54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1705

Definition 54.3.4 Let ε > 0 and for a sectorial operator as defined above, let the contourγε,φ be as shown next where the orientation is also as shown by the arrow, a being thecenter of the small circle.

Saφ

φ

The little circle has radius ε in the above contour but ε is not necessarily small.

Definition 54.3.5 For t ∈ S00(φ+π/2) the open sector shown in the following picture,

0S0(φ+π/2)

φ +π/2

define

S (t)≡ 12πi

∫γε,φ

eλ t (λ I−A)−1 dλ (54.3.11)

where ε is some positive number. Since the integrand is analytic, two different values for ε

give the same result in 54.3.11. The following picture shows S00(φ+π/2) and S0φ . Note how

the dotted line is at right angles to the solid line.

Sa(φ+π/2)

tφ π/2−φ

Also define S (0)≡ I.

I need to move A in and out of an integral.

Lemma 54.3.6 Let f (λ ) ,A f (y) be bounded and continuous on γ∗ε,φ and have values in

D(A). Then A∫

γε,φeλ t f (λ )dλ =

∫γε,φ

eλ tA f (λ )dλ provided t ∈ S00(φ+π/2). Also, for large

R, and ΓR the circle a+Reiθ with θ ∈ [π−φ ,π +φ ] , limR→∞

∫ΓR

eλ t f (λ )dλ = 0.