A metric on $\mathbb{R}^n$ such that $d(\lambda x, \lambda y)=|\lambda| d(x,y)$ which is not induced by a norm

Question

Let $V=\mathbb{R}^n$.
Let $d:V \times V\rightarrow \mathbb{R}$ a metric on $\mathbb{R}^n$.
Assume that for any $x,y\in V$ and $\lambda \in \mathbb{R}$, we have $d(\lambda x, \lambda y) = |\lambda|d(x,y)$.
Is $d$ necessarily induced by a norm?

Motivation: I've been thinking of $\pi$ and thought about why the ratio between a circles's circumference and its radius is constant. The proof is easy and is applicable to any norm. I think the "positive homogeneity" condition I posed on the metric above is enough for this ratio to be constant.

You did note that this ratio is normally $2\pi$, not? Doesn't render the question invalid, just sayin'. — Lord_Farin, Oct 20 '12 at 15:29
I did note that, thanks. I thought the formulation is clearer this way. — Gils, Oct 20 '12 at 15:31
This might fail for a non translation invariant metric (i.e. $d(x+c, y+c) \neq d(x,y)$. Have you checked that? — filmor, Oct 20 '12 at 15:40
See also: http://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm — Martin Sleziak, Oct 24 '14 at 08:48

Lord_Farin · Accepted Answer · 2012-10-20T16:16:26.497

5

The answer is no. You need translational invariance as well; then it's a pretty well-known theorem (see e.g. here).

As a counterexample when leaving out the translational invariance, consider:

$$d: \Bbb R^n \times \Bbb R^n \to \Bbb R_{\ge 0}: d (x,y)=\begin{cases} \|x\|+\|y\| & \text{if $x \ne y$}\\ 0 & \text{otherwise.} \end{cases}$$

This metric is sometimes referred to as the "metric of the French railway system", although there are similar metrics with the same name (cf. the comments).

edited Oct 20 '12 at 16:16

answered Oct 20 '12 at 15:40

Lord_Farin

17,743

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but some of our trains do have intermediate stops ! – mercio Oct 20 '12 at 15:53
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I suppose it could also be called the "metric of the French intercity service system locally around Paris", but that's such a mouthful. – Lord_Farin Oct 20 '12 at 15:55
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+1. I thought the SNCF distance $d$ was defined by $d(x,y)=|x-y|$ when $x$ and $y$ are colinear, and $d(x,y)=|x|+|y|$ otherwise. The intermediate stops again... – Did Oct 20 '12 at 16:13
@did: Apparently multiple versions exist; they both work, whatever their name. – Lord_Farin Oct 20 '12 at 16:15
@did’s version is also called the Paris metric. (I usually call it the hedgehog metric.) – Brian M. Scott Oct 20 '12 at 16:26

A metric on $\mathbb{R}^n$ such that $d(\lambda x, \lambda y)=|\lambda| d(x,y)$ which is not induced by a norm

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