Are all the $\ell^p$ metrics topologically equivalent for positive integer values of $p$?

Question

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?

Answer 1

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.

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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).

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

Answer 2

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.

Answer 3

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 $M$.