Show that $d(x,y)=|e^{-|x-y|}-1|$ is a metric, I already went throught the first 3 steps, but I'm having trouble with the triangular inequality, as far as i know is not possible to get $|e^{-|x-y|}-1|\leq |e^{-|x-z|}-1|+|e^{-|z-y|}-1|$ thanks for your help
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2Nice visualization https://www.desmos.com/calculator/6nry3chr9c, there are no points both red and green (since green shape is open) so the green equation should be $\ge k$. – zwim Dec 10 '21 at 21:32
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Nice way to see it, thanks – AlejandroL.g Dec 10 '21 at 22:16
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1Abstractly, the function $f(t) = 1 - e^{-|t|}$ is $0$ at $0$, and is increasing and concave on the non-negative reals. – Andrew D. Hwang Dec 11 '21 at 03:24
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Note that $e^x\le1$ for all $x\le0$.
$$\begin{aligned}&\quad |e^{-|x-z|}-1|+|e^{-|z-y|}-1| - (|e^{-|x-y|}-1|) \\ &= 1-e^{-|x-z|}+ 1 - e^{-|z-y|} - (1- e^{-|x-y|})\\ &= 1-e^{-|x-z|} - e^{-|z-y|} +e^{-|x-y|}\\ &\ge 1-e^{-|x-z|}- e^{-|z-y|} +e^{-(|x-z| + |z-y|)}\\ &=(1-e^{-|x-z|})(1 - e^{-|z-y|})\\ & \ge0 \end{aligned}$$ The first inequality holds since $|x-y|=|(x-z ) + (z-y)|\le |x-z| + |z-y|$ and the function $f(x)=e^x$ is an increasing function.
Apass.Jack
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