# Equivalence of Norms on Finite-Dimensional Spaces: Proof

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 $$0 with $$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

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 $$x$$.