Norms Equivalent to Absolute Value on Rational Numbers/Necessary Condition
Jump to navigation
Jump to search
Theorem
Let $\alpha \in \R_{> 0}$.
Let $\norm {\,\cdot\,}:\Q \to \R$ be the mapping defined by:
- $\forall x \in \Q: \norm x = \size x^\alpha$
where $\size x$ is the absolute value of $x$ in $\Q$.
Let $\norm {\,\cdot\,}$ be a norm on $\Q$.
Then:
- $\alpha \le 1$
Proof
The contrapositive is proved.
Let $\alpha > 1$.
The Norm Axiom $\text N 3$: Triangle Inequality is not satisfied:
\(\ds \norm {1 + 1}\) | \(=\) | \(\ds \size {1 + 1}^\alpha\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^\alpha\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 2\) | Power Function on Base Greater than One is Strictly Increasing | |||||||||||
\(\ds \) | \(=\) | \(\ds \size 1^\alpha + \size 1^\alpha\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm 1 + \norm 1\) |
By Rule of Transposition the result follows.
$\blacksquare$
Sources
- 2007: Svetlana Katok: p-adic Analysis Compared with Real: $\S 1.2$ Normed fields, Proposition $1.11$