54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1721

implies(−A)−1 x = 0

hence x = 0 so that (−A)−2 is also one to one. Similarly, (−A)−m is one to one for allpositive integers m.

From what was just shown, if (−A)−α x = 0 for α ∈ (0,1) , then

(−A)−1 x = (−A)−(1−α) (−A)−α x = 0

and so x = 0. This shows (−A)−α is one to one for all α ∈ [0,1] if is defined as (−A)0 ≡ I.What about α > 1? For such α, it is of the form m+ β where β ∈ [0,1) and m is a

positive integer. Therefore, if(−A)−(m+β ) x = 0

then(−A)−β

((−A)−m)x = 0

and so from what was just shown, ((−A)−m)x = 0

and now this implies x = 0 so that (−A)−α is one to one for all α ≥ 0.Consider 54.3.19. It was shown above that

(−A)−α (−A)−β = (−A)−(α+β )

Let x = (−A)−(α+β ) y. Then

x = (−A)−α (−A)−β y⊆ (−A)−α (−A)−β (H)⊆ (−A)−α (H) .

This proves 54.3.19. If α < β , (−A)−β (H)⊆ (−A)−α (H) .Now consider the problem of writing (−A)−α for α ∈ (0,1) in terms of A, not men-

tioning S (t) . By Proposition 19.14.5,

(λ I−A)−1 x =∫

∞

0e−λ tS (t)xdt

Then ∫∞

0λ−α (λ I−A)−1 dλ =

∫∞

0λ−α

∫∞

0e−λ tS (t)dtdλ

=∫

∞

0S (t)

∫∞

0λ−α e−λ tdλdt

=∫

∞

0S (t)

∫∞

0λ

β−1e−λ tdλdt

where β ≡ 1−α. Then using Lemma 54.3.15, this equals∫∞

0S (t)

∫∞

0µ

β−1t1−β e−µ t−1dµdt =∫

∞

0t−β S (t)

∫∞

0µ

β−1e−µ dµdt