Show the $d_p(x,y)$ is not a metric on $\mathbb{R}^2$

Question

---

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.

Answer 1

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$$d_p\bigl((1,0),(0,1)\bigr)=2^{1/p}>2.$$

Answer 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 $$\rho(f, g)=\int_{\Omega}|f-g|^p\,d\mu$$