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 of $M1$
So axiom $M1$ holds for $d_p$.

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:

So axiom $M2$ holds for $d_p$.

Proof of $M3$
So axiom $M3$ holds for $d_p$.

Proof of $M4$
So axiom $M4$ holds for $d_p$.

Also see

 * Taxicab Metric is Metric
 * Chebyshev Distance is Metric

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.