Consider the dual space of C_{0}(X) where X is a polish space in which the balls have compact closure. It will
turn out to be a space of measures. To show this, the following lemma will be convenient. Recall C_{0}
(X )
is
defined as follows.
Definition 23.6.1f ∈ C_{0}
(X )
means that for every ε > 0 there exists a compact set K such that
|f (x)|
< ε whenever x
∕∈
K. Recall the norm on this space is
||f||∞ ≡ ||f || ≡ sup{|f (x)| : x ∈ X }
The next lemma has to do with extending functionals which are defined on nonnegative functions to
complex valued functions in such a way that the extended function is linear. This exact process was used
earlier with the abstract Lebesgue integral. Basically, you can do it when the functional “desires to be
linear”.
Lemma 23.6.2Suppose λ is a mapping which has nonnegative real values which is defined on thenonnegative functions in C_{0}
(X)
such that
λ(af + bg) = aλ(f)+ bλ(g) (23.23)
(23.23)
whenever a,b ≥ 0 and f,g ≥ 0. Then there exists a unique extension of λ to all of C_{0}
(X )
, Λ such thatwhenever f,g ∈ C_{0}
(X )
and a,b ∈ ℂ, it follows
Λ (af + bg) = aΛ (f)+ bΛ(g).
If
|λ (f)| ≤ C ||f||∞ (23.24)
(23.24)
then
|Λf | ≤ C ∥f∥ , |Λf | ≤ λ(|f|)
∞
Proof: Here λ is defined on the nonnegative functions. First extend it to the continuous real valued
functions. There is only one way to do it and retain the map is linear. Let C_{0}(X; ℝ) be the real-valued
functions in C_{0}(X) and define
It is necessary to verify that Λ_{R}(cf) = cΛ_{R}(f) for all c ∈ ℝ. But (cf)^{±} = cf^{±},if c ≥ 0 while
(cf)^{+} = −c(f)^{−}, if c < 0 and (cf)^{−} = (−c)f^{+}, if c < 0. Thus, if c < 0,
It remains to verify the claim about continuity of Λ in case of 23.24. This is really pretty obvious
because f_{n}→ 0 in C_{0}
(X)
if and only if the positive and negative parts of real and imaginary parts also
converge to 0 and λ of each of these converges to 0 by assumption. What of the last claim that
Let L ∈ C_{0}(X)^{′}. Also denote by C_{0}^{+}(X) the set of nonnegative continuous functions defined on
X.
Definition 23.6.3Letting L ∈ C_{0}
(X )
^{′}, define for f ∈ C_{0}^{+}(X)
λ(f) = sup{|Lg| : |g| ≤ f}.
Note that λ(f) < ∞ because
|Lg|
≤
∥L∥
∥g∥
≤
∥L∥
||f|| for
|g|
≤ f. Isn’t this a lot like the total
variation of a vector measure? Indeed it is, and the proof that λ wants to be linear is also
similar to the proof that the total variation is a measure. This is the content of the following
lemma.
Lemma 23.6.4If c ≥ 0,λ(cf) = cλ(f),f_{1}≤ f_{2}implies λ
(f1)
≤ λ
(f2)
, and
λ(f1 + f2) = λ(f1) + λ(f2).
Also
0 ≤ λ (f ) ≤ ∥L∥||f ||∞
Proof: The first two assertions are easy to see so consider the third. For i = 1,2 and f_{i}∈ C_{0}^{+}
Let Λ be the unique linear extension of Theorem 20.7.8. It is just like defining the integral for functions
when you understand it for nonnegative functions. Then from the above lemma,
What follows is the Riesz representation theorem for C_{0}(X)^{′}.
Theorem 23.6.6Let L ∈ (C_{0}(X))^{′}for X a Polish space with closed balls compact. Then there exists afinite Radon measure μ and a function σ ∈ L^{∞}(X,μ) such that for all f ∈ C_{0}
(X )
,
∫
L(f) = fσd μ.
X
Furthermore,
μ (X) = ||L||, |σ| = 1 a.e.
and if
∫
ν(E) ≡ σdμ
E
then μ =
|ν|
.
Proof: From the above there exists a unique Radon measure μ such that for all f ∈ C_{c}
Since μ is both inner and outer regular, C_{c}(X) is dense in L^{1}(X,μ). (See Theorem 22.2.4 for more than is
needed.) Therefore L extends uniquely to an element of (L^{1}(X,μ))^{′},
^L
. By the Riesz representation
theorem for L^{1} for finite measure spaces, there exists a unique σ ∈ L^{∞}(X,μ) such that for all
f ∈ L^{1}