P-Product Metric on Real Vector Space is Metric/Proof 1
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Theorem
Let $\R^n$ be an $n$-dimensional real vector space.
Let $p \in \R_{\ge 1}$.
Let $d_p: \R^n \times \R^n \to \R$ be the $p$-product metric on $\R^n$:
- $\ds \map {d_p} {x, y} := \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{\frac 1 p}$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \R^n$.
Then $d_p$ is a metric.
Proof
Proof of Metric Space Axiom $(\text M 1)$
| \(\ds \map {d_p} {x, x}\) | \(=\) | \(\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - x_i}^p}^{\frac 1 p}\) | Definition of $d_p$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{i \mathop = 1}^n 0^p}^{\frac 1 p}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
So Metric Space Axiom $(\text M 1)$ holds for $d_p$.
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
It is required to be shown:
- $\map {d_p} {x, y} + \map {d_p} {y, z} \ge \map {d_p} {x, z}$
for all $x, y, z \in \R^n$.
Let:
- $(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
- $(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
- $(3): \quad \size {x_i - y_i} = r_i$
- $(4): \quad \size {y_i - z_i} = s_i$.
Thus we need to show that:
- $\ds \paren {\sum \size {x_i - y_i}^p}^{\frac 1 p} + \paren {\sum \size {y_i - z_i}^p}^{\frac 1 p} \ge \paren {\sum \size {x_i - z_i}^p}^{\frac 1 p}$
We have:
| \(\ds \map {d_p} {x, y} + \map {d_p} {y, z}\) | \(=\) | \(\ds \paren {\sum \size {x_i - y_i}^p}^{\frac 1 p} + \paren {\sum \size {y_i - z_i}^p}^{\frac 1 p}\) | 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 | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds \paren {\sum \paren {\size {x_i - y_i} + \size {y_i - z_i} }^p}^{\frac 1 p}\) | Definition of $r_i$ and $s_i$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum \size {x_i - z_i}^p}^{\frac 1 p}\) | Triangle Inequality for Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {d_p} {x, z}\) | Definition of $d_p$ |
So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_p$.
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
| \(\ds \map {d_p} {x, y}\) | \(=\) | \(\ds \paren {\sum \size {x_i - y_i}^p}^{\frac 1 p}\) | Definition of $d_p$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum \size {y_i - x_i}^p}^{\frac 1 p}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {d_p} {y, x}\) | Definition of $d_p$ |
So Metric Space Axiom $(\text M 3)$ holds for $d_p$.
$\Box$
Proof of Metric Space Axiom $(\text M 4)$
| \(\ds x\) | \(\ne\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists k \in \closedint 1 n: \, \) | \(\ds x_k\) | \(\ne\) | \(\ds y_k\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x_k - y_k\) | \(>\) | \(\ds 0\) | as $d_k$ fulfils Metric Space Axiom $(\text M 4)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\sum \size {x_k - y_k}^p}^{\frac 1 p}\) | \(>\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {d_p} {x, y}\) | \(>\) | \(\ds 0\) | Definition of $d_p$ |
So Metric Space Axiom $(\text M 4)$ holds for $d_p$.
$\blacksquare$