Logarithmic triangle inequality

Question

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.

score 1 · Accepted Answer · edited Sep 21 '17 at 11:43

1

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.

edited Sep 21 '17 at 11:43

Martin R

128,226

answered Sep 21 '17 at 10:32

ictibones

1,984

The same argument can be used to show that $d/(1+d)$ is a distance, that is done for example here: https://math.stackexchange.com/a/686821/42969. – Martin R Sep 21 '17 at 11:06
Unless I am mistaken, the term $\ln(1+d'(x,y)+1+d'(y,z))$ should be removed from the inequality chain. – Martin R Sep 21 '17 at 11:09
I insert that term to make clearer the following step (in which is used subadditivity). Is it incorrect or redundant? – ictibones Sep 21 '17 at 11:19
I think it is incorrect. $f(t) = \ln( 1+ t)$ is subadditive, not $f(t) = \ln t$. – Martin R Sep 21 '17 at 11:21
Btw, \ln t would make "ln" render like an operator. And $\ln[1+d'(x,y)+d'(y,z)]$ is there twice. – Martin R Sep 21 '17 at 11:23
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You're right, thanks, I think is correct now – ictibones Sep 21 '17 at 11:24

Logarithmic triangle inequality

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