P-Product Metrics on Real Vector Space are Topologically Equivalent

Theorem
Let $$A = \R^n$$ be an $n$-dimensional real vector space.

Let $$d_1, d_2, \ldots, d_\infty$$ be the generalized Euclidean metrics.

Then all of $$d_1, d_2, \ldots, d_\infty$$ are topologcally equivalent.

Proof
First we are going to show that:

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

Then we will have demonstrated Lipschitz equivalence between all of these metrics, from which topologcal equivalence follows.

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

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


 * First 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 $$\left({\sum_{i=1}^n \left|{x_i - y_i}\right|^r}\right)^{\frac 1 r} \ge \left({\sum_{i=1}^n \left|{x_i - y_i}\right|^{r+1}}\right)^{\frac 1 {r+1}}$$.

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 $$\left({\sum_{i=1}^{n-1} \left|{x_i - y_i}\right|^r}\right)^{\frac 1 r} \ge \left({\sum_{i=1}^{n-1} \left|{x_i - y_i}\right|^{r+1}}\right)^{\frac 1 {r+1}}$$, 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 $$f \left({r}\right) = \left({\sum_{i=1}^n s_i^r}\right)^{1/r}$$.

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

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

So:

$$ $$ $$ $$ $$ $$ $$

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

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

From Basic Properties of Natural Logarithm their logarithms are therefore negative.

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

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

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


 * Next we need to show that $$\forall r \in \N: n^{-\frac 1 {r+1}} d_{r+1} \left({x, y}\right) \ge n^{-\frac 1 r} d_r \left({x, y}\right)$$.

This is messier - please bear with it ...

In the same way as above, let $$\forall i \in \left[{1 \,. \, . \, n}\right]: s_i = \left|{x_i - y_i}\right|$$.

For similar reasons, we start with the assumption that $$\forall i \in \left[{1 \,. \, . \, n}\right]: s_i > 0$$.

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

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

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

So:

$$ $$ $$ $$ $$ $$ $$


 * Finally we need to note that $$\forall r \in \N: d_r \left({x, y}\right) \ge d_{\infty} \left({x, y}\right) \ge n^{-\frac 1 r} d_r \left({x, y}\right)$$.