Jerry's Digital Garden

Bibliography
/Part 3: Functional Analysis/Chapter 15: Order Structures/Section 15.1: Vector Lattices

Lemma 15.1.10.label Let $(E, \le)$ be a vector lattice and $x \in E$, then $|x| \ge 0$.

Proof. For any $x \in E$,

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

$\square$

Direct Backlinks

  • Section 15.1: Vector Lattices
  • Proposition 15.1.12: [V.1.1, SW99]
Powered by Spec

Jerry's Digital Garden

Bibliography

Direct Backlinks

  • Section 15.1: Vector Lattices
  • Proposition 15.1.12: [V.1.1, SW99]
Powered by Spec