For $\Bbb{R}^m$ metrics $d_1(v,w)=\sum|v_i-w_i|$, $d_2(v,w)=\sqrt{\sum(v_i-w_i)^2}$, why does checking $w=0$ suffice to show $d_1(v,w)\leq d_2(v,w)$?

Question

(Apologies for lack of information initially)

I had this proposition in my class earlier.

Let $v, w\in \mathbb{R}^m$, then with the $d_1$ and $d_2$ metrics

$$d_1(v,w)=\sum_{i=1}^m|v_i-w_i| \qquad d_2(v,w)=\sqrt{\sum_{i=1}^m(v_i-w_i)^2}$$

(with $v_i$ and $w_i$ the components of $v$ and $w$) we have the inequality $$d_1(v,w)\leq d_2(v,w)$$

The start of the proof was by assuming $w=0\in\mathbb{R}^m$ and then by considering the associated norms, $\|v\|_1=d_1(v,0)$, $\|v\|_2=d_2(v,0)$, which satisfy the inequality. The proof claims that since the norms satisfy the inequality, the metrics satisfy the inequality, as well.

I was wondering why is it allowed to just assume one of the vector to be the zero vector?

Answer 1

Both of these metrics are translation-invariant, so that $$ d(v,w) = d((v-w),(w-w)) = d(v-w,0). $$ That reduces the problem of proving this inequality to the problem of proving it in the case in which one of the vectors is $0.$