Proof. (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^{+} \in E^{b}$, and

then for any $x, y \in C$ and $\lambda \in \real$ with $\lambda \ge 0$,

Let

then $\Phi$ is a positive linear functional by Lemma 13.1.13.

For any $x \in C$, $\Phi(x) \ge \phi(x)$, so $\Phi \ge \phi$. On the other hand, let $\psi \in \hom(E; \real)$ such that $\psi \ge 0$ and $\psi \ge \phi$, then for each $x \in C$, $\psi(x) \ge \sup(\phi([0, x]))$. Therefore $\Phi = 0 \vee \phi$.

Since $0 \wedge \phi = -(0 \vee -\phi)$, $\phi = (0 \vee \phi) - (0 \vee -\phi)$ is a sum of two positive linear functionals, so $E^{b} = E^{+}$.

(2): For any $x, y \in E^{b}$, $x \wedge y = (x - y) \wedge 0 + y$ and $x \vee y = (x - y) \vee 0 + y$, so $E^{b}$ is a lattice.

(3): Let $A \subset E^{b}$ be order bounded, then since $E^{b}$ is a lattice, assume without loss of generality that $A$ is upward-directed under the canonical ordering. Let

then for any $\lambda \ge 0$ and $x \in C$, $\phi(\lambda x) = \lambda \phi(x)$. Let $x, y \in C$, then for any $f \in A$,

As this holds for all $f \in A$, $\phi(x + y) \le \phi(x) + \phi(y)$. On the other hand, for any $f, g \in A$, there exists $h \in A$ such that

Since such an $h \in A$ exists for all $f, g \in A$, $\phi(x + y) \ge \phi(x) + \phi(y)$.

By Lemma 13.1.13, the mapping

is a positive linear functional on $E$. By definition $\Phi \ge f$ for all $f \in A$. If $\psi \in \hom(E; \real)$ is a linear functional such that $\psi \ge f$ for all $f \in A$, then for any $x \in C$, $\psi(x) \ge \sup_{f \in A}f(x)$. Therefore $\Phi = \sup(A)$, and $E^{b}$ is order complete.

(3): Let $\phi \in E^{b}$ and $x \in E$ with $x \ge 0$, then by (1),

For any $u, v \in [0, x]$, $|u - v| \le x$, so

On the other hand, for any $y \in E$ with $|y| \le x$, $y^{+}, y^{-} \le x$, so

Therefore