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Jan 19, 2012 at 21:03 vote accept Larrieu
Jan 19, 2012 at 21:03 comment added Larrieu OK you are right. This counter example is difficult to find. Did you know it before, or you managed to find it so quickly ?
Jan 19, 2012 at 18:59 comment added Anton Petrunin $d((1,0),(0,\tfrac12))=f(1,\tfrac12)=1$. So each term in the last sum is 1...
Jan 19, 2012 at 18:50 comment added Larrieu I'm afraid your counter example doesn't work. I have not checked all the axioms (in particular the fact that $d$ is a distance) but the calculus of the last line is wrong. It should be replaced by : $$d(a.x;b.y)+d(a'.x;b'.y) = \frac{3}{2} + \frac{1}{2}+\frac{3}{4} > d((a+a').x;(b+b').y).$$ I'm sorry. Nevertheless, your method for finding a counter example is a good one, I've been looking this way for quite a long without any success.
Jan 19, 2012 at 18:38 vote accept Larrieu
Jan 19, 2012 at 18:38
Jan 19, 2012 at 18:38 vote accept Larrieu
Jan 19, 2012 at 18:38
Jan 19, 2012 at 16:31 history answered Anton Petrunin CC BY-SA 3.0