P-Product Metric on Real Vector Space is Metric

Theorem
$\newcommand{\dist} [2] {\left| {{#1} - {#2}} \right|}$ The generalized Euclidean metrics are metrics.

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

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

The generalized Euclidean metric is defined as follows:

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) $x_i - y_i = r_i$;
 * 4) $y_i - z_i = s_i$.

Then:


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

So we have to prove:


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

This is Minkowski's Inequality.

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

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

Then by the Triangle Inequality, $\left|{x_k - z_k}\right| \le \left|{x_k - y_k}\right| + \left|{y_k - z_k}\right|$.

But by the nature of the $\max$ operation, $\displaystyle \left|{x_k - y_k}\right| \le \max_{i=1}^n \left\{{\left|{x_i - y_i}\right|}\right\}$ and $\displaystyle \left|{y_k - z_k}\right| \le \max_{i=1}^n \left\{{\left|{y_i - z_i}\right|}\right\}$

Thus $\displaystyle \left|{x_k - y_k}\right| + \left|{y_k - z_k}\right| \le \max_{i=1}^n \left\{{\dist {x_i} {y_i}}\right\} + \max_{i=1}^n \left\{{\dist {x_i} {y_i}}\right\}$.

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

Alternative Proof
It can be noted that the generalized Euclidean metrics are special cases of the product space metrics $d_1, d_2, \ldots, d_{\infty}$.

Therefore, from Product Space Metrics are Metrics, it can be deduced that the generalized Euclidean metrics are metrics.

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

Hence the notation.