1704 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

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

)(λ I−A)

)−1. (54.3.10)

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 54.3.8 holds. Then from 54.3.9, 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

)k

exists and so for such b, the expression in 54.3.10 makes sense and equals

(λ I−A)−1(

I−B(λ I−A)−1)−1

and furthermore,∥∥∥∥(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 54.3.9. The followingexample shows that all the bounded operators are sectorial.

Example 54.3.3 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 above formcontained in the resolvent set of A. As to the estimate 54.3.4, let a be larger than 2∥A∥ andlet 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|