Analysis

24.9.3 Adjoints, Reflexive Banach Space

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].

Proof: Denote by

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

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}

Here is why 24.50 is so. For

∈G it follows that for all y ∈ D

∗ ∗ ∗ x (Ay) = A x (y)

and so for all

∈G,

− x∗(Ay )+ A∗x∗(y) = 0

which is what it means to say

∈^⊥. This shows

∗ ⊥ G(A ) ⊆ (τG (A))

To obtain the other inclusion, let

∈^⊥. This means that for all ∈G,

∗ ∗ − a (Ax )+ b (x) = 0

In other words, for all x ∈ D

,

|a∗(Ax)| ≤ ||b∗||||x||

which means by definition, a^∗∈ D

and A^∗a^∗ = b^∗. Thus ∈G.This shows the other inclusion.

Note that if V is any subspace of H × H,

(V⊥ )⊥ = V

and S^⊥ is always a closed subspace. Also τ and ⊥ commute. The reason for this is that

∈^⊥ means that

∗ ∗ − x (b)+ y (a) = 0

for all

∈ V and ∈ τ means ∈− = V ^⊥ so for all ∈ 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

were not closed it would still be the case that G is closed.

Why is D

dense? If it is not dense, then by a typical application of the Hahn Banach theorem, there exists y^∗∗∈ H^′′ such that y^∗∗ = 0 but y^∗∗≠0. Since H is reflexive, there exists y ∈ H such that x^∗ = 0 for all x^∗ ∈ D. Thus

∗ ⊥ ( ⊥)⊥ [y,0] ∈ G (A ) = (τG (A)) = τG(A )

and so

∈G which means y = A0 = 0, a contradiction. Thus D is indeed dense. Note this is where it was important to assume the space is reflexive. If you consider C it is not dense in L^∞ but if f ∈ L¹ satisfies ∫ ₀¹fgdm = 0 for all g ∈ C, then f = 0. Hence there is no nonzero f ∈ C^⊥.

Since A is a closed operator, meaning G

is closed in H ×H, it follows from the above formula that

and so ^∗ = A.

Now consider the final claim. First let y^∗∈ D

= D. Then letting x ∈ H be arbitrary,

∗ ( −1)∗ ∗ y (x) = (λI − A )(λI − A) y (x)

( ) = y∗ (λI − A)(λI − A )−1x

Since y^∗∈ D

and ⁻¹x ∈ D, this equals

∗ ∗ ( −1 ) (λI − A) y (λI − A) x

Now by definition, this equals

( −1)∗ ∗ ∗ (λI − A) (λI − A) y (x)

It follows that for y^∗∈ D

,

Next let y^∗∈ H^′ be arbitrary and x ∈ DIn going from the second to the third line, the first line shows ^∗y^∗∈ D and so the third line follows. Since D is dense, it follows

∗ ( −1)∗ (λI − A ) (λI − A ) = I (24.52)

(24.52)

Then 24.51 and 24.52 show λI − A^∗ is one to one and onto from D

t0 H^′ and

∗ −1 ( −1)∗ (λI − A ) = (λI − A) .

Finally, from the above,

( )n (( )∗)n (( )n)∗ (λI − A∗)−1 = (λI − A)−1 = (λI − A )−1 .

This proves the lemma.

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

Proof: First suppose S

is also a bounded semigroup, ≤ M. From Lemma 24.9.13 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 24.9.13 and the fact the adjoint of a bounded linear operator preserves the norm,

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 24.9.13, it follows that for x^∗∈ D and a suitable sequence ,

( −1)k ∗ − λnt∑∞ tk λ2n(λnI − A∗) ∗ T (t)x (y) = nli→m∞ e --------k!---------x (y) k=0

Therefore, since y is arbitrary, S^∗ = T on x ∈ D a dense set and this shows the two are equal. In particular, S^∗ is a semigroup because T is. 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

||S (t)|| ≤ M eαt.

Then consider the operator −αI + A and the bounded semigroup e^−αtS

. For x ∈ D

Thus −αI + A generates e^−αtS and it follows from the first part that −αI + A^∗ generates the semigroup e^−αtS^∗. Thus 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

−αt ∗ t → e S (t) x

is continuous. This proves the theorem.

class=”left” align=”middle”(U)24.10.  EXERCISES