Difference between revisions of "Norm"

Revision as of 18:43, 21 April 2015

An understanding of a norm is needed to proceed to linear isometries

A normed vector space is a vector space equipped with a norm $\|\cdot\|_V$ , it may be denoted $(V,\|\cdot\|_V,F)$

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

$\forall x\in V\ \|x\|\ge 0$
$\|x\|=0\iff x=0$
$\forall \lambda\in F, x\in V\ \|\lambda x\|=|\lambda|\|x\|$ where $|\cdot|$ denotes absolute value
$\forall x,y\in V\ \|x+y\|\le\|x\|+\|y\|$ - a form of the triangle inequality

Often parts 1 and 2 are combined into the statement

I don't like this

To get a metric space from a norm simply define $d(x,y)=\|x-y\|$

HOWEVER: It is only true that a normed vector space is a metric space also, given a metric we may not be able to get an associated norm.

Given a norm $\|\cdot\|_1$ and another $\|\cdot\|_2$ we say:

$\|\cdot\|_1$ is weaker than $\|\cdot\|_2$ if $\exists C> 0\forall x\in V$ such that $\|x\|_1\le C\|x\|_2$
$\|\cdot\|_2$ is stronger than $\|\cdot\|_1$ in this case

Given two norms and on a vector space $V$ we say they are equivalent if:

$\exists c,C\in\mathbb{R}\text{ with }c,C>0\ \forall x\in V:\ c\|x\|_1\le\|x\|_2\le C\|x\|_1$

We may write this as $\|\cdot\|_1\sim\|\cdot\|_2$ - this is an Equivalence relation

TODO: proof

Note also that if $\|\cdot\|_1$ is both weaker and stronger than $\|\cdot\|_2$ they are equivalent

Any two norms on $\mathbb{R}^n$ are equivalent
The norms $\|\cdot\|_{L^1}$ and $\|\cdot\|_\infty$ on $\mathcal{C}([0,1],\mathbb{R})$ are not equivalent.

Name	Norm	Notes
Norms on $\mathbb{R}^n$
1-norm	$\\|x\\|_1=\sum^n_{i=1}\|x_i\|$	it's just a special case of the p-norm.
2-norm	$\\|x\\|_2=\sqrt{\sum^n_{i=1}x_i^2}$	Also known as the Euclidean norm (see below) - it's just a special case of the p-norm.
p-norm	$\\|x\\|_p=\left(\sum^n_{i=1}\|x_i\|^p\right)^\frac{1}{p}$	(I use this notation because it can be easy to forget the $p$ in $\sqrt[p]{}$ )
$\infty-$ norm	$\\|x\\|_\infty=\sup(\{x_i\}_{i=1}^n)$	Also called $\infty-$ norm
Norms on $\mathcal{C}([0,1],\mathbb{R})$
$\\|\cdot\\|_{L^p}$	$\\|f\\|_{L^p}=\left(\int^1_0\|f(x)\|^pdx\right)^\frac{1}{p}$	NOTE be careful extending to interval $[a,b]$ as proof it is a norm relies on having a unit measure
$\infty-$ norm	$\\|f\\|_\infty=\sup_{x\in[0,1]}(\|f(x)\|)$	Following the same spirit as the $\infty-$ norm on $\mathbb{R}^n$
$\\|\cdot\\|_{C^k}$	$\\|f\\|_{C^k}=\sum^k_{i=1}\sup_{x\in[0,1]}(\|f^{(i)}\|)$	here $f^{(k)}$ denotes the $k^\text{th}$ derivative.
Induced norms
Pullback norm	$\\|\cdot\\|_U$	For a linear isomorphism $L:U\rightarrow V$ where V is a normed vector space

@@ Line 28: / Line 28: @@
 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>
+<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>
 We may write this as <math>\|\cdot\|_1\sim\|\cdot\|_2</math> - this is an [[Equivalence relation]]
@@ Line 37: / Line 37: @@
 *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==