Equivalent norms, and the dictionary with equivalent metrics

Let $V$ be a vector space over $\mathbb{R}$ (Vector space over a field) and let $M$ and $N$ be norms on $V$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). $M$ and $N$ are equivalent when there are reals $c > 0$ and $C > 0$ with

$c\,M(v) \;\le\; N(v) \;\le\; C\,M(v) \qquad \text{for every } v \in V .$

The constants are not part of the data and are not unique: any smaller $c$ and any larger $C$ serve as well.

This is an equivalence relation on the norms on $V$

• Reflexive: take $c = C = 1$.
• Symmetric: from $cM \le N \le CM$ and $c, C > 0$ one gets $C^{-1}N \le M \le c^{-1}N$, dividing by the positive constants (Inverses of positives are positive, and reciprocation reverses order).
• Transitive: if $cM \le N \le CM$ and $c'N \le P \le C'N$ then $c'c\,M \le P \le C'C\,M$, and $c'c > 0$, $C'C > 0$, a product of positives being positive.

The dictionary with equivalent metrics

Let $d_M(u,v) = M(u-v)$ and $d_N(u,v) = N(u-v)$ be the induced metrics (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Substituting $v := u - w$ in the displayed condition gives

$c\,d_M(u,w) \;\le\; d_N(u,w) \;\le\; C\,d_M(u,w) \qquad \text{for all } u, w \in V ,$

which is verbatim the Lipschitz equivalence of $d_M$ and $d_N$ in the sense of Topologically, uniformly and Lipschitz equivalent metrics on a set, with $\alpha = c$ and $\beta = C$. That is the strongest of the three tiers that item distinguishes: by Lipschitz equivalence implies uniform equivalence implies topological equivalence, Lipschitz equivalence implies uniform equivalence, which implies topological equivalence. So equivalent norms give

• the same open sets, hence the same closed sets, closures and interiors (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement);
• the same uniformly continuous maps into and out of $V$ (Uniform continuity of a map of metric spaces: one $\delta$ serving every point);
• the same convergent sequences with the same limits, and the same Cauchy sequences (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$, Cauchy sequence in a metric space).

The last line deserves its two-line verification, since it is used constantly below and is not literally a clause of Lipschitz equivalence implies uniform equivalence implies topological equivalence. If $d_M(v_k, v) \to 0$ then $0 \le d_N(v_k,v) \le C\,d_M(v_k,v)$, so given a rational $\varepsilon > 0$ an index beyond which $d_M(v_k,v) < \varepsilon/C$ serves for $d_N$; the converse uses $d_M \le c^{-1}d_N$ in the same way. The Cauchy statement is the same estimate applied to $d_N(v_k,v_l)$. In particular $(V, d_M)$ is complete if and only if $(V,d_N)$ is.

Naming. Many texts say strongly equivalent for what Topologically, uniformly and Lipschitz equivalent metrics on a set calls Lipschitz equivalent, and simply equivalent for what it calls topologically equivalent. As there, this library always writes the qualifier for metrics. For norms there is no fork to guard against: the condition displayed above is the only one anyone calls equivalence of norms, and it is always the Lipschitz-strength one.

• The converse of the dictionary fails. A metric on $V$ that is equivalent to a norm metric need not itself come from a norm: $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology turns any metric $d$ into $d' = \min\{d,1\}$, uniformly equivalent to $d$ and bounded, and a bounded metric on a nonzero vector space is not induced by any norm, since absolute homogeneity would make $d'(\lambda v, 0)$ unbounded in $\lambda$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). So "equivalent to a norm metric" is strictly weaker than "induced by an equivalent norm".

• Equivalence is a statement about a fixed vector space. Two norms on different spaces are never compared. On $\mathbb{R}^{n}$ with $n \ge 1$ the norms $\lVert\cdot\rVert_1$, $\lVert\cdot\rVert_2$ and $\lVert\cdot\rVert_\infty$ of The $p$-norms $\lVert x\rVert_p$ for rational $p \ge 1$, and $\lVert x\rVert_\infty$ are equivalent, with explicit constants proved in The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$; that every pair of norms on $\mathbb{R}^{n}$ is equivalent is For $n \ge 1$ all norms on $\mathbb{R}^n$ are equivalent. Neither statement survives to spaces that are not finite-dimensional, and the companion page carries the witness.

• Equivalence says nothing about the geometry. Equivalent norms have the same convergent sequences and the same open sets; they may still have quite different unit balls, and one of them may come from an inner product while the other does not. Each $\lVert\cdot\rVert_p$ is a norm on $\mathbb{R}^n$, and the induced metrics are exactly $d_1$, $d_2$ and $d_\infty$ of the published metric-spaces page records the metric identifications and nothing more.