22.8. GENERAL THEORY OF CONTINUOUS SEMIGROUPS 609

≤M2t∥∥Aµ x−Aλ x

∥∥≤M2t(∥∥Aµ x−Ax

∥∥+∥Ax−Aλ x∥)

Now by Lemma 22.8.9, the right side converges uniformly to 0 in t ∈ [0,T ] an arbitraryfinite interval. Denote that to which it converges S (t)x. Therefore, t→ S (t)x is continuousfor each x ∈ D(A) and also from 22.40,

∥S (t)x∥= limλ→∞

∥Sλ (t)x∥ ≤M ∥x∥

so that S (t) can be extended uniquely to a continuous linear map, still called S (t) defined onall of X which also satisfies ∥S (t)∥≤M since D(A) is dense in X . The uniform convergenceon [0,T ] implies t→ S (t) is continuous.

It remains to verify that A generates S (t) and for all x, limt→0+ S (t)x−x = 0. From theabove,

Sλ (t)x = x+∫ t

0Sλ (s)Aλ xds (22.43)

and so limt→0+ ∥Sλ (t)x− x∥ = 0. By the uniform convergence just shown, there exists λ

large enough that for all t ∈ [0,δ ] ,∥S (t)x−Sλ (t)x∥< ε. Then

lim supt→0+

∥S (t)x− x∥ ≤ lim supt→0+

(∥S (t)x−Sλ (t)x∥+∥Sλ (t)x− x∥)

≤ lim supt→0+

(ε +∥Sλ (t)x− x∥)≤ ε

It follows limt→0+ S (t)x = x because ε is arbitrary.Next, limλ→∞ Aλ x = Ax for all x ∈ D(A) by Lemma 22.8.9. Therefore, passing to the

limit in 22.43 yields from the uniform convergence

S (t)x = x+∫ t

0S (s)Axds

and by continuity of s→ S (s)Ax, it follows

limh→0+

S (h)x− xh

= limh→0

1h

∫ h

0S (s)Axds = Ax

Thus letting B denote the generator of S (t) , D(A) ⊆ D(B) and A = B on D(A) . It onlyremains to verify D(A) = D(B) .

To do this, let λ > 0 and consider the following where y ∈ X is arbitrary.

(λ I−B)−1 y = (λ I−B)−1((λ I−A)(λ I−A)−1 y

)Now (λ I−A)−1 y ∈ D(A)⊆ D(B) and A = B on D(A) and so

(λ I−A)(λ I−A)−1 y = (λ I−B)(λ I−A)−1 y

which implies,(λ I−B)−1 y =

(λ I−B)−1((λ I−B)(λ I−A)−1 y

)= (λ I−A)−1 y

Recall from Proposition 22.8.5, an arbitrary element of D(B) is of the form (λ I−B)−1 yand this has shown every such vector is in D(A) , in fact it equals (λ I−A)−1 y. HenceD(B)⊆ D(A) which shows A generates S (t) and this proves the first half of the theorem.