42.1.1 Weak Vector Valued Derivatives
In this section, several significant theorems are presented. Unless indicated otherwise, the
measure will be Lebesgue measure. First here is a lemma.
Lemma 42.1.1 Suppose g ∈ L1
where X is a Banach space. Then if
= 0 for all ϕ ∈ Cc∞
, then g
Proof: Let E be a measurable subset of
K ⊆ E ⊆ V ⊆
is compact, V
is open and m
Let K ≺ h ≺ V
as in the proof of
the Riesz representation theorem for positive linear functionals. Enlarging K
slightly and convolving with a mollifier, it can be assumed h ∈ Cc∞
Now let Kn ⊆ E ⊆ V n
Then from the above,
and the integrand of the last integral converges to 0 a.e. as n →∞ because
. By the dominated convergence theorem, this last integral
converges to 0. Therefore, whenever E ⊆
Since the endpoints have measure zero, it also follows that for any measurable E, the
above equation holds.
Now g ∈ L1
and so it is measurable. Therefore,
is separable. Let
be a countable dense subset and let E
denote the set of linear combinations of the form
is a rational point of F
and di ∈ D.
is countable. Denote by Y
the closure of E
is a separable closed subspace of X
which contains all the
values of g
Now let Sn ≡ g−1
This follows because if x ∈ Y
then in B
is a point of
x ∈ B
It follows that if each Sn
has measure zero, then g
= 0 for a.e.
Suppose then that for some n,
the set, Sn
has positive mesure. Then from what was
and so yn
= 0 which implies Sn
a contradiction to m
shows each Sn
has measure zero and so as just explained, g
= 0 a.e.
Definition 42.1.2 For f ∈ L1
, define an extension, f defined on
Definition 42.1.3 Also if f ∈ Lp
and h >
0, define for t ∈
for all h < b − a. Thus the map f → fh is continuous and linear on
. It is continuous because
The following lemma is on continuity of translation in Lp
Lemma 42.1.4 Let f be as defined in Definition ??. Then for f ∈ Lp
Proof: Regarding the measure space as
with Lebesgue measure, by Lemma
there exists g ∈ Cc
p < ε.
Here the norm is the norm in
is sufficiently small. This is because of the uniform continuity of g.
since ε >
0 is arbitrary, this proves the lemma. ■
Definition 42.1.5 Let f ∈ L1
. Then the distributional derivative in the sense
of X valued distributions is given by
Then f′∈ L1
if there exists h ∈ L1
such that for all ϕ ∈ Cc∞
Then f′ is defined to equal h. Here f and f′ are considered as vector valued distributions
in the same way as was done for scalar valued functions.
Lemma 42.1.6 The above definition is well defined.
Proof: Suppose both h and g work in the definition. Then for all ϕ ∈ Cc∞
Therefore, by Lemma 42.1.1, h
= 0 a.e.
The other thing to notice about this is the following lemma. It follows immediately
from the definition.
Lemma 42.1.7 Suppose f,f′∈ L1
. Then if
, it follows that
. This notation means the restriction to
Recall that in the case of scalar valued functions, if you had both f and its weak
derivative, f′ in L1
then you were able to conclude that f
is almost everywhere
equal to a continuous function, still denoted by f
In particular, you can define f
to be the initial value of this continuous function. It
turns out that an identical theorem holds in this case. To begin with here is the same sort
of lemma which was used earlier for the case of scalar valued functions. It says that if
= 0 where the derivative is taken in the sense of X
valued distributions, then f
Lemma 42.1.8 Suppose f ∈ L1
and for all ϕ ∈ Cc∞
Then there exists a constant, a ∈ X such that f
Proof: Let ϕ0 ∈ Cc∞
= 1 and define for ϕ ∈ Cc∞
Then ψϕ ∈ Cc∞
= ϕ −
It follows that for all ϕ ∈ Cc∞
and so by Lemma 42.1.1,
Theorem 42.1.9 Suppose f,f′ both are in L1
where the derivative is taken in
the sense of X valued distributions. Then there exists a unique point of X, denoted by
such that the following formula holds a.e. t.
Now consider ∫
Let Λ ∈ X′.
Then it is routine from approximating
with simple functions to verify
Now the ordinary Fubini theorem can be applied to obtain
separates the points of X,
Therefore, by Lemma 42.1.8
, there exists a constant, denoted as f
The integration by parts formula is also important.
Corollary 42.1.10 Suppose f,f′∈ L1
and suppose ϕ ∈ C1
. Then the
following integration by parts formula holds.
Proof: From Theorem 42.1.9
The interchange in order of integration is justified as in the proof of Theorem 42.1.9
There is an interesting theorem which is easy to present at this point.
Definition 42.1.11 Let
denote the functions f ∈ L2
whose weak derivative f′ is also in L2
Proposition 42.1.12 Let f ∈ H1
. Then f ∈ C0,
inclusion map is continuous.
Proof: First note that
It follows that
Then integrating by parts yields