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

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

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

Lemma 1
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.

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 $\alpha > 0$.