# Is this exotic function actually a metric?

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?

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.
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 $$d$$.
• @CiaPan it contradicts $\delta(x,y)\delta(y,z)\leq 2\delta(x,z)$. – user10354138 Jun 19 '19 at 21:44
• 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. – CiaPan Jun 19 '19 at 22:00