Equivalence of Norms on ℝⁿ

February 17, 2026 · Luciano Muratore

Equivalence of Norms on ℝⁿ (Using the ℓ∞ Norm)

In finite-dimensional vector spaces, all norms are equivalent.

In this article, we prove that any two norms on $\mathbb{R}^n$ are equivalent.
Instead of using the Euclidean norm as a reference, we use the $\ell_\infty$ norm.

Notation

For $x = (x_1, \dots, x_n) \in \mathbb{R}^n$, define

Its closed unit ball is the cube

Theorem (Equivalence of Norms)

Let $\|\cdot\|_a$ and $\|\cdot\|_b$ be two norms on $\mathbb{R}^n$.
Then there exist constants $c, C > 0$ such that for all $x \in \mathbb{R}^n$,

Consequently, the induced metrics

are bi-Lipschitz equivalent.

Proof

We divide the proof into three logically separate parts.

Part I — Compactness of the $\ell_\infty$ Unit Sphere

Consider the $\ell_\infty$ unit sphere:

Claim: $S_\infty$ is compact.

Closedness:
The map $x \mapsto \|x\|_\infty$ is continuous, and

so $S_\infty$ is closed.

Boundedness:
Since $S_\infty \subset [-1,1]^n$, and

is compact (finite product of compact intervals), it follows that

Part II — Comparison Between $\|\cdot\|_a$ and $\|\cdot\|_\infty$

Define a function

Since $S_\infty$ is compact and $g$ is continuous, it attains a minimum:

We have

because $\|u\|_a = 0$ would imply $u = 0$, but $0 \notin S_\infty$.

Global Inequality

Let $x \neq 0$ and define

Then $\|u\|_\infty = 1$, so $u \in S_\infty$.

By homogeneity,

Thus for all $x \in \mathbb{R}^n$,

Part III — Compactness of the $\|\cdot\|_a$ Unit Sphere

Define

Claim: $S_a$ is compact.

Closedness:
Since $x \mapsto \|x\|_a$ is continuous,

so it is closed.

Boundedness:
If $x \in S_a$, then $\|x\|_a = 1$, and using the previous inequality,

Hence

a compact cube.

Therefore,

Final Comparison with $\|\cdot\|_b$

Define

Since $S_a$ is compact, $f$ attains a minimum and maximum:

We have

because $\|u\|_b = 0 \Rightarrow u=0$, and $0 \notin S_a$.

Thus for all $u \in S_a$,

Final Step

Let $x \neq 0$ and define

Then $u \in S_a$, and by homogeneity,

Therefore,

For $x = 0$, the inequality is trivial.

This proves the equivalence of norms on $\mathbb{R}^n$.

Corollary — Bi-Lipschitz Equivalence of Metrics

Let

Then

and

Thus the identity map is bi-Lipschitz between
$(\mathbb{R}^n, d_a)$ and $(\mathbb{R}^n, d_b)$.

Logical Structure Summary

1. Use $\|\cdot\|_\infty$ to build a compact sphere $S_\infty$.
2. On $S_\infty$, the continuous map $u \mapsto \|u\|_a$ has a positive minimum $m_\infty$.
3. This gives the global bound $$\|x\|_\infty \le \frac{1}{m_\infty} \|x\|_a.$$
4. That implies $S_a$ is contained in a compact cube → hence compact.
5. On $S_a$, the continuous map $u \mapsto \|u\|_b$ has positive minimum and finite maximum.
6. This yields $$m\|x\|_a \le \|x\|_b \le M\|x\|_a.$$

Final Remark

This proof shows why finite dimensionality is essential:

• Compactness of spheres
• Continuity of norms
• Extreme value theorem

None of this survives automatically in infinite-dimensional spaces.

That is precisely why norm equivalence fails in general Banach spaces.