Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm

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Theorem

Let $R$ be a division ring.

Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.


Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:

$\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$


Then:

$\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$


Proof

Case 1

For all $x \in R: x \ne 0_R$, let $x$ satisfy $\norm x_1 \ge 1$.


Lemma 1
$\norm {\, \cdot \,}_1$ is the trivial norm.


By assumption, for all $x \in R, x \ne 0_R$, then $\norm x_2 \ge 1$.

Similarly $\norm {\, \cdot \,}_2$ is the trivial norm.

Hence $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equal.

For $\alpha = 1$ the result follows.

$\Box$


Case 2

Let $x_0 \in R$ such that $x_0 \ne 0_R$ and $\norm {x_0}_1 < 1$.

By assumption then $\norm {x_0}_2 < 1$.

Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$.

Then $\norm {x_0}_1 = \norm {x_0}_2^\alpha$.


As $\norm {x_0}_1, \norm {x_0}_2 < 1$:

$\log \norm {x_0}_1 < 0$
$\log \norm {x_0}_2 < 0$

So $\alpha > 0$.


Lemma 2
$\forall x \in R: \norm x_1 = \norm x_2^\alpha$

$\blacksquare$


Sources

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