17.1. THE RESOLVENT AND SPECTRAL RADIUS 421

Therefore, ∑∞k=0 Bk (I−B) = limn→∞ ∑

nk=0 Bk (I−B) = limn→∞

(I−Bn+1

)= I. As to the

estimate, if ∥x∥ ≤ 1,

∥∥∥(I−B)−1 x∥∥∥ =

∥∥∥∥∥(

∞

∑k=0

Bk

)(x)

∥∥∥∥∥=∥∥∥∥∥ ∞

∑k=0

Bk (x)

∥∥∥∥∥≤

∞

∑k=0

∥∥∥Bk∥∥∥≤ ∞

∑k=0∥B∥k =

11−∥B∥

■

The major result about resolvents is the following proposition. Note that the resolventhas values in L (X ,X) which is a Banach space.

Proposition 17.1.5 Let A ∈L (X ,X) for X a Banach space. Then the following hold.

1. ρ (A) is open

2. λ → (λ I−A)−1 is continuous

3. λ → (λ I−A)−1 is analytic

4. For |λ |> ∥A∥ ,∥∥∥(λ I−A)−1

∥∥∥≤ 1|λ |−∥A∥ and (λ I−A)−1 = ∑

∞k=0

Ak

λk+1

Proof: 1.) Let λ ∈ ρ (A) . Let |µ−λ |<∥∥∥(λ I−A)−1

∥∥∥−1. Then

µI−A = (µ−λ ) I +λ I−A = (λ I−A)[I− (λ −µ)(λ I−A)−1

](17.2)

Now∥∥∥(λ −µ)(λ I−A)−1

∥∥∥= |λ −µ|∥∥∥(λ I−A)−1

∥∥∥< 1 from the assumed estimate. Thus,

[I− (λ −µ)(λ I−A)−1

]−1=

∞

∑k=0

(λ −µ)k((λ I−A)−1

)k

and so from 17.2,

(µI−A)−1 =[I− (λ −µ)(λ I−A)−1

]−1(λ I−A)−1 (17.3)

=∞

∑k=0

(λ −µ)k((λ I−A)−1

)k+1

This shows ρ (A) is open.2.) Next consider continuity. This follows from Proposition 17.1.3.3.) From Theorem 14.8.1, it suffices to show that the function is differentiable. This

follows from the continuity and the resolvent equation.

(λ I−A)−1− (µI−A)−1

λ −µ=

(µ−λ )(µI−A)−1 (λ I−A)−1

λ −µ

= −(µI−A)−1 (λ I−A)−1

so taking the limit as λ → µ one obtains the derivative at µ is −((µI−A)−1

)2.