Lemma 16.1.12.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,
\[b \vee d - a \vee c \le |b - a| \vee |d - c| = |a - b| \vee |c- d|\]
$\square$
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