Could someone please help me prove that every metric on a linear space is induced by norm if and only if the metric is homogeneous and translation invariant? I found this answer about homogeneity but I'm not sure how to extend this to every metric. Thank you.

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    $\begingroup$ If $d(x,y)$ is translation invariant then $d(x,y)=d(x-y,0)$. Define $\left\|x\right\|=d(x,0)$ and check that it is a norm: $\left\|x\right\|=d(x,0)=0$ iff $x=0$. $\left\|ax\right\|=d(ax,0)=d(ax,a0)=|a|d(x,0)=|a|\left\|x\right\|$. Finally, $\left|x+y\right|=d(x+y,0)=d(x,-y)\leq d(x,0)+d(0,-y)=d(x,0)+d(y,0)=\left\|x\right|+\left\|y\right\|$. $\endgroup$ – orole Jan 15 '18 at 21:12
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    $\begingroup$ @orole: why post an answer in the comment section? $\endgroup$ – Martin Argerami Jan 16 '18 at 0:01

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