Norms on Finite-Dimensional Real Vector Space are Equivalent

Theorem
Norms on finite-dimensional real vector space are equivalent.

Proof
We will prove that all norms are equivalent to $\norm {\, \cdot \,}_2$.

By definition, two norms are equivalent on $\R^d$ iff:


 * $\forall \mathbf x \in \R^d : \exists m, M \in \R_{> 0} : m \norm {\mathbf x}_a \le \norm {\mathbf x}_b \le M \norm {\mathbf x}_a$

"Less or equal" condition
Let $\set{\mathbf e_1 \dots \mathbf e_n}$ be a standard basis in $\R^d$.

We have that each $\mathbf x \in \R^d$ is uniquely expressible as:


 * $\displaystyle \mathbf x = \sum_{i \mathop = 1}^d x_i \mathbf e_i$

where $x_i$ is a scalar.

Then:

By definition of norm, its image is non-negative real numbers: $M \in \R_{\ge 0}$.

Hence:


 * $\forall \mathbf x \in \R^d : \exists M \in \R_{\ge 0} : \norm {\mathbf x} \le M \norm {\mathbf x}_2$

Existance of $m$
Let $K := \set {\mathbf y \in \R^d : \norm {\mathbf y}_2 = 1}$.

By Unit Sphere is Closed in Normed Vector Space‎, $K$ is closed in $\struct {\R^d, \norm{\, \cdot \,}_2}$.

By definition, it is bounded in $\struct {\R^d, \norm{\, \cdot \,}_2}$.

Hence, by Heine-Borel theorem, it is a compact set.

By Norm on Vector Space is Continuous Function, the map $\norm {\, \cdot \,} : K \to \R_{\ge 0}$ is continuous from $\struct {K, \norm {\, \cdot \,}_2}$ to $\struct {\R_{\ge 0}, \size {\, \cdot \,}}$:


 * $\forall \mathbf y_1, y_2 \in K : \size {\norm {\mathbf y_1} - \norm {\mathbf y_2}} \le \norm {\mathbf y_1 - \mathbf y_2} \le M \norm {\mathbf y_1 - \mathbf y_2}_2$

By Weierstrass theorem, $\norm {\, \cdot \,} : K \to \R_{\ge 0}$ attains a minimum value $m$ for some $\mathbf y \in K$.

Suppose, $m = 0$.

Then $\norm {\mathbf y} = 0$.

By the norm axiom of positive definiteness, $\mathbf y = 0$.

But then $\mathbf y \notin K$.

Hence, $m \ne 0$.

So $m > 0$.

Furthermore:


 * $\forall \mathbf y \in \R^d : \norm {\mathbf y}_2 = 1 : \norm {\mathbf y} \ge m $

"Greater or equal" condition
Suppose, $\mathbf x = 0$.

Then we have equality.

Suppose, $\mathbf x \ne 0$.

Let $\displaystyle \mathbf y = \frac {\mathbf x} {\norm {\mathbf x}_2}$.

We have that $\norm {\mathbf y}_2 = 1$ and $\mathbf y \in K$.

Then:

This implies that:


 * $m \norm {\mathbf x}_2 \le \norm {\mathbf x}$

Therefore:


 * $\forall \mathbf x \in \R^d : \exists m, M : m \norm {\mathbf x}_2 \le \norm {\mathbf x} \le M \norm {\mathbf x}_2$

By definition, all norms are equivalent to $\norm{\, \cdot \,}_2$.

By Norm Equivalence is Equivalence, the equivalence relation $\norm {\, \cdot \,} \sim \norm {\, \cdot \,}_2$ is transitive.

Thus, all norms are equivalent.