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.

  • $\begingroup$ You did note that this ratio is normally $2\pi$, not? Doesn't render the question invalid, just sayin'. $\endgroup$ – Lord_Farin Oct 20 '12 at 15:29
  • $\begingroup$ I did note that, thanks. I thought the formulation is clearer this way. $\endgroup$ – Gils Oct 20 '12 at 15:31
  • $\begingroup$ This might fail for a non translation invariant metric (i.e. $d(x+c, y+c) \neq d(x,y)$. Have you checked that? $\endgroup$ – filmor Oct 20 '12 at 15:40
  • $\begingroup$ Ah, answered just now by Lord Farin … ;) $\endgroup$ – filmor Oct 20 '12 at 15:40
  • $\begingroup$ See also: math.stackexchange.com/questions/166380/… $\endgroup$ – Martin Sleziak Oct 24 '14 at 8:48

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).

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    $\begingroup$ but some of our trains do have intermediate stops ! $\endgroup$ – mercio Oct 20 '12 at 15:53
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    $\begingroup$ I suppose it could also be called the "metric of the French intercity service system locally around Paris", but that's such a mouthful. $\endgroup$ – Lord_Farin Oct 20 '12 at 15:55
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    $\begingroup$ +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... $\endgroup$ – Did Oct 20 '12 at 16:13
  • $\begingroup$ @did: Apparently multiple versions exist; they both work, whatever their name. $\endgroup$ – Lord_Farin Oct 20 '12 at 16:15
  • $\begingroup$ @did’s version is also called the Paris metric. (I usually call it the hedgehog metric.) $\endgroup$ – Brian M. Scott Oct 20 '12 at 16:26

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