13.1 Vector Lattices

Proof. (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.

(2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$.$\square$

Proof. For any $x \in E$,

Proof. (1): By Proposition 13.1.2,

(2): By (1),

By Proposition 13.1.2 and Lemma 13.1.10,

and

Since $x^{+}, x^{-} \ge 0$, $|x^{+}| = x^{+}$ and $|x^{-}| = x^{-}$, so by (1),

so $x^{+} \perp x^{-}$.

Now, let $y, z \in E$ with $y, z \ge 0$ such that $x = y - z$, then

and $x^{-} \le z$. If $y \perp z$, then $y - x^{+} \perp z - x^{-}$. Since $y - x^{+} = z - x^{-}$, $y - x^{+} = z - x^{-} = 0$, so $y = x^{+}$ and $z = x^{-}$.

(3): For any $\lambda > 0$, by (LO2),

(4): For any $x, y \in E$, by Proposition 13.1.2,

Likewise, $x^{-} + y^{-} \ge (x + y)^{-}$. Thus

(5): Let $x, y \in E$ with $x, y \ge 0$, then $[0, x] + [0, y] \subset [0, x + y]$. For any $z \in [0, x + y]$, let $u = z \wedge x$ and $v = z - u$, then

Proof. For any $\lambda \in \real$ with $\lambda \ge 0$,

Likewise, if $\lambda < 0$, then

For any $x, y \in E$, let $z = x + y$, then $z = z^{+} - z^{-} = x^{+} + y^{+} - x^{-} - y^{-}$. Thus

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