Theorem 17.3.5: Singer's Representation Theorem

Theorem 17.3.5 (Singer’s Representation Theorem). Let $X$ be an LCH space and $E$ be a normed space over $K \in \RC$. For each $\mu \in M_{R}(X; E^{*})$, let

\[I_{\mu}: C_{0}(X; E) \to K \quad \dpn{f, I_\mu}{C_0(X; E)}= \int \dpn{f, d\mu}{E}\]

then the map

\[M_{R}(X; E^{*}) \to C_{0}(X; E)^{*} \quad \mu \mapsto I_{\mu}\]

is an isometric isomorphism.

Proof [Hen96]. (Isometric): Let $\mu \in M_{R}(X; E^{*})$, then for any $f \in C_{0}(X; E)$,

\[|\dpn{f, I_\mu}{C_0(X; E)}| \le \int \norm{f}_{E} d|\mu| \le \norm{f}_{u} \cdot \norm{\mu}_{\text{var}}\]

so $\norm{I_\mu}_{C_0(X; E)^*}\le \norm{\mu}_{\text{var}}$.

On the other hand, let $\seqf{A_j}\subset \cb_{X}$ such that $\bigsqcup_{j = 1}^{n} A_{j} = X$ and $\eps > 0$.

By Proposition 17.1.3 applied to $|\mu|$, there exists $\seqf{K_j}$ compact such that for each $1 \le j \le n$, $K_{j} \subset A_{j}$ and $\norm{\mu(K_j) - \mu(A_j)}_{E^*}< \eps$. By outer regularity and Urysohn’s lemma, there exists $\seqf{\phi_j}\subset C_{c}(X; [0, 1])$ with disjoint support such that for each $1 \le j \le n$, $\norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)}< \eps$.

Let $\seqf{x_j}\subset \overline{B_E(0, 1)}$ such that for each $1 \le j \le n$, $\dpn{x_j, \mu(K_j)}{E}> \norm{\mu(K_j)}_{E^*}- \eps$. Define $\phi = \sum_{j = 1}^{n} x_{j} \phi_{j}$, then $\norm{\phi}_{u} \le 1$ and

\begin{align*}\abs{\sum_{j = 1}^n \norm{\mu(A_j)}_{E^*} - \int \dpn{\phi, d\mu}{E}}&\le \sum_{j = 1}^{n} \norm{\mu(A_j) - \mu(K_j)}_{E^*}\\&+ \sum_{j = 1}^{n} \norm{\phi_j - \one_{K_j}}_{L^1(|\mu|)}+ n\eps < 3n\eps\end{align*}

As such a $\phi \in C_{0}(X; E)$ exists for all $\eps > 0$ and $\seqf{A_j}$, $\norm{I_\mu}_{C_0(X; E)^*}\ge \norm{\mu}_{\text{var}}$. Therefore the map $\mu \mapsto I_{\mu}$ is isometric.

(Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak*-topology and

\[T: C_{0}(X; E) \to C_{0}(X \times B; K) \quad (Tf)(x, \phi) = \dpn{f(x), \phi}{E}\]

then $T$ is maps $C_{0}(X; E)$ continuously into a subspace of $C_{0}(X \times B; K)$.

Let $I \in C_{0}(X; E)^{*}$, then by the Hahn-Banach theorem, there exists $\ol{I}\in C_{0}(X \times B; K)^{*}$ such that $\ol I \circ T = I$. By Alaoglu’s Theorem, $B$ is a compact Hausdorff space. Therefore $X \times B$ is a LCH space by Proposition 4.20.10. By the Riesz Representation Theorem, there exists $\mu \in M_{R}(X \times B; K)$ such that for any $f \in C_{0}(X \times B; K)$,

\[\dpn{f, \ol I}{C_0(X \times B; K)}= \int_{X \times B}f d\mu\]

Now, let

\[\nu: \cb_{X} \to E^{*} \quad \dpn{y,\nu(A)}{E}= \int_{X \times B}\one_{A}(x) \cdot \dpn{y, \phi}{E}\mu(dx, d\phi)\]

then for each $A \in \cb_{X}$ and $y \in E$,

\[|\dpn{y,\nu(A)}{E}| \le \int_{X \times B}\one_{A}(x) \cdot \norm{y}_{E} |\mu|(dx, d\phi) \le \norm{y}_{E} \cdot |\mu|(A \times B)\]

As the above holds for all $y \in E$, $\norm{\nu(A)}_{E^*}\le |\mu|(A \times B)$. Moreover, for any pairwise disjoint sequence $\seq{A_n}\subset \cb_{X}$ and $A \in \cb_{X}$ such that $A = \bigsqcup_{n \in \nat}A_{n}$,

\[\limv{n}\norm{\nu(A) - \sum_{j = 1}^n \nu(A_n)}_{E^*}\le \limv{n}|\mu|\paren{\bigcup_{k > n}A_k \times B}= 0\]

so $\nu$ is a vector measure on $\cb_{X}$.

Since $\norm{\nu(A)}_{E^*}\le |\mu|(A \times B)$ for all $A \in \cb_{X}$, $|\nu|(A) \le |\mu|(A \times B)$ for all $A \in \cb_{X}$, and $\nu$ is a Radon measure by Lemma 17.1.6.

Finally, let $f \in C_{0}(X; K)$ and $y \in E$, then

\begin{align*}\int_{X} \dpn{y \cdot f, d\nu}{E}&= \int_{X \times B}f(x)\dpn{y, \phi}{E}\mu(dx, d\phi) \\&= \int_{X \times B}T(y \cdot f)(x, \phi)\mu(dx, d\phi) = \dpn{y \cdot f, I}{C_0(X; E)}\end{align*}

Therefore for any $f \in C_{0}(X; K) \otimes E$, $\int_{X} \dpn{f, d\nu}{E}= \dpn{f, I}{C_0(X; E)}$. By Proposition 4.21.3, $C_{0}(X; K) \otimes E$ is a dense subspace of $C_{0}(X; E)$, so

\[\int_{X} \dpn{f, d\nu}{E}= \dpn{f, I}{C_0(X; E)}\quad \forall f \in C_{0}(X; E)\]

$\square$

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