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:

$\forall x \in R:\norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$


Then:

$\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$

Proof

Case 1

Let for all $x \in R, x \neq 0_R$, satisfy $\norm{x}_1 \ge 1$.

Lemma 1

Then:

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


By assumption, for all $x \in R, x \neq 0_R$, 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 $x_0 \neq 0_R$ and $\norm{x_0}_1 \lt 1$.

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

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

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


Since $\norm{x_0}_1, \norm{x_0}_2 \lt 1$ then:

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

So $\alpha \gt 0$.

Lemma 2

Then:

$\forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$


$\blacksquare$

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