and take A = ‚ B R © N, B = M. Then A links B weakly. To see this, take =

B R © N, Q = ¯ . For u ∈ E, we write

u = v + w, v ∈ N, w ∈ M,

(10.31)

and take F to be the projection

Fu = Pu = v.

For any ¬nite-dimensional subspace S of E such that F S = {0}, take S0 = P S. Since

F| Q = I and M = F ’1 (0), we see by Theorem 10.2 that A links B weakly.

Example 2. We take M, N as in Example 1. Let w0 = 0 be an element of M, and take

= {v ∈ N : v ¤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

A

= ‚ Bδ © M, 0 < δ < R.

B

Then A links B weakly. Again, we may assume that w0 = 1. Let

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

˜

Then A = ‚ Q in N = N • {w0 }. If u is given by (10.31), we de¬ne

Fu = v + w w0 .

˜

Then F is a continuous map of E onto N , F| Q = I, and B = F ’1 (δw0 ). For any

¬nite-dimensional subspace S of E, take S0 = P S •{w0 }. We can now apply Theorem

10.2 to conclude that A links B weakly.

Example 3. Take M, N as before and let v 0 = 0 be an element of N. We write

N = {v 0 } • N . We take

= {v ∈ N : v ¤ R} ∪ {sv 0 + v : v ∈ N , s ≥ 0, sv 0 + v = R},

A

= {w ∈ M : w ≥ δ} ∪ {sv 0 + w : w ∈ M, s ≥ 0, sv 0 + w = δ},

B

where 0 < δ < R. Then A links B weakly. To see this, we let

Q = {sv 0 + v : v ∈ N , s ≥ 0, sv 0 + v ¤ R}

and reason as before. For simplicity, we assume that v 0 = 1, that E is a Hilbert

space, and that the splitting E = N • {v 0 } • M is orthogonal. If

u = v + w + sv 0 , v ∈ N , w ∈ M, s ∈ R,

(10.32)

10.3. Some examples 113

we de¬ne

= v + s+δ’ δ2 ’ w v0 , w ¤δ

2

F(u)

= v + (s + δ)v 0 , w > δ.

Note that F| Q = I while F ’1 (δv 0 ) is precisely the set B. For any ¬nite-dimensional

subspace S of E containing (δv 0 ), take S0 = F S • {δv 0 }. Hence, we can conclude via

Theorem 10.2 that A links B weakly.

Example 4. This is the same as Example 3 with A replaced by A = ‚ B R © N. The

¯

proof is the same with Q replaced by Q = B R © N.

Example 5. Let M, N be as in Example 1. Take A = ‚ Bδ © N, and let v 0 be any

element in ‚ B1 © N. Take B to be the set of all u of the form

u = w + sv 0 , w ∈ M,

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 links B weakly. To see this, take N = {v 0 } • N .

Then any u ∈ E can be written in the form (10.32). De¬ne

R0

F(u) = R0 ’ max w , |s ’ R0 | v0 + v

R

¯

and Q = Bδ © N. Then F ∈ C(E, N) and F| Q = I . Moreover, A = F ’1 (0). For any

¬nite-dimensional subspace S of E such that F S = {0}, take S0 = P S • {v 0 }. Hence,

A links B by Theorem 10.2.

Example 6. Let M, N be as in Example 1. Let v 0 be in ‚ B1 © N and write N =

¯

{v 0 } • N . Let A = ‚ Bδ © N, Q = Bδ © N, and

B = {w ∈ M : w ¤ R} ∪ {w + sv 0 : w ∈ M, s ≥ 0, w + sv 0 = R},

where 0 < δ < R. Then A links B weakly. To see this, write u = w + v + sv 0 ,

w ∈ M, v ∈ N , s ∈ R and take

F(u) = (c R ’ max{c w + sv 0 , |c R ’ s|})v 0 + v ,

where c = δ/(R ’ δ). Then F is the identity operator on Q, and F ’1 (0) = B. For any

¬nite-dimensional subspace S of E such that F S = {0}, take S0 = P S • {v 0 }. Apply

Theorem 10.2.

114 10. Weak Linking

10.4 Some applications

In this section we apply Theorem 10.2 to semilinear boundary-value problems. Let

be a domain in Rn and let A be a self-adjoint operator in L 2 ( ) having 0 in its resolvent

set.Thus, there is an interval (a, b) in its resolvent set satisfying a < 0 < b. Let f (x, t)

be a Carath´ odory function on — R such that

e

| f (x, t)| ¤ V (x)2 |t| + W (x)V (x), x∈ , t ∈ R,

(10.33)

and

f (x, t)/t ’ ±± (x) as t ’ ±∞,

(10.34)

where V, W ∈ L 2 ( ), and multiplication by V (x) > 0 is a compact operator from

D = D(|A|1/2) to L 2 ( ). Let

∞ a

M= N=

d E(»)D, d E(»)D,

’∞

b

where {E(»)} is the spectral measure of A. Then M, N are invariant subspaces for A

and D = M • N. If

(±+ u + ’ ±’ u ’ )vd x,

±(u, v) = ±(u) = ±(u, u),

(10.35)

then we assume that

±(v) ≥ (Av, v), v ∈ N,

(10.36)

(Aw, w) ≥ ±(w), w ∈ M.

(10.37)

We also assume that the only solution of

Au = ±+ u + ’ ±’ u ’

(10.38)

is u ≡ 0, where u ± = max{±u, 0}. We have

Theorem 10.5. Under the above hypotheses, there is at least one solution of