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I've got this pretty exotic metric of which I cannot seem to prove the triangle inequality. Given that I already have a metric $\delta$ on the unit ball in $\mathbb{R}^n$, I define a new metric $d(x,y)$ to be zero whenever $x=y$ and $\ln\dfrac{2}{\delta(x,y)}$ whenever this is not the case.

The first two properties are pretty straightforward. As for the triangular inequality, in the case that $x,y$ and $z$ are all different points (the other cases are trivial), I get this: $$\ln\dfrac{2}{\delta(x,y)} \leq \ln\dfrac{2}{\delta(x,z)}+\ln\dfrac{2}{\delta(z,y)} = \ln\dfrac{4}{\delta(x,z)\delta(z,y)} \Leftrightarrow \delta(x,z)\delta(z,y) \leq 2\delta(x,y)$$

I do have the extra condition that none of these three values are zero, nor do they exceed $1$. But now I'm stuck. I'm also not sure that this IS a metric indeed, yet I have failed to find a counterexample. The only other condition I have is that $\delta$ does satisfy the triangular inequality, hence $$\delta(x,y) \leq \delta(x,z)+\delta(z,y)$$

Who can help?

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Your can find a triangle whose sides in terms of Euclidean space's natural metric are arbitrarily close to 2, 2 and 0. Then their $\delta $ values would be close to 0 for the first two and arbitrarily large for the third one. Then the triangle inequality is not satisfied.

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There is a triangle $x,y,z$ in the ball such that $\delta(x,y)$ is arbitrarily close to $0$, and $\delta(x,z),\delta(y,z)\approx 1$. So that defeats your triangle inequality for $d$.

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  • $\begingroup$ Yes, I was starting to think it wouldn't hold. Thanks for the confirmation. $\endgroup$ – Werner Jun 17 '19 at 19:53
  • $\begingroup$ Zero, one and one doesn't seem to contradict the triangle inequality. It's simply a very thin isosceles triangle, isn't it? $\endgroup$ – CiaPan Jun 19 '19 at 21:41
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    $\begingroup$ @CiaPan it contradicts $\delta(x,y)\delta(y,z)\leq 2\delta(x,z)$. $\endgroup$ – user10354138 Jun 19 '19 at 21:44
  • $\begingroup$ Oh, yes, you're right. I have swapped the meanings of $d$ and $\delta$ in my mind. Now I see my answer is generally same as yours: one side arbitrarily short in the original metric becomes arbitrarily long in terms of a new function, proving it is not a metric. $\endgroup$ – CiaPan Jun 19 '19 at 22:00

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