Relation between P-Product Metric and Chebyshev Distance on Real Vector Space

Theorem

For $n \in \N$, let $\R^n$ be a Euclidean space.

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

Let $d_p$ be the $p$-product metric on $\R^n$.

Let $d_\infty$ be the Chebyshev distance on $\R^n$.

Then

$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1/p} \map {d_\infty} {x, y}$

By definition of the Chebyshev distance on $\R^n$, we have:

$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$

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

Let $j$ be chosen so that:

$\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n \size {x_i - y_i}$

Then:

$\ds \map {d_\infty} {x, y}$

$=$

$\ds \paren {\size {x_j - y_j}^p}^{1/p}$

$\ds $

$\le$

$\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{1/p}$

$\ds $

$=$

$\ds \map {d_p} {x, y}$

$\ds $

$\le$

$\ds \paren {n \size {x_j - y_j}^p}^{1/p}$

$\ds $

$=$

$\ds n^{1/p} \size {d_\infty} {x, y}$

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $3$