Prove the invariant property $d(u+w,v+w)=d(u,v)$ of quasi-norm

Question

Let $d(.,.)$ be a metric on a vector space $U$ such that $d(u,0)$ defines a quasi norm $\|.\|$, show that

$$d(u+w,v+w)=d(u,v)$$ for all $w\in U$

if $\|\alpha_n u\|\rightarrow 0$ as $\alpha_n\rightarrow 0$ and $ \|\alpha u_n \|\rightarrow 0$ as $u_n\rightarrow 0$.

Answer 1

Yep. I've been spending 3 hours trying to prove a false theorem.

Let $U = \Bbb R^2$ and $d$ be the French metro metric: $$d(\vec a,\vec b) = \begin{cases} |\vec a - \vec b| & \exists c \in \Bbb R: \vec a = c \vec b \\ |\vec a| + |\vec b| & \text{otherwise} \end{cases}$$

Now, the norm induced by the metric: $$\|\vec a\| = d(\vec a,0) = |\vec a|$$

Put $\vec u = \vec 0$, $\vec v = \vec i$, and $\vec w = \vec j$: $$d(\vec u+\vec w,\vec v + \vec w) = d(\vec i,\vec i+\vec j)= 1+\sqrt 2 \ne 1 = d(\vec 0,\vec i) = d(\vec u,\vec v)$$