Norms Equivalent to Absolute Value on Rational Numbers/Sufficient Condition
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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 $\Q$.
Then:
- $\alpha \le 1 \implies \norm {\,\cdot\,}$ is a norm on $\Q$
Proof
Suppose $\alpha \le 1$.
It is shown that $\norm {\,\cdot\,}$ satisfies the norm axioms $(\text N 1)$-$(\text N 3)$.
Norm Axiom $\text N 1$: Positive Definiteness
Let $x \in \Q$.
\(\ds \norm x = 0\) | \(\leadstoandfrom\) | \(\ds \size x^\alpha = 0\) | Definition of $\norm {\,\cdot\,}$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \size x = 0\) | Definition of $a$ to the power of $r$ for $a \in \R_{\ge 0}$ and $r \in \R_{> 0}$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds x = 0\) | Absolute Value is Norm and Norm Axiom $\text N 1$: Positive Definiteness |
$\Box$
Norm Axiom $\text N 2$: Multiplicativity
Let $x, y \in \Q$.
Then:
\(\ds \norm {x y}\) | \(=\) | \(\ds \size {x y}^\alpha\) | Definition of $\norm {\,\cdot\,}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\size x \size y}^\alpha\) | Absolute Value is Norm and Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \size x^\alpha \size {y}^\alpha\) | Power of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm x \norm y\) | Definition of $\norm {\,\cdot\,}$ |
$\Box$
Norm Axiom $\text N 3$: Triangle Inequality
Let $x, y \in \Q$.
Without loss of generality, let $\norm y < \norm x$.
If $\norm x = 0$ then $\norm y = 0$.
By Norm Axiom $\text N 1$: Positive Definiteness above:
- $x = y = 0$
Hence:
\(\ds \norm {x + y}\) | \(=\) | \(\ds \norm 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 + 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x + \norm y\) |
If $\norm x > 0$ then:
- $\norm x > 0 \leadstoandfrom \size x^\alpha > 0 \leadstoandfrom \size x > 0$
Hence:
\(\ds \norm {x + y}\) | \(=\) | \(\ds \size {x + y}^\alpha\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {\size x + \size y}^\alpha\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \size x^\alpha \paren {1 + \dfrac {\size y} {\size x} }^\alpha\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size x^\alpha \paren {1 + \dfrac {\size y} {\size x} }\) | Power Function on Base Greater than One is Strictly Increasing | |||||||||||
\(\ds \) | \(\le\) | \(\ds \size x^\alpha \paren {1 + \dfrac {\size y^\alpha} {\size x^\alpha} }\) | Power Function on Base between Zero and One is Strictly Decreasing | |||||||||||
\(\ds \) | \(\le\) | \(\ds \size x^\alpha + \size y^\alpha\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x + \norm y\) |
$\blacksquare$
Sources
- 2007: Svetlana Katok: p-adic Analysis Compared with Real: $\S 1.2$ Normed fields, Proposition $1.11$