Euclidean Space

Let $n\in \mathbb{N}$ and define

$${\mathbb{R}}^n=\{(x_1,\cdots ,x_n\mid x_j\in \mathbb{R}\}$$

where $\vec{x}$ or $\mathbf{x}$ is denoted as an element in it. We have some properties:

• Addition is kept from $\mathbb{R}$
• Multiplication by scalar. For $a\in \mathbb{R},\vec{x}\in {\mathbb{R}}^n$ we have $a\vec{x}=(a{x}_1,\cdots ,a{x}_n)$.
• We keep commutativity, associativity, distributive laws, etc.

Inner Product

Given two vectors,

$$\vec{x}\cdots \vec{y}=x_1{y}_1+x_2{y}_2=\sum _{i=1}^n{x}_i{y}_i$$

In particular,

$$\vec{x}\vec{x}=\sum _{i=1}^n{x}_i^2$$

Norm

We define the norm as:

$$||\vec{x}||:=\sqrt{\vec{x}\vec{x}}=\sqrt{\sum _{i=1}^n{x}_i^2}$$

Proposition

1. $||\vec{x}||>0$ unless $\vec{x}=\vec{0}$
2. $||\lambda \vec{x}||=|\lambda |\cdot ||\vec{x}||$ for $\lambda \in \mathbb{R}$
3. Cauchy-Schwarz Inequality: So, $|\vec{x}\vec{y}|\le ||\vec{x}||\cdots ||\vec{y}||$
4. Triangle Inequality: $||\vec{x}+\vec{y}||\le ||\vec{x}||+||\vec{y}||$
5. $||\vec{x}-\vec{y}||\le ||\vec{x}-\vec{z}||+||\vec{y}-\vec{z}||$

Proof:

We see that $1,2$ is trivial and $3$ is just Cauchy-Schwarz. Assume $4$. We prove $5$. We note that

$$x-y=(x-z)+(z-y)$$

so, with $4$, we get

$$\begin{array}{rl}||x-y||& \le ||x-z||+||z-y||\\ & =||x-z||+||-(y-z)||\\ & {=}_2||x-z||+|-1|\cdot ||y-z||\\ & =||x-z||+||y-z||\end{array}$$

Now, we prove $4$. Let $\vec{x}=(x_1,\cdots x_n)$ and $\vec{y}=(y_1,\cdots y_n)$. Because both sides are positive, it suffices to show

$$||\vec{x}+\vec{y}|{|}^2\le (||\vec{x}||+||\vec{y}||{)}^2$$

Then

$$\begin{array}{rl}||x+y|{|}^2& =\sum _{i=1}^n(x_i+y_i{)}^2\\ & =\sum _{i=1}^n\left(x_i^2+2x_i{y}_i+y_i^2\right)\\ & =\sum _{i=1}^n{x}_i^2+\sum x_i{y}_i+\sum y_i^2\\ & =||x|{|}^2+2xy+||y|{|}^2\end{array}$$

Now, by Cauchy-Schwarz, we have that $|xy|\le ||x||\cdot ||y||$, in which replacement of the middle term gives us:

$$\begin{array}{rl}& \le ||x|{|}^2+2||x||\cdot ||y||+||y|{|}^2\\ & =(||x||+||y||{)}^2\end{array}$$

And the proof is done.