1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Yes, you just need to check the three conditions. Two of them are rather immediate, the triangle inequality needs a good idea. $\endgroup$ – Daniel Fischer 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
9
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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 $y \in X$, $\alpha d_{1} (x, y) \leq d_{2} (x, y) \leq \beta d_{1} (x, y)$. $\endgroup$ – Mohammad W. Alomari May 29 '16 at 12:54
  • $\begingroup$ @Math1000 can you please do a favor of justifying the first inequality. $\endgroup$ – Urban PENDU May 29 '16 at 13:06
2
$\begingroup$

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 $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.