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Is there a way to classify which metrics defined on vector spaces can be induced by a norm? ie. there exists norm $n: X\to \mathbb{R}$ on the vector space $X$ such that the metric $d(x,y)=n(x-y)$.

I am considering metrics on vector spaces since a normed space is required to be a vector space. I know there are metrices (eg. the discrete metric) cannot be induced by a norm and I am wondering if there is a way to classify the metrices which could.

My attempt: metric $d$ on vector space $X$ on field $\mathbb{K}$ can be induced by a norm iff. for all $x,y\in X$, we get $d(\lambda x,\lambda y)=\lambda d(x,y)$ for all $\lambda \in \mathbb{K}$ and $d(0,x+y)\leq d(0,x)+d(0,y)$.

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  • $\begingroup$ I don't think that $d(\lambda x,y) = \lambda d(x,y)$ holds when $d$ is induced by norm (try putting $x=0$). For the classification, try rewriting the axioms of norm $n$ using the equality $d(x,y) = n(x-y)$ and see what comes out. $\endgroup$
    – KeeperOfSecrets
    Commented Jul 8 at 15:31
  • $\begingroup$ @KeeperOfSecrets Thanks for pointing out my mistake, I was using $d(x,y) = n(x-y)$ to figure out properties of $d$ but I was too sloopy. I thought the symmetry of $d$ means I can ignore one $\lambda $. $\endgroup$
    – HIH
    Commented Jul 8 at 15:39

1 Answer 1

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If $d$ is induced by a norm then $\|x\|=d(0,x)$ is that norm. So we immediately have the conditions:

  • $d(0,\lambda x)=\lambda d(0,x)$
  • $d(x,y)=d(0,y-x)$

From this, it follows that $\|\cdot\|$ satisfies the triangle inequality, as $$d(0,x+y)=d(-x,y) \le d(-x,0)+d(0,y) = d(0,x)+d(0,y)$$ And $\|\cdot\|$ must be positive definite, as $d$ is positive definite, hence these conditions are sufficient to show $\|\cdot\|$ is a norm. And $$d(x,y)=d(0,y-x)$$ guarantees that $\|\cdot\|$ induces $d$. So these are necessary and sufficient conditions.

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