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

180

15.

Weak Sandwich Pairs

B is bounded in E

, vk

→

u weakly in E and (10.5) implies that

Since ˆ

(15.19)

(G (vk ), h(u)) → (G (u), h(u)) ≥ 2δ

B be the set B with the inherited

in view of (15.14). This contradicts (15.17). 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., [77]).

(15.20)

Y (u) =

ψτ (

)

(

u

τ ),

u

˜ .

u

h

B

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

(15.21)

Y (u) ≤

ψτ (u) h(uτ ) ≤ 1

and

=

≥

ˆ

(G (u), Y (u))

τ

τ

(15.22)

ψ

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

))

δ,

u

B.

For u

¯

∩ ˆ

, let σ (t)u be the solution of

E

(15.23)

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

0

−

Next, we note that if u

¯ ∩ ˆ

σ (t)u

δ

E, we cannot have

B and G(σ (t)u) > b

for 0 ≤ t ≤ T : for by (15.23), (15.22),

(15.24)

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

(15.25)

G(σ (T )u) − G(u) ≤ −δT = −(a0 − 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

B. To see this, note that

t

(15.26)

u − σ (t)u = zt (u) :=

0

Y (σ (s)u)ds.

By (15.21),

(15.27)

zt (u) ≤ t.

Consequently,

(15.28)

σ (t)u ≤ u + t < R.

15.2. Weak sandwich pairs

181

Thus,

σ ( )

u

ˆ

¯

∩

ˆ

, there is a t

≥

0 such

t

B. We can now conclude that for each u

E

that σ (s)u exists for 0 ≤ s ≤ t and G(σ (t)u) ≤ b0 − δ. Let

¯

∩

ˆ

(15.29)

u

= {

≥

0 : G

(σ ( )

u

) ≤

b

0 − δ},

u

T

: inf t

t

E.

σˆ (t)u =

σ (t)u,

0 ≤ t ≤ Tu,

σ (Tu )u,

Tu ≤ t ≤ T,

¯

∩

ˆ

¯

\

ˆ

ˆ

for u

ˆ

=

u, 0

≤

t

≤

E. For u

E, define σ (t)u

T . Then σ (t)u is continuous

in (u, t), and

ˆ

≤

−

(15.30)

G

T

u

b

δ,

u

.

(σ (

)

)

0

¯

Let

=

ˆ

v

¯

≤

≤

(15.31)

ϕ(v, t)

0

t

T.

Fσ (t)v,

,

¯

×

[0

,

T ] to N. Let

Then ϕ is a continuous map of

}

K

= {

(u, t)

: u

= ˆ

¯

,

t

[0,

T ]

.

σ (t)v, v

Q

Then K

is a compact subset of

E

×

R

t

be any sequence in K

˜

. To see this, let (uk ,

k )

.

Then u

k = σ ( k )vk ,

where

vk

¯ .

Since Q is bounded, there is a subsequence such

t

Q

that vk → v0

weakly in E and t

k → t0 in [0, T ]. Since

Q is convex and bounded,

v0

is

− |

→

¯

in Q and

|

ˆ

E

×

R

we have

¯

vk

v0 w

0. Since σ (t) is continuous in

˜

,

0

ˆ

uk = σˆ (tk )vk σˆ (t0)v0 = u0 K .

Each u

(

u

0)

˜

(

u

0)

B has a neighborhood W

in E and a finite-dimensional subspace S

such that Y

(

u

)

(

u

0) for u

W (u0)

ˆ .

Since

σ (

)

u is continuous in

(

u

,

t

),

for each

S

∩

B

ˆ

t

t

K

E

R and a finite-dimensional

(u0,

0)

, there are a neighborhood W (u0, t0)

˜ ×

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

t

ˆ

E

u

ˆ ,

(15.32)

zˆt (u) := u − σˆ (t)u =

0 Y (σ (s)u)ds,

0,

u

ˆ .

E

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.

, we have z

By hypothesis, there is a finite-dimensional subspace S0

= {0} of N containing p such

that F(v

− ˆt (v)) 0

v

¯

∩

0

¯ ∩

0 ×

[0, T ]

z

S for all

S . We note that

ϕ(u, t) maps

S

into S0. For t in [0, T ], let ϕt (v) = ϕ(v, t). Then

(15.33)

ϕt (v) = p,

v ∂ ( ∩ S0) = ∂ ∩ S0, 0 ≤ t ≤ T.

To see this, note that if v ∂ , then

v − p ≤ v − Fσˆ (t)v + Fσˆ (t)v − p .

182

15. Weak Sandwich Pairs

Hence,

(15.34)

Fσˆ (t)v − p > K T + δ − t K > 0,

v ∂ , 0 ≤ t ≤ T

since

t

(s)v

Fσ (t)v

−

v

≤

K

σ

ds

≤

K t.

ˆ

0

ˆ

(15.35)

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

=

Hence, there is a v

ˆ

=

ˆ

F

−1( p)

B.

such that Fσ (T )v

p. Consequently, σ (T )v

(15.36)

uk → u weakly in E,

there exists a renamed subsequence such that

(15.37)

J (uk ) → J (u) weakly in F.

We have

(15.38)

J (u)−1 ≤ C, u E,

(15.39)

−∞ < b0

:= J B

≤ a0

:= J A

< ∞.

inf G

sup G

Let

G1(u) = G( J u),

u E.

15.2. Weak sandwich pairs

183

J (uk )

→

J (u) weakly

J (uk )

→

J

(u).

;

(15.40)

−∞ < b0

:= J B

= J u J B

(

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 weak sandwich pair, there is a sequence {hk } E such that

(15.41)

G1(hk )

→

c,

b0

≤

c

≤

a0

, G (hk )

→

0.

1

If we set uk = J hk , this becomes

(15.42)

G(uk ) → c,

b0 ≤ c ≤ a0,

G (uk ) J (hk ) → 0.

0 ≤ ϕ(t) ≤ 1, ϕ(0) = 1,

and

ϕ(t) = 0, |t| ≥ 1.

Let

(15.43) F(v + w + sv0) = v + [s + δ − δϕ( w 2/δ2)]v0,

v N, w M, s R.

Then A = N = N {v0}, B = F−1(δv0) forms a weak sandwich pair.

Proof. Define

J (v + w + sv0) = v + w + [s − δ + δϕ( w 2/δ2)]v0,

v N, w M, s R.

(15.44)

f (x, v, w) = ∂ F/∂ v, g(x, v, w) = ∂ F/∂ w

are also Carath´eodory functions satisfying

(15.45)

| f (x, v, w)| + |g(x, v, w)| ≤ C0(|v| + |w| + 1), v, w R,

and

(15.46)

f (x, t y, tz)/t → α+(x)v+ − α−(x)v− + β+(x)w+ − β−(x)w−,

(15.47)

g(x, t y, tz)/t → γ+(x)v+ − γ−(x)v− + δ+(x)w+ − δ−(x)w−

(15.48)

Av = − f (x, v, w)

(15.49)

Bw = g(x, v, w).

(15.50)

− Av = α+v+ − α−v− + β+w+ − β−w−,

(15.51)

Bw = γ+v+ − γ−v− + δ+w+ − δ−w−

is v = w = 0. Our first result is

Theorem 15.8. Assume

(15.52)

2F(x, s, 0) ≥ −λ02 − W1(x),

x , t R,

and

(15.53)

2F(x, 0, t) ≤ μ0t2 + W2(x),

x , t R,

(15.54)

u 2D = ( Av, v) + (Bw, w), u = (v, w) D.

We define

(15.55)

G(u) = b(w) − a(v) − 2

F(x, v, w)dx, u D,

where

(15.56)

a(v) = ( Av, v),

b(w) = (Bw, w).

15.3. Applications

185

Then G C1(D, R) and

(15.57)

(G (u), h)/2 = b(w, h2) − a(v, h1) − ( f (u), h1) − (g(u), h2),

(15.58)

G (u) = 0.

(15.59)

D = M N.

If we define

(15.60)

Lu = 2(−v, w), u = (v, w) D,

then L is a self-adjoint, bounded operator on D. Also,

(15.61)

G (u) = Lu + c0(u),

where

(15.62)

c0(u) = −( A−1 f (u), B−1g(u))

(15.63)

G(0, w) ≥ b(w) − μ0 w 2

−

W2(x) dx,

(0, w) M.

Thus,

(15.64)

M

≥ −

(

x

)

dx

≡

b

0.

inf G

W

On the other hand, (15.52) implies

(15.65)

G(v, 0) ≤ −a(v) + λ0 v 2 +

W1(x) dx,

(v, 0) N.

Thus,

(15.66)

sup G ≤

W1

(x) dx ≡ a0.

N

(15.67)

ρk2 = a(vk ) + b(wk ) ≤ C,