54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1713

φ(A u|u| ,

u|u| )

λ

In the picture the angle which is a little larger than φ is φ′. Let λ be as shown with

|argλ | ≤ π−φ′. Then from the picture and trigonometry, if u ∈ D(A) ,

|λ |sin(φ′−φ

)<

∣∣∣∣λ −(Au|u|

,u|u|

)∣∣∣∣and so |u| |λ |sin(φ ′−φ) <

∣∣∣(λu−Au, u|u|

)∣∣∣ ≤ ∥(λ I−A)u∥ . Hence for all λ such that

|argλ | ≤ π−φ′ and u ∈ D(A) ,

|u|<(

1sin(φ ′−φ)

)1|λ ||(λ I−A)u| ≡ M

|λ ||(λ I−A)u|

Thus (λ I−A) is one to one on S0,φ ′ and if λ ∈ r (A) , then∥∥∥(λ I−A)−1∥∥∥< M|λ |

.

By assumption 0 ∈ r (A). Now if |µ| is small, (µI−A)−1 must exist because it equals((µA−1− I

)A)−1 and for |µ|<

∥∥A−1∥∥ ,(µA−1− I

)−1 ∈L (H,H) since the infinite series

∞

∑k=0

(−1)k (µA−1)k

converges and must equal to(µA−1− I

)−1. Therefore, there exists µ ∈ S0,φ ′ such that

µ ̸= 0 and µ ∈ r (A). Also if µ ̸= 0 and µ ∈ S0,φ ′ , then if |λ −µ| < |µ|M ,(λ I−A)−1 must

exist because

(λ I−A)−1 =[(

(λ −µ)(µI−A)−1− I)(µI−A)

]−1

where((λ −µ)(µI−A)−1− I

)−1exists because

∥∥∥(λ −µ)(µI−A)−1∥∥∥= |λ −µ|

∥∥∥(µI−A)−1∥∥∥< |µ|

M· M|µ|

= 1.

It follows that if S ≡{

λ ∈ S0,φ ′ : λ ∈ r (A)}, then S is open in S0,φ . However, S is also

closed because if λ = limn→∞ λ n where λ n ∈ S, then if λ = 0, it is given λ ∈ S. If λ ̸= 0,then for large enough n, |λ −λ n| < |λ n|

M and so λ ∈ S. Since S0,φ ′ is connected, it followsS = S0,φ ′ .