432 CHAPTER 17. SPECTRAL THEORY OF LINEAR MAPS

Proof: I need to consider (λ I− (A+B))−1 . This equals((I−B(λ I−A)−1

)(λ I−A)

)−1. (17.18)

The issue is whether this makes any sense for all λ ∈ Sbφ for some b∈R. Let b > a be verylarge so that if λ ∈ Sbφ , then 17.16 holds. Then from 17.17, it follows that for ∥x∥ ≤ 1,∥∥∥B(λ I−A)−1 x

∥∥∥ ≤ ε

∥∥∥A(λ I−A)−1 x∥∥∥+K

∥∥∥(λ I−A)−1 x∥∥∥

≤ εC+K/ |λ −a|

and so if b is made sufficiently large and λ ∈ Sbφ , then for all ∥x∥ ≤ 1,∥∥∥B(λ I−A)−1 x∥∥∥≤ εC+K/ |λ −a|< r < 1

Therefore, for such b,(

I−B(λ I−A)−1)−1

=∑∞k=0

(B(λ I−A)−1

)kexists and so for such

b, the expression in 17.18 makes sense and equals (λ I−A)−1(

I−B(λ I−A)−1)−1

andfurthermore, ∥∥∥∥(λ I−A)−1

(I−B(λ I−A)−1

)−1∥∥∥∥≤ M|λ −a|

11− r

≤ M′

|λ −b|

by adjusting the constants because M|λ−a|

|λ−b|1−r is bounded for λ ∈ Sbφ . ■

In finite dimensions, this kind of thing just shown always holds. There you have D(A)is the whole space typically and B will satisfy such an inequality in 17.17. The followingexample shows that all the bounded operators are sectorial.

Example 17.4.5 If A ∈L (H,H) , then A is sectorial.

The spectrum σ (A) is bounded by ∥A∥ and so there is clearly a sector of the aboveform contained in the resolvent set of A. As to the estimate 17.12, let a be larger than 2∥A∥and let Saφ be contained in the resolvent set. Then for λ ∈ Saφ , |λ |> 2∥A∥ and so∥∥∥(λ I−A)−1

∥∥∥= |λ |−1

∥∥∥∥∥(

I− Aλ

)−1∥∥∥∥∥≤ |λ |−1

∥∥∥∥∥ ∞

∑k=0

(Aλ

)k∥∥∥∥∥≤ |λ |−1 2

Now for λ ∈ Saφ ,∣∣∣λ−a

λ

∣∣∣≤M for some constant M and so∥∥∥(λ I−A)−1

∥∥∥≤ 2M|λ−a| Thus this

theory includes the case of ordinary differential equations in Rp. You might note that bothA and −A will be sectorial in this case.

Definition 17.4.6 For a sectorial operator as defined above, let the contour γ be asshown next where the orientation is also as shown by the arrow, a being the center of thecircle having radius d.

Saφγd,φ

φ