Norms on Finite-Dimensional Real Vector Space are Equivalent

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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 $\R^d$ if and only if:

$ \exists m, M \in \R_{> 0} : \forall \mathbf x \in \R^d : 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:

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

where $x_i$ is a scalar.


Then:

\(\ds \norm {\sum_{i \mathop = 1}^d x_i \mathbf e_i}\) \(\le\) \(\ds \sum_{i \mathop = 1}^d \norm {x_i \mathbf e_i}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^d \size {x_i} \norm {\mathbf e_i}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \sqrt {\paren {\sum_{i \mathop = 1}^d \size {x_i} \norm {\mathbf e_i} }^2 }\)
\(\ds \) \(\le\) \(\ds \sqrt {\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2} \sqrt {\sum_{j \mathop = 1}^d \size {x_j}^2 }\) Cauchy's Inequality
\(\ds \) \(\le\) \(\ds \sqrt {\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2} \norm {\mathbf x}_2\) Definition of $p$-norm
\(\ds \) \(=\) \(\ds M \norm {\mathbf x}_2\) where we define $\ds M := \sqrt {\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2}$

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

But $M$ is independent of $\mathbf x$.

Hence:

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

$\Box$


Existence of $m$

Let $K := \set {\mathbf y \in \R^d : \norm {\mathbf y}_2 = 1}$ be a unit sphere in $\struct {\R^d, \norm {\, \cdot \,}_2}$.

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

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

Hence, by the Heine-Borel theorem, $K$ 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 $\struct {\R_{\ge 0}, \size {\, \cdot \,}}$:

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

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


Suppose $m = 0$.

Then:

\(\ds \norm {\mathbf y}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \mathbf y\) \(=\) \(\ds 0\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \leadsto \ \ \) \(\ds \mathbf y\) \(\notin\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds m\) \(\ne\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds m\) \(>\) \(\ds 0\)


Furthermore:

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

$\Box$


"Greater or equal" condition

Suppose $\mathbf x = 0$.

Then we have equality.

Suppose $\mathbf x \ne 0$.

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

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

Then:

\(\ds m\) \(\le\) \(\ds \norm {\mathbf y}\)
\(\ds \) \(=\) \(\ds \norm {\frac {\mathbf x} {\norm {\mathbf x}_2} }\)
\(\ds \) \(=\) \(\ds \frac {\norm {\mathbf x} } {\norm {\mathbf x}_2}\)

This implies that:

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

But once again, $m$ is independent of $\mathbf x$.

Therefore:

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

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.

$\blacksquare$


Sources

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