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  1. Let $f:\mathbb{R}\rightarrow \mathbb{R}$. What conditions ensure that $d(x,y)=|f(x)-f(y)|$ defines a metric on $\mathbb{R}$

  2. Let $g:[0,\infty) \to \mathbb{R}$. What conditions on $g$ ensure that $\phi(x,y) = g(|x-y|)$ defines a metric on $\mathbb{R}$.

For the first one I think the only condition is that it must be injective to ensure that there is only one kernel satisfying that is is positive definite with $d(x,y)=0$ iff $x = y$. The other two conditions for being metric seem to hold regardless of $f$.

For the second one I again think it needs to be injective, but I think it would also have to be surjective so we don't have a situation such as |5-4|=|4-5|=|1-0| etc.

Are the above two correct and can anyone spot anything else I have missed out? Thanks

naslundx
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user132548
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  • Have a look at : http://math.stackexchange.com/questions/155001/condition-on-function-f-mathbbr-rightarrow-mathbbr-so-that-a-b-mapsto?rq=1 – pitchounet Mar 02 '14 at 18:50
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    For the second: If $g$ is increasing and sublinear, then $\phi$ satisfies the triangle inequality. – J.R. Mar 02 '14 at 18:53

1 Answers1

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  1. is correct: only the injectivity of $f$ is needed.

  2. Here, $g$ need not be injective. For example, $g(t)=\min(t,1)$ is a perfectly fine metric transform: the function $g\circ d $ satisfies the triangle inequality for any metric $d$. The property you need is subadditivity:

$$g(a+b)\le g(a)+g(b)$$

A nice sufficient condition: every concave increasing function mapping $0$ to $0$ is subadditive, which is easy to verify after rewriting the property as

$$\frac{g(a+b)-g(b)}{a}\le \frac{g(a)-g(0)}{a}$$ and considering the slopes of secant lines.