# The metric comes from a norm if and only if satisfies the following:

Let $E$ a vector space over $\mathbb{R}$ and $d$ a metric in $E$. We say $d$ comes from a norm if there exist a norm $||.||$ in $E$ such that $d(x,y) = ||x-y||, \forall x,y \in E$. Prove that $d$ comes from a norm if and only if $d$ satisfies:

(i) $d(x+z,y+z) = d(x,y) , \forall x,y,z \in E$

(ii) $d(\lambda x,\lambda y) = |\lambda|d(x,y) , \forall x,y \in E$

My atttempt: I think I did the right implications right, but my problem is the left. Here is what I did:

(i) $(\Rightarrow)$ $\exists ||.|| \rightarrow \mathbb{R} st. d(x,y) = ||x-y|| \forall x,y \in E$

Then $d(x+z,y+z) = ||x+z-y-z|| = ||x-y|| = d(x,y)$

(ii) $(\Rightarrow)$ $\exists ||.|| \rightarrow \mathbb{R} st. d(x,y) = ||x-y|| \forall x,y \in E$

Then $d(\lambda x, \lambda y) = || \lambda x - \lambda y|| = || \lambda (x-y)|| = |\lambda| ||x-y|| = |\lambda| d(x,y).$

For the left implications I believe I need to create a norm, but I'm lacking creativity, as I usually do in analysis. Can someone help me? Also, can someone please confirm that "metric comes from a norm" is the standard term for what I'm saying? I did a free translation from portuguese (métrica provém de uma norma). Thanks.

• Intuitively the norm of something is its distance from $0$ so if you want a norm coming from a distance expect $||x||=d(x,0)$ – user290300 Aug 21 '16 at 18:31

For the backward direction, you need to define a norm. Given the distance function $d$, try defining $$\|x\| := d(x,0).$$
Intuitively the norm of something is its distance from $0$. So if you want a norm coming from a distance, expect $||x||=d(x,0)$.