Difference between revisions of "Norm"

Revision as of 21:58, 11 July 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 22:16, 11 July 2015 (view source) Alec (Talk | contribs) m Newer edit →
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	* <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
−		+	==Terminology==
		+	Such a vector space equipped with such a function is called a [[Normed space|normed space]]
	==Induced metric==		==Induced metric==
	To get a [[Metric space|metric space]] from a norm simply define:		To get a [[Metric space|metric space]] from a norm simply define:

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Revision as of 22:16, 11 July 2015

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

Definition

A norm on a vector space $(V,F)$ (where $F$ is either $\mathbb{R}$ or $\mathbb{C}$) is a function

$$\|\cdot\|:V\rightarrow\mathbb{R}$$ such that^[1][2]:

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

Often parts 1 and 2 are combined into the statement

• $$\|x\|\ge 0\text{ and }\|x\|=0\iff x=0$$ so only 3 requirements will be stated.

I don't like this

Terminology

Such a vector space equipped with such a function is called a normed space

Induced metric

To get a metric space from a norm simply define:

• $$d(x,y):=\|x-y\|$$ and $d$ is indeed a metric^[2]

(See Subtypes of topological spaces for more information, this relationship is very important in Functional analysis)

Weaker and stronger norms

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

Equivalence of norms

Given two norms

$$\|\cdot\|_1$$ and $$\|\cdot\|_2$$ on a vector space $V$ we say they are equivalent if:

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

Note also that if

$$\|\cdot\|_1$$ is both weaker and stronger than $$\|\cdot\|_2$$ they are equivalent

Examples

• 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.

Common norms

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 - 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 sup-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

Examples

• Euclidean norm

References

1. Jump up ↑ Analysis - Part 1: Elements - Krzysztof Maurin
2. ↑ ^{Jump up to: 2.0 2.1} Functional Analysis - George Bachman and Lawrence Narici