5.35)	ψ(d(u, B )) ≤ G (u)
would hold for all u in the set
(5.36)	Q = {u E : b0 − 3δ ≤ G(u) ≤ a + 3δ}.

By reducing δ if necessary, we can find θ < 1, T > 0 such that

(5.37)	a0 − b0 + δ < θ T, T ≤	R+α
		ψ(s) ds.
		δ+α

Thus, by Lemma 4.3, if u(t) is the solution of (4.6) with ρ(t) = 1/ψ(t), γ = 1, t0 = 0, and u0 = R, then

Let		u(t) ≥ δ, t [0, T ].
(5.38)	Q0	= {u Q : b0	− 2δ ≤ G(u) ≤ a + 2δ},
(5.39)	Q1	= {u Q : b0	− δ ≤ G(u) ≤ a + δ},
and
(5.40)	Q2 = E\Q0, η(u) = d(u, Q2)/[d(u, Q1) + d(u, Q2)].

46 5. The Method Using Flows

As before, we note that η satisfies (5.4). There is a locally Lipschitz continuous map

ˆ

= {

u

E : G (u)

=

0

}

into itself such that

Y (u) of E

ˆ

(5.41)

≤

1,

G

≤

(G (u), Y (u)),

Y (u)

θ

(u)

u

E

(cf., e.g., [120]). Let σ (t) be the flow generated by

(5.42)

=

W (u) = −η(u)Y (u)ρ(d(u, B )),

˜

where ρ(τ )

1/ψ(τ ). Since

≤

ρ(d(u, B ))

≤ ˜ =

W (u)

ρ(

u

)

1/ψ(

u

) and is

locally Lipschitz continuous, σ (t) exists for all t R+ in view of Theorem 4.5. Since

					t
(5.43)	σ (t)u − u =				0	W (σ (τ )u)dτ,
we have						t
								)) dr.
σ (t)u		−	σ (s)u	≤			ρ(d(σ (r )u, B
					s
If v B , we have
h(s) = d(σ (s)u, B ) ≤ σ (s)u − v ≤ σ (t)u − v +									t	)) dr,
								s	ρ(d(σ (r )u, B
which implies
							t
(5.44)		h(s) ≤ h(t) + s					ρ(h(r )) dr.
We also have
(5.45) dG(σ (t)u)/dt	= (G (σ ), σ ) = −η(σ )(G (σ ), Y (σ ))ρ(d(σ, B ))

≤−θ η(σ ) G (σ ) ρ(d(σ, B ))

≤−θ η(σ )ψ(d(σ, B ))ρ(d(σ, B ))

=−θ η(σ )

in view of (5.35) and (5.41). Now suppose u Ea+δ is such that there is a t1 [0, T ] for which σ (t1)u / Q1. Then

G(σ (t1)u) < b0 − δ,

since we cannot have G(σ (t1)u) > a + δ for such u by (5.45). But this implies

(5.46) G(σ (T )u) < b0 − δ.

On the other hand, if σ (t)u Q1 for all t [0, T ], then

T

G(σ (T )u) ≤ G(u) − θ dt ≤ a − θ T < b0 − δ

0

5.5. Theorem 2.21

by (5.37). Thus, (5.46) holds for u definition (2.6) of a, there is a K K

47

Ea+δ and, in particular, for any u A. By the such that

(5.47)					sup G < a + (δ/2).
					K
Note that σ (t)u = v				B for u A and t [0, T ]. For v				B , this follows from
(5.44) and the fact that
			h(t) = d(σ (t)u, B ) ≥ u(t) ≥ δ, t [0, T ],
in view of Lemma 4.8. If v B\B , we have, by (5.45),
		G(σ (t)u) ≤ a − θ					t
						0	η(σ (τ )u)dτ < a,	t > 0,	¯
unless η(σ (τ )	u	) =	0 for 0		≤ τ ≤ t. But this would mean that η(σ (τ )u)					2	in view
									Q

of (5.4). But then we would have G(u) ≤ b0 − 2δ since we cannot have G(u) ≥ a + 2δ. Thus,

G(σ (t)u) < a.

This shows that

(5.48)

B ∩ σ (t) A = φ,

0 ≤ t ≤ T.

Taking b

=

a

+ (δ/

),

(2.14) tells us that

there is a

˜

K

such that

2

K

(5.49)

˜

σ (t) A σ (T )[Eb K ].

K

t [0,T ]

K

∩

B

= φ,

contradicting the fact that A links

But (5.46), (5.47), and (5.48) imply that ˜

B [mm].

We also give the proof of Theorem 2.22.

Proof. Again, we may assume that a = a0. We interchange A and B and consider the

functional

˜ (

) = −

(

).

Then

G u

G u

˜0

=

B

˜

= −

B

= −

0 < ∞

a

sup G

inf G

b

and

˜0

= A

˜

= − A

= −

0 > −∞.

b

inf G

sup G

a

Moreover,

R+β

a

b

=

a

0 −

b

t

dt

˜0

− ˜0

0

<

β

ψ(

)

,

where

R ≤ d = d( A , B).

	48							5. The Method Using Flows
	Since	= {			˜	˜	}
	A		u					,
					A : G(u) < a0

we can apply Theorem 2.21 to conclude that for each δ > 0, there is a u E such that

(5.50)	˜0 −		≤ ˜	≤ ˜0	+		˜	(u)		< ψ(d(u, A )).
	b	δ	G(u)	a		δ,	G

This implies (2.42).

Theorems 2.24 and 2.19 follow from the same arguments used in the proof of Theorem 2.4 (in the case of Theorem 2.24 we substitute −G for G). Now we give the proof of Theorem 2.18.

Proof. If the conclusion of the theorem were not true, then there would exist ε > 0 and ψ(r ) such that

(5.51)		G	(u)	≥	ψ( u )
whenever
(5.52)	|G(u) − b| ≤ 3ε.

Let

Q = {u E : |G(u) − b| ≤ 2ε},

Q1 = {u E : |G(u) − b| ≤ ε},

Q2 = B\Q,

and

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

Note that η(u) is locally Lipschitz continuous and satisfies

	η(u) = 1,	u Q1,
(5.53)	η(u) = 0,	u ¯ 2,
		Q
	η(u) (0, 1),	otherwise.

Let Y (u) be a locally Lipschitz continuous pseudo-gradient for G satisfying

(5.54)	G	(u)Y (u)	≥	α		G	(u) ,	Y (u)	≤	1,

for some α > 0. Let W (u) = η(u)Y (u). Then W (u) is a locally Lipschitz continuous mapping of E into itself such that

(5.55)	G (u)W (u) ≥ 0, W (u) ≤ 1.
Let σ (t)u be the solution of the initial-value problem
(5.56)	dσ (t)u/dt = W (σ (t)u), σ (0)u = u.

5.5. Theorem 2.21

49

The solution of (5.56) exists for every u E and t ≥ 0 by Theorem 4.5. By (5.55),

(5.57)			σ (t)u − u ≤ t,
and by (5.54) and (5.56),
(5.58)	dG(σ (t)u)/dt	=	G (σ (t)u)W (σ (t)u)		≥	αη(σ (t)u)		G	(σ (t)u) .
This implies that
(5.59)	G(σ (t1)u) ≤ G(σ (t2)u), 0 ≤ t1 < t2.
By hypothesis,
(5.60)			G(u) ≥ b, u ∂ ω0.
Let
(5.61)			M =	sup u
				u ∂ω0
and let T be given by
(5.62)			T = d(0, ∂ ω0)/2.
Take ε > 0 so small that
(5.63)			2ε < α	T +M
				ψ(t) dt.
			M
By (5.57) and (5.62),
(5.64)	σ (t)u = 0, u ∂ ω0, 0 ≤ t ≤ T.
If u ∂ ω0, but u Q1, we must have
(5.65)			G(u) > b + ε
since we cannot have
(5.66)			G(u) < b − ε
by (5.60). Thus, by (5.59),
(5.67)	G(σ (t)u) ≥ G(u) > b + ε,			u ∂ ω0, u Q1, 0 ≤ t ≤ T.

On the other hand, if u ∂ ω0 ∩ Q1, let t1 be the largest number not greater than T such that σ (t)u Q1 for 0 ≤ t ≤ t1. If t1 < T, then for δ > 0 sufficiently small,

G(σ (t1 + δ)u) ≥ G(σ (t1)u) ≥ b

50						5. The Method Using Flows
and since σ (t1 + δ)u Q1, we must have
G(σ (t1 + δ)u) > b + ε.
Consequently,
(5.68)	G(σ (T )u) > b + ε.
If t1 = T , then by (5.51), (5.53), (5.57), (5.58), (5.61),and (5.63),
					T	G (σ (t)u)
G(σ (T )u)	−	G(u)	≥	0				dt
					T
			≥ α		0	ψ( σ (t)u ) dt
			≥ α		T	ψ( u + t) dt
					0
					T
			≥ α		0	ψ(M + t) dt
			= α		T +M
					M	ψ(r )dr

> 2ε.

Thus, (5.68) holds as well in this case by (5.60). Consequently, (5.68) holds for all u ∂ ω0. Let ωT be the set of points σ (T, u) where u ω0. Then ωT is a bounded, open set in E with ∂ ωT consisting of those points of the form σ (T )u, u ∂ ω0 by (5.57) and the continuous dependence of σ (T )u on u. Since 0 is in Q2 and η ≡ 0 there, we see that σ (T )0 = 0 by the uniqueness of solutions of (5.56). Thus, 0 = σ (T )0 ωT . Thus, ∂ ωT K, and

(5.69) G(u) > b + ε, u ∂ ωT ,

by (5.68). But (5.69) contradicts (2.27). This completes the proof.

In proving Theorem 2.20, we merely replace G by −G and follow the proof of Theorem 2.18.

Chapter 6

Finding Linking Sets

6.1 Introduction

As we saw in Chapter 3, there are several sufficient conditions that imply that a set A links a set B in the sense of Definition 2.9. Our goal is to find all subsets that link according to this definition. At the present, we are very close to achieving this goal.

In the present chapter we define two relationships that are close to each other. The stronger one is sufficient for linking, while the weaker one is necessary. We use the following maps.

Definition 6.1. We shall say that a map ϕ : E → E is of class if it is a homeo-

morphism onto E and both ϕ, ϕ−1 are bounded on bounded sets. If, furthermore, ϕ, ϕ−1 C1(E; E), we shall say that ϕ U .

Definition 6.2. We shall say that a bounded set A is chained to a set B if A ∩ B = φ and

(6.1)	x		B ϕ(x) ≤ sup ϕ(x) , ϕ	U .
	inf
			x A

Definition 6.3. We shall say that a bounded set A is strongly chained to a set B if A ∩ B = φ and (6.1) holds for every ϕ .

Definition 6.4. For A E, we define
U ( A)	=	ϕ	−1(BR) : ϕ		U , R > sup ϕ(u)		,
K

u A

where

BR = {u E : u < R}.

M.Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_6, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009