# P-Product Metric is Metric

## Theorem

Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be metric spaces.

Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.

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

Let $d_p: \mathcal A \times \mathcal A \to \R$ be the $p$-product metric on $\mathcal A$:

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

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

Then $d_p$ is a metric.

## Proof

### Proof of $M1$

 $\displaystyle d_p \left({x, x}\right)$ $=$ $\displaystyle \left({\sum_{i \mathop = 1}^n \left({d_{i'} \left({x_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}$ Definition of $d_p$ $\displaystyle$ $=$ $\displaystyle \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}$ as $d_{i'}$ fulfils axiom $M1$ $\displaystyle$ $=$ $\displaystyle 0$

So axiom $M1$ holds for $d_p$.

$\Box$

### Proof of $M2$

Let:

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

Thus we need to show that:

$\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p} \ge \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}$

We have:

 $\displaystyle d_p \left({x, y}\right) + d_p \left({y, z}\right)$ $=$ $\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}$ 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 $\displaystyle$ $=$ $\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right) + d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}$ Definition of $r_i$ and $s_i$ $\displaystyle$ $\ge$ $\displaystyle \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}$ as $d_{i'}$ fulfils axiom $M2$ $\displaystyle$ $=$ $\displaystyle d_p \left({x, z}\right)$ Definition of $d_p$

So axiom $M2$ holds for $d_p$.

$\Box$

### Proof of $M3$

 $\displaystyle d_p \left({x, y}\right)$ $=$ $\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p}$ Definition of $d_p$ $\displaystyle$ $=$ $\displaystyle \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}$ as $d_i$ fulfils axiom $M3$ $\displaystyle$ $=$ $\displaystyle d_p \left({y, x}\right)$ 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 \ \$ $\displaystyle d_k \left({x_k, y_k}\right)$ $>$ $\displaystyle 0$ as $d_k$ fulfils axiom $M4$ $\displaystyle \implies \ \$ $\displaystyle \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}$ $>$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle d_p \left({x, y}\right)$ $>$ $\displaystyle 0$ Definition of $d_p$

So axiom $M4$ holds for $d_p$.

$\blacksquare$

## Comment on notation

It can be shown that:

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

That is:

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

Hence the notation.