Vector space/K/Finite-dimensional/Norms equivalent/Fact/Proof

Proof

We use fact. The norm and the topology depend only on the underlying real vector space; therefore, we may assume

𝕂 = ℝ,

For a basis $v_{1}, \dots, v_{n} \in V$ , there exists an isomorphism

φ : ℝ^{n} ⟶ V

with $e_{i} \mapsto v_{i}$ . Under an isomorphism $φ$ , by setting

‖ u ‖^{'} := ‖ φ (u) ‖

we get a norm on $ℝ^{n}$ . Hence, we may directly consider the case $V = ℝ^{n}$ . We compare now an arbitrary norm on $ℝ^{n}$ with the maximum norm and with the Euclidean norm; we know by example that these two norms are equivalent to each other. Let $v = \sum_{i = 1}^{n} a_{i} e_{i}$ . Because of

\begin{matrix} ‖ v ‖ & = ‖ \sum_{i = 1}^{n} a_{i} e_{i} ‖ \\ \leq \sum_{i = 1}^{n} ‖ a_{i} e_{i} ‖ \\ = \sum_{i = 1}^{n} | a_{i} | \cdot ‖ e_{i} ‖ \\ \leq n \cdot max (‖ e_{i} ‖, i = 1, \dots, n) ‖ v ‖_{m a x}, \end{matrix}

sufficiently small $‖ - ‖_{m a x}$ -open balls are contained in $‖ - ‖$ -open balls. Therefore, the topology of the maximum norm is as fine as the topology of any other norm. To prove the converse, we consider the identity

ℝ^{n} ⟶ ℝ^{n},

where the topology on the left-hand side is given by the Euclidean norm (or maximum norm), and on the right-hand side by the norm. The reasoning so far shows that this mapping is continuous. The Euclidean $1$ -sphere $S$ on the left is compact due to the theorem of Heine-Borel

and, according to fact, $S$ is also compact with respect to the norm $‖ - ‖$ . We denote this set by $S^{'}$ . Since $ℝ^{n}$ is a Hausdorff-space with every norm, it follows by exercise that $S^{'}$ is closed. Since the origin does not belong to $S^{'}$ , there exists a

δ > 0

such that

U (0, δ) \cap S^{'} = \emptyset

(the open ball with respect to $‖ - ‖$ ). For $v \neq 0$ we obtain, because of $\frac{v}{‖ v ‖_{E u c}} \in S = S^{'}$ , the estimate

‖ \frac{v}{‖ v ‖_{E u c}} ‖ \geq δ .

Therefore,

‖ v ‖_{E u c} \leq \frac{1}{δ} ‖ v ‖ .

To fact