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Is the metric $$d(u,v)=\frac{||u-v||_v}{1+||u-v||_v}$$ induced by a norm?

My attempt at an answer:

Suppose that it was then, there would be a norm $||.||_m$ such that $$||u-v||_m=\frac{||u-v||_v}{1+||u-v||_v}$$ therefore $$||u-0||_m=\frac{||u-0||_v}{1+||u-0||_v}$$ would have to be norm but $$||\lambda u||_m= \frac{\lambda||u||_v}{1+\lambda||u||_v} \neq \frac{\lambda||u||_v}{1+||u||_v}=\lambda||u||_m$$ so $||.||_m$ isn't a norm and therefore no norm can induce this metric.

user66733
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Gottfried
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1 Answers1

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That's it.

Basically, if you have a metric on a vector space defined via a norm, the metric must satisfy $d(x+z,y+z)=d(x,y)$ and $d(\lambda x,\lambda y)=|\lambda|d(x,y)$.

This should also be enough - if $d$ is a metric satisfying the above two conditions then $d(x-y,0)$ is a norm.

Thomas Andrews
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  • I have never seen those condition before but have almost accidentally ended up using them. Is the logic correct that for a metric to be induced by a norm $d(x,0)$ must be a norm? Is that in some way equivalent to what you have said? – Gottfried Sep 24 '13 at 13:23
  • Yes. Basically, my first condition with $z=-y$ means that $d(x,y)=d(x-y,0)$, which must be the norm of $x-y$. – Thomas Andrews Sep 24 '13 at 13:27
  • Sorry to get bogged down on this but I have one more question. From what you've said I can see that if $d(x,0)$ is not a norm then $d(x,y)$ was not induced by a norm but I can't tell if you're saying the reverse hold as well. That is if, $d(x,0)$ is a norm does it imply that $d(x,y)$ was induced by a norm. – Gottfried Sep 24 '13 at 13:43
  • Only if also $d(x,y)=d(x-y,0)$. @Gottfried – Thomas Andrews Sep 24 '13 at 13:46
  • See also: http://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm – Martin Sleziak Oct 24 '14 at 08:45