Equivalence of Definitions of Non-Archimedean Vector Space Norm
Theorem
Let $\struct {R, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_R$.
Let $X$ be a vector space over $R$, with zero $0_X$.
The following definitions of the concept of Non-Archimedean Vector Space Norm are equivalent:
Definition 1
A norm $\norm {\,\cdot\,} $ on $X$ is non-Archimedean if and only if $\norm {\, \cdot \,}$ satisfies the axiom:
| \((\text N 4)\) | $:$ | Ultrametric Inequality: | \(\ds \forall x, y \in X:\) | \(\ds \norm {x + y} \) | \(\ds \le \) | \(\ds \max \set {\norm x, \norm y} \) |
Definition 2
A non-Archimedean norm on $X$ is a mapping from $X$ to the non-negative reals:
- $\norm {\, \cdot \,}: X \to \R_{\ge 0}$
satisfying the non-Archimedean norm axioms:
| \((\text N 1)\) | $:$ | Positive Definiteness: | \(\ds \forall x \in X:\) | \(\ds \norm x = 0 \) | \(\ds \iff \) | \(\ds x = 0_R \) | |||
| \((\text N 2)\) | $:$ | Positive Homogeneity: | \(\ds \forall x \in X, \lambda \in R:\) | \(\ds \norm {\lambda x} \) | \(\ds = \) | \(\ds \norm {\lambda}_R \times \norm x \) | |||
| \((\text N 4)\) | $:$ | Ultrametric Inequality: | \(\ds \forall x, y \in X:\) | \(\ds \norm {x + y} \) | \(\ds \le \) | \(\ds \max \set {\norm x, \norm y} \) |
Proof
Definition 1 implies Definition 2
Let $\norm {\,\cdot\,} : X \to \R_{\ge 0}$ be a norm on a division ring satisfying:
| \((\text N 4)\) | $:$ | Ultrametric Inequality: | \(\ds \forall x, y \in X:\) | \(\ds \norm {x + y} \) | \(\ds \le \) | \(\ds \max \set {\norm x, \norm y} \) |
It remains only to show that $\norm {\,\cdot\,}$ satisfies $(\text N 1)$ and $(\text N 2)$.
This follows from the definition of a norm on a division ring.
$\Box$
Definition 2 implies Definition 1
Let $\norm{\,\cdot\,} : X \to \R_{\ge 0}$ satisfy the non-Archimedean norm axioms: $(\text N 1)$, $(\text N 2)$ and $(\text N 4)$.
To show that $\norm{\,\cdot\,}$ is a norm on a division ring satisfying $(\text N 4)$, it remains to show that $\norm{\,\cdot\,}$ satisfies:
| \((\text N 3)\) | $:$ | Triangle Inequality: | \(\ds \forall x, y \in X:\) | \(\ds \norm {x + y} \) | \(\ds \le \) | \(\ds \norm x + \norm y \) |
Let $x, y \in X$.
Without loss of generality, suppose $\norm x \le \norm y$.
From non-Archimedean norm axiom $(\text N 1)$ : Positive Definiteness:
- $0 \le \norm x$
Then:
| \(\ds \norm{x + y}\) | \(\le\) | \(\ds \max \set {\norm x, \norm y}\) | Non-Archimedean Norm Axiom $(\text N 4)$ : Ultrametric Inequality | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm y\) | as $\norm x \le \norm y$ by assumption | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm x + \norm y\) | as $0 \le \norm x$ |
The result follows.
$\blacksquare$