A well-known result from functional analysis is that all norms on a finite-dimensional space $X$ are equivalent. The source [1] proves it by showing that every norm on a finite-dimensional space is equivalent to the Euclidean norm $||⋅||_2$.

  • First, Werner assumes that $\dim(X) = n$ and that $||\cdot||$ is an arbitrary norm on $X$. By setting $K := \max\{ ||e_1||, \cdots, ||e_n||\}$, where $\{e_1, \cdots, e_n\}$ shall be a basis of $X$, he shows with the Cauchy-Schwartz inequality that $$||x|| \leq K\sqrt{n}||x||_2 \quad \forall x\in X.$$

  • Now Werner goes on and defines the compact set $S:=\{x\in X \mid ||x||_2 = 1\} \subset X$. Since $S$ is compact and since $||\cdot||$ defines a continuous function, we know that $||\cdot||$ takes its minimum $m > 0$ on $S$. Now he states that since $x\cdot ||x||_2^{-1}\in S \ \forall x\in X\backslash \{0\}$, it follows that $$m||x||_2 \leq ||x|| \quad \forall x\in X.$$

Question: I know the following the definition for the equivalence of norms (taken from Definition I.2.3 of [1]):

Two norms $||\cdot||$ and $|||\cdot||||$ on a vector space $X$ are called equivalent fif there are two numbers $0<m\leq M$ with $$m||x|| \leq |||x||| \leq M||x|| \quad \forall x\in X$$

Applied to our case, this would mean we have to show that $m\leq K\sqrt{n}$, but I'm not sure on how to go about this, I'm afraid.

[1] Dirk Werner. Funktionalanalysis. Springer. $8$th edition


2 Answers 2


Your well-known result is true for finite-dimensional normed vector spaces over any complete valued field, not just over $\mathbf R$ and $\mathbf C$. The proof relies on completeness of the scalar field instead of local compactness, which may not be true over certain complete $p$-adic fields like $\mathbf C_p$. See Definition 1.3 (a more conceptual definition for equivalence of norms), Theorem 2.1, and Theorem 3.2 here.


Well, $m\|x\|_2\le\|x\|\le K\sqrt n\|x\|_2$ is already proved.

It readily implies $m\le K\sqrt n$: just plug in any nonzero $x$.


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