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"Proof of the existence of the invariant measure
We divide the proof into four steps:
Step 1. Let q > 1, α ∈ (1
q,1], denote
Mαφ =R1
0(1 − s)α−1S(1 − s)φ(s)ds
ˆ
Mαφ =R1
0(1 − s)α−1ˆS(1 − s)φ(s)ds
then operator Mαandˆ
Mαare compact.
Proof. For ? ∈ (0,1), φ ∈ Lq(0,1;H), define
(3.43)
M?
αφ =
Z1−?
0
(1 − s)α−1S(1 − s)φ(s)ds
Then
(3.44)
M?
αφ = S(?)
Z1−?
0
(1 − s)α−1S(1 − ? − s)φ(s)ds
By Proposition 3.7, we obtain that M?
αis a compact operator.
By the H¨ older inequality, we have
(3.45)
|Mαφ − M?
αφ| = |
Z1
1−?
(1 − s)α−1S(1 − s)φ(s)ds|
≤ (
Z1
1−?
(1 − s)(α−1)q0kS(1 − s)kq0ds)1/q0(
Z1
1−?
|φ(s)|q
Hds)1/q
≤ M(
?(α−1)q0+1
(α − 1)q0+ 1)1/q0|φ|Lq(0,1;H)
where (α − 1)q0+ 1 =
αq−1
q−1> 0 and M=sups∈[0,1]kS(s)k. Then M?
α→ Mαas ? → 0 in
the operator norm so that Mαis compact. Employing Proposition 3.8, the proof ofˆ
Mαis
analogous.
As a consequence, for every r > 0 and α ∈ (1
q,1] ,
(3.46) K(r) = {S(1)u0+ M1φ1+ˆ
M(t)φ2+ Mαφ3: |u0|H< r,|φi|Lq([0,1];H)< r,i = 1,2,3}
is relatively compact in H."