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I would probably get this answer eventually, I pose this question because of the time I have spent looking for why it is not.

The metric is:

$$d_p(x,y)=(|x_1-y_1|^p+|x_2-y_2|^p)^\frac{1}{p}$$

for $p\in(0,1)$

I've just realised I actually missinterpreted it, I read it as $|x_1-x_2|$ ...

A contradiction would be easy to find in that case (Using $d(x,z)\le d(x,y)+d(y,z)$)

edit: It's even easier, consider (1,1) and (2,2), the metric is 0 but d(x,y)=0 $\iff x=y$ - thus a contradiction.

If the question were:

$$d_p(x,y)=(|x_1-x_2|^p+|y_1-y_2|^p)^\frac{1}{p}$$

Is that a metric? I think it is but I've had little luck proving it.

Alec Teal
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  • It may be helpful: http://math.stackexchange.com/questions/278420/minkowskis-inequality-for-0p1 – Stephen Dedalus Jan 13 '14 at 14:30
  • @Baranovskiy I misread a similar question, I want to know if what I read it as initially is a metric. – Alec Teal Jan 13 '14 at 14:31
  • Check $p=\frac{1}{2}$ and points $p_1=(4,2), p_2=(2,4), p_3=(0,8)$ for reverse triangle inequality. – Stephen Dedalus Jan 13 '14 at 14:49
  • Or any other linary indepedendent triple. – Stephen Dedalus Jan 13 '14 at 15:17
  • It would help if we knew whether $x=(x_1,x_2)$ and $y=(y_1,y_2)$, or on the other hand is the notation $x=(x_1,y_1)$ and $y=(x_2,y_2)$ [to me the latter would be a strange choice of notation.] Anyway it should be pth root of sum of pth powers of absolute differences for first and second coordinates, to avoid notational convention misunderstanding. – coffeemath Jan 13 '14 at 16:47

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Hint: What is $d_p(x,y)$ when $x=(x_1,x_2)=(1,1)$ and $y=(y_1,y_2)=(2,2)$?

Remark: $x\neq y$ ;)

yo'
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  • Oh, you answered yourself. Now it seems to me the question makes a little if any sense ;) – yo' Jan 13 '14 at 14:35
  • I should probably remove that, you see at the bottom where it says "If the question were" - that's what I want answering – Alec Teal Jan 13 '14 at 14:36
  • I answered that one. It is not, with the counter-example I give. – yo' Jan 13 '14 at 14:37
  • $(1^p+1^p)^\frac{1}{p}$ - wait, the point x = $(x_1,y_1)$ is what I meant, there is a lot of ambiguity here. grr! – Alec Teal Jan 13 '14 at 14:40