17.4. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 435

Proof: Since S (t) does not depend on d > 0, we can take d = 1/ |t| . Then the circu-

lar part of the contour is λ = a+ 1|t|e

iθ ,dλ = i|t|e

iθ dθ . Then eλ t = e(

a+ 1|t| e

iθ)(|t|(eiarg t))

=

eateei(θ+arg(t)). Then on the circle which is part of γ the contour integral equals

12π

∫π−φ

φ−π

eateei(θ+arg(t))((

a+1|t|

eiθ)

I−A)−1 1|t|

eiθ dθ

≤ e2π

∫π−φ

φ−π

∣∣eat ∣∣ M∣∣∣ 1|t|e

iθ∣∣∣ 1|t|

dθ =Me2π

∫π−φ

φ−π

∣∣eat ∣∣dθ < Me∣∣eat ∣∣≡ M̂eaRe(t) (17.25)

where M̂ does not depend on t. The estimate 17.12 was used to obtain the inequality. Whatabout the rest of the contour defining S (t)? Letting arg(w) be chosen as either π − φ or3π

2 −φ , λ = yw+a where y = |λ −a| . Then on either straight segment we have

12πi

∫∞

1/|t|e(yw+a)t ((yw+a) I−A)−1 wdy, |w|= 1

and the bottom is similar. Thus, from Lemma 17.4.8 and the resolvent estimate 17.12 wecan dominate these two by an expression of the form

1π

∫∞

1/|t|

∣∣eat ∣∣My

e−δ ry|t|dy =Mπ

∣∣eat ∣∣ 1|t|

∫∞

1e−δ ru |t|du

=Mπ

∣∣eat ∣∣∫ ∞

1e−δ rudu≤ M

π

1δ r

∣∣eat ∣∣Taking u = y |t|. This together with 17.25 gives 17.22. In particular, ∥S (t)∥e−at is boundedfor t ∈ [0,∞). As noted in Lemma 17.4.9. It was important that |arg t| ≤ r <

(π

2 −φ).

Now let x ∈ D(A) . From 17.14,

eλ t

λ(λ −A)−1 Ax+

eλ t

λx = eλ t (λ I−A)−1 x (17.26)

On the circular part of the contour, λ = a+ 1|t|e

iθ . The contour integral is of the form

∫π−φ

φ−π

eateei(θ+arg(t)) 1a+ 1

|t|eiθ

((a+

1|t|

eiθ)

I−A)−1

Axi|t|

eiθ dθ

which is dominated by

e∣∣eat ∣∣∫ π−φ

φ−π

1∣∣∣a+ 1|t|e

iθ∣∣∣ M∣∣∣ 1|t|e

iθ∣∣∣ ∥Ax∥ 1

|t|≤ eatM̂ ∥Ax∥

∫π−φ

φ−π

|t||a |t|+ eiθ |

dθ

which converges to 0 as t→ 0. On the other part of the contour, λ = yw+a where |w|= 1,arg(w) = π−φ ,y > 1/ |t|, either of the straight segments are of the form

eat

2πi

∫∞

1/|t|eywt 1

yw+a((yw+a) I−A)−1 wAxdy