Difference between revisions of "Norm"

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An understanding of a norm is needed to proceed to [[Linear isometry|linear isometries]]
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{{:Norm/Heading}}
 
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==Normed vector spaces==
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A normed vector space is a [[Vector space|vector space]] equipped with a norm <math>\|\cdot\|_V</math>, it may be denoted <math>(V,\|\cdot\|_V,F)</math>
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__TOC__
 
==Definition==
 
==Definition==
A norm on a [[Vector space|vector space]] {{M|(V,F)}} is a function <math>\|\cdot\|:V\rightarrow\mathbb{R}</math> such that:
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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{{rAPIKM}}<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} }}, it may commonly be {{M|\mathbb{R} }} but it does not ''need'' 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
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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
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I don't like this (inline with the [[Doctrine of monotonic definition]])
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==Properties==
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* [[The norm of a space is a uniformly continuous map with respect to the topology it induces]] - {{M|\Vert\cdot\Vert:X\rightarrow\mathbb{R} }} is a [[uniformly continuous]] map.
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==Terminology==
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Such a vector space equipped with such a function is called a [[Normed space|normed space]]<ref name="APIKM"/>
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==Relation to various [[subtypes of topological spaces]]==
 +
The reader should note that:
 +
* Every [[inner product]] induces a ''norm'' and
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* Every ''norm'' induces a [[metric]]
 +
These are outlined below
 +
===Relation to [[inner product]]===
 +
Every [[inner product]] {{M|\langle\cdot,\cdot\rangle:V\times V\rightarrow(\mathbb{R}\text{ or }\mathbb{C})}} induces a ''norm'' given by:
 +
* {{M|1=\Vert x\Vert:=\sqrt{\langle x,x\rangle} }}
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{{Todo|see [[inner product#Norm induced by|inner product (norm induced by)]] for more details, on that page is a proof that {{M|\langle x,x\rangle\ge 0}}, this needs its own page with a proof.}}
  
==Norms may define a metric space==
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===Metric induced by a norm===
To get a [[Metric space|metric space]] from a norm simply define <math>d(x,y)=\|x-y\|</math>
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To get a [[Metric space|metric space]] from a norm simply define<ref name="FA"/><ref name="APIKM"/>:
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* <math>d(x,y):=\|x-y\|</math>
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(See [[Subtypes of topological spaces]] for more information, this relationship is very important in [[Functional analysis]])
 +
{{Todo|Move to its own page and do a proof (trivial)}}
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 +
==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>
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* <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|vector space]] {{M|V}} 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>
 +
 
 +
{{Begin Theorem}}
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Theorem: This is an [[Equivalence relation]] - so we may write this as <math>\|\cdot\|_1\sim\|\cdot\|_2</math>
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{{Begin Proof}}
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{{Todo|proof}}
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{{End Proof}}
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{{End Theorem}}
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Note also that if <math>\|\cdot\|_1</math> is both weaker and stronger than <math>\|\cdot\|_2</math> they are equivalent
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===Examples===
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*Any two norms on <math>\mathbb{R}^n</math> are equivalent
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*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==
 
==Common norms==
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| 2-norm
 
| 2-norm
 
|<math>\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}</math>
 
|<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.
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| Also known as the [[Euclidean norm]] - it's just a special case of the p-norm.
 
|-
 
|-
 
| p-norm
 
| p-norm
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| <math>\infty-</math>norm
 
| <math>\infty-</math>norm
 
|<math>\|x\|_\infty=\sup(\{x_i\}_{i=1}^n)</math>
 
|<math>\|x\|_\infty=\sup(\{x_i\}_{i=1}^n)</math>
|Also called <math>\infty-</math>norm<br/>
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|Also called sup-norm<br/>
 
|-
 
|-
 
!colspan="3"|Norms on <math>\mathcal{C}([0,1],\mathbb{R})</math>
 
!colspan="3"|Norms on <math>\mathcal{C}([0,1],\mathbb{R})</math>
 
|-
 
|-
 
| <math>\|\cdot\|_{L^p}</math>
 
| <math>\|\cdot\|_{L^p}</math>
| <math>\|f\|_{L^p}=\left(\int^1_0|f(x)|^pdx\right)</math>
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| <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
 
| '''NOTE''' be careful extending to interval <math>[a,b]</math> as proof it is a norm relies on having a unit measure
 
|-
 
|-
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|For a [[Linear map|linear isomorphism]] <math>L:U\rightarrow V</math> where V is a normed vector space
 
|For a [[Linear map|linear isomorphism]] <math>L:U\rightarrow V</math> where V is a normed vector space
 
|}
 
|}
 
==Equivalence of norms==
 
Given two norms <math>\|\cdot\|_1</math> and <math>\|\cdot\|_2</math> on a [[Vector space|vector space]] {{M|V}} we say they are equivalent if:
 
 
<math>\exists c,C\in\mathbb{R}\ \forall x\in V:\ c\|x\|_1\le\|x\|_2\le C\|x\|_1</math>
 
 
We may write this as <math>\|\cdot\|_1\sim\|\cdot\|_2</math> - this is an [[Equivalence relation]]
 
{{Todo|proof}}
 
===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.
 
  
 
==Examples==
 
==Examples==
 
* [[Euclidean norm]]
 
* [[Euclidean norm]]
{{Definition|Linear Algebra}}
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
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{{Normed and Banach spaces navbox}}
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{{Metric spaces navbox}}
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{{Topology navbox}}
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{{Definition|Linear Algebra|Topology|Metric Space|Functional Analysis}}
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[[Category:Exemplary pages]]
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[[Category:First-year friendly]]

Latest revision as of 20:33, 9 April 2017

Norm
[ilmath]\Vert\cdot\Vert:V\rightarrow\mathbb{R}_{\ge 0} [/ilmath]
Where [ilmath]V[/ilmath] is a vector space over the field [ilmath]\mathbb{R} [/ilmath] or [ilmath]\mathbb{C} [/ilmath]
relation to other topological spaces
is a
contains all
Related objects
Induced metric
  • [ilmath]d_{\Vert\cdot\Vert}:V\times V\rightarrow\mathbb{R}_{\ge 0}[/ilmath]
  • [ilmath]d_{\Vert\cdot\Vert}:(x,y)\mapsto\Vert x-y\Vert[/ilmath]
Induced by inner product
  • [ilmath]\Vert\cdot\Vert_{\langle\cdot,\cdot\rangle}:V\rightarrow\mathbb{R}_{\ge 0}[/ilmath]
  • [ilmath]\Vert\cdot\Vert_{\langle\cdot,\cdot\rangle}:x\mapsto\sqrt{\langle x,x\rangle}[/ilmath]
A norm is a an abstraction of the notion of the "length of a vector". Every norm is a metric and every inner product is a norm (see Subtypes of topological spaces for more information), thus every normed vector space is a topological space to, so all the topology theorems apply. Norms are especially useful in functional analysis and also for differentiation.

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]:

  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 (inline with the Doctrine of monotonic definition)

Properties

Terminology

Such a vector space equipped with such a function is called a normed space[1]

Relation to various subtypes of topological spaces

The reader should note that:

These are outlined below

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], this needs its own page with a proof.



Metric induced by a norm

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: Move to its own page and do a proof (trivial)



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

  1. 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]!
  2. The other mistake books make is saying explicitly that the field of a vector space needs to be [ilmath]\mathbb{R} [/ilmath], it may commonly be [ilmath]\mathbb{R} [/ilmath] but it does not need to be [ilmath]\mathbb{R} [/ilmath]

References

  1. 1.0 1.1 1.2 Analysis - Part 1: Elements - Krzysztof Maurin
  2. 2.0 2.1 Functional Analysis - George Bachman and Lawrence Narici
  3. Functional Analysis - A Gentle Introduction - Volume 1, by Dzung Minh Ha
  4. Real and Abstract Analysis - Edwin Hewitt & Karl Stromberg