Generalized Euclidean Metrics 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:

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

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 $\displaystyle d_r \left({x, y}\right) = \left({\sum_{i=1}^n \left|{x_i - y_i}\right|^r}\right)^{\frac 1 r}$.

• 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:

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

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:

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

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

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

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:

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

So:

	\(\displaystyle \)	\(\displaystyle D_r \left({\left({\sum_{i=1}^n s_i^r}\right)^{1/r} }\right)\)	\(=\)	\(\displaystyle \frac 1 r \left({\sum_{i=1}^n s_i^r}\right)^{\frac 1 r - 1} \left({\sum_{i=1}^n s_i^r \ln s_i}\right) - \frac {\left({\sum_{i=1}^n s_i^r}\right)^{1/r} \ln \left({\sum_{i=1}^n s_i^r}\right)} {r^2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({\sum_{i=1}^n s_i^r}\right)^{1/r} } r \left({\frac {\sum_{i=1}^n s_i^r \ln s_i} {\sum_{i=1}^n s_i^r} - \frac {\ln \left({\sum_{i=1}^n s_i^r}\right)} r}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({\sum_{i=1}^n s_i^r}\right)^{1/r} } r \left({\frac {r \left({\sum_{i=1}^n s_i^r \ln s_i}\right) - \left({\sum_{i=1}^n s_i^r}\right) \ln \left({\sum_{i=1}^n s_i^r}\right)} {r \left({\sum_{i=1}^n s_i^r}\right)} }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({r \left({\sum_{i=1}^n s_i^r \ln s_i}\right) - \left({\sum_{i=1}^n s_i^r}\right) \ln \left({\sum_{i=1}^n s_i^r}\right)}\right)\)	\(\displaystyle \)	where $\displaystyle K = \frac {\left({\sum_{i=1}^n s_i^r}\right)^{1/r} } {r^2 \left({\sum_{i=1}^n s_i^r}\right)} > 0$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({\sum_{i=1}^n s_i^r \ln \left({s_i^r}\right) - \left({\sum_{i=1}^n s_i^r}\right) \ln \left({\sum_{i=1}^n s_i^r}\right)}\right)\)	\(\displaystyle \)	Logarithms of Powers
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({\sum_{j=1}^n \left({s_j^r \left({\ln \left({s_j^r}\right) - \ln \left({\sum_{i=1}^n s_i^r}\right)}\right)}\right)}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({\sum_{j=1}^n \left({s_j^r \ln \left({\frac {s_j^r} {\sum_{i=1}^n s_i^r} }\right)}\right)}\right)\)	\(\displaystyle \)

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

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

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

So:

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

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

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

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

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

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:

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

So:

	\(\displaystyle \)	\(\displaystyle D_r \left({\left({\frac {\sum_{i=1}^n s_i^r} n}\right)^{1/r} }\right)\)	\(=\)	\(\displaystyle \frac 1 r \left({\frac {\sum_{i=1}^n s_i^r} n}\right)^{\frac 1 r - 1} \left({\frac {\sum_{i=1}^n s_i^r \ln s_i} n}\right) - \frac {\left({\frac {\sum_{i=1}^n s_i^r} n}\right)^{1/r} \ln \left({\frac {\sum_{i=1}^n s_i^r} n}\right)} {r^2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 r \left({\frac {\sum_{i=1}^n s_i^r} n}\right)^{\frac 1 r} \left({\frac {\sum_{i=1}^n s_i^r \ln s_i} {\sum_{i=1}^n s_i^r} - \frac {\ln \left({\frac {\sum_{i=1}^n s_i^r} n}\right)} r}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 r \left({\frac {\sum_{i=1}^n s_i^r} n}\right)^{\frac 1 r} \left({\frac {r \left({\sum_{i=1}^n s_i^r \ln s_i}\right) - \left({\sum_{i=1}^n s_i^r}\right) \ln \left({\frac {\sum_{i=1}^n s_i^r} n}\right)} {r \left({\sum_{i=1}^n s_i^r}\right)} }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({r \left({\sum_{i=1}^n s_i^r \ln s_i}\right) - \left({\sum_{i=1}^n s_i^r}\right) \ln \left({\frac {\sum_{i=1}^n s_i^r} n}\right)}\right)\)	\(\displaystyle \)	where $\displaystyle K = \frac 1 {r^2 \left({\sum_{i=1}^n s_i^r}\right)} \left({\frac {\sum_{i=1}^n s_i^r} n}\right)^{\frac 1 r} > 0$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({\sum_{i=1}^n s_i^r \ln \left({s_i^r}\right) - \left({\sum_{i=1}^n s_i^r}\right) \ln \left({\frac {\sum_{i=1}^n s_i^r} n}\right)}\right)\)	\(\displaystyle \)	Logarithms of Powers
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({\sum_{j=1}^n \left({s_j^r \left({\ln \left({s_j^r}\right) - \ln \left({\frac {\sum_{i=1}^n s_i^r} n} \right)} \right)} \right)} \right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle K \left({\sum_{j=1}^n \left({s_j^r \ln \left({\frac {n s_j^r} {\sum_{i=1}^n s_i^r} }\right)}\right)}\right)\)	\(\displaystyle \)

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

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Example $2.4.5$