54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1707

Lemma 54.3.7 The above definition is well defined for t ∈ S00(φ+π/2). Also there is a con-

stant Mr such that∥S (t)∥ ≤Mreat

for every t ∈ S00(φ+π/2) such that |arg t| ≤ r <

(π

2 −φ). If Sr is the sector just described, t

such that |arg t| ≤ r <(

π

2 −φ), then for any x ∈ H,

limt→0,t∈Sr

S (t)x = x (54.3.12)

Also, for |arg t| ≤ r <(

π

2 −φ)∥AS (t)∥ ≤Mr

∣∣eat ∣∣ 1|t|

+Nr∣∣eat ∣∣ |a| (54.3.13)

Proof: In the definition of S (t)

S (t)≡ 12πi

∫γε,φ

eλ t (λ I−A)−1 dλ

Since S (t) does not depend on ε, we can take ε = 1/ |t| . Then the circular part of the

contour is λ = a + 1|t|e

iθ . Then eλ t = e(

a+ 1|t| e

iθ)(|t|(eiarg t))

= eateei(θ+arg(t)). Then on the

circle which is part of γε,φ the contour integral equals

12π

∫π−φ

φ−π

eateei(θ+arg(t))((

a+1|t|

eiθ)

I−A)−1 1|t|

eiθ dθ

Now ∣∣∣eei(θ+arg(t))∣∣∣= ∣∣∣ecos(θ+arg t)+isin(θ+arg t)

∣∣∣≤ e

and by assumption, the norm of the integrand is no larger than eeat M1/|t|

1|t| and so the norm of

this integral is dominated by

eeatM2π

∫π−φ

φ−π

dθ =eeatM

2π(2π−2φ)≤ eatM

where M is independent of t.Now consider the part of the contour used to define S (t) which is the top line segment.

λ = yw+a where arg(w) = π−φ ,y > 1/ |t|. This part of the contour integral equals

12πi

∫∞

1/|t|e(yw+a)t ((yw+a) I−A)−1 wdy

Then from the resolvent estimate 54.3.4, the norm of this is dominated by

eat 12πi

∫∞

1/|t|eyeiargw(|t|eiarg t)M

ydy = eat 1

2πi

∫∞

1/|t|e|t|yei(argw+arg t) M

ydy