P-Product Metric on Real Vector Space is Metric

Theorem

Let $\R^n$ be an $n$-dimensional real vector space.

Let $p \in \R_{\ge 1}$.

Let $d_p: \R^n \times \R^n \to \R$ be the $p$-product metric on $\R^n$:

$\displaystyle d_p \left({x, y}\right) := \left({\sum_{i \mathop = 1}^n \left \vert {x_i - y_i} \right \vert^p}\right)^{\frac 1 p}$

where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \R^n$.

Then $d_p$ is a metric.

Proof 1

Proof of $M1$

 $\displaystyle d_p \left({x, x}\right)$ $=$ $\displaystyle \left({\sum_{i \mathop = 1}^n \left \vert {x_i - x_i} \right \vert^p}\right)^{\frac 1 p}$ Definition of $d_p$ $\displaystyle$ $=$ $\displaystyle \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}$ $\displaystyle$ $=$ $\displaystyle 0$

So axiom $M1$ holds for $d_p$.

$\Box$

Proof of $M2$

It is required to be shown:

$d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$

for all $x, y, z \in \R^n$.

Let:

$(1): \quad z = \left({z_1, z_2, \ldots, z_n}\right)$
$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
$(3): \quad \left\vert{x_i - y_i}\right\vert = r_i$
$(4): \quad \left\vert{y_i - z_i}\right\vert = s_i$.

Thus we need to show that:

$\displaystyle \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p} \ge \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}$

We have:

 $\displaystyle d_p \left({x, y}\right) + d_p \left({y, z}\right)$ $=$ $\displaystyle \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p}$ Definition of $d_p$ $\displaystyle$ $=$ $\displaystyle \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}$ $\displaystyle$ $\ge$ $\displaystyle \left({\sum \left({r_i + s_i}\right)^p}\right)^{\frac 1 p}$ Minkowski's Inequality for Sums $\displaystyle$ $\ge$ $\displaystyle \left({\sum \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)^p}\right)^{\frac 1 p}$ Definition of $r_i$ and $s_i$ $\displaystyle$ $=$ $\displaystyle \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}$ Triangle Inequality for Real Numbers $\displaystyle$ $=$ $\displaystyle d_p \left({x, z}\right)$ Definition of $d_p$

So axiom $M2$ holds for $d_p$.

$\Box$

Proof of $M3$

 $\displaystyle d_p \left({x, y}\right)$ $=$ $\displaystyle \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p}$ Definition of $d_p$ $\displaystyle$ $=$ $\displaystyle \left({\sum \left\vert{y_i - x_i}\right\vert^p}\right)^{\frac 1 p}$ $\displaystyle$ $=$ $\displaystyle d_p \left({y, x}\right)$ Definition of $d_p$

So axiom $M3$ holds for $d_p$.

$\Box$

Proof of $M4$

 $\displaystyle x$ $\ne$ $\displaystyle y$ $\displaystyle \implies \ \$ $\displaystyle \exists k \in \left[{1 \,.\,.\, n}\right]: x_k$ $\ne$ $\displaystyle y_k$ $\displaystyle \implies \ \$ $\displaystyle x_k - y_k$ $>$ $\displaystyle 0$ as $d_k$ fulfils axiom $M4$ $\displaystyle \implies \ \$ $\displaystyle \left({\sum \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}$ $>$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle d_p \left({x, y}\right)$ $>$ $\displaystyle 0$ Definition of $d_p$

So axiom $M4$ holds for $d_p$.

$\blacksquare$

Proof 2

This is an instance of $p$-Product Metric is Metric.

$\blacksquare$

Comment on notation

It can be shown that:

$\displaystyle d_\infty \left({x, y}\right) = \lim_{p \to \infty} d_p \left({x, y}\right)$

That is:

$\displaystyle \lim_{p \to \infty} \left({\sum_{i \mathop = 1}^n \left \vert {x_i - y_i} \right \vert^p}\right)^{\frac 1 p} = \max_{i \mathop = 1}^n \left\{{\left \vert {x_i - y_i} \right \vert}\right\}$

Hence the notation.