User:Dfeuer/General Euclidean Metrics are Topologically Equivalent

Proof
First we are going to show that:


 * $\displaystyle d_1 \left({x, y}\right) \ge d_2 \left({x, y}\right) \ge \cdots \ge d_r \left({x, y}\right) \ge \cdots \ge d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x, y}\right)$

Then we can demonstrate Lipschitz equivalence between all of these metrics, from which topological equivalence follows.

Let $r \in \N: r \ge 1$.

Let $d_r$ be the metric defined as $\displaystyle d_r \left({x, y}\right) = \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{\frac 1 r}$.


 * First we wish to show that $d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x, y}\right)$.

By the definitions of $d_\infty$ and $d_1$, this is equivalent to showing that


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

But this holds trivially.


 * Now we wish to show that that $\forall r \in \N: d_r \left({x, y}\right) \ge d_{r+1} \left({x, y}\right)$.

That is, that:
 * $\displaystyle \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{\frac 1 r} \ge \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^{r+1}}\right)^{\frac 1 {r+1}}$

Let $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i = \left|{x_i - y_i}\right|$.

Suppose $s_k = 0$ for some $k \in \left[{1 \,.\,.\, n}\right]$.

Then the problem reduces to the equivalent one of showing that:
 * $\displaystyle \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^r}\right)^{\frac 1 r} \ge \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^{r+1}}\right)^{\frac 1 {r+1}}$

that is, of reducing the index by $1$.

Note that when $n = 1$, from simple algebra $d_r \left({x, y}\right) = d_{r+1} \left({x, y}\right)$.

So, let us start with the assumption that $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i > 0$.

Let $\displaystyle f \left({r}\right) = \left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}$.

Let $\displaystyle u = \sum_{i \mathop = 1}^n s_i^r, v = \frac 1 r$.

From Derivative of Powers of Functions‎, $D_r \left({u^v}\right) = v u^{v-1} D_r \left({u}\right) + u^v \ln u D_r \left({v}\right)$

Here:
 * $\displaystyle D_r \left({u}\right) = \sum_{i \mathop = 1}^n s_i^r \ln s_i$ from Derivative of Exponential Function and Sum Rule for Derivatives
 * $\displaystyle D_r \left({v}\right) = - \frac 1 {r^2}$ from Power Rule for Derivatives

So:

$K > 0$ because all of $s_i, r > 0$.

For the same reason, $\displaystyle \forall j: \frac{s_j^r} {\sum_{i \mathop = 1}^n s_i^r} < 1$.

From Logarithm of 1 is 0 and Logarithm is Strictly Increasing and Concave, their logarithms are therefore negative.

So:
 * $\displaystyle D_r \left({\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}}\right) < 0$

So, from Derivative of Monotone Function, it follows that (given the conditions on $r$ and $s_i$) $\displaystyle \left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}$ is decreasing.

Hence $\forall r \in \N: d_r \left({x, y}\right) \ge d_{r+1} \left({x, y}\right)$.

Since $\forall r \in \N: d_r \left({x, y}\right) \ge d_{r+1} \left({x, y}\right)$:
 * $\forall r \in \N: n^{-1} d_r \left({x, y}\right) \ge n^{-1} d_{r+1} \left({x, y}\right)$.

Since $d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x, y}\right)$, we see by transitivity that:
 * $\forall r \in \N: n^{-1} d_\infty \left({x, y}\right) \ge n^{-1} d_r \left({x, y}\right)$.

Thus:
 * $\forall r \in \N: d_r \left({x,y}\right) \ge n^{-1} d_\infty \left({x, y}\right) \ge n^{-1} d_r \left({x, y}\right)$

Therefore, $d_r$ and $d_\infty$ are Lipschitz equivalent for each $r$.