Difference between revisions of "Norm"

Revision as of 03:22, 8 March 2015

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

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 (see below) - 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 [math]\infty-[/math]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)[/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.

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}\ \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

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.

@@ Line 62: / Line 62: @@
 ==Examples==
-===The Euclidean Norm===
+* [[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.
 {{Definition|Linear Algebra}}