Metric Defines Norm iff it Preserves Linear Structure

Theorem
Let $V$ be a vector space over a valued field $k$.

Let $d : V \times V \to k$ be a metric on $V$.

Then the function $\| v \| := d(v,0)$ is a norm on $V$ if and only if for all $x,y,z \in V$, $\lambda \in k$:


 * 1. $d(x+z,y+z) = d(x,y)$ (homogenity or translation invariance)


 * 2. $d(\lambda x, \lambda y) = | \lambda | d(x,y)$ (the enlargement property)

Proof
Suppose first that $d$ satisfies the hypotheses 1. and 2..

Since $d(u,v) \geq 0$ for all $u,v \in V$, $\| u \| = d(u,0) \geq 0$ for all $u \in V$.

Moreover if $\| u \| = 0$ then $d(u,0) = 0$, so $u = 0$.

Now let $\lambda \in K$, $u \in V$.

Then, using the enlargement property of $d$:

Finally if $u,v \in V$, then we have