1734 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

The first of these equations shows D((−B)β

)⊆ D

((−A)β

)and the second turns the

inclusion around. Thus they are equal as claimed.Next consider the case where β = 1. In this case

(A−B)B−α

is bounded on D(A) and so(A−B)B−α B−1+α

is also bounded on D(A) . But this equals

(A−B)B−1.

Thus AB−1 is bounded on D(A) . Similarly you can show

(B−A)A−1

is bounded which implies BA−1 is bounded on D(A). This proves the theorem.

Definition 54.3.26 Let A be sectorial for the sector Sa,φ . Let b > a so that A−bI is secto-rial for S−δ ,φ where δ = b−a. Then for each α ∈ [0,1] , define a norm on D

((bI−A)α

)≡

Hα by||x||

α≡∣∣∣∣(bI−A)α x

∣∣∣∣The {Hα}α∈[0,1] is called a scale of Banach spaces.

Proposition 54.3.27 The Hα above are Banach spaces and they decrease in α. Further-more, if bi > a for i = 1,2 then the two norms associated with the bi are equivalent.

Proof: That the Hα are decreasing was shown above in Theorem 54.3.17. They areBanach spaces because (bI−A)α is a closed mapping which is also one to one.

It only remains to verify the claim about the equivalence of the norms. Let b2 > b1 > a.Then if α ∈ (0,1) ,

((b1I−A)− (b2I−A))(b2I−A)−α

= (b1−b2)(b2I−A)−α ∈L (H,H)

and so by Theorem 54.3.25, for each β ∈ [0,1] ,

D((b1I−A)β

)= D

((b2I−A)β

)so the spaces, Hβ are the same for either choice of b > a. Also from this theorem,

(b1I−A)β (b2I−A)−β , (b2I−A)β (b1I−A)−β

are both bounded on D(A) . Therefore, for x ∈ Hβ∣∣∣∣∣∣(b1I−A)β x∣∣∣∣∣∣ =

∣∣∣∣∣∣(b1I−A)β (b2I−A)−β (b2I−A)β x∣∣∣∣∣∣

≤ C∣∣∣∣∣∣(b2I−A)β x

∣∣∣∣∣∣