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A metric cone $C$ is a nonempty metric space (whose distance is denoted $d$) together with a map $\cdot\colon \mathbf{R}\times C \mapsto C$ satisfying these axioms:

  • $a\cdot(b\cdot x) = (ab)\cdot x$ for all reals $a$ and $b$, and all $x$ in $C$,

  • $d(a\cdot x;a\cdot y) = \vert a\vert d(x;y)$ for all real $a$ and all $x$ and $y$ in $C$,

  • $d(a\cdot x;b\cdot x) = \vert a-b\vert d(x;0)$ for all reals $a$ and $b$, and all $x$ in $C$, where $0$ denote $0\cdot x$ for any $x \in C$ (this definition is independent of the choice of $x$),

  • $d((a+a')\cdot x;(b+b')\cdot y) \leq d(a\cdot x;b\cdot y) + d((a'\cdot x;b'\cdot y)$ for all reals $a$, $b$, $a'$, and $b'$, and all $x$ and $y$ in $C$.

Those structures arise in my work about metric vector spaces. In some situations, I manage to prove the three first axioms, but not the fourth. So my question is:

Is the fourth axiom independent from the three other ones?

I have looked for a counterexample, and for a demonstration of the last axiom from the previous, but I failed in both attempts.

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up vote 5 down vote accepted

Yes, it is independent.

Consider function $f\colon \mathbb R^2\to \mathbb R_\ge$ defened the following way: $$f(x,y)=f(y,x)$$ and if $|y|\le |x|$ then $$f(x,y)=|x|+\min\{|y|,|y-\tfrac12{\cdot} x|\}.$$ Note that one can define a metric on the union of two coordinate lines $C=\{(x,y)\in \mathbb R^2\mid x{\cdot}y=0\}$ the following way: $$d((x,0),(y,0))=d((0,x),(0,y))=|x-y|$$ and $$d((x,0),(0,y))=f(x,y).$$

It is straightforward to check that your first axioms hold for $d$. Take $x=(1,0)$, $y=(0,1)$, $a=1$, $b=\tfrac12$, $a'=\tfrac12$ and $b'=1$. Then $$d((a+a')\cdot x;(b+b')\cdot y)=\tfrac32+\tfrac34>2.$$

On the other hand $$d(a\cdot x;b\cdot y) + d((a'\cdot x;b'\cdot y)=2.$$

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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. – Larrieu Jan 19 '12 at 18:50
$d((1,0),(0,\tfrac12))=f(1,\tfrac12)=1$. So each term in the last sum is 1... – Anton Petrunin Jan 19 '12 at 18:59
OK you are right. This counter example is difficult to find. Did you know it before, or you managed to find it so quickly ? – Larrieu Jan 19 '12 at 21:03

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