Consider the dual space of C_{0}(X) where X is a locally compact Hausdorff space. It will turn
out to be a space of measures. To show this, the following lemma will be convenient. Recall
this space is defined as follows.
Definition 25.6.1f ∈ C_{0}
(X )
means that for every ε > 0 there exists acompact set K such that
|f (x)|
< ε whenever x
∕∈
K. Recall the norm on this spaceis
||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 25.6.2Suppose λ is a mapping which has nonnegative values which is definedon the nonnegative functions in C_{0}
(X)
such that
λ(af + bg) = aλ(f)+ bλ(g) (25.19)
(25.19)
whenever a,b ≥ 0 and f,g ≥ 0. Then there exists a unique extension of λ to all of C_{0}
(X )
, Λ
such that whenever f,g ∈ C_{0}
(X )
and a,b ∈ ℂ, it follows
Λ (af + bg) = aΛ (f)+ bΛ(g).
If
|λ (f)| ≤ C ||f|| (25.20)
∞
(25.20)
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
( + − ) ( +) ( − ) + −
ΛR(f) = ΛR f − f = ΛR f − ΛR f = λf − λf
for f ∈ C_{0}(X; ℝ). This is the only thing possible if Λ_{R} is to be linear. Is
Λ_{R}
equivalently, since Λ_{R} = λ on nonnegative functions and λ tries to be linear,
( + − − ) ( − + +)
λ (f + g) + f + g = λ (f + g) + f + g
But this is so because
(f + g)
^{+} + f^{−} + g^{−} =
(f + g)
^{−} + f^{+} + g^{+}.
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 25.20. 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 25.6.3Letting 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 25.6.4If c ≥ 0,λ(cf) = cλ(f),f_{1}≤ f_{2}implies λf_{1}≤ λf_{2}, 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.
Therefore, Λ is a positive linear functional on C_{0}(X). In particular, it is a positive linear
functional on C_{c}
(X )
. Thus there are now two linear continuous mappings L,Λ which are
defined on C_{0}
(X )
. The above shows that in fact
∥Λ∥
≤
∥L∥
. Also, from the definition of
Λ
|Lg| ≤ λ (|g|) = Λ(|g|) ≤ ∥Λ ∥∥g∥∞
so in fact,
∥L ∥
≤
∥Λ ∥
showing that these two have the same operator norms.
∥L ∥ = ∥Λ∥ (25.22)
(25.22)
By Theorem 21.6.2 on Page 1490, there exists a unique measure μ such that
∫
Λf = fdμ
X
for all f ∈ C_{c}(X). This measure is inner regular on all open sets and on all measurable sets
having finite measure. In fact, it is actually a finite measure.
Lemma 25.6.5Let L ∈ C_{0}
(X)
^{′}as above. Then letting μ be the Radon measure justdescribed, it follows μ is finite and
μ(X ) = ∥Λ∥ = ∥L∥
Proof:First of all, it was observed above in 25.22 that
∥Λ∥
=
∥L∥
.
Now X is an open set and so
μ (X) = sup {μ(K ) : K ⊆ X }
and so letting K ≺ f ≺ X for one of these K, it also follows
What follows is the Riesz representation theorem for C_{0}(X)^{′}.
Theorem 25.6.6Let L ∈ (C_{0}(X))^{′}for X a locally compact Hausdorf space. Thenthere exists a finite Radon measure μ and a function σ ∈ L^{∞}(X,μ) such that for allf ∈ C_{0}
(X )
,
∫
L(f) = fσdμ.
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}
(X )
,
∫
Λf = fdμ
X
Then for f ∈ C_{c}
(X )
,
∫
|Lf | ≤ λ(|f|) = Λ (|f|) = |f|dμ = ||f|| 1 .
X L (μ)
Since μ is both inner and outer regular thanks to it being finite, C_{c}(X) is dense
in L^{1}(X,μ). (See Theorem 15.2.4 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}