I was working through some exercises from a Functional Analysis book, and am stumped by a question which seems like it should be easy enough:
let $d$ be a metric on a vector space $E$, satisfying the following condition (1): $$ \forall x,y \in E \hspace{5pt} \forall \lambda \in \mathbb{R} : d(x,x+\lambda y) = |\lambda|d(x,x+y) $$ prove that $\|x\| := d(x,0)$ defines a norm.
To prove this I already checked three of the four properties of a metric: trivially we have $\|x\| \geq 0$ for all $x$, with equality if and only if $x=0$. We can also see that
$$ \|\lambda x\| = d(\lambda x, 0) = d(0, 0+\lambda x) = |\lambda| d(x,0) = |\lambda|\|x\| $$
as expected. However, I got stuck trying to prove that $\|x+y\| \leq \|x\| + \|y\|$, that is, proving that for a metric $d$ satisfying (1) we have $d(x+y,0) \leq d(x,0)+d(y,0)$.
I have searched this site as well as others for necessary conditions for a metric to induce a norm, hoping to find a similar proof. I did find that for a metric to induce a norm through $d(x,0)$, the metric must be translation invariant as well as homogeneous: see here. As such, these two conditions are necessary, but I am left wondering why they would be sufficient: one comment mentions that it 'turns out that these conditions are sufficient', but does not elaborate. Most of the other discussions here seem to be about why these conditions are necessary, so further search didn't help me. Finally, when I am convinced that these conditions are actually sufficient, I would then still have to show that the condition given by the exercise implies homogeneity as well as translation invariance.
It feels like I am missing something obvious. Could someone help here?
