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 $\R^d$ 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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces