P-Product Metric on Real Vector Space is Metric

Theorem
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 M1, M3 and M4 of the conditions for being a metric are satisfied. So all we need to do is check M2.

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

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 $\displaystyle d_\infty \left({x, y}\right) = \max_{i=1}^n \left\{{\left \vert {x_i - y_i} \right \vert}\right\}$.

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

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

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

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

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

Hence the notation.