24.9.2 Adjoints, Hilbert Space
In Hilbert space, there are some special things which are true.
Definition 24.9.9 Let A be a densely defined closed operator on H a real Hilbert
space. Then A ^{∗} is defined as follows.
∗
D (A ) ≡ {y ∈ H : |(Ax,y)| ≤ C |x|}
Then since D
is dense, there exists a unique element of H denoted by A ^{∗} y such
that
for all x ∈ D
.
Lemma 24.9.10 Let A be closed and densely defined on D
⊆ H, a Hilbert space.
Then A ^{∗} is also closed and densely defined. Also ^{∗} =
A. In addition to this, if
^{−1} ∈ℒ , then ^{−1} ∈ℒ and
(( ) ) ( )
(λI − A )− 1 n ∗ = (λI − A∗)−1 n
Proof: Denote by
an ordered pair in
H × H. Define
τ :
H × H → H × H
by
Then the definition of adjoint implies that for G
equal to the graph of
B,
G(A∗) = (τG (A))⊥ (24.47)
(24.47)
In this notation the inner product on H × H with respect to which ⊥ is defined is given
by
([x,y],[a,b]) ≡ (x,a)+ (y,b).
Here is why this is so. For
∈G it follows that for all
y ∈ D
∗ ∗
([x,A x],[− Ay,y]) = − (Ay,x)+ (y,A x) = 0
and so
∈ ^{⊥} which shows
To obtain the other inclusion, let
∈ ^{⊥} . This means that for all
x ∈ D ,
In other words, for all x ∈ D
,
and so
≤ C for all
x ∈ D which shows
a ∈ D and
for all x ∈ D
. Therefore, since
D is dense, it follows
b =
A ^{∗} a and so
∈G .
This shows the other inclusion.
Note that if V is any subspace of the Hilbert space H × H,
and S ^{⊥} is always a closed subspace. Also τ and ⊥ commute. The reason for this is that
∈ ^{⊥} means that
for all
∈ V and
∈ τ means
∈ V ^{⊥} so for all
∈ V,
which says the same thing. It is also clear that τ ∘ τ has the effect of multiplication by
− 1.
It follows from the above description of the graph of A ^{∗} that even if G
were not closed
it would still be the case that
G is closed.
Why is D
dense? Suppose
z ∈ D ^{⊥} . Then for all
y ∈ D so that
∈G , it follows
∈G ^{⊥} =
^{⊥} =
τ G but this
implies
and so z = − A 0 = 0. Thus D
must be dense since there is no nonzero vector in
D ^{⊥} .
Since A is a closed operator, meaning G
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)
and so
^{∗} =
A.
Now consider the final claim. First let y ∈ D
=
D . Then letting
x ∈ H be
arbitrary,
( ( )∗ )
x, (λI − A )(λI − A)−1 y
( ) ( ( )∗ )
(λI − A )(λI − A)−1x,y = x, (λI − A )−1 (λI − A ∗) y
Thus
( )∗ ( )∗
(λI − A )(λI − A)−1 = I = (λI − A)−1 (λI − A ∗) (24.48)
(24.48)
on D
. Next let
x ∈ D =
D and
y ∈ H arbitrary.
( − 1 ) ( ( −1)∗ )
(x,y) = (λI − A ) (λI − A)x,y = (λI − A )x, (λI − A) y
Now it follows
|( ( ) ) |
|| (λI − A )x, (λI − A)−1 ∗y ||
≤ for any
x ∈ D and
so
( )
(λI − A )−1 ∗ y ∈ D (A∗)
Hence
( ∗ ( −1)∗ )
(x,y) = x,(λI − A ) (λI − A) y .
Since x ∈ D
is arbitrary and
D is dense, it follows
( −1)∗
(λI − A∗) (λI − A ) = I (24.49)
(24.49)
From 24.48 and 24.49 it follows
( ) ∗
(λI − A ∗)−1 = (λI − A )− 1
and
is one to one and onto with continuous inverse. Finally, from the
above,
( ∗ −1)n (( −1)∗)n (( −1)n)∗
(λI − A ) = (λI − A) = (λI − A ) .
This proves the lemma.
With this preparation, here is an interesting result about the adjoint of the generator of a
continuous bounded semigroup. I found this in Balakrishnan [4 ] .
Theorem 24.9.11 Suppose A is a densely defined closed operator which
generates a continuous semigroup, S
. Then A ^{∗} is also a closed densely defined
operator which generates S ^{∗} and S ^{∗} is also a continuous semigroup.
Proof: First suppose S
is also a bounded semigroup,
≤ M . From Lemma
24.9.10 A ^{∗} is closed and densely defined. It follows from the Hille Yosida theorem, Theorem
24.9.7 that
||( − 1)n|| M
| (λI − A ) | ≤ λn
From Lemma 24.9.10 and the fact the adjoint of a bounded linear operator preserves the
norm,
| | | |
M-- ||(( −1)n)∗|| ||(( −1)∗)n||
λn ≥ | (λI − A ) | = | (λI − A) |
||( ∗−1)n ||
= | (λI − A ) |
and so by Theorem
24.9.7 again it follows
A ^{∗} generates a continuous semigroup,
T which
satisfies
≤ M. I need to identify
T with
S ^{∗} . However, from the proof of
Theorem
24.9.7 and Lemma
24.9.10 , it follows that for
x ∈ D and a suitable sequence
,
( ( )k )
| ∑∞ tk λ2n(λnI − A ∗)− 1 |
(T (t)x,y) = ( nli→m∞e−λnt --------k!---------x,y)
k=0
( (( ) ) ∗ )
∞ tk λ2 (λ I − A)−1 k
|| − λnt∑ -----n---n------------ ||
= lnim→∞ (e k! x,y)
k=0
( ( ( ( −1)k) ) )
| | ∑∞ tk λ2n(λnI − A) | |
= lnim→∞ |(x, e− λnt|( --------------------- |)y|)
k=0 k!
∗
= (x,S (t)y) = (S (t)x,y).
Therefore, since
y is arbitrary,
S ^{∗} =
T on
x ∈ D a dense set and this
shows the two are equal. This proves the proposition in the case where
S is also
bounded.
Next only assume S
is a continuous semigroup. Then by Proposition
24.9.5 there exists
α > 0 such that
Then consider the operator − αI + A and the bounded semigroup e ^{−αt} S
. For
x ∈ D
e−αhS(h)x-−-x- ( − αhS(h)x-−-x e−-αh −-1-)
hli→m0+ h = hl→im0+ e h + h x
= − αx + Ax
Thus
− αI +
A generates
e ^{−αt} S and it follows from the first part that
− αI +
A ^{∗} generates
e ^{−αt} S ^{∗} . Thus
e− αhS∗(h)x− x
− αx + A∗x = lhi→m0+ --------------
( h∗ −αh )
= lim e−αhS-(h)x-−-x+ e---−-1x
h→0+ h h
S∗(h)x-− x
= − αx + hli→m0+ h
showing that
A ^{∗} generates
S ^{∗} . It follows from Proposition
24.9.5 that
A ^{∗} is closed and
densely defined. It is obvious
S ^{∗} is a semigroup. Why is it continuous? This also follows
from the first part of the argument which establishes that
is continuous. This proves the theorem.