1732 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

54.3.4 A Scale Of Banach SpacesNext I will present an important and interesting theorem which can be used to prove equiv-alence of certain norms.

Theorem 54.3.25 Let A,B be sectorial for S−a,φ where −a < 0 and suppose D(A) =D(B) . Also suppose

(A−B)(−A)−α ,(A−B)(−B)−α

are both bounded on D(A) for some α ∈ (0,1). Then for all β ∈ [0,1] ,

(−A)β (−B)−β ,(−B)β (−A)−β

are both bounded on D(A) = D(B). Also D((−A)β

)= D

((−B)β

).

Proof: First of all it is a good idea to verify (A−B)(−A)−α ,(A−B)(−B)−α makesense on D(A) . If x ∈D(A) , then why is (−A)−α x ∈D(A)? Here is why. Since x ∈D(A) ,

x = (−A)−1 y

for some y ∈ H. Then

(−A)−α x = (−A)−α (−A)−1 y = (−A)−1 (−A)−α y ∈ D(A) .

The case of (A−B)(−B)−α is similar.Next for β ∈ (0,1) and λ > 0, use 54.3.25 to write∣∣∣∣∣∣(−A)β (λ I−A)−1 x

∣∣∣∣∣∣≤ C

∣∣∣∣∣∣(−A)(λ I−A)−1 x∣∣∣∣∣∣β ∣∣∣∣∣∣(λ I−A)−1 x

∣∣∣∣∣∣1−β

≤ C∣∣∣∣∣∣(−A)(λ I−A)−1

∣∣∣∣∣∣β ∣∣∣∣∣∣(λ I−A)−1∣∣∣∣∣∣1−β

||x||

≤ C∣∣∣∣∣∣I−λ (λ I−A)−1

∣∣∣∣∣∣β M

(λ +δ )1−β||x||

≤ C(

1+λ

(λ +δ )

)β M

(λ +δ )1−β||x|| ≡ C

(λ +δ )1−β||x|| (54.3.29)

where −a <−δ < 0 where C denotes a generic constant. Similarly, for all β ∈ (0,1) ,∣∣∣∣∣∣(−B)β (λ I−B)−1 x∣∣∣∣∣∣≤ C

(λ +δ )1−β||x|| (54.3.30)

Now from Theorem 54.3.17 and letting β ∈ (0,1) ,

(−B)−β − (−A)−β =sin(πβ )

π

∫∞

0λ−β((λ I−B)−1− (λ I−A)−1

)dλ

1732 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS54.3.4 A Scale Of Banach SpacesNext I will present an important and interesting theorem which can be used to prove equiv-alence of certain norms.Theorem 54.3.25 Let A,B be sectorial for S_a where —a <0 and suppose D(A) =D(B). Also suppose(A—B)(—A) “,(A—B) (—B) ©are both bounded on D(A) for some a € (0,1). Then for all B € [0,1],(—A)P (—B) P ,(—B)? (—a) *are both bounded on D(A) = D(B). Also D ((-4)*) =D ((-8)*) .Proof: First of all it is a good idea to verify (A — B) (—A) * ,(A—B) (—B) * makesense on D(A). If x € D(A), then why is (—A) “x € D(A)? Here is why. Since x € D(A),for some y € H. Then(-A) 7x = (—A)-%(—A) ty = (A)! (A) “y € D(A).The case of (A —B)(—B) “ is similar.Next for B € (0,1) and A > 0, use 54.3.25 to write| (-ayP (ar—A) x||< ¢||(-a) aay ts|)? ||ar—ayts||!z ¢lf-war-ayfljar-a 7< ne ce< (1+ gts) teal = (54.3.29)where —a < —6 < 0 where C denotes a generic constant. Similarly, for all B € (0,1),<I——__, ||x(A+6)'?Now from Theorem 54.3.17 and letting B € (0,1),\(-8)? (Al—B)"' || < (54.3.30)(—B) 8 —(-a) 8 = ma) ("4 ((ar—B)"'—(ar—ay) an