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

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20

3. Examples of Minimax Systems

and

ψ(t)u = u, t [0, 1], u A,

then ψ(1)K K. He calls such a collection a homotopy-stable family with extended boundary A. He proves

Theorem 3.7. If K is a homotopy-stable family with extended closed boundary A, B is a closed subset of E satisfying (1.1), and G is a C1-functional that satisfies

sup G a inf G,

A B

where the quantity a is given by (2.6), then there is a PS sequence satisfying (1.4).

We shall prove

Theorem 3.8. Under the same assumptions, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence satisfying (2.9). In particular, there is a Cerami sequence satisfying (1.5).

Proof. It is easy to show that such a collection K is indeed a minimax system. Indeed, if K K and ϕ ( A), then

ψ(t)u = tϕ(u) + (1 t)u

satisfies the stipulations above, and consequently ϕ(K ) K. Since each of the members of K is compact, it follows that every C1-functional is Lipschitz continuous on

some set Kρ defined by (2.25) for ρ > 0 sufficiently small. Moreover, if σ (t) C(R+ × E, E) is such that σ (0) = I, one has S(K ) K, where

S(u) = σ (d(u, A))u, u E.

This follows from the fact that S(t)u = σ (td(u, A))u is in C([0, 1] × E, E) and satisfies

S(0)u = u, u E,

and

S(t)u = u, t [0, 1], u A.

Consequently, S(K ) = S(1)K K by hypothesis. It now follows that the conclusion of Theorem 2.14 holds.

Note. It is not required to have a satisfy

a inf G.

B

If it does, one obtains additional information as described in [74].

3.5. Examples of linking sets

21

3.5 Examples of linking sets

We now discuss various subsets of a Banach space E with respect to linking. First we have

Proposition 3.9. [122] Let A, B be two closed, bounded subsets of E such that E\ A is path connected. If A links B [hm], then B links A [hm].

The next proposition gives a very useful method of checking the linking of two sets.

Proposition 3.10. [122] Let F be a continuous map from E to Rn, and let Q E be such that F0 = F|Q is a homeomorphism of Q onto the closure of a bounded open subset of Rn. If p , then F01(∂ ) links F1( p) [hm].

Proposition 3.11. [122] If H is a homeomorphism of E onto itself and A links B [hm], then H A links H B [hm].

The following examples were given in [122].

Example 1. Let B be an open set in E, and let A consist of two points e1, e2 with

e

1

 

 

B. Then A links

B [hm].

B links A [hm] as well if

B is bounded.

 

 

B and e2 / ¯

 

 

 

Example 2. Let M, N be closed subspaces such that dim N < and E = M N. Let

(3.9)

BR = {u E : u < R}

and take A = BR N,

B = M. Then A links B [hm].

Example 3. We take M, N as in Example 2. 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 and B link each other [hm].

Example 4. 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 [hm].

Example 5. This is the same as Example 4 with A replaced by A = BR N.

Example 6. Let M, N be as in Example 2. 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,

22

3. Examples of Minimax Systems

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 and B link each other [hm].

Example 7. Let M, N be as in Example 2.

Let v0 be in B1 N and write N =

{v0} N . Let

N

,

Q

B

N

,

A = Bδ

 

 

= ¯δ

 

and

 

 

 

 

 

 

B = {w M : w R} {w + sv0 : w M, s 0, w + sv0 = R}, where 0 < δ < R. Then A and B link each other [hm].

Example 8. Let M, N be closed subspaces of E, one of which is finite-dimensional,

and such that

E = M N.

If

BR := {u E : u < R}, then M BR links N [hm] for each R > 0.

Example 9. Let M, N be closed subspaces of E such that

E = M N,

with one of them being finite-dimensional. Let w0 be an element of M\{0}, and let 0 < δ < r < R. Take

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

B= Br M, 0 < δ < r < R.

Then A and B link each other [hm].

Example 10. Let M, N be closed subspaces of E such that

E = M N,

with one of them being finite-dimensional. Let w0 be an element of M\{0}, and let 0 < r < R,

A= {w M : w = R},

B= {v N : v r } {u = v + sw0 : v N, s 0, u = r }.

Then A links B [hm].

3.5. Examples of linking sets

23

Example 11. Let M, N be as in Example 2. 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)s = 0

(b)s = 2R0

where 0 < δ < R0. Then A links B [hm].

Example 12. Let M, N be as in Example 2. 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 > 0,

where 0 < δ < . Then A links B [hm].

Example 13. Let M be a closed subspace of a Hilbert space E with complement N {v0}, where v0 is an element in E having unit norm, and let δ be any positive number. Let ϕ(t) C1(R) be such that

0 ϕ(t) 1, ϕ(0) = 1,

and

ϕ(t) = 0, |t| ≥ 1.

Let

F(v + w + sv0) = w + [s + δ δϕ( v 22)]v0, v N, w M, s R.

Assume that one of the subspaces M, N is finite-dimensional. Take

A = [M {v0}] BR

and

B = {v + r v0 : v N, r = δϕ( v 22)}.

Then A links B [hm] provided 0 < δ < R.

Proposition 3.12. [122] If K is any subset of a bounded open set E, then

links K [hm].

24

3. Examples of Minimax Systems

3.6 Various geometries

We now apply the theorems of the preceding sections to various geometries in Banach space. As before, we assume that G C1(E, R) and that ψ satisfies the hypotheses of Theorem 2.4.

Theorem 3.13. Assume that there is a δ > 0 such that

(3.10)

G(0) α G(u),

u Bδ ,

and that there are a R0 < and a ϕ0 B1 such that

(3.11)

G( Rϕ0) γ ,

R > R0.

Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a

sequence {uk } E such that

 

 

 

 

 

G

 

 

 

 

 

 

 

 

 

(3.12)

 

G(uk )

c, α

c

γ ,

(uk )/ψ(

 

uk )

0.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Proof. We take A = {0, Rϕ0}, B = Bδ . Then A

= { Rϕ0}. Note that a given by

(2.6) is finite for each R since

 

R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a

0 r R G(r ϕ0).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

max

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

≤ ≤

 

 

 

 

 

 

 

 

 

 

We apply Theorem 2.24. We note that in each case

 

 

 

 

 

 

 

 

(3.13)

 

 

 

 

 

 

aR γ ,

R > 0,

 

 

 

 

 

 

 

since the mapping

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(3.14)

 

 

 

 

 

 

 

(s)u = (1 s)u

 

 

 

 

 

 

 

 

(which is in ) satisfies

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(3.15)

 

 

 

 

G( (s)u) γ , 0 s 1,

u A.

 

 

 

This implies (3.13). We replace

ψ(

)

 

 

˜

=

ψ(t

+

δ), which also satisfies the

 

t

 

with ψ (t)

 

 

 

hypotheses of Theorem 2.4. By Theorem 2.24, we can find a sequence satisfying

(3.16)

α

(1/ k)

G(uk )

γ

+

(1/ k), G

(uk )/ψ (d(uk , B))

0.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

˜

 

 

This implies (3.12), since u d(u, B) + δ.

 

 

 

 

 

 

 

 

 

Theorem 3.14. Let M, N be closed subspaces of E such that

 

 

 

(3.17)

 

 

 

 

E = M N,

 

M = E, N = E,

 

 

 

with

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(3.18)

 

 

 

 

 

dim M < or

dim N < .

 

 

 

 

3.6. Various geometries

25

Let G C1(E, R) be such that

 

(3.19)

G(v) γ , v ∂ BR N, R > R0,

and

 

 

(3.20)

 

G(w) α, w M.

Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {uk } E such that

(3.21) G(uk ) c, α c γ , G (uk )/ψ(d(uk , M)) 0.

Proof. This time we take A and B as in Example 2 above. Thus, A links B [hm]. Again, aR given by (2.6) is finite for each R since

a

R

B N G(u).

 

max

 

 

u ¯ R

Again we see that we can apply Theorem 2.24 to conclude that the desired sequence exists.

Theorem 3.15. Let M, N be as in Theorem 3.14, and let G C1(E, R) satisfy

(3.22)

G(v) α,

v N,

(3.23)

G(w) α,

w ∂ Bδ M,

(3.24)

G(sw0 + v) γ ,

s 0, v N, sw0 + v = R > R0,

for some w0 B1 M, where 0 < δ < R0. Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {uk } E such that (3.12) holds.

Proof. Here we take A, B as in Example 3 above. Thus, A and B link each other [hm].

Here

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

Again, for each R, the quantity a given by (2.6) is finite since

aR max G,

Q

where

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

We now apply Theorem 2.24 to conclude that the desired sequence exists.

Theorem 3.16. Let M, N be as in Theorem 3.14, and let v0 B1 N. Take N = {v0} N . Let G C1(E, R) be such that

(3.25)

G(v) γ ,

v ∂ BR N, R > R0,

(3.26)

G(w) α,

w M, w δ,

(3.27)

G(sv0 + w) α,

s 0, w M, sv0 + w = δ,

where 0 < δ < R0. Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {uk } E such that (3.12) holds.

26 3. Examples of Minimax Systems

Proof. We take A, B as in Example 5 above. Thus, A links B [hm]. As before, we note that aR < for each R. Hence, (3.12) holds by Theorem 2.24.

3.7 A sandwich theorem

We now discuss a very useful theorem that allows one to consider functionals that are bounded from below on one subspace and bounded from above on another with no correlation between the bounds. This provides such functionals with the same advantages as those that are semibounded. One drawback is the requirement that one of the subspaces be finite-dimensional. This condition will be removed in Chapter 15 if we assume that the functional satisfies more than the mere continuity of its derivative.

Theorem 3.17. Let N be a closed subspace of a Hilbert space E and let M = N . Assume that at least one of the subspaces M, N is finite-dimensional. Let G be a C1-functional on E such that

(3.28)

m0

:= w M

G

(w) = −∞

 

 

inf

 

 

 

 

 

 

and

 

 

 

 

(3.29)

m1 := sup G(v) = ∞.

 

 

v N

 

 

Let ψ(t) be a positive, nonincreasing, locally Lipschitz continuous function on [0, ) such that (2.8) holds. Then there are a constant c R and a sequence {uk } E such that

(3.30)

G(uk )

c, m0

c

m1, G (uk )/ψ(

 

Puk

)

0,

 

 

 

 

 

 

 

where P is the (orthogonal) projection onto M.

Proof. We may assume dim N < ; otherwise, we can consider G in place of G. Let A be the set BR N, and take B = M, where R > 0 is arbitrary. Then A links B [hm] by Example 2 above. We now apply Theorem 2.21. Note that a0 m1, m0 = b0, and (2.31) holds for R sufficiently large. We also note that

aR sup G m1

BR N

by taking (s)u = (1 s)u, u E. Hence, by Theorem 2.21, for each δ > 0, there is a u E such that

m0 δ G(u) m1 + δ, G (u) < ψ(d(u, B )).

Since this is true for each δ > 0, we obtain the desired conclusion.

3.7. A sandwich theorem

27

An immediate consequence is

Corollary 3.18. Under the hypotheses of Theorem 3.17, there are a constant c R and a sequence {uk } E such that

(3.31)

G(uk ) c, m0 c m1, (1 + Puk )G (uk ) 0,

where P is the (orthogonal) projection onto M.

The following is a consequence of Theorem 2.23.

Theorem 3.19. Under the hypotheses of Theorem 3.17, for any sequence { Rk } R+ such that Rk → ∞, there are a constant c R and a sequence {uk } E such that

(3.32)

G(u

)

c, m

0

c

m

, ( R

u

)

G

(u )

m1 m0

.

 

 

k

 

 

 

1

 

k +

k

 

k

ln(4/3)

Proof. We may assume dim N < . Let Ak be the set BRk N, and take Bk = M. Then, for each k, Ak links Bk [hm] by Example 2 above. We now apply Theorem 2.23. Note that αk = Rk and ak0 m1, m0 = bk0. Take

ψ (t)

=

m1 m0

.

 

k

[2Rk + t] ln(4/3)

Since Rk + d(u, Bk ) u , we see that (3.32) holds for each k.

The following is another consequence of Theorem 2.23.

Theorem 3.20. Let N be a closed subspace of a Hilbert space E, and let M = N . Assume that at least one of the subspaces M, N is finite-dimensional. Let G be a C1-functional on E such that

(3.33)

m0

:= v N w M G(v + w) = −∞

 

 

sup inf

 

and

 

:= w M v N

 

(3.34)

m1

(v + w) = ∞.

 

 

inf sup G

 

Then, for any ε > 0 and any sequence { Rk } R+ such that Rk → ∞, there are a constant c R and a sequence {uk } E such that

(3.35)

G(uk ) c,

m0 ε c m1 + ε,

 

( R

u )

G (u )

m1 m0 + 3ε

.

 

 

 

 

k + k

k

ln(5/4)

Proof. We may assume dim N such that

m0 ε < inf

w M

< . Let ε > 0 be given. Then there is a uε = vε + wε

G(vε + w), sup G(v + wε ) < m1 + ε.

v N

28

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3. Examples of Minimax Systems

Note that αk

= Rk and ak0 m1, m0 = bk0. Take

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ (t)

 

 

 

 

m1 m0 + 3ε

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

= [3Rk + t] ln(5/4)

 

 

 

 

 

 

 

 

Note that

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2Rk

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m1 m0 + 2ε <

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Rk

 

 

ψk (t) dt.

 

 

 

 

 

We now apply Theorem 2.23. Thus, there is a sequence such that

 

 

 

 

 

 

 

 

 

G(uk + uε ) c, m0 ε c m1 + ε,

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m1 m0 + 3ε

 

 

 

 

 

 

[3Rk

+

d(uk , B

)]

 

G

(uk

+

u )

.

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

k

 

 

 

 

ε

 

ln(5/4)

 

 

 

 

 

Since Rk

d(u, B

k

, we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

)

 

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2Rk

+

d(uk , B

)

Rk

 

 

 

uk

 

 

 

u

ε +

uk

 

u

+

uk

 

 

 

 

 

 

 

k

 

 

 

+

 

 

ε

 

when Rk uε . Let hk = uk + uε . Then we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

G(hk ) c, m0 ε c m1 + ε,

 

 

 

 

 

 

 

 

 

 

( Rk

 

 

hk )

 

G

(hk )

m1 m0 + 3ε

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

ln(5/4)

 

 

 

 

 

Since ε was arbitrary, we see that (3.35) holds.

Here are some consequences.

Theorem 3.21. Let G be a C1-functional on E such that

(3.36)

a0 = sup G < .

 

E

If ψ satisfies the hypotheses of Theorem 2.4, then there is a sequence {uk } E such that

(3.37)

G(uk )

a0,

G

(uk )/ψ( uk

)

0.

 

 

 

 

 

 

 

 

The same holds if

 

 

 

 

 

 

 

 

 

(3.38)

 

 

b

0 =

E

 

> −∞,

 

 

 

 

 

inf G

 

 

 

with

 

 

 

 

 

 

 

 

 

(3.39)

G(uk ) b0,

G (uk )/ψ( uk ) 0.

3.8. Notes and remarks

29

Proof. We refer to Theorem 2.24. We take a sequence of points such that G(vk ) >

a0 (1/ k) and a sequence { Rk } such that Rk

> 2 vk and

 

Rk

2βk

1

ck = 2 k

 

+

0, k → ∞,

2βk

ψ(t) dt

where βk = vk . We then take

ψk (t) = ck ψ(t + βk )

in Theorem 2.24. Then (2.53) holds. We used the fact that BRk +βk links {vk } for each k (Theorem 3.12). The conclusion follows since

u d(u, vk ) + βk .

In the second case, we replace G with G.

Corollary 3.22. If (3.36) holds, then there is a sequence satisfying

(3.40)

G(uk )

a0,

(1

+

uk

)G

(uk )

0.

 

 

 

 

 

 

 

 

If (3.38) holds, there is a sequence satisfying

 

 

 

 

 

(3.41)

G(uk )

b0,

(1

+

uk

)G

(uk )

0.

 

 

 

 

 

 

 

 

3.8 Notes and remarks

The results of Section 3.2 are from [136], [114], and [120] (cf. also [122]). Sections 3.5 and 3.6 are from [122]. The results of Section 3.7 come from [143], [109], [108], [129], and [132].

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