Difference between revisions of "Norm"
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==Examples== | ==Examples== | ||
===The Euclidean Norm=== | ===The Euclidean Norm=== | ||
+ | {{Todo|Migrate this norm to its own page}} | ||
The Euclidean norm is denoted <math>\|\cdot\|_2</math> | The Euclidean norm is denoted <math>\|\cdot\|_2</math> | ||
Revision as of 17:29, 7 March 2015
Contents
Definition
A norm on a vector space [ilmath](V,F)[/ilmath] is a function [math]\|\cdot\|:V\rightarrow\mathbb{R}[/math] such that:
- [math]\forall x\in V\ \|x\|\ge 0[/math]
- [math]\|x\|=0\iff x=0[/math]
- [math]\forall \lambda\in F, x\in V\ \|\lambda x\|=|\lambda|\|x\|[/math] where [math]|\cdot|[/math] denotes absolute value
- [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
TODO: Migrate this norm to its own page
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.