P-Product Metric on Real Vector Space is Metric

Theorem
The generalized Euclidean metric is a metric.

Proof
The generalized Euclidean metric is as follows:

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

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

Consider the Euclidean metric $$d \left({x, y}\right) = \left({\sum_{i=1}^n \left({x_i - y_i}\right)^2}\right)^{\frac 1 2}$$ on $$\reals^n$$.

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:

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

So we have to prove:

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

This is Minkowski's Inequality.

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

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

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, $$\left|{x_k - y_k}\right| \le \max_{i=1}^n \left\{{\left|{x_i - y_i}\right|}\right\}$$ and $$\left|{y_k - z_k}\right| \le \max_{i=1}^n \left\{{\left|{y_i - z_i}\right|}\right\}$$

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

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 $$d_\infty \left({x, y}\right) = \lim_{r \to \infty} d_r \left({x, y}\right)$$.

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

Hence the notation.