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In complex numbers $\Bbb{C}=\{z=x+iy:x,y\in \Bbb{R}\}$ $$d(z_1,z_2)=|z_1-z_2|$$ $$p(z_1,z_2)=\frac{2.|z_1-z_2|}{{\sqrt{1+|z_1|^2}}.{\sqrt{1+|z_2|^2}}}$$

Show that $p$ and $d$ aren't equivalent using

$$\exists M,m>0 \ \ \text{such that} \ \ m.d \le p \le M.d.$$

I showed $p \le 2d$ but I can't find $m.d \le p$.

Hacemat
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  • I wrote wrong thanks for warning – Hacemat Feb 10 '21 at 18:26
  • It looks impossible that this so-called "chordal metrics" $p$ is "strongly equivalent" to the usual distance in $\mathbb{C}$. See math.stackexchange.com/q/2087814 and https://en.wikipedia.org/wiki/Equivalence_of_metrics – Jean Marie Feb 10 '21 at 22:25
  • I don't understand why your instructor has proposed you an erroneous result to be proven... He/she should have said you to establish that these distances are equivalent in a weaker sense... (see Wikipedia article). – Jean Marie Feb 11 '21 at 12:22
  • This was an exam question she prepared 60 quenstions we solved 5 question. I pass the lesson but I try to solve all question. Probably she wrote wrong and didn't noticed. – Hacemat Feb 11 '21 at 20:33

1 Answers1

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Note that there is no $m > 0$ such that $m d \leq p.$ Indeed, if this were the case, then we would have that $m \vert z_1 - z_2 \vert \leq \frac{2 \vert z_1 - z_2 \vert}{\sqrt{1 + \vert z_1 \vert^2} \sqrt{1 + \vert z_2 \vert^2}}$ for all $z_1, z_2 \in \mathbb{C}.$ The inequality is of course trivial when $z_1 = z_2,$ so assuming $z_1 \neq z_2,$ we would get that $m \leq \frac{2}{\sqrt{1 + \vert z_1 \vert^2} \sqrt{1 + \vert z_2 \vert^2}}$ for all $z_1, z_2 \in \mathbb{C}$ with $z_1 \neq z_2.$ In particular, let $z_1^n := 0,$ $z_2^n := n$ for all $n \in \mathbb{N}.$ Then $m \leq \frac{2}{\sqrt{n^2 + 1}}$ for all $n \in \mathbb{N},$ which yields the contradiction $0 < m \leq 0.$ Hence, these two metrics are not equivalent.

Actually Fritz
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