Difference between revisions of "Norm"
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==Definition== | ==Definition== | ||
| − | A norm on a [[Vector space|vector space]] {{M|(V,F)}} (where {{M|F}} is either {{M|\mathbb{R} }} or {{M|\mathbb{C} }}) is a function <math>\|\cdot\|:V\rightarrow\mathbb{R}</math> such that<ref name="KMAPI">Analysis - Part 1: Elements - Krzysztof Maurin</ref><ref name="FA">Functional Analysis - George Bachman and Lawrence Narici</ref><ref name="FAAGI">Functional Analysis - A Gentle Introduction - Volume 1, by Dzung Minh Ha</ref><sup>{{Highlight|See warning <ref group="Note">A lot of books, including the brilliant Analysis | + | A norm on a [[Vector space|vector space]] {{M|(V,F)}} (where {{M|F}} is either {{M|\mathbb{R} }} or {{M|\mathbb{C} }}) is a function <math>\|\cdot\|:V\rightarrow\mathbb{R}</math> such that<ref name="KMAPI">Analysis - Part 1: Elements - Krzysztof Maurin</ref><ref name="FA">Functional Analysis - George Bachman and Lawrence Narici</ref><ref name="FAAGI">Functional Analysis - A Gentle Introduction - Volume 1, by Dzung Minh Ha</ref>{{RRAAAHS}}<sup>{{Highlight|See warning notes:<ref group="Note">A lot of books, including the brilliant [[Books:Analysis - Part 1: Elements - Krzysztof Maurin|Analysis - Part 1: Elements - Krzysztof Maurin]] referenced here state ''explicitly'' that it is possible for {{M|\Vert\cdot,\cdot\Vert:V\rightarrow\mathbb{C} }} they are wrong. I assure you that it is {{M|\Vert\cdot\Vert:V\rightarrow\mathbb{R}_{\ge 0} }}. Other than this the references are valid, note that this is 'obvious' as if the image of {{M|\Vert\cdot\Vert}} could be in {{M|\mathbb{C} }} then the {{M|\Vert x\Vert\ge 0}} would make no sense. What ordering would you use? The [[canonical]] ordering used for the product of 2 spaces ({{M|\mathbb{R}\times\mathbb{R} }} in this case) is the [[Lexicographic ordering]] which would put {{M|1+1j\le 1+1000j}}!</ref><ref group="Note">The other mistake books make is saying explicitly that the [[field of a vector space]] needs to be {{M|\mathbb{R} }}</ref>}}</sup>: |
# <math>\forall x\in V\ \|x\|\ge 0</math> | # <math>\forall x\in V\ \|x\|\ge 0</math> | ||
# <math>\|x\|=0\iff x=0</math> | # <math>\|x\|=0\iff x=0</math> | ||
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# <math>\forall x,y\in V\ \|x+y\|\le\|x\|+\|y\|</math> - a form of the [[Triangle inequality|triangle inequality]] | # <math>\forall x,y\in V\ \|x+y\|\le\|x\|+\|y\|</math> - a form of the [[Triangle inequality|triangle inequality]] | ||
| − | Often parts 1 and 2 are combined into the statement | + | 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. | * <math>\|x\|\ge 0\text{ and }\|x\|=0\iff x=0</math> so only 3 requirements will be stated. | ||
| − | I don't like this | + | I don't like this (inline with the [[Doctrine of monotonic definition]]) |
| + | |||
==Terminology== | ==Terminology== | ||
Such a vector space equipped with such a function is called a [[Normed space|normed space]]<ref name="KMAPI"/> | Such a vector space equipped with such a function is called a [[Normed space|normed space]]<ref name="KMAPI"/> | ||
Revision as of 10:31, 8 January 2016
An understanding of a norm is needed to proceed to linear isometries.
A norm is a special case of metrics. See Subtypes of topological spaces for more information
Contents
Definition
A norm on a vector space [ilmath](V,F)[/ilmath] (where [ilmath]F[/ilmath] is either [ilmath]\mathbb{R} [/ilmath] or [ilmath]\mathbb{C} [/ilmath]) is a function [math]\|\cdot\|:V\rightarrow\mathbb{R}[/math] such that[1][2][3][4]See warning notes:[Note 1][Note 2]:
- [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 (inline with the Doctrine of monotonic definition)
Terminology
Such a vector space equipped with such a function is called a normed space[1]
Relation to inner product
Every inner product [ilmath]\langle\cdot,\cdot\rangle:V\times V\rightarrow(\mathbb{R}\text{ or }\mathbb{C})[/ilmath] induces a norm given by:
- [ilmath]\Vert x\Vert:=\sqrt{\langle x,x\rangle}[/ilmath]
TODO: see inner product (norm induced by) for more details, on that page is a proof that [ilmath]\langle x,x\rangle\ge 0[/ilmath] - I cannot think of any complex norms!
Induced metric
To get a metric space from a norm simply define[2][1]:
- [math]d(x,y):=\|x-y\|[/math]
(See Subtypes of topological spaces for more information, this relationship is very important in Functional analysis)
TODO: Some sort of proof this is never complex
Weaker and stronger norms
Given a norm [math]\|\cdot\|_1[/math] and another [math]\|\cdot\|_2[/math] we say:
- [math]\|\cdot\|_1[/math] is weaker than [math]\|\cdot\|_2[/math] if [math]\exists C> 0\forall x\in V[/math] such that [math]\|x\|_1\le C\|x\|_2[/math]
- [math]\|\cdot\|_2[/math] is stronger than [math]\|\cdot\|_1[/math] in this case
Equivalence of norms
Given two norms [math]\|\cdot\|_1[/math] and [math]\|\cdot\|_2[/math] on a vector space [ilmath]V[/ilmath] we say they are equivalent if:
[math]\exists c,C\in\mathbb{R}\text{ with }c,C>0\ \forall x\in V:\ c\|x\|_1\le\|x\|_2\le C\|x\|_1[/math]
Theorem: This is an Equivalence relation - so we may write this as [math]\|\cdot\|_1\sim\|\cdot\|_2[/math]
TODO: proof
Note also that if [math]\|\cdot\|_1[/math] is both weaker and stronger than [math]\|\cdot\|_2[/math] they are equivalent
Examples
- Any two norms on [math]\mathbb{R}^n[/math] are equivalent
- The norms [math]\|\cdot\|_{L^1}[/math] and [math]\|\cdot\|_\infty[/math] on [math]\mathcal{C}([0,1],\mathbb{R})[/math] are not equivalent.
Common norms
| Name | Norm | Notes |
|---|---|---|
| Norms on [math]\mathbb{R}^n[/math] | ||
| 1-norm | [math]\|x\|_1=\sum^n_{i=1}|x_i|[/math] | it's just a special case of the p-norm. |
| 2-norm | [math]\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}[/math] | Also known as the Euclidean norm - it's just a special case of the p-norm. |
| 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]) |
| [math]\infty-[/math]norm | [math]\|x\|_\infty=\sup(\{x_i\}_{i=1}^n)[/math] | Also called sup-norm |
| Norms on [math]\mathcal{C}([0,1],\mathbb{R})[/math] | ||
| [math]\|\cdot\|_{L^p}[/math] | [math]\|f\|_{L^p}=\left(\int^1_0|f(x)|^pdx\right)^\frac{1}{p}[/math] | NOTE be careful extending to interval [math][a,b][/math] as proof it is a norm relies on having a unit measure |
| [math]\infty-[/math]norm | [math]\|f\|_\infty=\sup_{x\in[0,1]}(|f(x)|)[/math] | Following the same spirit as the [math]\infty-[/math]norm on [math]\mathbb{R}^n[/math] |
| [math]\|\cdot\|_{C^k}[/math] | [math]\|f\|_{C^k}=\sum^k_{i=1}\sup_{x\in[0,1]}(|f^{(i)}|)[/math] | here [math]f^{(k)}[/math] denotes the [math]k^\text{th}[/math] derivative. |
| Induced norms | ||
| Pullback norm | [math]\|\cdot\|_U[/math] | For a linear isomorphism [math]L:U\rightarrow V[/math] where V is a normed vector space |
Examples
Notes
- ↑ A lot of books, including the brilliant Analysis - Part 1: Elements - Krzysztof Maurin referenced here state explicitly that it is possible for [ilmath]\Vert\cdot,\cdot\Vert:V\rightarrow\mathbb{C} [/ilmath] they are wrong. I assure you that it is [ilmath]\Vert\cdot\Vert:V\rightarrow\mathbb{R}_{\ge 0} [/ilmath]. Other than this the references are valid, note that this is 'obvious' as if the image of [ilmath]\Vert\cdot\Vert[/ilmath] could be in [ilmath]\mathbb{C} [/ilmath] then the [ilmath]\Vert x\Vert\ge 0[/ilmath] would make no sense. What ordering would you use? The canonical ordering used for the product of 2 spaces ([ilmath]\mathbb{R}\times\mathbb{R} [/ilmath] in this case) is the Lexicographic ordering which would put [ilmath]1+1j\le 1+1000j[/ilmath]!
- ↑ The other mistake books make is saying explicitly that the field of a vector space needs to be [ilmath]\mathbb{R} [/ilmath]
