Normed Vector Space Requires Multiplicative Norm on Division Ring

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Theorem

Let $R$ be a normed division ring with a submultiplicative norm $\norm {\, \cdot \,}_R$.

Let $V$ be a vector space that is not a trivial vector space.

Let $\norm {\, \cdot \,}: V \to \R_{\ge 0}$ be a mapping from $V$ to the positive real numbers satisfying the vector space norm axioms.

Then $\norm {\, \cdot \,}_R$ is a multiplicative norm.

That is:

$\forall r, s \in R: \norm {r s}_R = \norm r_R \norm s_R$


Proof

Since $V$ is not a trivial vector space:

$\exists \mathbf v \in V: \mathbf v \ne 0$

By Norm axiom (N1):

$\norm {\mathbf v} > 0$


Let $r, s \in R$:

\(\displaystyle \norm {r s}_R \norm {\mathbf v}\) \(=\) \(\displaystyle \norm {\paren {r s} \mathbf v}\) Norm axiom (N2)
\(\displaystyle \) \(=\) \(\displaystyle \norm {r \paren {s \mathbf v} }\) Vector Space Axiom $V \, 7$: Associativity with Scalar Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \norm r_R \norm {s \mathbf v}\) Norm axiom (N2)
\(\displaystyle \) \(=\) \(\displaystyle \norm r_R \norm s_R \norm {\mathbf v}\) Norm axiom (N2)


By dividing both sides of the equation by $\norm {\mathbf v}$ then:

$\norm {r s}_R = \norm r_R \norm s_R$


The result follows.

$\blacksquare$