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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.

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    See also: http://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm2014-10-24

1 Answers 1

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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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    @did’s version is also called the [Paris metric](http://en.wikipedia.org/wiki/Hedgehog_space#Paris_metric). (I usually call it the *hedgehog metric*.)2012-10-20