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If d is a metric on a (finite dimensional) Banach Space and there exist norms $\|-\|_1$ and $\|-\|_2$ and constants $C_1,C_2 \in [1,\infty)$ satisfying:

\begin{equation} C_1\|x-y\|_1 \leq d(x,y) \leq C_2\|x-y\|_2, \end{equation} then is $d$ normable?

By normable I mean there exists some norm $\|\cdot \|$ such that $\|x,y\|=d(x,y)$.

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    $\begingroup$ Sorry, what do you mean by normable? $\endgroup$ – user14717 May 16 '15 at 19:56
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    $\begingroup$ Presumably expressible as a norm? It would need to be translation invariant, at least. I don't have my Rudin handy at the moment. $\endgroup$ – copper.hat May 16 '15 at 20:00
  • $\begingroup$ Yes, by normable I mean there exists some norm $\|\|$ such that $\|x,y\|=d(x,y)$. $\endgroup$ – AIM_BLB May 16 '15 at 20:40
  • $\begingroup$ The metric $d$ defines the same topology as any norm on the Banach space $X$, but if $d$ is not translation-invariant, then there does not exist a norm $\left\|\cdot\right\|$ such that $\left\|x-y\right\|=d(x,y)$. $\endgroup$ – Matt Rosenzweig May 16 '15 at 20:54
  • $\begingroup$ ok, but how does that play a role in our (dis)proof? $\endgroup$ – AIM_BLB May 16 '15 at 21:06
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First of all: since all norms on a finite-dimensional space are comparable, the condition could be simpler stated as $$ C_1\|x-y\| \leq d(x,y) \leq C_2\|x-y\| $$ where $\|\cdot \|$ is a norm of our choice, e.g., Euclidean.

Second: the answer is negative, for example $$d(x,y) = |x-y|+\min(|x-y|,1)$$ is a translation-invariant metric on $\mathbb{R}$ that satisfies $$|x-y|\le d(x,y)\le 2|x-y|$$ but is not given by any norm.

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  • $\begingroup$ It might be helpful to mention that $d$ is not given by a norm since $d$ fails the homogeneity condition. $\endgroup$ – Matt Rosenzweig May 17 '15 at 18:09

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