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 $\displaystyle \mathcal M = \left({\prod_{i=1}^n \left({A_{i'}, d_{i'}}\right), d_n}\right)$, where $d_n$ is defined as:

where:
 * $\displaystyle x = \left({x_1, x_2, \ldots, x_n}\right) \in \prod_{i=1}^n A_{i'}$

and:
 * $\displaystyle y = \left({y_1, y_2, \ldots, y_n}\right) \in \prod_{i=1}^n A_{i'}$

These product spaces are metrics.

Proof for p = 1
This is the taxicab metric, which has been proved to be a metric.

Proof for p = 2, 3, ...
It is easy to see that conditions M0, M1 and M2 of the conditions for being a metric are satisfied. So all we need to do is check M3.

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

Then:


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

As all the $d_{i'}$ are metrics, $d_{i'} \left({x_i, y_i}\right) \ge 0$ and $d_{i'} \left({y_i, z_i}\right) \ge 0$.

Thus $r_i = \left|{r_i}\right|$ and $s_i = \left|{s_i}\right|$, and of course $r_1 + s_i = \left|{r_i + s_i}\right|$.

This reduces our condition for being a metric space to:


 * $\displaystyle \left({\sum \left|{r_i}\right|^p}\right)^{\frac 1 p} + \left({\sum \left|{s_i}\right|^p}\right)^{\frac 1 p} \ge \left({\sum \left|{r_i + s_i}\right|^p}\right)^{\frac 1 p}$

This is Minkowski's Inequality.

Proof for Infinite Case
We have that:
 * $\displaystyle d_\infty \left({x, y}\right) = \max_{i=1}^n \left\{{d_{i'} \left({x_i, y_i}\right)}\right\}$

Let $k \in \left[{1 \,. \, . \, n}\right]$ such that $\displaystyle d_{k'} \left({x_k, y_k}\right) = d_\infty \left({x, z}\right) = \max_{i=1}^n \left\{{d_{i'} \left({x_i, y_i}\right)}\right\}$.

Then by the definition of a metric:
 * $d_{k'} \left({x_k, z_k}\right) \le d_{k'} \left({x_k, y_k}\right) + d_{k'} \left({y_k, z_k}\right)$

But by the nature of the $\max$ operation:
 * $\displaystyle d_{k'} \left({x_k, y_k}\right) \le \max_{i=1}^n \left\{{d_{i'} \left({x_i, y_i}\right)}\right\}$ and $d_{k'} \left({y_k, z_k}\right) \le \max_{i=1}^n \left\{{d_{i'} \left({y_i, z_i}\right)}\right\}$

Thus:
 * $d_{i'} \left({x_i, y_i}\right) + d_{k'} \left({y_k, z_k}\right) \le \max_{i=1}^n \left\{{d_{i'} \left({x_i, y_i}\right)}\right\} + \max_{i=1}^n \left\{{d_{i'} \left({y_i, z_i}\right)}\right\}$

Hence $d_\infty \left({x, z}\right) \le d_\infty \left({x, y}\right) + d_\infty \left({y, z}\right)$.

Comment on notation
It can be shown that:
 * $\displaystyle d_\infty \left({x, y}\right) = \lim_{r \to \infty} d_r \left({x, y}\right)$

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

Hence the notation.