19.2.2 Taking A Closed Operator Out Of The Integral
Now let X and Y be separable Banach spaces and suppose A : D
(A)
⊆ X → Y be a
closed operator. Recall this means that the graph of A,
G(A ) ≡ {(x,Ax ) : x ∈ D (A )}
is a closed subset of X × Y with respect to the product topology obtained from the
norm
||(x,y)|| = max (||x||,||y||).
Thus also G
(A )
is a separable Banach space with the above norm. You can also consider
D
(A)
as a separable Banach space having the graph norm
||x||D (A) ≡ max(||x||,||Ax||) (19.2.17)
(19.2.17)
which is isometric to G
(A)
with the mapping, θx ≡
(x,Ax )
.
Lemma 19.2.7A closed subspace of a reflexive Banach space is reflexive.
Proof:Consider the following diagram in which Y is a closed subspace of the
reflexive space, X.
i∗∗ 1-1
Y ′′ ∗→ X ′′
Y′ i←onto X′
Y →i X
This diagram follows from theorems on adjoints presented earlier.
Now let y^{∗∗}∈ Y^{′′}. Is y^{∗∗} = J
(y)
for some y ∈ Y ? From the above, there exists
y ∈ X such that for all x^{∗}, i^{∗∗}y^{∗∗}
(x∗)
= x^{∗}
(y)
. Is y ∈ Y ? If it is not in Y
then there exists x^{∗}∈ X^{′} such that x^{∗}
(y)
≠0 but x^{∗}
(Y )
= 0. Then i^{∗}x^{∗} = 0.
Hence
∗∗ ∗ ∗ ∗∗ ∗∗ ∗ ∗
0 = y (i x ) = i y (x ) = x (y) ⁄= 0,
a contradiction. Hence y ∈ Y . Thus
y∗∗ (i∗x∗) = i∗∗y∗∗(x∗) = x∗ (y) = x∗(iy) = i∗x∗(y), y ∈ Y
Now i^{∗} is onto. and so this says y^{∗∗}
∗
(y )
= y^{∗}
(y)
for all y^{∗}∈ Y^{′}. In other words,
y^{∗∗} = J
(y)
. ■
Corollary 19.2.8Suppose Y is a reflexive Banach space and X is a Banach spacesuch that there exists a continuous one to one mapping, g : X → Y such that g
(X)
is a closed subset of Y. Then X is reflexive.
Proof: By the open mapping theorem, g
(X )
and X are homeomorphic since g^{−1}
must also be continuous. Therefore, since g
(X )
is reflexive, it follows X is also.
■
Lemma 19.2.9Suppose V and W are reflexive Banach spaces and that V is adense subset of W in the topology of W. Then i^{∗}W^{′}is a dense subset of V^{′}wherehere i is the inclusion map of V into W.
Proof:First note that i^{∗} is one to one. If i^{∗}w^{∗} = 0 for w^{∗}∈ W^{′}, then this means
that for all v ∈ V,
i∗w (v) = w∗ (v) = 0
and since V is dense in W, this shows w^{∗} = 0.
Consider the following diagram
′′ i∗∗ ′′
V →∗ W
V′ i← W ′
V →i W
in which i is the inclusion map. Next suppose i^{∗}W^{′} is not dense in V^{′}. Then there exists
v^{∗∗}∈ V^{′′} such that v^{∗∗}≠0 but v^{∗∗}
∗ ′
(i W )
= 0. It follows from V being reflexive, that
v^{∗∗} = Jv_{0} where J is the James map from V to V^{′′}for some v_{0}∈ V . Thus for every
w^{∗}W^{′},
and since W^{′} separates the points of W, it follows v_{0} = 0 which contradicts v^{∗∗}≠0.■
Note that in the proof, only V reflexive was used.
This lemma implies an easy corollary.
Corollary 19.2.10Let E and F be reflexive Banach spaces and let A be a closedoperator, A : D
(A)
⊆ E → F. Suppose also that D
(A )
is dense in E. Then makingD
(A)
into a Banach space by using the above graph norm given in 19.2.17, it followsthat D
(A)
is a Banach space and i^{∗}E^{′}is a dense subspace of D
(A)
^{′}.
Proof: First note that E × F is a reflexive Banach space and G
(A)
is a closed
subspace of E ×F so it is also a reflexive Banach space. Now D
(A )
is isometric to G
(A)
and so it follows D
(A )
is a dense subspace of E which is reflexive. Therefore, from
Lemma 19.2.9 the conclusion follows. ■
With this preparation, here is another interesting theorem. This one is about taking
outside the integral a closed linear operator as opposed to a continuous linear
operator.
Theorem 19.2.11Let X,Y be separable Banach spaces and letA : D
(A )
⊆ X → Y be a closed operator where D
(A)
is a denseseparable subset of X with respect to the graph norm on D
(A)
describedabove^{1}.Suppose also that i^{∗}X^{′}is a dense subspace of D
(A )
^{′}where D
(A)
is a Banach spacehaving the graph norm described in 19.2.17. Suppose that
(Ω, ℱ,μ)
is a σ finite measurespace and x : Ω → X is strongly measurable and it happens that x
(s)
∈ D
(A)
for alls ∈ Ω. Then x is strongly measurable as a mapping into D
(A)
. Also Ax is stronglymeasurable as a map into Y and if
∫ ∫
||x (s)||dμ, ||Ax (s)||dμ < ∞, (19.2.18)
Ω Ω
(19.2.18)
then
∫
Ω x(s)dμ ∈ D (A) (19.2.19)
(19.2.19)
and
∫ ∫
A Ω x(s)dμ = Ω Ax (s)dμ. (19.2.20)
(19.2.20)
Proof: First of all, consider the assertion that x is strongly measurable into D
(A)
.
Letting f ∈ D
(A )
^{′} be given, there exists a sequence,
{gn}
⊆ i^{∗}X^{′} such that g_{n}→ f in
D
(A)
^{′}. Therefore,
s → gn(x (s))
is measurable by assumption and
gn (x (s)) → f (x (s))
which shows that s → f
(x (s))
is measurable. By the Pettis theorem, it follows
s → x(s)
is strongly measurable as a map into D
(A )
.
It follows from Theorem 19.2.5 there exists a sequence of simple functions,
{x }
n
of
the form
m∑n n
xn(s) = akXEnk (s),xn(s) ∈ D (A ),
k=1
which converges strongly and pointwise to x
(s)
in D
(A )
. Thus
x (s) → x(s),Ax (s) → Ax (s) ,
n n
which shows s → Ax
(s)
is stongly measurable in Y as claimed.
It remains to verify the assertions about the integral. 19.2.18 implies x is Bochner
integrable as a function having values in D
(A)
with the norm on D
(A)
described above.
Therefore, by Theorem 19.2.5 there exists a sequence of simple functions {y_{n}} having
values in D
(A)
,
∫
lim ||yn − ym ||D(A)dμ = 0,
m,n→∞ Ω
y_{n}
(s)
converging pointwise to x
(s)
,
||yn(s)|| ≤ 2||x (s)||
D(A) D (A)
and
∫
lim ||x (s)− yn (s)|| ds = 0.
n→ ∞ Ω D(A)
Therefore,
∫ ∫ ∫
yn (s) dμ ∈ D (A), yn(s)dμ → x(s)dμ in X,
Ω Ω Ω
and since y_{n} is a simple function and A is linear,
∫ ∫ ∫
A y (s)dμ = Ay (s)dμ → Ax (s)dμ in Y.
Ω n Ω n Ω
It follows, since A is a closed operator, that
∫
x(s)dμ ∈ D (A)
Ω
and
∫ ∫
A x(s)dμ = Ax (s)dμ. ■
Ω Ω
Here is another version of this theorem which has different hypotheses.
Theorem 19.2.12Let X and Y be separable Banach spaces and let A : D
(A)
⊆ X → Ybe a closed operator. Also let
(Ω,ℱ,μ )
be a σ finite measure space and let x : Ω → X beBochner integrable such that x
(s)
∈ D
(A )
for all s. Also suppose Ax is Bochnerintegrable. Then
∫ ∫
Axd μ = A xdμ
and∫xdμ ∈ D
(A)
.
Proof: Consider the graph of A,
G(A ) ≡ {(x,Ax ) : x ∈ D (A )} ⊆ X × Y.
Then since A is closed, G
(A)
is a closed separable Banach space with the norm
||(x,y)||
≡ max
(||x||,||y||)
. Therefore, for g^{∗}∈ G
(A)
^{′}, one can apply the Hahn Banach
theorem and obtain
∗ ∗
(x ,y )
∈
(X × Y )
^{′} such that g^{∗}
(x,Ax )
=
∗ ∗
(x (x),y (Ax))
. Now it
follows from the assumptions that s →
∗ ∗
(x (x(s)),y (Ax(s)))
is measurable with values
in G
(A)
. It is also separably valued because this is true of G
(A)
. By the Pettis theorem,
s →
(x (s),A (x(s)))
must be strongly measurable. Also ∫
||x(s)||
+
||A (x(s))||
dμ < ∞
by assumption and so there exists a sequence of simple functions having values in
G
(A)
,
{(xn (s),Axn (s))}
which converges to
(x (s),A (s))
pointwise such that
∫
||(xn,Axn) − (x,Ax )||
dμ → 0 in G
(A )
. Now for simple functions is it routine to
verify that