22.8. GENERAL THEORY OF CONTINUOUS SEMIGROUPS 617

C ([0,1]) it is not dense in L∞ ([0,1]) but if f ∈ L1 ([0,1]) satisfies∫ 1

0 f gdm = 0 for allg ∈C ([0,1]) , then f = 0. Hence there is no nonzero f ∈C ([0,1])⊥.

Since A is a closed operator, meaning G (A) is closed in H ×H, it follows from theabove formula that

G((A∗)∗

)=

(τ

((τG (A))⊥

))⊥=(

τ (τG (A))⊥)⊥

=((−G (A))⊥

)⊥=(G (A)⊥

)⊥= G (A)

and so (A∗)∗ = A.Now consider the final claim. First let y∗ ∈ D(A∗) = D(λ I−A∗) . Then letting x ∈ H

be arbitrary,

y∗ (x) =((λ I−A)(λ I−A)−1

)∗y∗ (x) = y∗

((λ I−A)(λ I−A)−1 x

)Since y∗ ∈D(A∗) and (λ I−A)−1 x ∈D(A) , this equals (λ I−A)∗ y∗

((λ I−A)−1 x

). Now

by definition, this equals((λ I−A)−1

)∗(λ I−A)∗ y∗ (x). It follows that for y∗ ∈ D(A∗) ,

((λ I−A)−1

)∗(λ I−A)∗ y∗ =

((λ I−A)−1

)∗(λ I−A∗)y∗ = y∗ (22.50)

Next let y∗ ∈ H ′ be arbitrary and x ∈ D(A)

y∗ (x) = y∗((λ I−A)−1 (λ I−A)x

)=((λ I−A)−1

)∗y∗ ((λ I−A)x)

= (λ I−A)∗((λ I−A)−1

)∗y∗ (x)

In going from the second to the third line, the first line shows((λ I−A)−1

)∗y∗ ∈ D(A∗)

and so the third line follows. Since D(A) is dense, it follows

(λ I−A∗)((λ I−A)−1

)∗= I (22.51)

Then 22.50 and 22.51 show λ I−A∗ is one to one and onto from D(A∗) to H ′ and

(λ I−A∗)−1 =((λ I−A)−1

)∗.

Finally, from the above,((λ I−A∗)−1

)n=((

(λ I−A)−1)∗)n

=((

(λ I−A)−1)n)∗

. Thisproves the lemma.

With this preparation, here is an interesting result about the adjoint of the generator ofa continuous bounded semigroup.

Theorem 22.8.17 Suppose A is a densely defined closed operator which generatesa continuous semigroup, S (t) . Then A∗ is also a closed densely defined operator whichgenerates S∗ (t) and S∗ (t) is also a continuous semigroup.