13.2 Banach Lattices
Definition 13.2.1 (Banach Lattice). Let $(E, \normn{\cdot}_{E})$ be a Banach space over $\real$ and $\le$ be a partial order on $E$, then $(E, \normn{\cdot}_{E}, \le)$ is a Banach lattice if for any $x, y \in E$, $|x| \le |y|$ implies that $\normn{x}_{E} \le \normn{x}_{E}$.
Proposition 13.2.2. Let $(E, \normn{\cdot}_{E}, \le)$ be a Banach lattice, then:
For any $x \in E$, $\normn{\ |x|\ }_{E} = \normn{x}_{E}$.
For any $x \in E$, $\normn{x^+}_{E}, \normn{x^-}_{E} \le \normn{x}_{E}$.
For any $x \in E$ and positive linear functional $\phi \in E^{*}$,
\[\norm{\phi}_{E^*}= \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x \ge 0, \norm{x}_E = 1}\]
Proof. (1): Let $x \in E$, then $|\ |x|\ | = |x|$, so $\normn{\ |x|\ }_{E} = \normn{x}_{E}$.
(2): Let $x \in E$, then $|x| = x^{+} + x^{-}$, so
(3): Since $|x| = x \vee (-x)$,
Proposition 13.2.3. Let $(E, \normn{\cdot}_{E}, \le)$ be a Banach lattice and $(E^{*}, \normn{\cdot}_{E^*}, \le)$ be its dual, equipped with the canonical ordering, then $E^{*}$ is a Banach lattice.
Proof. Let $x, y \in E$ and $z \in [x, y]$, then
so $[x, y]$ is bounded by $\norm{x}_{E} \vee \norm{y}_{E}$. Thus $E^{*} \subset E^{b}$.
Let $\phi \in E^{*}$, then by Proposition 13.1.14, $0 \vee \phi$ exists in $E^{b}$. For any $x \in E$ with $x \ge 0$,
so for arbitrary $x \in E$,
and $\norm{0 \vee \phi}_{E^*}\le \norm{\phi}_{E^*}$. Therefore $E^{*}$ is a vector lattice.
Now, let $x \in E$ with $\norm{x}_{E} = 1$, then
On the other hand, by Proposition 13.1.14, for any $x \in E$ with $x \ge 0$ and $\norm{x}_{E} = 1$,
so $\norm{\ |\phi|\ }_{E^*}\le \norm{\phi}_{E^*}$. Therefore for any $\phi, \psi \in E$ with $|\phi| \le |\psi|$,