Difference between revisions of "Norm"

Revision as of 03:04, 8 March 2015

Definition

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

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

Common norms

The 1-norm

$$\|x\|_1=\sum^n_{i=1}|x_i|$$ - it's just a special case of the p-norm.

The 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.

The 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 $$\sqrt[p]{}$$ )

The supremum-norm

Also called $$\infty-$$ norm
$$\|x\|_\infty=\sup(\{x_i\}_{i=1}^n)$$

Equivalence of norms

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

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

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

TODO: proof

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.

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.