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Assume that (X,d) is a metric space.

Is there a simple/smart way (e.g. using some convexity argument) to show shat: $d^{*}(x,y) = ln(1+\frac{d(x,y)}{1+d(x,y)})$ is a distance, esp. that it does satisfy the triangle inequality?

Thus far I can only manage to prove it by using exp and brutally expanding/simplifying.

Thanks, J.

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Lemma: Given a concave function $f:[0, +\infty] \rightarrow \mathbb{R}$ s.t. $f(0)=0$ then $f$ is subadditive i.e. $\forall x,y \space f(x+y) \le f(x)+f(y)$.

You should also know that given a distance $d$, the function $d'=\frac{d}{1+d}$ is a distance; then is straightforward: take arbitrarily $x,y,z \in X$

$$ d^*(x,z)=\ln[1+d'(x,z)]\le \ln[1+d'(x,y)+d'(y,z)] \\ \stackrel{subadditivity}{\le} \ln[1+d'(x,y)]+\ln[1+d'(y,z)]=d^*(x,y)+d^*(y,z) $$


Proof of lemma: https://en.wikipedia.org/wiki/Concave_function point 6 of "properties" section.

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  • $\begingroup$ The same argument can be used to show that $d/(1+d)$ is a distance, that is done for example here: math.stackexchange.com/a/686821/42969. $\endgroup$ – Martin R Sep 21 '17 at 11:06
  • $\begingroup$ Unless I am mistaken, the term $\ln(1+d'(x,y)+1+d'(y,z))$ should be removed from the inequality chain. $\endgroup$ – Martin R Sep 21 '17 at 11:09
  • $\begingroup$ I insert that term to make clearer the following step (in which is used subadditivity). Is it incorrect or redundant? $\endgroup$ – Leonardo Vannini Sep 21 '17 at 11:19
  • $\begingroup$ I think it is incorrect. $f(t) = \ln( 1+ t)$ is subadditive, not $f(t) = \ln t$. $\endgroup$ – Martin R Sep 21 '17 at 11:21
  • $\begingroup$ Btw, \ln t would make "ln" render like an operator. And $\ln[1+d'(x,y)+d'(y,z)]$ is there twice. $\endgroup$ – Martin R Sep 21 '17 at 11:23

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