1706 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

Proof: On one of the straight lines making up the contour, we have λ = a+ yw where|w| = 1,y ≥ ε . Then eλ t = eat+ywt = eatey|t|ei(argw+arg t)

= eatey(cos(argw+arg t)+isin(argw+arg t)).Now for t ∈ S0

0(φ+π/2),

argw+ arg t >π

2

and so cos(argw+ arg t)< 0. Therefore,∣∣eλ t∣∣≤ |eat |e−y|t|δ t where δ t > 0. Thus this part

of the contour integral is of the form∫

∞

εeat+ytw f (a+ yw)wdy and∥∥∥∥∫ ∞

Reat+ytw f (a+ yw)wdy

∥∥∥∥≤C∣∣eat ∣∣∫ ∞

Re−y|t|δ t dy < η

if R is large enough. Now consider A∫ R

ε

g(y)︷ ︸︸ ︷eat+ytw f (a+ yw)wdy. There is a sequence of

Riemann sums converging to the integral, {S (g,Pn)} as ∥Pn∥ → 0 for Pn a partition. Eachof these sums is in the D(A) . Then

S (g,Pn)→∫ R

ε

g(y)dy,AS (g,Pn) = S (Ag,Pn)→∫ R

ε

Ag(y)dy

Since A is a closed operator,∫ R

εg(y)dy ∈ D(A) and A

(∫ Rε

g(y)dy)=∫ R

εAg(y)dy. Now∫ R

εg(y)dy ∈ D(A) and limR→∞

∫ Rε

g(y)dy =∫

∞

εg(y)dy while

limR→∞

A∫ R

ε

g(y)dy = limR→∞

∫ R

ε

Ag(y)dy =∫

∞

ε

Ag(y)dy

Since A is closed,∫

∞

εAg(y)dy = A

∫∞

εg(y)dy and

∫∞

εg(y)dy ∈ D(A). The other straight

line is similar. As to the circular part, it is easier because it is not an improper integral. Theargument for taking A on the inside is similar, approximating with Riemann sums and thenpassing to a limit.

It remains to consider the other claim. On the circle, λ = a+Reiθ so dλ = Rieiθ dθ

and ∫ΓR

eλ t f (λ )dλ =∫

π+φ

π−φ

eat+|t|R(cos(θ+arg t)+isin(θ+arg t)) f (λ )Rieiθ dθ

Now for t ∈ S00(φ+π/2) and θ as indicated, θ + arg t > π

2 and θ + arg t < 3π

2 and so themagnitude of the above integral is no more than an expresson of the form

∣∣eat ∣∣C∫ π+φ

π−φ

e−|t|Rδ Rdθ

which clearly converges to 0 as R→ ∞.Because of this lemma, I will move A into and out of the integrals which occur in what

follows. Also, it is possible to approximate contour integrals over γε,φ with closed contoursand use the Cauchy integral formula.

Next is consideration of the above definition along with estimates.