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 $\|x,y\|=d(x,y)$.

  • 2
    $\begingroup$ Sorry, what do you mean by normable? $\endgroup$ – user14717 May 16 '15 at 19:56
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.