P-Product Metric is Metric

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Theorem

Let $M_{1'} = \struct {A_{1'}, d_{1'} }, M_{2'} = \struct {A_{2'}, d_{2'} }, \ldots, M_{n'} = \struct {A_{n'}, d_{n'} }$ be metric spaces.

Let $\ds \AA = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.

Let $p \in \R_{\ge 1}$.


Let $d_p: \AA \times \AA \to \R$ be the $p$-product metric on $\AA$:

$\ds \map {d_p} {x, y} := \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p}$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.


Then $d_p$ is a metric.


Proof

Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map {d_p} {x, x}\) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, x_i} }^p}^{\frac 1 p}\) Definition of $d_p$
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^n 0^p}^{\frac 1 p}\) as $d_{i'}$ fulfills Metric Space Axiom $(\text M 1)$
\(\ds \) \(=\) \(\ds 0\)

So Metric Space Axiom $(\text M 1)$ holds for $d_p$.

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

Let:

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

Thus we need to show that:

$\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^p}^{\frac 1 p}$


We have:

\(\ds \map {d_p} {x, y} + \map {d_p} {y, z}\) \(=\) \(\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p}\) Definition of $d_p$
\(\ds \) \(=\) \(\ds \paren {\sum r_i^p}^{\frac 1 p} + \paren {\sum s_i^p}^{\frac 1 p}\)
\(\ds \) \(\ge\) \(\ds \paren {\sum \paren {r_i + s_i}^p}^{\frac 1 p}\) Minkowski's Inequality for Sums
\(\ds \) \(=\) \(\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} + \map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p}\) Definition of $r_i$ and $s_i$
\(\ds \) \(\ge\) \(\ds \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^p}^{\frac 1 p}\) as $d_{i'}$ fulfills Metric Space Axiom $(\text M 2)$: Triangle Inequality
\(\ds \) \(=\) \(\ds \map {d_p} {x, z}\) Definition of $d_p$

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_p$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

\(\ds \map {d_p} {x, y}\) \(=\) \(\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p}\) Definition of $d_p$
\(\ds \) \(=\) \(\ds \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^p}^{\frac 1 p}\) as $d_i$ fulfills Metric Space Axiom $(\text M 3)$
\(\ds \) \(=\) \(\ds \map {d_p} {y, x}\) Definition of $d_p$

So Metric Space Axiom $(\text M 3)$ holds for $d_p$.

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

\(\ds x\) \(\ne\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \exists k \in \closedint 1 n: \, \) \(\ds x_k\) \(\ne\) \(\ds y_k\)
\(\ds \leadsto \ \ \) \(\ds \map {d_k} {x_k, y_k}\) \(>\) \(\ds 0\) as $d_k$ fulfills Metric Space Axiom $(\text M 4)$
\(\ds \leadsto \ \ \) \(\ds \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^p}^{\frac 1 p}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \map {d_p} {x, y}\) \(>\) \(\ds 0\) Definition of $d_p$

So Metric Space Axiom $(\text M 4)$ holds for $d_p$.

$\blacksquare$


Also see

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