54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1729

and now writing in what γ is in terms of θ yields

Γ(θ)∣∣∣∣(−A)γ x

∣∣∣∣≤C(

1β −α

)((ηβ−α

)1−θ

(1−θ)

∣∣∣∣∣∣(−A)β x∣∣∣∣∣∣+ (ηβ−α

)−θ

θ

∣∣∣∣(−A)α x∣∣∣∣)

Letting λ = ηβ−α , it follows

Γ(θ)∣∣∣∣(−A)γ x

∣∣∣∣≤C(

1β −α

)(λ

1−θ

(1−θ)

∣∣∣∣∣∣(−A)β x∣∣∣∣∣∣+ λ

−θ

θ

∣∣∣∣(−A)α x∣∣∣∣)

then let

λ =

∣∣∣∣(−A)α x∣∣∣∣∣∣∣∣∣∣(−A)β x∣∣∣∣∣∣

which is obtained from minimizing the expression on the right in the above. then placingthis in the inequality yields

Γ(θ)∣∣∣∣(−A)γ x

∣∣∣∣≤ C

(1

β −α

)(||(−A)α x||∣∣∣∣∣∣(−A)β x

∣∣∣∣∣∣)1−θ

(1−θ)

∣∣∣∣∣∣(−A)β x∣∣∣∣∣∣

+

(||(−A)α x||∣∣∣∣∣∣(−A)β x

∣∣∣∣∣∣)−θ

θ

∣∣∣∣(−A)α x∣∣∣∣

=C(

1β −α

)(1

(1−θ)+

1θ

)∣∣∣∣(−A)α x∣∣∣∣1−θ

∣∣∣∣∣∣(−A)β x∣∣∣∣∣∣θ

and this proves the proposition.Note that the constant is not bounded as θ → 1.Here is another interesting result about compactness.

Proposition 54.3.22 Let A be sectorial for S−a,φ where −a < 0. Then the following areequivalent.

1. (−A)−α is compact for all α > 0.

2. S (t) is compact for each t > 0.