Vector Norms
Definition. A norm is mapping $|\cdot|:\mathbb{F}^m \to \mathbb{R}$ such that
- $|x| \ge 0$ for every $x \in \mathbb{F}^m$ with $|x| = 0$ if and only if $x = 0$.
- $|x + y| \le |x| + |y|$ for all $x, y\in \mathbb{F}^m$.
$|\alpha x| = \alpha |x|$ for all $\alpha\in\mathbb{F}$ and $x \in \mathbb{F}^m$.
| Remark. An obvious result that follows immediately from the triangle inequality and homogenity is that $|\alpha x + \beta y| \le | \alpha | |x| + | \beta | |y|$. |
Remark. In numerical linear algebra, we are not too interested in working with abstract norms. Rather, we are most interested in working with the $p$-norms.
Definition. The $p$-norm (with $p \ge 1$) $|\cdot|_p : \mathbb{F}^m \to \mathbb{R}$ is defined by
\[\|x\|_p = (|x_1|^p + \cdots + |x_m|^p)^{1/p}.\]Definition. The $\infty$-norm $|\cdot|_\infty:\mathbb{F}^m \to \mathbb{R}$ is defined as
\[\|x\|_\infty = \max_i |x_i|.\]| Remark. It is intuitively easy to see that the $\infty$-norm is a pointwise limit of $p$-norm as $p \to \infty$ as $|x|_p$ is dominated by $\max_i | x_i | $. To rigorously justify this, consider letting $M = \max_i | x_i | $. Then |
as $p \to \infty$.
Remark. The most important $p$-norms are the $1$-norm, $2$-norm, and $\infty$-norm.
Proposition. Let $p, q\ge 1$ and $\frac{1}{p} + \frac{1}{q} = 1$.
- $\langle x, y \rangle \le |x|_p |y|_p$
- $\langle x, y \rangle \le |x|_2 |y|_2$
- $|x|_p \ge |x|_q$ if $1 \le p \le q \le \infty$.
Proof. TODO
Definition. Let $|\cdot|$ be a norm on $\mathbb{F}^m$ with $W \in \mathbb{F}^{m\times m}$ invertible. Then
\[\|x\|_W := \|Wx\|\]is the weighted norm induced by $|\cdot|$ with weight $W$. In particular, if the norm is the $p$-norm $|\cdot|_p$, then we write
\[\|x\|_{W,p} := \|Wx\|_p.\]Similarly, if the norm is the $\infty$-norm $|\cdot|_\infty$, then
\[\|x\|_{W,\infty} = \|Wx\|_\infty.\]Remark. Notice that invertibility of $W$ is required. If it wasn’t it would have a nontrivial nullspace and we contradict the definition of a norm.