54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1723

There exists a constant C independent of α ∈ [0,1) such that for x ∈ D(A) and ε > 0,∣∣∣∣(−A)α x∣∣∣∣≤ ε ||(−A)x||+Cε

−α/(1−α) ||x|| (54.3.24)

There exists a constant C′ independent of α ∈ [0,1] such that for x ∈ D(A) ,∣∣∣∣(−A)α x∣∣∣∣≤C′ ||(−A)x||α ||x||1−α (54.3.25)

The formula 54.3.25 is called an interpolation inequality.

Proof: It is obvious (−A)α is densely defined because its domain is at least as large asD(A) which was assumed to be dense. It is a closed operator because if xn ∈ D

((−A)α

)and

xn→ x, (−A)α xn→ y,

then(−A)−α xn→ (−A)−α x, xn = (−A)−α (−A)α xn→ (−A)−α y

and so(−A)−α y = x

showing x ∈ D((−A)α

)and y = (−A)−α x. Thus (−A)α is closed and densely defined.

Let−δ >−a where the sector for A was S−a,φ ,a> 0. Then recall from Corollary 54.3.8there is a constant, N such that

||(−A)S (t)|| ≤ Nt

e−δ t

What about∣∣∣∣(−A)α S (t)

∣∣∣∣? First note that for α ∈ [0,1) this at least makes sense becauseS (t) maps into D(A). For any α > 0,

S (t)(−A)−α = (−A)−α S (t)

follows from the definiton of (−A)−α . Therefore,

(−A)α S (t)(−A)−α = S (t) . (54.3.26)

Note this implies that on D((−A)α

),

(−A)α S (t) = S (t)(−A)α .

Also(−A)−1 S (t) = S (t)(−A)−1 = S (t)(−A)−α (−A)−(1−α)

and soS (t) = (−A)S (t)(−A)−α (−A)−(1−α)

From 54.3.26 it follows

(−A)α S (t) = (−A)(−A)α S (t)(−A)−α (−A)−(1−α)

= (−A)S (t)(−A)−(1−α) (54.3.27)