Difference between revisions of "Norm"

Revision as of 17:27, 7 March 2015

Definition

A norm on a vector space [ilmath](V,F)[/ilmath] is a function [math]\|\cdot\|:V\rightarrow\mathbb{R}[/math] such that:

1. [math]\forall x\in V\ \|x\|\ge 0[/math]
2. [math]\|x\|=0\iff x=0[/math]
3. [math]\forall \lambda\in F, x\in V\ \|\lambda x\|=|\lambda|\|x\|[/math] where [math]|\cdot|[/math] denotes absolute value
4. [math]\forall x,y\in V\ \|x+y\|\le\|x\|+\|y\|[/math] - a form of the triangle inequality

Often parts 1 and 2 are combined into the statement

• [math]\|x\|\ge 0\text{ and }\|x\|=0\iff x=0[/math] so only 3 requirements will be stated.

I don't like this

Common norms

The 1-norm

[math]\|x\|_1=\sum^n_{i=1}|x_i|[/math] - it's just a special case of the p-norm.

The 2-norm

[math]\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}[/math] - Also known as the Euclidean norm (see below) - it's just a special case of the p-norm.

The p-norm

[math]\|x\|_p=\left(\sum^n_{i=1}|x_i|^p\right)^\frac{1}{p}[/math] (I use this notation because it can be easy to forget the [math]p[/math] in [math]\sqrt[p]{}[/math])

The supremum-norm

Also called [math]\infty-[/math]norm
[math]\|x\|_\infty=\sup(\{x_i\}_{i=1}^n)[/math]

Examples

The Euclidean Norm

The Euclidean norm is denoted [math]\|\cdot\|_2[/math]

Here for [math]x\in\mathbb{R}^n[/math] we have:

[math]\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}[/math]

Proof that it is a norm

TODO: proof

Part 4 - Triangle inequality

Let [math]x,y\in\mathbb{R}^n[/math]

[math]\|x+y\|_2^2=\sum^n_{i=1}(x_i+y_i)^2[/math] [math]=\sum^n_{i=1}x_i^2+2\sum^n_{i=1}x_iy_i+\sum^n_{i=1}y_i^2[/math] [math]\le\sum^n_{i=1}x_i^2+2\sqrt{\sum^n_{i=1}x_i^2}\sqrt{\sum^n_{i=1}y_i^2}+\sum^n_{i=1}y_i^2[/math] using the Cauchy-Schwarz inequality

[math]=\left(\sqrt{\sum^n_{i=1}x_i^2}+\sqrt{\sum^n_{i=1}y_i^2}\right)^2[/math] [math]=\left(\|x\|_2+\|y\|_2\right)^2[/math]

Thus we see: [math]\|x+y\|_2^2\le\left(\|x\|_2+\|y\|_2\right)^2[/math], as norms are always [math]\ge 0[/math] we see:

[math]\|x+y\|_2\le\|x\|_2+\|y\|_2[/math] - as required.