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

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