578 CHAPTER 19. HILBERT SPACES

Proof: First note D(A) ̸= /0. In fact 0∈D(A). It follows from Theorem 19.14.3 that forall λ larger than α , one can define a Laplace transform, R(λ )x≡

∫∞

0 e−λ tS (t)xdt ∈ X .Herethe integral is the ordinary improper Riemann integral. I claim each of these R(λ )x for λ

large is in D(A) .S (h)

∫∞

0 e−λ tS (t)xdt−∫

∞

0 e−λ tS (t)xdth

Using the semigroup property and changing the variables in the first of the above integrals,this equals

=1h

(eλh

∫∞

he−λ tS (t)xdt−

∫∞

0e−λ tS (t)xdt

)=

1h

((eλh−1

)∫ ∞

0e−λ tS (t)xdt− eλh

∫ h

0e−λ tS (t)xdt

)Then it follows that the limit as h→ 0 exists and equals

λR(λ )x− x = limh→0+

S (h)R(λ )x−R(λ )xh

≡ A(R(λ )x) (19.14.80)

and R(λ )x ∈ D(A) as claimed. Hence

x = (λ I−A)R(λ )x. (19.14.81)

Since x is arbitrary, this shows that for λ > α , λ I−A is onto. Also, if x ∈D(A) , you couldapproximate with Riemann sums and pass to a limit and obtain

1h(R(λ )S (h)x−R(λ )x) =

∫∞

0e−λ tS (t)

S (h)x− xh

dt

Then, passing to a limit as h→ 0 using the dominated convergence theorem, one obtains

limh→0

1h(R(λ )S (h)x−R(λ )x) =

∫∞

0S (t)e−λ tAxdt ≡ R(λ )Ax

Also, S (h) commutes with R(λ ) . This follows in the usual way by approximating withRiemann sums and taking a limit. Thus for x ∈ D(A)

λR(λ )x− x = limh→0+

S (h)R(λ )x−R(λ )xh

= limh→0+

R(λ )S (h)x−R(λ )xh

= R(λ )Ax

and so, for x ∈ D(A) ,

x = λR(λ )x−R(λ )Ax = R(λ )(λ I−A)x (19.14.82)

which shows that for λ > α,(λ I−A) is one to one on D(A). Hence from 19.14.80and 19.14.82, (λ I−A) is an algebraic isomorphism from D(A) onto X . Also R(λ ) =