P-Product Metric is Metric

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 of
So holds for $d_p$.

Proof of
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:

So holds for $d_p$.

Proof of
So holds for $d_p$.

Proof of
So holds for $d_p$.

Also see

 * Taxicab Metric is Metric
 * Chebyshev Distance is Metric


 * Chebyshev Distance is Limit of P-Product Metric, that is: $\ds \map {d_\infty} {x, y} = \lim_{p \mathop \to \infty} \map {d_p} {x, y}$