I'm attempting to verify a statement made by the author of a textbook:
Given a metric d on a vector space $X$, setting $||x|| = d(x,0)$ defines a norm on $X$ if:
1) $d(x,y) = d(x+z,y+z) $
2) $d(ax,ay) = |a|d(x,y) $
for any vectors $x,y,z$ and scalar $a$.
Here is what I've tried:
Suppose the metric $d$ is such that assigning $d(x,0)$ to every $x$ is a valid norm, where $||x|| := d(x,0)$.
Then we can use the norm ||,|| to define a new metric d' as follows:
$d'(x,y) := ||x-y||$.
Then we have $ d'(x+z,y+z) = ||(x+z)-(y+z)|| = ||x-y|| = d'(x,y)$. And $d'(ax,ay) = ||ax-ay|| = ||a(x-y)|| = |a|||x-y|| = |a|d'(x,y$).
So I've showed this metric $d'$ satisfies 1) and 2). and I know that $d'(x,0) = ||x|| = d(x,0)$. But it is not clear to me that $d=d'$, and I don't know how to proceed/show that $d$ must satisfy 1) and 2).
The converse is false. Consider the metric $d$ on $\mathbb{R}$ given by $$ d(x,y) = \begin{cases} \lvert{x\rvert}+\lvert{y\rvert} &\text{if } x\neq y\\ 0 &\text{if } x=y \end{cases} $$ Clearly, $d(x,0) = \lvert x\rvert$ defines a norm (the usual norm on the reals). On the other hand, property (1) is not satisfied. To see this, take for instance $x=2, y=3, z=-1$.