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.

Euclidean Space

Vector Space

Note: $(x)=(x_1,\dots ,x_k)$ is a vector.

An inner product $⟨\vec{x},\vec{y}⟩:{\mathbb{R}}^k\times {\mathbb{R}}^k\to \mathbb{R}$ for $\vec{x},\vec{y}\in {\mathbb{R}}^k$ is defined as

$$⟨\vec{x},\vec{y}⟩=\vec{x}\cdot \vec{y}=x_1{y}_1+x_2{y}_2+\cdots +x_k{y}_k$$

The norm (length) of a vector is $||\vec{x}||=\sqrt{⟨\vec{x},\vec{x}⟩}$. For ${\mathbb{R}}^k$, we have:

1. $⟨\vec{x},\vec{y}⟩=⟨\vec{y},\vec{x}⟩$
2. $⟨a\vec{x}+b\vec{y},\vec{z}⟩=a⟨\vec{x},\vec{z}⟩+b⟨\vec{y},\vec{z}⟩$
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. $||\vec{x}-\vec{y}||\ge |||\vec{x}||-||\vec{y}|||$

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.