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

14.3. Local estimates

169

Lemma 14.6. If (14.16) holds, then for each ρ > 0 sufficiently small, there is a positive ε such that

(14.50)	G(v + y) ≤ −ε v 2, v N −1, y E(λ ), v + y ≤ ρ.
Proof. Since E(λ ) L∞( ), there is a ρ > 0 such that
(14.51)	y D ≤ ρ implies |y(x)| ≤ δ/2, y E(λ ),

where δ is the constant given in (14.16). Let w = v + y, where v N −1, y E(λ ). If

(14.52)			w D ≤ ρ					and	|w(x)| ≥ δ,
then
(14.53)		δ ≤ |w(x)| ≤ |v(x)| + |y(x)| ≤ |v(x)| + δ/2.
Consequently, (14.52) implies
(14.54)					|y(x)| ≤ δ/2 ≤ |v(x)|
and
(14.55)					|w(x)| ≤ 2|v(x)|.
By (14.10),
(14.56)				|F(x, t)| ≤ C(t2 + |t|).
Thus,
(14.57)	G(w)	≤	w 2D	− λ				w2dx + C		(w2 + |w|)dx
							|w|<δ		|w|>δ
		≤ w 2D − λ w 2 + C							(w2 + |w|)dx
									|w|>δ
		≤ v 2D − λ v 2 + 4C (1 + δ−1)								v2dx
									2|v |>δ
		≤ v 2D − λ v 2 + C							|v|σ dx
		≤	1 −	λ			+ C	v σD−2 v 2D ,
				λ	−1

where σ > 2. If we take ρ sufficiently small, this implies (14.50).

Lemma 14.7. If (14.20) holds, then for each ρ > 0 sufficiently small, there is an ε > 0 such that

(14.58) G(w + y) ≥ ε w 2, w M , y E(λ ), w + y ≤ ρ.

170	14. Type (II) Regions

The proof of Lemma 14.7 is similar to that of Lemma 14.6 and is omitted.

Lemma 14.8. Let θ be a continuous map from M −1 to N −1 satisfying (14.32), and define

(14.59)

H (v + w) = v + w + θ (w), v N −1, w M −1.

If (14.16) holds and G is given by (14.48), then we have the following alternative: Either (a) for each ρ > 0 sufficiently small, there is an ε > 0 such that

(14.60)	G(H v) ≤ −ε, v N ∩ ∂ Bρ ,
or (b)	there is a y E(λ )\{0} such that
(14.61)	Ay = λ y = f (x, y).

Proof. For v N , write v = v + y, where v N −1 and y E(λ ). Then

(14.62) H v = v + θ (y), H v ≤ C v .

Thus, by Lemma 14.6, for each ρ > 0 sufficiently small, there is an ε > 0 such that

(14.63) G(H v) ≤ −ε v + θ (y) 2, v N ∩ Bρ .

In particular, G(H v) ≤ 0 for such v. Assume that option (a) does not hold. Then there

is a sequence vk = vk	+ yk , vk N −1, yk E(λ ) such that
(14.64)	G(H vk) → 0, vk = ρ.

Since H vk = vk + θ (uk ), we see from (14.63) that vk + θ (yk ) → 0. Moreover, there is a renamed subsequence such that yk → y0 in E(λ ). By continuity, θ (yk ) → θ (y0)

and vk → −θ (y0). If y0 = 0, then vk → 0, contradicting the fact that vk = ρ. Thus,
y0 = 0 and H vk → y0. This implies
(14.65)	G(y0) = 0.
If ρ > 0 is such that (14.51) holds, then (14.16) implies that
(14.66)	λ y0(x)2 ≤ 2F(x, y0(x)),	x .
But (14.65) says
(14.67)	{λ y0(x)2 − 2F(x, y0(x))}dx = 0.
This together with (14.66) implies that
(14.68)	λ y0(x)2 ≡ F(x, y0(x)),	x .
Let ζ (x) be any function in C0∞( ). Then, for t > 0 sufficiently small,
(14.69)	t−1λ [(y0 + tζ )2 − y02] ≤ t−12[F(x, y0 + tζ ) − F(x, y0)].

14.3. Local estimates		171
Taking the limit as t → 0, we have
(14.70)	λ y0(x)ζ (x) ≤ f (x, y0)ζ (x),	x .
From this we conclude that
(14.71)	λ y0(x) ≡ f (x, y0(x)),	x .

Since y0 E(λ ), this implies that y0 satisfies (14.61).

Lemma 14.9. Let τ be a continuous map from N to M such that (14.41) holds and define

(14.72)

H (v + w) = v + w + τ (v),

v N , w M .

If F(x, t) satisfies (14.20), then the following alternative holds: Either

(a) for each

ρ > 0 sufficiently small, there is an ε > 0 such that

(14.73)

G(H w) ≥ ε,

w M −1 ∩ ∂ Bρ ,

or (b)

there is a y E(λ )\{0} such that (14.61) holds.

Proof. The proof is similar to that of Lemma 14.8. If w = w + y, w

M , y E(λ ),

then H w = w + τ (y), H w D ≤ C w D and

(14.74)

G(H w) ≥ ε w + τ (y) 2,

w M −1 ∩ Bρ ,

for ρ sufficiently small by Lemma 14.7. If (a) does not hold and wk

= wk + yk , wk

M , yk

E(λ ), wk

∂ Bρ , and G(H wk ) → 0, then wk

+ τ (yk )

→ 0, yk →

y1, τ (yk ) → τ (y1), and wk → −τ (y1) for a renamed subsequence. If y1

= 0, then

wk → 0, contradicting the fact that wk D = ρ. Now H wk → y1 and

(14.75)

G(y1) = 0.

If ρ is sufficiently small, (14.51) holds and

(14.76)

2F(x, y1(x)) ≤ λ y1(x)2,

x ,

by (14.20). This together with (14.75) implies that

(14.77)

2F(x, y1(x)) ≡ λ y1(x)2,

x .

This and (14.20) imply that

(14.78)

t−12[F(x, y1

+

tζ )

−

F(x, y1)]

≤

t

−1λ [(y1

+

tζ )2

−

y2

]

1

for t small and ζ

C

0

∞( ). This yields

(14.79)

f (x, y1) ≡ λ y1(x),

x

and y1 is a solution of (14.61).

172	14. Type (II) Regions

14.4 The solutions

We can now present the proof of Theorems 14.2 and 14.3. We begin with the proof of Theorem 14.2.

Proof. We shall study G given by (14.48) and look for solutions of (14.49). By (14.13) we have for w M −1, v0, v1 N −1, v = v1 − v0,

(14.80)	1	(G (w + v1) − G (w + v0), v)
	2
		= v 2D − ( f (w + v1) − f (w + v0), v)
		≤	{λ −1v2 − [ f (x, w + v1) − f (x, w + v0)]v}dx.

The right-hand side will be negative unless v ≡ 0 in view of (14.13). Thus, there is a

unique solution ˆ of

θ (w)

(14.81)

G

θ (w))

=

sup

G

(w

+ v)

(w + ˆ

v N −1

ˆ

(Lemma 13.5). Moreover, θ is continuous and satisfies (14.32). Define

(14.82)

H (v

+

w)

=

v

+ w +

ˆ

v

N

, w

M

−1,

θ (w),

which is continuous as well. Let

(14.83)

A

R = [M ∩ ¯ R

]

{

u

= w +

sy

0

:

w

,

s

≥

0

, =

R

}

B

M

u

where y0 is the element of E(λ )\{0} given in Lemma 14.4. By Example 7 of Section 3.5, AR links N ∩ ∂ Bρ [hm] whenever 0 < ρ < R. In view of Proposition 3.11, H AR links H [N ∩ ∂ Bρ ] [hm]. Let θ be the map described in Section 14.2 satisfying (14.32)–(14.36). Then, by (14.81),

(14.84) G(w

+

sy

0

+ ˆ

+

0

))

≥

G(w

+

sy

0 +

θ (w

+

0

))

θ (w

sy

sy

=I (w + sy0 + θ (w + sy0), a, b)

−2 P(x, w + sy0 + θ (w + sy0))dx

	≥	−	W1(x) dx
	≡
		−B1
for w M , s ≥ 0. Thus,
(14.85)	G(u) ≥ −B1,		u H AR.

14.4. The solutions

173

We can do a bit better on that portion of AR contained in M . In fact, (14.16) implies

(14.86)

G(w

+

sy

θ (w))

≥

G

2

−

2

F

x

dx

0 + ˆ

(w) = w D

(

, w)

≥w 2D − λ +1 w 2

≥0

for w M . On the other hand, we see from Lemma 14.8 that

(14.87)

G(u) ≤ −ε < 0, u H [N ∩ ∂ Bρ ],

for ρ > 0 sufficiently small unless there is a y E(λ )\{0} satisfying (14.61). But in the latter case the theorem is proved. Thus, we may assume that (14.87) holds. We can now apply Theorem 3.15 to conclude that there is a sequence {uk } D such that

(14.88)	G(uk ) → c,	−∞ < c ≤ −ε,					G (uk ) → 0.
Thus
(14.89)	uk 2D − 2		F(x, uk )dx → c
and
(14.90)	(uk , v)D − ( f (uk ), v) → 0,							v D.
If
(14.91)		ρk = uk D → ∞,
let u˜k = uk /ρk . Then u˜k D		= 21, and there is a renamed subsequence such that
u˜ k → u˜ weakly in D, strongly in L ( ), and a.e. in . By (14.56),
(14.92)	|F(x, uk )|/ρk2 ≤ C(u˜k2 + |u˜k |/ρk )
and the right-hand side converges to Cu˜2 in L1( ). Moreover,
(14.93)	2F(x, uk )/ρ2	b(u+)2	+	a(u−)2 as k				→ ∞			a.e. in .
	k →	˜			˜
Hence,
(14.94)	2 F(x, uk )dx/ρ2				b u	+ 2	+		a u	− 2.
			k →		˜				˜

But the left-hand side of (14.94) converges to 1 by (14.89). Thus, u˜ ≡ 0. Now, by (14.10),

(14.95)	|	f (x, u	k	) /ρ		C( u	ρ	−1)
				|	k ≤	| ˜k | +		k

174

14. Type (II) Regions

and the right-hand side converges to C|u˜| in L2( ). Since

(14.96)

f (x, u

)/ρ

k →

bu

+

au− a.e. in as k

→ ∞

,

k

˜

−

˜

we have

(14.97)

(u, v)D

=

(bu+

−

au−, v), v

D,

˜

˜

˜

by (14.90). This says that u˜ satisfies (14.3). Since (a, b)

, we must have u˜ ≡

0, contradicting the conclusion reached earlier. Thus, (14.91) cannot hold. But then standard arguments imply that (14.1) has a solution u satisfying G(u) = c ≤ −ε. This shows that u ≡ 0, and the proof is complete.

Proof of Theorem 14.3. In this case we take

v

≥

, =

},

(14.98)

A

R = [N −1 ∩

¯ R

]

{

u

= v +

sy

1

:

−1,

s

0

R

B

N

u

where y1 is the element of E(λ )\{0} given by Lemma 14.5. By (14.18), there is a continuous map τˆ from N to M such that (14.41) holds (for τˆ) and

(14.99)	G(v + τˆ(v)) = w M G(v + w).
	inf
If
(14.100)	H (v + w) = v + w + τˆ(v), v N , w M ,

then H is a homeomorphism of D onto itself. Thus, H AR links H [N 0 < ρ < R (Example 7 of Section 3.5 and Proposition 3.11). Let τ described in Section 14.2 satisfying (14.41)–(14.45). Then we have

∩ ∂ Bρ ] for be the map

G(v + sy1 + τˆ (v + sy1)) ≤ G(v + sy1 + τ (v + sy1))

=I (v + sy1 + τ (v + sy1), a, b)

−2 P(x, v + sy1 + τ (v + sy1))dx

	≤
		B1
for v N −1, s ≥ 0. This implies
(14.101)	G(u) ≤ B1, u H AR.
Moreover, we have
	G(v + τˆ(v)) ≤ G(v) = v 2D − 2		F(x, v) dx
	≤ v 2D − λ −1 v 2 ≤ 0, v N −1.
Also, Lemma 14.9 implies that
(14.102)	G(w) ≥ ε > 0,	w H [M −1 ∩ ∂ Bρ ],

14.5. Notes and remarks

175

for ρ > 0 sufficiently small unless there is a y E(λ )\{0} satisfying (14.61). The latter option implies the conclusion of the theorem. We may therefore assume that (14.102) holds. We can now apply Theorem 3.15 to conclude that there is a sequence {uk } D such that

(14.103) G(uk ) → c, ε ≤ c < ∞, G (uk ) → 0.

We now follow the proof of Theorem 14.2 to conclude that there is a solution of (14.1) satisfying G(u) = c ≥ ε. This implies that u ≡ 0, and the proof is complete.

14.5 Notes and remarks

Since the work of Fuˇc´ık , many authors have studied other problems of the form (14.1) when (14.2) holds (cf., e.g.,[29]–[30], [52], [54], [55], [58], [56], [63], [64], [72], [73], [80], [81], [82], [90], [105], [122]–[116] and the references quoted in them. In [72], [73], [33], [30] the problem (14.1) was considered for (a, b) Q when λ is a simple eigenvalue. If ϕ is a corresponding eigenfunction, then it was shown in [72] that for each s R, there are a unique function us and a unique constant Cs (a, b) such that

(14.104)	Aus = bus+ − aus− + Cs (a, b)ϕ,		(us , ϕ) = s,
with
	Cs = Cs (a, b) =	sC1,	s ≥ 0
	=	sC−1,	s < 0.

Clearly, (a, b) iff C1C−1 = 0. The region where C1 > 0, C−1 > 0 is below both curves, the region C1 < 0, C−1 < 0 is above the curves, and the region C1C−1 < 0 is between the curves. It is in this sense that problems for regions of type (II) were considered by these authors for simple eigenvalues.

Theorems 14.2 and 14.3 were given in [116].

Chapter 15

Weak Sandwich Pairs

15.1 Introduction

In Chapter 7 we discussed the situation in which one cannot find linking sets that separate the functional, i.e., satisfy (7.1). Are there sets such that the opposite of (7.1) will imply (3.31)? More precisely, are there sets A, B such that (7.3) implies that there is a sequence satisfying (7.4)? This was answered in the affirmative in Chapter 7. Such pairs exist. This has led to

Definition 15.1. We say that a pair of subsets A, B of a Banach space E forms a sandwich, if, for any G C1(E, R), inequality (7.3) implies the existence of a PS sequence (7.4).

It follows from Theorem 3.17 that M, N form a sandwich pair if one of them is finite-dimensional. (Note that m0 ≤ m1.) This is a severe drawback in many applications.

The purpose of the present chapter is to find a counterpart of sandwich pairs that deals with the case when both sets in the pair are infinite-dimensional. In order to do this, we required weak-to-weak continuous differentiability of the functional as we did in Theorem 15.2. We call such pairs weak sandwiches.

The purpose of this chapter is to solve systems of equations of the form

(15.1)	Av = f (x, v, w)
(15.2)	Bw = g(x, v, w),

where A, B are linear partial differential operators. The variational approach to solving such a system is to study a functional G(u) chosen so that the system is equivalent to

(15.3)

G (u) = 0.

(In very many cases, such a functional can be found.) The sandwich theorem, Theorem 3.17, is very useful in dealing with equations or systems for which the corresponding functional is semibounded in one of the directions only on a subspace of finite

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_15, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

178	15. Weak Sandwich Pairs

dimension. However, there are many systems for which this is not the case. On the other hand, the theorem is probably not true if both subspaces are infinite-dimensional. In the present chapter we shall show that the theorem is indeed true if we require more than mere continuous differentiability of the functional. The requirement we have chosen is present in many applications. It is the weak-to-weak continuous differentiability defined in Chapter 10 (cf. Definition 10.1). For such functionals, we have

Theorem 15.2. Let N be a closed subspace of a Hilbert space E and let M = N . Let G be a weak-to-weak continuously differentiable functional on E such that

(15.4)	m	0	:= w M	G	(w) = −∞
			inf
and
(15.5)	m1 := sup G(v) = ∞.
			v N

Then there are a constant c R and a sequence {uk } E such that (15.6) G(uk ) → c, m0 ≤ c ≤ m1, quG (uk ) → 0.

We shall prove Theorem 15.2 in the next section, where we introduce weak sandwich pairs. Applications will be given in Section 15.3.

15.2 Weak sandwich pairs

We now introduce the corresponding definition for the case when both sets A, B are infinite-dimensional.

Definition 15.3. We shall say that a pair of subsets A, B of a Banach space E forms a weak sandwich pair if, for any weak-to-weak continuously differentiable G C1(E, R), the inequality

(15.7)	−∞ <	b	0	:= B	≤	a	0	=	A	< ∞
				inf G				:	sup G
implies that there is a sequence {uk } satisfying									G (uk ) → 0.
(15.8)	G(uk ) → c,			b0 ≤ c ≤ a0,
We have

Theorem 15.4. Let E be a separable Hilbert space, let N be a closed subspace of E, and let p be any point of N. Let F be a Lipschitz continuous map of E onto N such that F|N = I,

(15.9) F(g) − F(h) ≤ K g − h , g, h E,

and, 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

(15.10) v S0, w S F(v + w) S0.

Then A = N, B = F−1( p) form a weak sandwich pair.

15.2. Weak sandwich pairs

179

Proof. Assume that the theorem is false. Let G be a weak-to-weak continuously differentiable functional on E satisfying (15.7), where A, B are the subsets of E specified in the theorem, such that there is no sequence satisfying (15.8). Then there is a positive number δ > 0 such that

(15.11)

G (u) ≥ 2δ

whenever u belongs to the set

(15.12)

ˆ

= {

u

E : b

0 −

2

δ ≤

(

u

) ≤

a

0 +

2

δ}.

E

G

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

(15.13) |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

k=1

k2

˜

ˆ

=

G (u)/

We denote E equipped with this norm by E. For u

E, let h(u)

G (u) .

Then, by (15.11),

(15.14)

(G (u), h(u))

≥

2δ,

u

ˆ

E.

Let

T = (a0 − b0 + 4δ)/δ,

(15.15)

BR = {u E : u < R},

R = sup u + T,

ˆ

=

∩

ˆ ,

¯ R

B

B

E

where is a bounded, open subset of N containing the point p such that

(15.16)

ρ (∂ , p) > K T + δ,

and

ρ

is the distance in E

.

For each u

B

E neighborhood W

(

u

)

of u such

ˆ , there is an

˜

that

∩

ˆ

(15.17)

(G (v), h(u)) > δ,

v

W (u)

B.

Otherwise there would be a sequence {vk }

B such that

ˆ

(15.18)

|vk − u|w → 0

and

(G (vk ), h(u)) ≤ δ.