Difference between revisions of "Norm"

Revision as of 03:22, 8 March 2015

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

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 $\mathcal{C}([0,1],\mathbb{R})$
$\\|\cdot\\|_{L^p}$	$\\|f\\|_{L^p}=\left(\int^1_0\|f(x)\|^pdx\right)$	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.

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

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

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

TODO: proof

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.

@@ 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}}