Normed, Metric and Topological spaces

A norm on a vector space is a generalised notion of “length” of a vector.

definition Norms

A norm on a vector space $X$ is a map $||\cdot ||:X\to {\mathbb{R}}^{+}$ such that

• $||x||=0$ if and only if $x=0$
• $||\lambda x||=|\lambda |||x||$ for every $\lambda \in \mathbb{R},x\in X$ (“homogeneity”) and
• $||x+y||\le ||x||+||y||$ for every $x,y\in X$ (the triangle inequality)

example 2.2

In the vector space ${\mathbb{R}}^n$, for $x=(x_1,...,x_n)$ define
$||x||=({\sum }_{j=1}^n|x_j{|}^2{)}^{\frac{1}{2}}$
The “standard norm” or “Euclidean norm.”

Lets check this is a norm. If $x=0$, the $||x||=0$, and if $||x||=0$ then $|x_j|=0$ for every $j$, i.e. $x=0$, giving us bullet point 1.

For bullet point 2, we have
$||\lambda x||=({\sum }_{j=1}^n|\lambda x_j{|}^2{)}^{\frac{1}{2}}=({\sum }_{j=1}^n|\lambda {|}^2|x_j{|}^2{)}^{\frac{1}{2}}=|\lambda |||x||,$
as required.

The triangle inequality proof is non-examinable, but I will show this below.
Let $x\cdot y$ denote the “dot product”: $x\cdot y={\sum }_{j=1}^n{x}_j{y}_j$. We have

$||x+y|{|}^2=(x+y)\cdot (x+y)=||x|{|}^2+2x\cdot y+||y|{|}^2$
$le||x|{|}^2+2||x||||y||+||y|{|}^2$
$le(||x||+||y||{)}^2$
Using the inequality $|x\cdot y|\le ||x||||y||$

definition Normed spaces

If $X$ is a vector space and $\Vert \cdot \Vert$ is a norm on $X$, the pair $(X,\Vert \cdot \Vert )$ is a normed space.

Many spaces have a “standard norm”, so for exampled ${\mathbb{R}}^n$ is usually ${\mathbb{R}}^n$ with the Euclidean norm, or it is clear that we are working with a particular norm. Therefore, we might talk about “the normed space $X$” rather than the normed space $(X,\Vert \cdot \Vert )$ for brevity.

definition Convexity

Let $X$ be a vector space. A subset $K$ of $X$ is convex if whenever $x,y\in K$ and $0\le \lambda \le 1$ we have $\lambda x+(1-\lambda )y\in K$.
Informally, a set is convex if the line segment joining any two points in the set is entirely contained in the set.

lemma

In any normed space $(X,\Vert \cdot \Vert )$, the closed unit ball $\overline{{\mathcal{B}}_X}$ is convex

- Proof $x,y\in \overline{{\mathcal{B}}_X}$ then $\Vert x\Vert \le 1$ and $\Vert y\Vert \le 1$. Then, for $0<\lambda <1$, $\Vert \lambda x+(1-\lambda )y\Vert \le |\lambda |\Vert x\Vert +|1-\lambda |\Vert y\Vert le\lambda +(1-\lambda )=1$ So $\lambda x+(1-\lambda )y\in \overline{{\mathcal{B}}_X}$

If