Does the function $d: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ given by: $$d(x,y)= \frac{\lvert x-y\rvert} {1+{\lvert x-y\rvert}}$$ define a metric on $\mathbb{R}^n?$

How do you go about proving this? Do I need to just show that it satisfies the three conditions to be a metric? If so how do I show them?

  • 2
    $\begingroup$ Yes, you just need to check the three conditions. Two of them are rather immediate, the triangle inequality needs a good idea. $\endgroup$ May 28 '16 at 13:50
  • $\begingroup$ I do not think the function you give is well-defined. If $| \cdot |$ does denote the absolute value, how would you compute $| x - y |$ when $x, y \in \mathbb{R}^n$? $\endgroup$
    – user342207
    May 28 '16 at 13:51
  • 1
    $\begingroup$ @user342207 I think this is the standard Euclidean norm $\endgroup$
    – M10687
    May 28 '16 at 13:52

In general, if $(E,d)$ is a metric space, then $d':=\frac d{1+d}$ is a metric. The triangle inequality is the only nontrivial property here. If $x,y,z\in E$, then $d(x,z)\leqslant d(x,y)+d(y,z)$, so so monotonicity of the map $t\mapsto\frac t{1+t}$ on $[0,\infty)$ yields \begin{align} d'(x,z) &= \frac{d(x,z)}{1+d(x,z)}\\ &\leqslant \frac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}\\ &\leqslant \frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}\\ &=d'(x,y)+d'(y,z). \end{align}

  • $\begingroup$ Is there a definition that states if (E,d) is a metric space, then d' is a metric? I don't think I have come across this concept and would like to look further into it $\endgroup$
    – user323388
    May 29 '16 at 12:37
  • $\begingroup$ is the d' suppose to represent a derivative or is it just a notation? $\endgroup$
    – user323388
    May 29 '16 at 12:44
  • $\begingroup$ @user323388 See here: math.stackexchange.com/questions/987602/… $d'$ - notation. I edited the post to use $:=$ to make it more clear. $\endgroup$
    – Math1000
    May 29 '16 at 12:44
  • $\begingroup$ @user323388: a metric $d_1$ is equivalent to $d_2$ iff for each $x \in X$, there exist positive constants $\alpha$ and$\beta$ such that, for every point $y \in X$, $\alpha d_{1} (x, y) \leq d_{2} (x, y) \leq \beta d_{1} (x, y)$. $\endgroup$ May 29 '16 at 12:54
  • $\begingroup$ @Math1000 can you please do a favor of justifying the first inequality. $\endgroup$ May 29 '16 at 13:06

This first conditions holds trivially, to prove the third condition (triangle inequality) consider the function $f(t)=\frac{t}{1+t}$, $t>0$ and check the monotonicity of $f$.


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