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
\[\normn{x}_{E} = \normn{x^+ + x^-}_{E} \ge \normn{x^+}_{E}, \normn{x^-}_{E}\]
(3): Since $|x| = x \vee (-x)$,
\[\norm{\phi}_{E^*}= \sup_{\substack{x \in E \\ \norm{x}_E = 1}}\dpn{x, \phi}{E}\le \sup_{\substack{x \in E \\ \norm{x}_E = 1}}\dpn{|x|, \phi}{E}\le \sup_{\substack{x \in E \\ x \ge 0 \\ \norm{x}_E = 1}}\dpn{x, \phi}{E}\le \norm{\phi}_{E^*}\]
$\square$