1

Question:

For $0\lt p\lt 1$ and $x, y \in\mathbb{R}^2$ with $x=(\xi_1, \xi_2), y = (\eta_1, \eta_2)$, let $d :\mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R} $

Be defined as $d_p(x, y) = (|\xi_1, − \eta_1|^p + |\xi_2 - \eta_2|^p)^{1/p}$. Show that $d_p$ is not a metric on $\mathbb{R}^2$.

My attempt:

$d_p(x, y) = (|\xi_1 − \eta_1|^p + |\xi_2 - \eta_2|^p)^{1/p}$

Note: Minkowski Inequality for $0\lt p\lt1$:

$(\sum_{k=1}^n |x_k+y_k|^p)^{1/p}\ge(\sum_{k=1}^n |x_k|^p)^{1/p}+(\sum_{k=1}^n |y_k|^p)^{1/p}$

$\Rightarrow d_p(x,y)\ge(|\xi_1|^p+|\xi_2|^p)^{1/p}+(|\eta_1|^p+|\eta_2|^p)^{1/p}$

Note: I know this is not true for all $a$, but it's the best I've got :/ $|x|\ge|x-a| \Rightarrow |x|^p\ge|x-a|^p \Rightarrow |x|^p+|y|^p\ge|x-a|^p+|y-b|^p$

So... $d_p(x,y)\ge(|\xi_1|^p+|\xi_2|^p)^{1/p}+(|\eta_1|^p+|\eta_2|^p)^{1/p}\ge(|\xi_1-\alpha_1|^p+|\xi_2-\alpha_2|^p)^{1/p}+(|\eta_1-\alpha_1|^p+|\eta_2-\alpha_2|^p)^{1/p}$

$=d_p(x,z)+d_p(y,z)$.

Hence the Triangle Inequality Axiom is not satisfied, therefore $d_p(x,y)$ is not a metric on $\mathbb{R}^2$.

Summary

I don't think the use of the Minkowski inequality is necessary when tackling this problem but this is the best I could come up with today. I've just started a Functional Analysis course and I'm just getting to grips with some of the problems before any worksheets are due. Any help/alternative solutions would be greatly appreciated.

George Cooper
  • 635
  • 1
    You really just need to find three point $x,y,s$ with $d_p(x,z)>d_p(x,y)+d_p(y,z).$ – Thomas Andrews Sep 27 '21 at 20:07
  • 1
    $d_p(x,y)\geq d_p(x,z)+d_p(y,z)$ does not contradict being a metric directly, but you can deduce it is not a metric from there if your metric space contains at least three distinct points. – Thomas Andrews Sep 27 '21 at 20:09
  • Okay, are you referring to the case where the LHS and RHS could be equal? – George Cooper Sep 27 '21 at 20:10
  • Would it be best to choose three distinct points that do not satisfy the inequality? @ThomasAndrews – George Cooper Sep 27 '21 at 20:13
  • I haven’t followed the entire argument, but if $x=y,$ the inequality can’t be true. – Thomas Andrews Sep 27 '21 at 20:14
  • That’s the easiest way. No reason to figure out all cases $x,y,z$ which fail the triangle inequality, when one triple is enough. And I don’t think your inequality is correct. – Thomas Andrews Sep 27 '21 at 20:18
  • Oh yeah, I'm it's awfully incorrect, as I've made many false assumptions. I was in the routine of manipulating the inequalities to acquire the desired result. Thank you for directing me in the correct direction :) – George Cooper Sep 27 '21 at 20:23

2 Answers2

4

You can simply use that fact that$$d_p\bigl((1,0),(0,0)\bigr)+d_p\bigl((0,0),(0,1)\bigr)=1+1=2,$$but$$d_p\bigl((1,0),(0,1)\bigr)=2^{1/p}>2.$$

José Carlos Santos
  • 427,504
2

Just to expand on the counter example of José Carlos Santos, when $0<p<1$ $$\rho_p((x,y),(x',y'))=|x-x'|^p+|y-y'|^p$$ is a complete metric on $\mathbb{R}^n$. That this is a metric follows from $$(a+b)\leq a^p+b^p, \qquad a,b\geq0$$ which in turn can be established from the inequality $(1+x)^p\leq 1+ x^p$, $x\geq0$. The metric $\rho_p$ is translation invariant, that is $\rho_p(\boldsymbol{u}+\boldsymbol{z},\boldsymbol{v}+\boldsymbol{z})=\rho_p(\boldsymbol{u},\boldsymbol{v})$; however, $\rho_p$ is not homogeneous: $\rho_p(c\boldsymbol{u},c\boldsymbol{v})=c^p\rho_p(\boldsymbol{u},\boldsymbol{v})$.

More generally, for any measure space $(\Omega,\mathscr{F},\mu)$, the space $L_p(\mu)$ (modulo identification of almost surely functions) of measurable functions $f$ such that $\int_\Omega|f|^p\,d\mu<\infty$ is a complete metric space under the metric $$\rho(f, g)=\int_{\Omega}|f-g|^p\,d\mu$$

Mittens
  • 39,145