Here the adjoint of a generator of a semigroup is considered. I will show that the adjoint of
the generator generates the adjoint of the semigroup in a reflexive Banach space. This is
about as far as you can go although a general but less satisfactory result is given in Yosida
[32].
Definition 24.9.12Let A be a densely defined closed operator on H a real Banachspace. Then A^{∗}is defined as follows.
∗ ∗ ′ ∗
D (A ) ≡ {y ∈ H : |y (Ax)| ≤ C ||x|| for all x ∈ D (A)}
Then since D
(A)
is dense, there exists a unique element of H^{′}denoted by A^{∗}y suchthat
∗ ∗ ∗
A (y )(x) = y (Ax )
for all x ∈ D
(A )
.
Lemma 24.9.13Let A be closed and densely defined on D
(A)
⊆ H, a Banach space.Then A^{∗}is also closed and densely defined. Also
(A∗)
^{∗} = A. In addition to this, if
(λI − A)
^{−1}∈ℒ
(H,H )
, then
(λI − A ∗)
^{−1}∈ℒ
(H ′,H ′)
and
(( ) ) ( )
(λI − A )− 1 n ∗ = (λI − A∗)−1 n
Proof: Denote by
[x,y]
an ordered pair in H × H. Define τ : H × H → H × H
by
τ [x,y] ≡ [− y,x]
A similar notation will apply to H^{′}×H^{′}. Then the definition of adjoint implies that for G
(B)
equal to the graph of B,
∗ ⊥
G(A ) = (τG (A)) (24.50)
(24.50)
For S ⊆ H × H, define S^{⊥} by
{[a∗,b∗] ∈ H ′ ×H ′ : a∗(x)+ b∗(y) = 0 for all [x,y] ∈ S}
If S ⊆ H^{′}× H^{′} a similar definition holds.
∗ ∗ ∗ ∗
{[x,y] ∈ H × H : a (x)+ b (y) = 0 for all [a ,b ] ∈ S}
and S^{⊥} is always a closed subspace. Also τ and ⊥ commute. The reason for this is that
∗ ∗
[x ,y ]
∈
(τV)
^{⊥} means that
∗ ∗
− x (b)+ y (a) = 0
for all
[a,b]
∈ V and
[x∗,y∗]
∈ τ
( )
V ⊥
means
[− y∗,x∗]
∈−
( )
V⊥
= V^{⊥} so for all
[a,b]
∈ V,
− y∗(a)+ x∗(b) = 0
which says the same thing. It is also clear that τ ∘τ has the effect of multiplication by −1. If
V ⊆ H^{′}× H^{′}, the argument for commuting ⊥ and τ is similar.
It follows from the above description of the graph of A^{∗} that even if G
(A )
were not closed
it would still be the case that G
(A∗)
is closed.
Why is D
(A ∗)
dense? If it is not dense, then by a typical application of the
Hahn Banach theorem, there exists y^{∗∗}∈ H^{′′} such that y^{∗∗}
(D (A∗))
= 0 but y^{∗∗}≠0.
Since H is reflexive, there exists y ∈ H such that x^{∗}
(y)
= 0 for all x^{∗}∈ D
(A ∗)
.
Thus
∗ ⊥ ( ⊥)⊥
[y,0] ∈ G (A ) = (τG (A)) = τG(A )
and so
[0,y]
∈G
(A)
which means y = A0 = 0, a contradiction. Thus D
(A∗)
is indeed dense.
Note this is where it was important to assume the space is reflexive. If you consider C
([0,1])
it is not dense in L^{∞}
([0,1])
but if f ∈ L^{1}
([0,1])
satisfies ∫_{0}^{1}fgdm = 0 for all g ∈ 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 the above
formula that
( ∗ ∗) ( ( ⊥ ))⊥ ( ⊥ )⊥
G (A ) = τ (τG (A )) = τ (τG (A ))
( ⊥ )⊥ ( ⊥) ⊥
= (− G (A )) = G(A) = G (A)