1712 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

Now the following diagram might help in drawing some interesting conclusions.

The first iterated integral equals 0. This can be seen from the above picture. By Lemma54.3.6, the inner integral taken over γε,φ is essentially equal to the integral over the closedcontour in the above picture provided the radius of the part of the large circle in the aboveclosed contour is large enough. This closed contour integral equals 0 by the Cauchy integraltheorem. The second iterated integral equals

12πi

∫γε,φ

(λ I−A)−1 eλ teλ sdλ = S (t + s)

from the Cauchy integral formula. This verifies the semigroup identity.4.) is done in Lemma 54.3.7 which also includes 5.) when you let t be positive and real.

54.3.1 The Numerical Range

In Hilbert space, there is a useful easy to check criterion which implies an operator issectorial.

Definition 54.3.9 Let A be a closed densely defined operator A : D(A)→H for H a Hilbertspace. The numerical range is the following set.

{(Au,u) : u ∈ D(A)}

Also recall the resolvent set, r (A) consists of those λ ∈C such that (λ I−A)−1 ∈L (H,H) .Thus, to be in this set λ I−A is one to one and onto with continuous inverse.

Proposition 54.3.10 Suppose the numerical range of A,a closed densely defined operatorA : D(A)→ H for H a Hilbert space is contained in the set

{z ∈ C : |arg(z)| ≥ π−φ}

where 0 < φ < π/2 and suppose A−1 ∈L (H,H) ,(0 ∈ r (A)). Then A is sectorial with thesector S0,φ ′ where π/2 > φ

′ > φ . Here arg(z) is the angle which is between −π and π .

Proof: Here is a picture of the situation along with details used to motivate the proof.