17.3 Vector Lattices

Definition 17.3.1 (Vector Lattice).label Let $(E, \le)$ be an ordered vector space, then $E$ is a vector lattice if for any $x, y \in E$, $x \vee y = \sup\bracs{x, y}$ and $x \wedge y = \inf\bracs{x, y}$ exist.

Definition 17.3.2 (Absolute Value).label Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| = x \vee -x$ is the absolute value of $x$.

Lemma 17.3.3.label Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| \ge 0$.

Proof. For any $x \in E$,

\[2|x| = 2(x \vee (-x)) \ge x + -x = 0\]

$\square$

Definition 17.3.4 (Disjoint).label Let $(E, \le)$ be a vector lattice and $x, y \in E$, then $x$ and $y$ are disjoint, denoted $x \perp y$, if $|x| \wedge |y| = 0$.

Lemma 17.3.5.label Let $E$ be a vector lattice and $a, b, c, d \in E$, then

(1)
$a + b = a \vee b + a \wedge b$.
(2)
$b \le a + |b - a|$.
(3)
If $c \ge 0$, then $(a + c) \vee b \le a \vee b + c$.
(4)
$|a \vee c - b \vee c| \le |a - b|$.
(5)
$|a \vee c - b \vee d| \le |a - b| \vee |c- d|$

Proof. (1): By translation invariance,

\begin{align*}x \vee y + x \wedge y - x - y&= 0 \vee (y - x) + (x - y) \wedge 0 \\&= 0 \vee (y - x) - 0 \vee (x - y) = 0\end{align*}

(2): $b = a + b - a \le a + |b - a|$.

(3): Since $c \ge 0$,

\[(a + c) \vee b = (a + c) \vee (b - c + c) = a \vee (b - c) + c \le a \vee b + c\]

(4): By (2) and then (3),

\begin{align*}a \vee c - b \vee c&\le (b + |a - b|) \vee c - b \vee c \\&\le b \vee c + |a - b| - b \vee c \le |a - b|\end{align*}

Similarly, $b \vee c - a \vee c \le |b - a| = |a - b|$.

(5): Finally,

\begin{align*}a \vee c - b \vee d&\le (b + |a - b|) \vee (d + |c- d|) - b \vee d \\&\le (b + |a - b| \vee |c- d|) \vee (d + |a - b| \vee |c- d|) - b \vee d \\&= b \vee d + |a - b| \vee |c- d| - b \vee d \\&\le |a - b| \vee |c- d|\end{align*}

Similarly,

\[b \vee d - a \vee c \le |b - a| \vee |d - c| = |a - b| \vee |c- d|\]

$\square$

Proposition 17.3.6.label Let $(E, \le)$ be a vector lattice, then:

(1)
Let $x \in E$, $x^{+} = x \vee 0 \ge 0$, and $x^{-} = -(x \wedge 0) \ge 0$, then $x = x^{+} - x^{-}$ and $|x| = x^{+} + x^{-}$. Moreover, $(x^{+}, x^{-})$ are the unique disjoint non-negative elements of $E$ such that $x = x^{+} - x^{-}$.

For any $x, y \in E$ and $\lambda \in \real$,

(2)
$|\lambda x| = |\lambda| \cdot |x|$
(3)
$|x + y| \le |x| + |y|$.

Finally, for any $x, y \in E$ with $x, y \ge 0$,

(4)
$[0, x] + [0, y] = [0, x + y]$.

Proof, [V.1.1, SW99]. (1): By Lemma 17.3.5,

\[x = x + 0 = x \vee 0 + x \wedge 0 = x^{+} - x^{-}\]

By Proposition 17.1.3 and Lemma 17.3.3,

\[x^{+} + x^{-} = x \vee 0 + (-x \vee 0) = x \vee -x \vee 0 = |x|\]

and

\[x^{+} \vee x^{-} = x \vee 0 \vee (-x) \vee 0 = |x|\]

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

\[x^{+} \wedge x^{-} = x^{+} + x^{-} - x^{+} \vee x^{-} = 0\]

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

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

\[x^{+} = x \vee 0 = (y - z) \vee 0 \le y \vee 0 = y\]

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 $z = x^{-}$.

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

\[|\lambda x| = (\lambda x) \vee (-\lambda x) = \lambda (x \vee -x) = \lambda |x|\]

(3): For any $x, y \in E$, by Proposition 17.1.3,

\begin{align*}x^{+} + y^{+}&= (x \vee 0) + (y \vee 0) = x \vee y \vee (x + y) \vee 0 \\&\ge (x + y) \vee 0 = (x + y)^{+}\end{align*}

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

\[|x+y| = (x+y)^{+} + (x + y)^{-} \le x^{+} + y^{+} + x^{-} + y^{-} = |x| + |y|\]

(4): 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

\[v = z - z \wedge x = z + (-z \vee -x) = (0 \vee z - x) \le z - x \le y\]

$\square$

Lemma 17.3.7.label Let $(E, \le)$ be a vector lattice, $C = \bracs{x \in E|x \ge 0}$ and $\phi: C \to [0, \infty)$ such that:

(1)
For any $x \in C$ and $\lambda \in \real$ with $\lambda \ge 0$, $\phi(\lambda x) = \lambda \phi(x)$.
(2)
For any $x, y \in C$, $\phi(x + y) = \phi(x) + \phi(y)$.

then the mapping

\[\Phi: E \to \real \quad x \mapsto \phi(x^{+}) - \phi(x^{-})\]

is a positive linear functional on $E$.

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

\begin{align*}\Phi(\lambda x)&= \phi((\lambda x)^{+}) - \phi((\lambda x)^{-}) \\&= \lambda\phi(x^{+}) - \lambda\phi(x^{-}) = \lambda\Phi(x)\end{align*}

Likewise, if $\lambda < 0$, then

\begin{align*}\Phi(\lambda x)&= \phi((\lambda x)^{+}) - \phi((\lambda x)^{-}) \\&= -\lambda\phi(x^{-}) + \lambda\phi(x^{+}) = \lambda\Phi(x)\end{align*}

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

\begin{align*}z^{+} + x^{-} + y^{-}&= z^{-} + x^{+} + y^{+} \\ \phi(z^{+}) + \phi(x^{-}) + \phi(y^{-})&= \phi(z^{-}) + \phi(x^{+}) + \phi(y^{+}) \\ \phi(z^{+}) - \phi(z^{-})&= \phi(x^{+}) - \phi(x^{-}) + \phi(y^{+}) - \phi(y^{-}) \\ \Phi(z)&= \Phi(x) + \Phi(y)\end{align*}

$\square$

Proposition 17.3.8.label Let $(E, \le)$ be an ordered vector space. If for any $x, y \in E$ with $x, y \ge 0$, $[0, x] + [0, y] = [0, x + y]$, then:

(1)
$E^{b} = E^{+}$.
(2)
The order bound dual $E^{b}$ equipped with its canonical ordering is a vector lattice.
(3)
$E^{b}$ is order complete.
(4)
For any $\phi \in E^{b}$ and $x \in E$ with $x \ge 0$,
\[|\phi|(x) = \sup\bracs{\phi(y)|y \in E, |y| \le x}\]

Proof, [V.1.4, SW99]. (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^{+} \in E^{b}$, and

\[\Phi^{+}: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))\]

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

\begin{align*}\Phi^{+}(\lambda x + y)&= \sup(f([0, \lambda x + y])) = \sup(f(\lambda [0,x]) + f([0, y])) \\&= \lambda \sup(f([0, x])) + \sup(f([0, y])) = \lambda \Phi^{+}(x) + \Phi^{+}(y)\end{align*}

Let

\[\Phi: E \to \real \quad x \mapsto \Phi^{+}(x^{+}) - \Phi^{+}(x^{-})\]

then $\Phi$ is a positive linear functional by Lemma 17.3.7.

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

\[\phi: C \to [0, \infty) \quad x \mapsto \sup_{f \in A}f(x)\]

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

\[f(x + y) = f(x) + f(y) \le \phi(x) + \phi(y)\]

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

\[h(x + y) = h(x) + h(y) \ge f(x) + g(y)\]

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

By Lemma 17.3.7, the mapping

\[\Phi: E \to \real \quad x \mapsto \phi(x^{+}) - \phi(x^{-})\]

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),

\begin{align*}|\phi|(x)&= \sup(\phi([0, x])) + \sup(-\phi([0, x])) \\&= \sup\bracs{\phi(u) - \phi(v)| u, v \in [0, x]}\end{align*}

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

\[|\phi|(x) \le \sup\bracs{\phi(y)|y \in E, |y| \le x}\]

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

\[\phi(y) = \phi(y^{+}) - \phi(y^{-}) \le \sup\bracs{\phi(u) - \phi(v)| u, v \in [0, x]}= |\phi|(x)\]

Therefore

\[|\phi|(x) = \sup\bracs{\phi(y)|y \in E, |y| \le x}\]

$\square$

Jerry's Digital Garden

17.3 Vector Lattices

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