1722 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

= Γ(1−α)∫

∞

0tα−1S (t)dt = Γ(α)Γ(1−α)(−A)−α

=

(∫ 1

0xα−1 (1− x)−α dx

)(−A)−α =

π

sin(πα)(−A)−α

and so this gives the formula

(−A)−α =sin(πα)

π

∫∞

0λ−α (λ I−A)−1 dλ .

This proves 54.3.20.

Definition 54.3.18 For α ≥ 0, define (−A)α on D((−A)α

)≡ (−A)−α (H) by

(−A)α ≡((−A)−α

)−1

Note that if α,β > 0, then if x ∈ D((−A)α+β

),

(−A)α+β x =((−A)−(α+β )

)−1x =(

(−A)−α (−A)−β)−1

x = (−A)β (−A)α x. (54.3.21)

Next let β > α > 0 and let x ∈ D((−A)β

). Then from what was just shown,

(−A)α (−A)β−α x = (−A)β x

and so(−A)β−α x = (−A)−α (−A)β x

If x ∈D((−A)β

), does it follow that (−A)−α x ∈D

((−A)β

)? Note x = (−A)−β y and so

(−A)−α x = (−A)−α (−A)−β y = (−A)−(α+β ) y ∈ D((−A)α+β

).

Therefore, from 54.3.21,

(−A)β−α x = (−A)β−α (−A)α((−A)−α x

)= (−A)β (−A)−α x.

Theorem 54.3.19 The definition of (−A)α is well defined and (−A)α is densely definedand closed. Also for any α > 0,∣∣∣∣(−A)α S (t)

∣∣∣∣≤ Cα

δ

1tα

e−δ t (54.3.22)

where −δ > −a. Furthermore, Cα is bounded as α → 0+ and is bounded on compactintervals of (0,∞). Also for α ∈ (0,1) and x ∈ D

((−A)α

),

||(S (t)− I)x|| ≤ C1−α

αδtα∣∣∣∣(−A)α x

∣∣∣∣ (54.3.23)