568 CHAPTER 19. HILBERT SPACES
where θ n (h) ∈ (0,h). Then taking a limit as h→ 0 and using the dominated convergencetheorem, the limit of the difference quotient is
∞
∑n=1
Antn−1
(n−1)!= A
∞
∑n=1
An−1tn−1
(n−1)!= A
∞
∑n=0
(At)n!
n
Thus ∑∞n=0
(At)n!
nsatisfies the differential equation. It clearly satisfies the initial condition.
Hence it equals S (t).Note that as a consequence of the above argument showing that T and S are the same, it
follows that T (t)A=AT (t) so one obtains that if the generator is a bounded linear operator,then the semigroup commutes with this operator.
When dealing with differential equations, one of the best tools is Gronwall’s inequality.This is presented next.
Theorem 19.12.6 Suppose u is nonnegative, continuous, and real valued and that
u(t)≤C+∫ t
0ku(s)ds, k ≥ 0
Then u(t)≤Cekt .
Proof: Let w(t)≡∫ t
0 ku(s)ds. Then
w′ (t) = ku(t)≤ kC+ kw(t)
and so w′ (t)− kw(t)≤ kC which implies ddt
(e−ktw(t)
)≤ kCe−kt . Therefore,
e−ktw(t)≤Ck∫ t
0e−ksds =Ck
(1k− 1
ke−kt
)so w(t)≤C
(ekt −1
). From the original inequality, u(t)≤C+w(t)≤C+Cekt−C =Cekt .
19.13 Fractional Powers of OperatorsLet A ∈L (X ,X) for X a Hilbert space, A = A∗. We want to define Aα for α ∈ (0,1) insuch a way that things work as they should provided that (Ax,x)≥ 0.
If A ∈ (0,∞) we can get A−α for α ∈ (0,1) as
A−α ≡ 1Γ(α)
∫∞
0e−Atta−1dt
Indeed, you change the variable as follows letting u = At,∫∞
0e−Atta−1dt =
∫∞
0e−u( u
A
)a−1 1A
du
=∫
∞
0e−uuα−1A1−α 1
Adu = A−α
Γ(α)
Next we need to define e−At for A ∈L (X ,X).