Norms Equivalent to Absolute Value on Rational Numbers

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Theorem

Let $\alpha \in \R_{\gt 0}$.

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:

$\norm{\,\cdot\,}$ is a norm on $\Q$ if and only if $\,\,\alpha \le 1$


Proof

Necessary Condition

The contrapositive is proved.

Let $\alpha \gt 1$.

The norm axiom (N3) (Triangle Inequality) is not satisfied:

\(\displaystyle \norm {1 + 1}\) \(=\) \(\displaystyle \size {1 + 1}^\alpha\)
\(\displaystyle \) \(=\) \(\displaystyle 2^\alpha\)
\(\displaystyle \) \(\gt\) \(\displaystyle 2\) Real Power Function on Base Greater than One is Strictly Increasing
\(\displaystyle \) \(=\) \(\displaystyle \size{1}^\alpha + \size{1}^\alpha\)
\(\displaystyle \) \(=\) \(\displaystyle \norm{1} + \norm{1}\)

By Rule of Transposition the result follows.

$\Box$

Sufficient Condition

Suppose $\alpha \le 1$.

It is shown that $\norm{\,\cdot\,}$ satisfies the norm axioms (N1)-(N3).

(N1) Positive Definiteness

Let $x \in \Q$.

\(\displaystyle \norm {x} = 0\) \(\iff\) \(\displaystyle \size {x}^\alpha = 0\) Definition of $\norm{\,\cdot\,}$
\(\displaystyle \) \(\iff\) \(\displaystyle \size {x} = 0\) Definition of $a$ to the power of $r$ for $a \in \R_{\ge 0}$ and $r \in \R_{\gt 0}$
\(\displaystyle \) \(\iff\) \(\displaystyle x = 0\) Absolute Value is Norm and norm axiom (N1) (Positive Definiteness)

$\Box$

(N2) Multiplicativity

Let $x, y \in \Q$.

Then:

\(\displaystyle \norm {x y}\) \(=\) \(\displaystyle \size {x y}^\alpha\) Definition of $\norm{\,\cdot\,}$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\size {x} \size {y} }^\alpha\) Absolute Value is Norm and norm axiom (N2) (Multiplicativity)
\(\displaystyle \) \(=\) \(\displaystyle \size {x}^\alpha \size {y}^\alpha\) Power of product
\(\displaystyle \) \(=\) \(\displaystyle \norm {x} \norm {y}\) Definition of $\norm{\,\cdot\,}$

$\Box$

(N3) Triangle Inequality

Let $x, y \in \Q$.

Without loss of generality let $\norm y \lt \norm x$.


If $\norm x = 0$ then $\norm y = 0$.

By (N1) above, $x = y = 0$.

Hence:

\(\displaystyle \norm{x + y}\) \(=\) \(\displaystyle \norm 0\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)
\(\displaystyle \) \(=\) \(\displaystyle 0 + 0\)
\(\displaystyle \) \(=\) \(\displaystyle \norm x + \norm y\)


If $\norm x \gt 0$ then:

$\norm x \gt 0 \iff \size {x}^\alpha \gt 0 \iff \size x \gt 0$

Hence:

\(\displaystyle \norm{x + y}\) \(=\) \(\displaystyle \size{x + y}^\alpha\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren{\size x + \size y}^\alpha\) Norm axiom (N3) (Triangle Inequality)
\(\displaystyle \) \(=\) \(\displaystyle \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }^\alpha\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }\) Real Power Function on base Greater than One is Strictly Increasing
\(\displaystyle \) \(\le\) \(\displaystyle \size {x}^\alpha \paren{1 + \dfrac {\size {y}^\alpha} {\size {x}^\alpha} }\) Real Power Function on base between Zero and One is Strictly Decreasing
\(\displaystyle \) \(\le\) \(\displaystyle \size {x}^\alpha + \size {y}^\alpha\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {x} + \norm {y}\)

$\blacksquare$

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