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

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

Let $r, t \in \R_{\ge 1}$.

Let $d_r$, $d_t$ and $d_\infty$ be general Euclidean metrics on $\R^n$.

Then $d_r$, $d_t$ and $d_\infty$ are topologically equivalent.

Inequalities for Infinite Case
Without loss of generality, assume that $r \le t$. First, we show that $d_t \left({x, y}\right) \ge d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x, y}\right)$.

By definition of $d_\infty$, we have $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$.

With $i'$ chosen so $\left\vert{x_{i'} - y_{i'} }\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$, it follows that:


 * $\displaystyle d_t \left({x, y}\right) = \left({ \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert^t }\right)^{1/t} \ge \left({ \left\vert{x_{i'} - y_{i'} }\right\vert^t }\right)^{1/t} = \left\vert{x_{i'} - y_{i'} }\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert} = d_\infty \left({x, y}\right)$

The second inequality holds trivially, as:


 * $\displaystyle n \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert} \ge  \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$