Explain why $|x - y|^2 $ is not a metric.

Question

Define $d(x,y) = |x - y|^2$ . Explain why d is not a metric.

I have seen that the first two properties of a metric hold, so naturally it's left to prove that the triangle inequality does not hold. But after some algebra, I get stuck at the following:

$2|y|(|x| + |z|)-2|x||z|- 2|y|^2<= 0 $

Am I missing something obvious here? I am not sure how to prove that this inequality doesn't hold from here.

Its R x R. Sorry I couldn't figure out how to format that part on this site @MyGlasses — Hoodtingz, Feb 21 '17 at 05:39
For what it's worth, on $\Bbb{R}$, we have $d(0,4)=16$ and $d(0,2)=d(2,4)=4$, so we don't have $d(0,4)\le d(0,2)+d(2,4)$. Probably works on $\Bbb{R}^2$ too: just look at $(0,0)$, $(0,2)$, and $(0,4)\in \Bbb{R}^2$. — ForgotALot, Feb 21 '17 at 05:40

score 2 · Answer 1 · answered Feb 21 '17 at 05:41

2

On $\mathbb{R}$ we have: $$d(0,1)=1 $$ and $$d(0,\frac{1}{2})=d(\frac{1}{2},1)=\frac{1}{4}$$ under your proposed metric. Hence $$d(0,1)>d(0,\frac{1}{2}) +d(\frac{1}{2},1). $$

Note that this extends to $\mathbb{R}^2$ by simply taking the second coordinate zero.

answered Feb 21 '17 at 05:41

Leon Sot

1,293

Is there a way to generalize this? – Hoodtingz Feb 21 '17 at 05:44
Or I suppose providing a counterexample is proof enough – Hoodtingz Feb 21 '17 at 05:45
Well, the same argument works for any normed vector space. – Leon Sot Feb 21 '17 at 05:45
Excellent. thank you – Hoodtingz Feb 21 '17 at 05:50

Explain why $|x - y|^2 $ is not a metric.

1 Answers1

Linked