Difference between revisions of "Norm"
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==Examples== | ==Examples== | ||
===The Euclidean Norm=== | ===The Euclidean Norm=== | ||
+ | {{Todo|Migrate this norm to its own page}} | ||
The Euclidean norm is denoted <math>\|\cdot\|_2</math> | The Euclidean norm is denoted <math>\|\cdot\|_2</math> | ||
Revision as of 17:29, 7 March 2015
Contents
[hide]Definition
A norm on a vector space (V,F) is a function ∥⋅∥:V→R such that:
- ∀x∈V ∥x∥≥0
- ∥x∥=0⟺x=0
- ∀λ∈F,x∈V ∥λx∥=|λ|∥x∥ where |⋅| denotes absolute value
- ∀x,y∈V ∥x+y∥≤∥x∥+∥y∥ - a form of the triangle inequality
Often parts 1 and 2 are combined into the statement
- ∥x∥≥0 and ∥x∥=0⟺x=0 so only 3 requirements will be stated.
I don't like this
Common norms
The 1-norm
∥x∥1=n∑i=1|xi| - it's just a special case of the p-norm.
The 2-norm
∥x∥2=√n∑i=1x2i - Also known as the Euclidean norm (see below) - it's just a special case of the p-norm.
The p-norm
∥x∥p=(n∑i=1|xi|p)1p (I use this notation because it can be easy to forget the p in p√)
The supremum-norm
Also called ∞−norm
∥x∥∞=sup
Examples
The Euclidean Norm
TODO: Migrate this norm to its own page
The Euclidean norm is denoted \|\cdot\|_2
Here for x\in\mathbb{R}^n we have:
\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}
Proof that it is a norm
TODO: proof
Part 4 - Triangle inequality
Let x,y\in\mathbb{R}^n
\|x+y\|_2^2=\sum^n_{i=1}(x_i+y_i)^2 =\sum^n_{i=1}x_i^2+2\sum^n_{i=1}x_iy_i+\sum^n_{i=1}y_i^2 \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 using the Cauchy-Schwarz inequality
=\left(\sqrt{\sum^n_{i=1}x_i^2}+\sqrt{\sum^n_{i=1}y_i^2}\right)^2 =\left(\|x\|_2+\|y\|_2\right)^2
Thus we see: \|x+y\|_2^2\le\left(\|x\|_2+\|y\|_2\right)^2, as norms are always \ge 0 we see:
\|x+y\|_2\le\|x\|_2+\|y\|_2 - as required.