I'm currently studying the proof of Theorem 6.19 from Real and Complex Analysis by Walter Rudin, which concerns the Riesz-Markov-Kakutani Theorem. I'd like to verify my understanding as well as some minor questions. $\vert \mu \vert$ is the total variation of the measure $\mu$.
Theorem 6.19
If $X$ is a locally compact Hausdorff space, then every bounded linear functional $\Phi$ on $C_0(X)$ is represented by a unique regular complex Borel measure $\mu$, in the sense that
$$
\tag{1}
\Phi f=\int_X f d \mu
$$
for every $f \in C_0(X)$. Moreover, the norm of $\Phi$ is the total variation of $\mu$ :
$$
\tag{2}
\|\Phi\|=|\mu|(X) .
$$
Proof
Consider a given bounded linear functional $\Phi$ on $C_0(X)$. Assume $\|\Phi\|=1$, without loss of generality. We shall construct a positive linear functional $\Lambda$ on $C_c(X)$, such that
$$
\tag{4}
|\Phi(f)| \leq \Lambda(|f|) \leq\|f\| \quad\left(f \in C_c(X)\right),
$$
where $\|f\|$ denotes the supremum norm.
Once we have this $\Lambda$, we associate with it a positive Borel measure $\lambda$, as in Theorem 2.14. The conclusion of Theorem 2.14 shows that $\lambda$ is regular if $\lambda(X)<\infty$. Since
$$
\lambda(X)=\sup \left\{\Lambda f: 0 \leq f \leq 1, f \in C_c(X)\right\}
$$
and since $|\Lambda f| \leq 1$ if $\|f\| \leq 1$, we see that actually $\lambda(X) \leq 1$.
We also deduce from (4) that
$$
\tag{5}
|\Phi(f)| \leq \Lambda(|f|)=\int_X|f| d \lambda=\|f\|_1 \quad\left(f \in C_c(X)\right) .
$$
The last norm refers to the space $L^1(\lambda)$. Thus $\Phi$ is a linear functional on $C_c(X)$ of norm at most 1 , with respect to the $L^1(\lambda)$-norm on $C_c(X)$. There is a normpreserving extension of $\Phi$ to a linear functional on $L^1(\lambda)$, and therefore Theorem 6.16 (the case $p=1$ ) gives a Borel function $g$, with $|g| \leq 1$, such that
$$
\tag{6}
\Phi(f)=\int_X f g d \lambda \quad\left(f \in C_c(X)\right) .
$$
Each side of (6) is a continuous functional on $C_0(X)$, and $C_c(X)$ is dense in $C_0(X)$. Hence (6) holds for all $f \in C_0(X)$, and we obtain the representation (1) with $d \mu=g d \lambda$.
Since $\|\Phi\|=1,(6)$ shows that
$$
\tag{7}
\int_X|g| d \lambda \geq \sup \left\{|\Phi(f)|: f \in C_0(X),\|f\| \leq 1\right\}=1 .
$$
We also know that $\lambda(X) \leq 1$ and $|g| \leq 1$. These facts are compatible only if $\lambda(X)=1$ and $|g|=1$ a.e. [ $\lambda]$. Thus $d|\mu|=|g| d \lambda=d \lambda$, by Theorem 6.13, and
$$
|\mu|(X)=\lambda(X)=1=\|\Phi\|,
$$
which proves (2).
Questions:
- Shouldn't it be $C_{c}$ instead of $C_{0}$ in the statement:
Each side of (6) is a continuous functional on $C_{0}(X)$
- Is the norm-preserving extension a result of the Hahn-Banach Theorem?
There is a normpreserving extension of $\Phi$ to a linear functional on $L^1(\lambda)$
- Is the extension of equation (6) to $C_{0}(X)$ a consequence of the BLT-Theorem? (See definition at the end)
Each side of (6) is a continuous functional on $C_0(X)$, and $C_c(X)$ is dense in $C_0(X)$. Hence (6) holds for all $f \in C_0(X)$,
-
By theorem 6.16 ( dual of $L^{1}(\lambda)$ is isometric isomorph to $L^{\infty}$), $g \in L^{\infty}$. The definition $d \mu=g d \lambda$ is only valid if $g \in L^{1}$ .
-
Am I correct in understanding that equation (7) is derived as described below?
My Thoughts:
-
I think this follows from $L^{\infty}(\lambda) \subset L^{1}(\lambda)$ which is valid since $\lambda$ is a finite measure.
-
By equation (6), for all $f \in C_{0}(X)$ with $\vert \vert f \vert \vert \leq 1$ the inequality
$$\vert \Phi(f) \vert = \vert \int_{X} fg d \lambda \vert \leq \int_{X} \vert fg \vert d \lambda \leq \vert \vert f \vert \vert \int_{X} \vert g \vert d \lambda \leq \int_{X} \vert g \vert d \lambda$$
holds and hence
$$ \int_{X} \vert g \vert d \lambda \geq \sup\{ \Phi(f): f \in C_{0}(X),\; \vert \vert f \vert \vert \leq 1 \}:=\vert \vert \Phi \vert \vert =1$$
Theorems
Theorem 2.14
Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$.
Then there exists a $\sigma$-algebra $\mathcal{M}$ in $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $\mathcal{M}$ which represents $\Lambda$ in the sense that
$$\Lambda f=\int_X f d \mu$$ for every $f \in C_c(X)$
and which has the following additional properties:
- $\mu(K)<\infty$ for every compact set $K \subset X$.
- For every $E \in \mathcal{M}$, we have
$
\mu(E)=\inf \{\mu(V): E \subset V, V\, \text{ is open }\} .
$ - The relation
$
\mu(E)=\sup \{\mu(K): K \subset E, K\; \text{ is compact }\}
$
holds for every open set $E$, and for every $E \in \mathcal{M}$ with $\mu(E)<\infty$. - If $E \in \mathcal{M}, A \subset E$, and $\mu(E)=0$, then $A \in \mathcal{M}$.
Theorem 5.16 (Hahn-Banach)
If $M$ is a subspace of a normed linear space $X$ and if $f$ is $a$ bounded linear functional on $M$, then $f$ can be extended to a bounded linear functional $F$ on $X$ so that $\|F\|=\|f\|$.
B.L.T. Theorem
Suppose that $Z$ is a normed space, $Y$ is a Banach space, and $\mathcal{S} \subset Z$ is a dense linear subspace of $Z$. If $T: \mathcal{S} \rightarrow Y$ is a bounded linear transformation (i.e. there exists $C<\infty$ such that $\|T z\| \leq C\|z\|$ for all $z \in \mathcal{S}$ ), then $T$ has a unique extension to an element of $L(Z, Y)$.
Theorem 6.12
Let $\mu$ be a complex measure on a $\sigma$-algebra $\mathfrak{M}$ in $X$. Then there is a measurable function $h$ such that $|h(x)|=1$ for all $x \in X$ and such that
$$
d \mu=h d|\mu| .
$$
Theorem 6.13
Suppose $\mu$ is a positive measure on $\mathfrak{M}, g \in L^1(\mu)$, and
$$
\lambda(E)=\int_E g d \mu \quad(E \in \mathfrak{M}) .
$$
Then
$$
|\lambda|(E)=\int_E|g| d \mu \quad(E \in \mathfrak{M})
$$
Theorem 6.16
Suppose $1 \leq p<\infty, \mu$ is a $\sigma$-finite positive measure on $X$, and $\Phi$ is a bounded linear functional on $L^p(\mu)$. Then there is a unique $g \in L^q(\mu)$, where $q$ is the exponent conjugate to $p$, such that
$$
\Phi(f)=\int_X f g d \mu \quad\left(f \in L^p(\mu)\right) .
$$
Moreover, if $\Phi$ and $g$ are related as in (1), we have
$$
\|\Phi\|=\|g\|_q .
$$
Best Answer
First and third questions: No to both. The proof does the following: given two functions $f,g: A \rightarrow B$, they show that they are equal by showing
In particular, both sides of (6) are already defined on $C_0(X)$. We just don’t know that they’re equal.
Second question: Hahn-Banach is probably overkill (but not incorrect) in this case. I’d say it follows from BLT since $C_c(X)$ is dense in $L^1(\lambda)$ for the $L^1$ norm.
Digression
I think the result above is classical, but here’s a sketch of proof if you’re unconvinced: because $\lambda$ is finite, you can restrict to the case where $X$ is (Hausdorff) compact. Consider the collection of Borel subsets $A \subset X$ such that $1_A$ is in the closure of $C(X)$. It contains the closed subsets (outer regularity+sort of Urysohn) and is a monotone class (formal), so it’s the whole $\sigma$-algebra. Now every $L^1$ function is the limit of a finite linear combination of indicators of Borel subset of $X$.
If you don’t want to use that $C_c(X)$ is dense in $L^1(X)$, you can apply BLT to replace $C_c$ with its closure in $L^1$ then Hahn-Banach to extend to all of $L^1$.
Fourth and fifth questions: your thoughts are correct in both cases.