I was asked to prove that $d(x,y) =\max_i{\lvert x_i - y_i \rvert }$ is a distance function in $\mathbb{R}^2$. I've got myself stuck with proving the triangle inequality.

Can someone give me an hint about how to proceed?

  • 2
    $\begingroup$ $\vert x_{i}-y_{i} \vert \leq \vert x_{i}-z_{i} \vert + \vert z_{i}-y_{i} \vert$ by triangular inequality. $\endgroup$ – jibounet Jul 24 '14 at 15:52
  • $\begingroup$ Thanks for the hint! Someone answered with the complete solution, but you centered the target of the question, I've really appreciated this! Thumbs up for you :-) $\endgroup$ – Filippo De Bortoli Jul 24 '14 at 16:18

$$ |x_i - z_i|\leq |x_i - y_i| + |y_i - z_i| \quad\forall i \qquad\implies $$

$$ \max_i|x_i - z_i|\leq \max_i\left(|x_i - y_i| + |y_i - z_i|\right) \leq \max_i|x_i - y_i| + \max_i |y_i - z_i| $$

  • $\begingroup$ Great! Solutions are always easy, but sometimes we are not able to see them at first glance. Thank you very much! $\endgroup$ – Filippo De Bortoli Jul 24 '14 at 16:16

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