1724 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

Then with this formula,∣∣∣∣(−A)α S (t)∣∣∣∣ =

∣∣∣∣∣∣(−A)S (t)(−A)−(1−α)∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣ 1Γ(1−α)

∫∞

0s1−α (−A)S (t + s)ds

∣∣∣∣∣∣∣∣≤ N

Γ(1−α)

∫∞

0

s1−α

(t + s)e−δ (s+t)ds

=N

Γ(1−α)

∫∞

t

(u− t)1−α

ue−δuds

≤ NΓ(1−α)

∫∞

t

(1− t

u

)1−α 1uα

e−δuds

≤ NΓ(1−α)

1tα

∫∞

te−δuds =

NΓ(1−α)δ

1tα

e−δ t

≡ Cα

δ

1tα

e−δ t .

this establishes the formula when α ∈ [0,1). Next suppose α = m, a positive integer.

||AmS (t)|| =

∣∣∣∣∣∣∣∣AmS( t

m

)m∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣(AS( t

m

))m∣∣∣∣∣∣∣∣≤ N

tm mm.

This is why the above inequality holds.If α,β > 0, ∣∣∣∣∣∣Aα+β S (t)

∣∣∣∣∣∣ =∣∣∣∣∣∣Aα+β S

( t2

)S( t

2

)∣∣∣∣∣∣=

∣∣∣∣∣∣Aα S( t

2

)Aβ S

( t2

)∣∣∣∣∣∣≤ Cα

tα

Cβ

tβe−2δ t =

Ctα+β

e−δ t

Suppose now that α > 0. Thenα = m+β

where β ∈ [0,1). Then from what was just shown,∣∣∣∣∣∣Am+β S (t)∣∣∣∣∣∣≤ C

tm+βe−δ t .

Next consider 54.3.23. First note that whenever α > 0,

(−A)−α S (s) = S (s)(−A)−α