17.3 Dual of $C_{0}$

Proof. (1), (2): Since $C_{0}(X; \real)$ is a Banach lattice, $C_{0}(X; \real)^{*}$ is also a Banach lattice by Proposition 13.2.3. Therefore there exists positive linear functionals $I^{+}, I^{-} \in C(X; \real)^{*}$ such that $I = I^{+} - I^{-}$ and $I^{+} \perp I^{-}$.

(3): By Proposition 13.2.2.$\square$

Proof. Let $\mu, \nu \in M_{R}(X; E)$, then for any $A \in \cb_{X}$, $|\mu + \nu|(A) \le |\mu|(A) + |\nu|(A)$. Let $\eps > 0$, then by outer regularity and Proposition 17.1.4, there exists $K \subset A$ compact and $U \in \cn^{o}(A)$ such that $(|\mu| + |\nu|)(A \setminus K), (|\mu| + |\nu|)(U \setminus A) < \eps$. Therefore $|\mu + \nu|$ is regular on all Borel sets, and hence Radon.$\square$

Proof [Theorem 7.17, Fol99]. Let $f \in C_{0}(X; \complex)$, then

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

On the other hand, let $\seqf{A_j}\subset \cb_{X}$ such that $X = \bigsqcup_{j = 1}^{n}A_{j}$. Let $\eps > 0$, then by Proposition 17.1.3 applied to $|\mu|$, there exists compact sets $\seqf{K_j}$ such that for each $1 \le j \le n$, $K_{j} \subset A_{j}$ and $|\mu(K_{j}) - \mu(A_{j})| < \eps$. 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 $\phi = \sum_{j = 1}^{n} \ol{\sgn(\mu(K_j))}\phi_{j}$, then $\norm{\phi}_{u} \le 1$ and

so

As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)}\ge \sum_{j = 1}^{n} |\mu(A_{j})|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)}\ge |\mu|(X)$. Therefore the map $\mu \mapsto I_{\mu}$ is isometric.

Finally, let $I \in C_{0}(X; \complex)^{*}$, then there exists bounded linear functionals $I_{r}, I_{i} \in C_{0}(X; \real)^{*}$ such that for any $f \in C_{0}(X; \real)$,

By Lemma 17.3.1, there exists bounded positive linear functionals $I_{r}^{+}, I_{r}^{-}, I_{i}^{+}, I_{i}^{-}$ such that for any $f \in C_{0}(X; \real)$,

Thus by the Riesz Representation Theorem, there exists finite Radon measures $\mu_{r}^{+}, \mu_{r}^{-}, \mu_{i}^{+}, \mu_{i}^{-} \in M_{R}(X; \complex)$ such that for any $f \in C_{0}(X; \real)$,

Let $\mu = \mu_{r}^{+} - \mu_{r}^{-} + i\mu_{i}^{+} - i\mu_{i}^{-}$, then $I = I_{\mu}$, and the map $\mu \mapsto I_{\mu}$ is surjective.$\square$

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

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

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

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 $f \in C_{0}(X \times B; K)$,

Now, let

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

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 $A = \bigsqcup_{n \in \nat}A_{n}$,

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

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