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/Part 3: Functional Analysis/Chapter 13: Order Structures/Section 13.1: Vector Lattices

Lemma 13.1.10. 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 13.1: Vector Lattices
  • Proposition 13.1.12
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Jerry's Digital Garden

Bibliography

Direct Backlinks

  • Section 13.1: Vector Lattices
  • Proposition 13.1.12
Powered by Spec