1730 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

Proof: First suppose (−A)−α is compact for all α > 0. Then

Γ(α)(−A)−α =∫ t

0sα−1S (s)ds+

∫∞

tsα−1S (s)ds

=tα

αS (t)−

∫ t

0

sα

αAS (s)ds+ sα−1S (s)A−1|∞t

−(α−1)∫

∞

tsα−2S (s)A−1ds

Now ∣∣∣∣∣∣∣∣ sα

αAS (s)

∣∣∣∣∣∣∣∣≤Csα−1

α

and so the second integral satisfies∣∣∣∣∣∣∣∣∫ t

0

sα

αAS (s)ds

∣∣∣∣∣∣∣∣≤Ctα

α2

Γ(α)(−A)−α = O(

tα

α2

)+

tα

αS (t)

−tα−1A−1S (t)− (α−1)∫

∞

tsα−2S (s)dsA−1

It follows that for t > 0, and ε > 0 given,

S (t) =

(tα

α− tα−1

)−1 (Γ(α)(−A)−α

+(α−1)∫

∞

tsα−2S (s)dsA−1 +O

(tα

α2

))=

(tα

α− tα−1

)−1 (Γ(α)(−A)−α

+(α−1)∫

∞

tsα−2S (s)dsA−1

)+O

(1α

)= Nα +O

(1α

).

where Nα is a compact operator. Now let B be a bounded set in H, ||x|| ≤M for all x∈B andlet η > 0 be given. Then choose α large enough that

∣∣∣∣O( 1α

)∣∣∣∣< η

4+4M . Then there existsa η/2 net, {Nα xn}N

n=1 for Nα (B) . Then consider {S (t)xn}Nn=1 . For x ∈ B, there exists xn

such that ||Nα xn−Nα x||< η/2. Then

||S (t)x−S (t)xn|| ≤ ||S (t)x−Nα x||+ ||Nα x−Nα xn||+ ||Nα xn−S (t)xn||