Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent

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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:

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


Then for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ if and only if $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$


Proof

Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_1$.

Let $\epsilon > 0$ be given.


Since $\sequence {x_n}$ is a Cauchy sequence then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 < \epsilon^\alpha$

Then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon^\alpha$

Hence:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 < \epsilon$


So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$


It follows that for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

$\Box$


Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_2$.

Let $\epsilon > 0$ be given.


Since $\sequence {x_n}$ is a Cauchy sequence then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$

Then:

$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon$

Hence:

$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_1 < \epsilon$


So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$


It follows that for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$


The result follows.

$\blacksquare$


Sources

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