# Test for normability of a metric on a Banach space

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:

$$C_1\|x-y\|_1 \leq d(x,y) \leq C_2\|x-y\|_2,$$ then is $d$ normable?

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

• Sorry, what do you mean by normable? – user14717 May 16 '15 at 19:56
• Presumably expressible as a norm? It would need to be translation invariant, at least. I don't have my Rudin handy at the moment. – copper.hat May 16 '15 at 20:00
• Yes, by normable I mean there exists some norm $\|\|$ such that $\|x,y\|=d(x,y)$. – AIM_BLB May 16 '15 at 20:40
• 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 $\left\|x-y\right\|=d(x,y)$. – Matt Rosenzweig May 16 '15 at 20:54
• ok, but how does that play a role in our (dis)proof? – AIM_BLB May 16 '15 at 21:06

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