P-Product Metric is Metric

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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

Proof of $M1$

\(\displaystyle d_p \left({x, x}\right)\) \(=\) \(\displaystyle \left({\sum_{i \mathop = 1}^n \left({d_{i'} \left({x_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}\) as $d_{i'}$ fulfils axiom $M1$
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So axiom $M1$ holds for $d_p$.

$\Box$


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:

\(\displaystyle d_p \left({x, y}\right) + d_p \left({y, z}\right)\) \(=\) \(\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}\)
\(\displaystyle \) \(\ge\) \(\displaystyle \left({\sum \left({r_i + s_i}\right)^p}\right)^{\frac 1 p}\) Minkowski's Inequality for Sums
\(\displaystyle \) \(=\) \(\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right) + d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}\) Definition of $r_i$ and $s_i$
\(\displaystyle \) \(\ge\) \(\displaystyle \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}\) as $d_{i'}$ fulfils axiom $M2$
\(\displaystyle \) \(=\) \(\displaystyle d_p \left({x, z}\right)\) Definition of $d_p$

So axiom $M2$ holds for $d_p$.

$\Box$


Proof of $M3$

\(\displaystyle d_p \left({x, y}\right)\) \(=\) \(\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p}\) Definition of $d_p$
\(\displaystyle \) \(=\) \(\displaystyle \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}\) as $d_i$ fulfils axiom $M3$
\(\displaystyle \) \(=\) \(\displaystyle d_p \left({y, x}\right)\) Definition of $d_p$

So axiom $M3$ holds for $d_p$.

$\Box$


Proof of $M4$

\(\displaystyle x\) \(\ne\) \(\displaystyle y\)
\(\displaystyle \implies \ \ \) \(\displaystyle \exists k \in \left[{1 \,.\,.\, n}\right]: x_k\) \(\ne\) \(\displaystyle y_k\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_k \left({x_k, y_k}\right)\) \(>\) \(\displaystyle 0\) as $d_k$ fulfils axiom $M4$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}\) \(>\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_p \left({x, y}\right)\) \(>\) \(\displaystyle 0\) Definition of $d_p$

So axiom $M4$ holds for $d_p$.

$\blacksquare$


Also see


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.