Norm induces a metric (and conversely)

Proposition. Let $(X,\|\cdot\|)$ be a normed vector space . Define

Then $d$ is a metric on $X$ (hence $(X,d)$ is a metric space).

Conversely, let $X$ be a vector space equipped with a metric $d$ satisfying:

• translation invariance: $d(x+z,y+z)=d(x,y)$ for all $x,y,z\in X$,
• absolute homogeneity: $d(\alpha x,\alpha y)=|\alpha|\,d(x,y)$ for all scalars $\alpha$ and all $x,y\in X$.

Then $\|x\|:=d(x,0)$ defines a norm on $X$, and $d(x,y)=\|x-y\|$.

Context. This identifies norms as exactly the translation-invariant, homogeneous metrics on vector spaces.

Proof sketch. For the forward direction, metric axioms follow from the norm axioms; in particular the triangle inequality for $d$ is $\|x-z\|\le \|x-y\|+\|y-z\|$. For the converse, check the norm axioms directly from the two metric properties and the metric triangle inequality.