$d$ is a metric space on $X\not=\{0\}$, obtained from a norm. $d'(x,y)=d(x,y)+1$. Show $d'$ cannot be obtained from a norm.

Question

If $d$ is a metric on a vector space $X\not=\{0\}$ which is obtained from a norm, and $d'$ is defined by $d'(x,x)=0$, $d'(x,y)=d(x,y)+1, (x\not=y)$, show that $d'$ cannot be obtained from a norm.

I've proven that if $X\not=\{0\}$ is a discrete metric space cannot be obtained from a norm, but I don't know how to go about proving the above.

Any help would be welcomed.

Answer 1

It isn't homogeneous. If $d$ and $d'$ were obtained from norms you would have $$ |\alpha| \|x-y\|' = d'(\alpha x,\alpha y) = d(\alpha x,\alpha y) + 1 = |\alpha| ||x-y|| + 1.$$ What happens as $\alpha \to 0$?

Answer 2

A metric induced by a norm on a nontrivial vector space is always surjective as a map $d : X\times X\to [0,\infty)$. This is not the case for your metric (why?).