A related concept is that of weak derivatives. Let Ω ⊆ ℝ^{n} be an open set. A distribution on Ω
is defined to be a linear functional on C_{c}^{∞}
(Ω)
, called the space of test functions. The space of
all such linear functionals will be denoted by D^{∗}
(Ω )
. Actually, more is sometimes done here.
One imposes a topology on C_{c}^{∞}
(Ω)
making it into a topological vector space, and
when this has been done, D^{′}
(Ω )
is defined as the dual space of this topological
vector space. To see this, consult the book by Yosida [32] or the book by Rudin
[29].
Example: The space L_{loc}^{1}
(Ω)
may be considered as a subset of D^{∗}
(Ω )
as follows.
∫
f (ϕ) ≡ f (x)ϕ (x)dx
Ω
for all ϕ ∈ C_{c}^{∞}
(Ω )
. Recall that f ∈ L_{loc}^{1}
(Ω)
if fX_{K}∈ L^{1}
(Ω )
whenever K is compact.
Lemma 17.2.5 makes this identification possible. Recall that this said that if
f ∈ L_{loc}^{1}
(ℝn)
and for all ϕ ∈ C_{c}^{∞}
(Ω )
∫
fϕdmn = 0
then f
(x)
= 0 for a.e. x.
Example:δ_{x}∈D^{∗}
(Ω)
where δ_{x}
(ϕ)
≡ ϕ
(x)
.
It will be observed from the above two examples and a little thought that D^{∗}
(Ω )
is truly
enormous. We shall define the derivative of a distribution in such a way that it agrees with
the usual notion of a derivative on those distributions which are also continuously
differentiable functions. With this in mind, let f be the restriction to Ω of a smooth function
defined on ℝ^{n}. Then D_{xi}f makes sense and for ϕ ∈ C_{c}^{∞}
(Ω)
∫ ∫
D f (ϕ) ≡ D f (x)ϕ (x)dx = − fD ϕdx = − f (D ϕ).
xi Ω xi Ω xi xi
Motivated by this, here is the definition of a weak derivative.
Definition 17.4.1For T ∈D^{∗}
(Ω)
D T (ϕ) ≡ − T (D ϕ).
xi xi
Of course one can continue taking derivatives indefinitely. Thus,
( )
DxixjT ≡ Dxi Dxj T
and it is clear that all mixed partial derivatives are equal because this holds for the functions
in C_{c}^{∞}
(Ω )
. Thus one can differentiate virtually anything, even functions that may be
discontinuous everywhere. However the notion of “derivative” is very weak, hence the name,
“weak derivatives”.
Example: Let Ω = ℝ and let
{ 1 if x ≥ 0,
H (x) ≡ 0 if x < 0.
Then
∫ ′
DH (ϕ) = − H (x)ϕ (x)dx = ϕ(0) = δ0(ϕ).
Note that in this example, DH is not a function.
What happens when Df is a function?
Theorem 17.4.2Let Ω =
(a,b)
and suppose that f and Df are both in L^{1}
(a,b)
.Then f is equal to a continuous function a.e., still denoted by f and
∫
f (x) = f (a)+ xDf (t)dt.
a
The proof of Theorem 17.4.2 depends on the following lemma.
Lemma 17.4.3Let T ∈D^{∗}
(a,b)
and suppose DT = 0. Then there exists a constant Csuch that
∫ b
T (ϕ) = C ϕdx.
a
Proof:T
(Dϕ)
= 0 for all ϕ ∈ C_{c}^{∞}
(a,b)
from the definition of DT = 0. Let
∞ ∫ b
ϕ0 ∈ C c (a,b), a ϕ0 (x)dx = 1,
and let
( )
∫ x ∫ b
ψϕ(x) = [ϕ(t)− ϕ (y) dy ϕ0 (t)]dt
a a
for ϕ ∈ C_{c}^{∞}
(a,b)
. Thus ψ_{ϕ}∈ C_{c}^{∞}
(a,b)
and
( ∫ b )
D ψϕ = ϕ− ϕ(y)dy ϕ0.
a
Therefore,
( )
∫ b
ϕ = D ψϕ + ϕ (y)dy ϕ0
a
and so
( ∫ b ) ∫ b
T (ϕ) = T (D ψϕ)+ ϕ(y)dy T(ϕ0) = T (ϕ0)ϕ(y)dy.
a a
Let C = Tϕ_{0}. This proves the lemma.
Proof of Theorem 17.4.2 Since f and Df are both in L^{1}
(a,b)
,
∫
b
Df (ϕ)− a Df (x)ϕ (x) dx = 0.
Consider
∫ (⋅)
f (⋅)− Df (t)dt
a
and let ϕ ∈ C_{c}^{∞}
(a,b)
.
( ∫ (⋅) )
D f (⋅) − Df (t)dt (ϕ)
a
∫ b ∫ b( ∫ x )
≡ − f (x)ϕ′(x)dx+ Df (t) dt ϕ′(x)dx
a a a
∫ b∫ b
= Df (ϕ) + Df (t)ϕ′(x)dxdt
∫ a t
= Df (ϕ) − bDf (t)ϕ(t)dt = 0.
a
By Lemma 17.4.3, there exists a constant, C, such that
( ∫ (⋅) ) ∫ b
f (⋅)− Df (t) dt (ϕ) = C ϕ(x)dx
a a
for all ϕ ∈ C_{c}^{∞}
(a,b)
. Thus
∫ b ( ∫ x )
{ f (x)− Df (t)dt − C}ϕ(x)dx = 0
a a
for all ϕ ∈ C_{c}^{∞}
(a,b)
. It follows from Lemma 17.2.5 in the next section that
dt, if Df is interpreted as a weak derivative.
Somehow, this is the way it ought to be. It follows from the fundamental theorem of calculus
that f^{′}
(x)
exists for a.e. x in the classical sense where the derivative is taken in the sense of a
limit of difference quotients and f^{′}
(x)
= Df
(x)
. This raises an interesting question. Suppose
f is continuous on [a,b] and f^{′}
(x)
exists in the classical sense for a.e. x. Does it follow
that
∫ x
f (x) = f (a)+ f ′(t)dt?
a
The answer is no. You can build such an example from the Cantor function which is
increasing and has a derivative a.e. which equals 0 a.e. and yet climbs from 0 to 1. Thus this
function is not recovered from integrating its classical derivative. Thus, in a sense weak
derivatives are more agreeable than the classical ones.