Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2

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:

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

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

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

Then:

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

Proof

Since $\norm {x_0}_1 < 1$ then $\norm {x_0}_2 < 1$ and:

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

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

Hence $\alpha > 0$

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

Lemma 1

Then:

$\forall y \in R: \norm y_1 > 1 \iff \norm y_2 > 1$

Lemma 2

Then:

$\forall y \in R:\norm y_1 = 1 \iff \norm y_2 = 1$

Let $x \in R, x \ne 0_R$.

Case 1

Let $\norm x_1 = 1$.

By Lemma 2 then:

$\norm x_2 = 1$

Hence:

$\norm x_1 = 1 = 1^\alpha = \norm x_2^\alpha$

$\Box$

Case 2

Let $\norm x_1 \ne 1$.

By Lemma 2 then:

$\norm x_2 \ne 1$

Let $\beta = \dfrac {\log \norm x_1 } {\log \norm x_2 }$.

Then $\norm x_1 = \norm x_2^\beta$.

Lemma 3

$\alpha = \beta$

Hence $\norm x_1 = \norm x_2^\alpha$.

$\blacksquare$