Why is $d(x,0)$ not a norm?

Question

If $\|x\|$ is a norm, then we can define $d(x,y):=\|x-y\|$ and it will be a metric. Now, if $d$ is a metric, why is $\|x\|:= d(x,0)$ not a norm? I think it fail for the sub-linearity, but I don't see how.

Answer 1

Consider $$d(x,y)=\lvert x-y\rvert^3.$$ You can prove that this is a metric. However, if you define $\lVert x\rVert:=d(x,0),$ then in general, for $x\neq 0,$ we will have $$\lVert\alpha x\rVert\neq\lvert\alpha\rvert\lVert x\rVert,$$ with a few exceptions.

Answer 2

Homogeneity is not verified for the discrete metric for example. You can check the answer at here

Answer 3

Take for example $d$ as the discrete distance, then for $x\not=0$ and $|\lambda|\not=0,1$, we $$1=d(\lambda x,0)=\|\lambda x\|\not=|\lambda|\|x\|=|\lambda|d(x,0)=|\lambda|$$ and the absolute homogeneity property does not hold.