54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1709

Which converges to 0 as t→ 0 in the sector |arg t| ≤ r <(

π

2 −φ). Also note that for t close

to 0, the contour contains 0 on the left. Similarly the integral over the other straight lineconverges to 0 as t→ 0 in that sector.

S (t)x = ε (t)+1

2πi

∫γ1/|t|,φ

eλ t

λxdλ , lim

t→0+ε (t) = 0

Now approximate γ1/|t|,φ with a closed contour having a large circular arc of radius R suchthat the resulting bounded contour ΓR has 0 on its inside and∣∣∣∣∣ 1

2πi

∫γ1/|t|,φ

eλ t

λxdλ − 1

2πi

∫ΓR

eλ t

λxdλ

∣∣∣∣∣< η (R) .

By the Cauchy integral formula, this shows that

S (t)x = ε (t)+η (R)+ x, limt→0+

ε (t) = 0 = limR→∞

η (R)

So let R→ ∞ and obtain S (t)x = ε (t)+ x and now let t → 0 to obtain S (t)x→ x. By thefirst part, ∥S (t)∥ is bounded for small t in that sector so it follows that for any x ∈ H,

∥S (t)x− x∥ ≤ ∥S (t)x−S (t)y∥+∥S (t)y− y∥+∥y− x∥≤ C∥x− y∥+∥S (t)y− y∥

Choosing ∥x− y∥ small enough for y∈D(A) , the above is no more than ε/2+∥S (t)y− y∥and the second term converges to 0 from what was just shown. Hence, for all x ∈ H,

limt→0

S (t)x = x

where t is in the sector |arg t| ≤ r <(

π

2 −φ).

Now for |arg t| ≤ r <(

π

2 −φ), AS (t) = 1

2πi∫

γε,φeλ tA(λ I−A)−1 dλ . From 54.3.5 this

is1

2πi

∫γε,φ

eλ t(−I +λ (λ I−A)−1

)dλ

As above, let ε = 1/ |t| . On the circle, λ = a+ 1|t|e

iθ and as above, this is

∫π−φ

φ−π

eateei(θ+arg(t))

(−I +

(a+

1|t|

eiθ)((

a+1|t|

eiθ)

I−A)−1

)i|t|

eiθ dθ

As before, because of the choice of t, the above is dominated by

e∣∣eat ∣∣∫ π−φ

φ−π

1+M

∣∣∣a+ 1|t|e

iθ∣∣∣

1/ |t|

 1|t|

dθ = e∣∣eat ∣∣M ∫

π−φ

φ−π

(1+∣∣∣a |t|+ eiθ

∣∣∣) 1|t|

dθ

= e∣∣eat ∣∣M ∫

π−φ

φ−π

(1|t|

+

∣∣∣∣a+ eiθ 1|t|

∣∣∣∣)dθ

≤ e∣∣eatM

∣∣2π2|t|

+Me∣∣eat ∣∣ |a|2π (54.3.15)