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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8If $d(x,y)$ 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, $\left|x+y\right|=d(x+y,0)=d(x,-y)\leq d(x,0)+d(0,-y)=d(x,0)+d(y,0)=\left|x\right|+\left|y\right|$. – orole Jan 15 '18 at 21:12
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5@orole: why post an answer in the comment section? – Martin Argerami Jan 16 '18 at 00:01