Metric not induced by any norm

Question

I'm stuck with two problems, both of them involving metric not induced by any norm.

1) $d_p:\mathbb{R}^2 \times \mathbb{R}^2 \longrightarrow \mathbb{R}$

$d_p(x,y) = \left\{ \begin{array}{ll}||x||_2+||y||_2 & \mbox{if } x \neq y \\ 0 & \mbox{if } x=y \end{array} \right.$

I guess this is the French Railways Metric.

2) $d_r:X \times X \longrightarrow \mathbb{R}$, where $(X, ||\cdot ||)$ is a normed space:

$d_p(x,y) = min\{1, ||x-y||\}$

Well, for point 1) I supossed that there's a norm $|| \cdot ||^*:\mathbb{R}^2 \longrightarrow \mathbb{R}$ that $||x||^*=d_p(x,0)$ and I try to use the properties of the metric and prove that $||\cdot||^*$ is not a norm. I readed that the problem usually is prove $||\alpha x||^* = |\alpha| ||x||^*$ but:

• If $\alpha x = 0$ then:

$||\alpha x||^* = d_p(\alpha x,0)=0=|\alpha| ||x||_2 =|\alpha|d_p(x,0) = |\alpha| ||x||^*$

• If $\alpha x \neq 0$ then:

$||\alpha x||^* = d_p(\alpha x,0)=||\alpha x||_2 = |\alpha| ||x||_2=|\alpha|d_p(x,0) = |\alpha| ||x||^*$

Where's my mistake?

Answer 1

Indeed it is not a nonzero norm. Let its define a norm: $$||2x||^*=d_p(2x,0)=d_p(\frac{n-1}{n}x,\frac{n+1}{n}x)$$ for any natural number $n$. Now since $||\cdot||^*$ is a norm so the function $d_p$ is continuous and $\lim_{n\rightarrow \infty}(\frac{n-1}{n}x,\frac{n+1}{n}x)=(x,x)$, therefore $$||2x||^*= \lim_{n\rightarrow \infty} d_p(\frac{n-1}{n}x,\frac{n+1}{n}x)=d_p(x,x)=0;$$ $$||x||^*=0$$ for all $x$.

Hint. If $c>1$ and $r\neq 0$ be reals then $$d_p(\frac{rc-r}{2c}x,\frac{rc+r}{2c}x)=||rx||_2=||rx||^*$$