1
$\begingroup$

In a course on Metric spaces, we saw that the $\ell^1$, $\ell^2$ and $\ell^\infty$ metrics are topologically equivalent.

To clarify what I mean by $\ell^1$, $\ell^2$, $\ell^\infty$:

Definition ($\ell^p$, $\ell^\infty$ Metrics). $~~$ Suppose $(M_1, d_1), (M_2, d_2), \dots, (M_n, d_n)$ are metric spaces, let $M = M_1 \times M_2 \times \cdots \times M_n$, and let $p$ be a positive integer. We define the metrics $\ell^p$ and $\ell^\infty$ on $M$ by $$ \ell^p(\mathbf x, \mathbf y) = \left(\sum_{i=1}^n \left(d_i(x_i, y_i)\right)^p \right)^{1/p} \qquad \text{and} \qquad \ell^\infty(\mathbf x, \mathbf y) = \max_{i = 1, \dots, n} d_i(x_i,y_i),$$ for any $\mathbf x = (x_1, x_2, \dots, x_n), \mathbf y = (y_1, y_2, \dots, y_n)\in M$.

Now is it not the case that all $\ell^p$ metrics are topologically equivalent for $p \in \{1,2,3,\dots\}$? Or did we simply stick to $p = 1,2, \infty$ in the course because they're the most often encountered?

$\endgroup$
0
$\begingroup$

Yes, the metrics $d_p(\mathbf x,\mathbf y):=(\sum_{i=1}^nd_i(x_i,y_i)^p)^{\frac 1 p}$ defined on the product of metric spaces $\prod_{i=1}^n M_i$ are equivalent for $p\in[1,\infty]$.

Let's inspect, for the moment, the $\mathbb R^n$ space with $p$-norm. We'll be needing an intermediate result of the following.

Lemma: Let $\lVert x\rVert_p:=(|x_1|^p+\dots+|x_n|^p)^{\frac 1 p}$ be the $p$-norm on $\mathbb R^n$. For all $x\in\mathbb R^n$, we have $$ \lim_{p\rightarrow\infty}\lVert x\rVert_p=\max_{i\in\mathbb N_n} |x_i|^p=:\lVert x \rVert_\infty $$

Proof. $$(\lVert x \rVert_\infty)^p=\big(\max_{i\in\mathbb N_n}|x_i|\big)^p=\max_{i\in\mathbb N_n}|x_i|^p\leq\underbrace{|x_1|^p}_{\leq \lVert x \rVert_\infty{}^p}+\dots+\underbrace{|x_n|^p}_{\leq \lVert x \rVert_\infty{}^p}=(\lVert x\rVert_p)^p\leq n(\lVert x \rVert_\infty)^p, $$ and if we take the $p$-th root of this expression, we get \begin{align} \tag{1}\label{1} \lVert x \rVert_\infty \leq \lVert x \rVert_p \leq n^{\frac 1 p}\lVert x \rVert_\infty, \end{align} a result that we'll need to prove the equivalence of metrics. Taking the limit as $p\rightarrow \infty$ of the last equation, we get the desired result.


Proposition: The metrics $d_p$ defined on $\prod_{i=1}^nM_i$ are equivalent for $p\in[1,\infty]$.

Proof. It suffices to check that $d_p$ metric is equivalent to $d_\infty$ metric for some $p\in[1,\infty)$. Let $\mathbf x, \mathbf y \in \prod_{i=1}^nM_i$. From \eqref{1} above it follows that $$ d_\infty(\mathbf x,\mathbf y)\leq d_p(\mathbf x, \mathbf y)\leq n^{\frac 1 p}d_\infty(\mathbf x,\mathbf y) \text{ for all }p\in[1,\infty] $$ because $d_i(x_i,y_i)\in \mathbb R$ for all $i\in\mathbb N_n$. We need to check that $d_p$ metric is finer than $d_\infty$ and also the reverse, to see that they are really equivalent. It is equivalent to say for a metric $d$ to be finer than $d'$ on some set $M$ and to say that $\mathrm{id}_M:(M,d)\rightarrow (M,d')$ is continuous, so that's what we'll check.

  1. We're proving $d_p$ is finer than $d_\infty$. Let $\varepsilon>0$ and pick $\delta = \varepsilon$. Then if $d_p(\mathbf x,\mathbf y)<\delta$ it follows $d_\infty(\mathbf x,\mathbf y)<\varepsilon$, since $d_\infty(\mathbf x,\mathbf y)\leq d_p(\mathbf x,\mathbf y)<\varepsilon$. Thus $\mathrm{id}_M$ is uniformly continuous and also continuous.
  2. Similarly, just pick $\delta=\frac {\varepsilon}{n^{1/p}}$ for any $\varepsilon>0$.

It follows that the metrics $d_p$ and $d_\infty$ are equivalent (and even uniformly equivalent, as the identity map is uniformly continuous in both directions).

$$\tag*{$\blacksquare$}$$

$\endgroup$
1
$\begingroup$

It is the case that the $\ell^p$-metrics on the product space are all topologically equivalent (for $1 \leq p \leq \infty$, without necessarily even having $p \in \mathbb{N}$).

To see this note that the $\ell^p$ norms on $\mathbb{R}^n$ are equivalent norms since all norms on finite dimensional spaces are equivalent. So for any $1 \leq p,q \leq \infty$ there are constants $c,C$ such that for all $z$ in $\mathbb{R}^n$, $c \| z \|_p \leq \| z \|_q \leq C \| z \|_p$. Applying this to the vectors $z = (d_i(x_i,y_i))_{i=1}^n \in \mathbb{R}^n$ gives the equivalence of metrics you want.

$\endgroup$
1
$\begingroup$

All of your metrics are obtained by the following procedure: we have a fixed map $d: M \times M \to \mathbb R^n$ given by $$ d((x_1, \dots, x_n), (y_1, \dots, y_n)) = (d_1(x_1, y_1), \dots, d_n(x_n, y_n)). $$ Now you follow that by the $\ell^p$ norm on $\mathbb R^n$ to get your $\ell^p$ metric on $M$. Check that all the metrics are equivalent on $M$ because all the $\ell^p$ norms are equivalent on $\mathbb R^n$. In fact, you could replace the $\ell^p$ norm by any norm whatever on $\mathbb R^n$ and get an equivalent metric on $M$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.