Euclidean Space
Definition 1. $\forall n \in \mathbb{N}$, let $$\mathbb{R}^n := \{ \textbf{x} \,|\, \textbf{x} = (x_1, ..., x_n) \, \text{where} \, x_i \in \mathbb{R} (i = 1, 2, ..., n)\}.$$ For $\textbf{x} = (x_1, ..., x_n)$ and $\textbf{y} = (y_1, ..., y_n) \in\mathbb{R}^n$, we define the coordinatewise operations: $$\textbf{x} + \textbf{y} = (x_1 + y_1, ..., x_n + y_n) \\ \alpha \textbf{x} = (\alpha x_1, ..., \alpha x_n)$$ so that $\textbf{x} + \textbf{y}, \alpha \textbf{x} \in \mathbb{R}^n$.
이러한 정의 하에, $\mathbf{0} = (0, ..., 0)$을 영벡터로 선언하면 $\mathbb{R}^n$은 실수 집합 $\mathbb{R}$에 대한 vector space가 된다는 것을 쉽게 보일 수 있다.
Euclidean metric
이러한 집합 $\mathbb{R}^n$은 다음과 같이 정의되는 metric을 가진다.
Definition 2. In $\mathbb{R}^n$, the metric $d$ is defined for each $n$ by
$$d(x, y) := (\sum_{i=1}^{n} (x_i - y_i)^2)^{\frac{1}{2}} \\ \text{where} \,\, \mathbf{x} = (x_1, ..., x_n), \mathbf{y} = (y_1, ..., y_n) \in \mathbb{R}^n$$
여기서 metric의 notation은 $d_n$ 이 되어서 각 $n$ 에 대해서 다른 metric임을 명시해야 하지만 편의상 간단하게 $d$ 라고만 표기하자. 이렇게 정의되는 metric을 가지는 집합 $\mathbb{R}^n$은 metric space가 된다.
Definition 3.
(a) The inner product of $\textbf{x}$ and $\textbf{y}$ is defined by $$\textbf{x} \cdot \textbf{y} := \sum_{i=1}^{n} x_iy_i$$
(b) The norm (or length) $||\mathbf{x}||$ of $\textbf{x}$ is defined by $$||\textbf{x}|| := d(\mathbf{x}, \mathbf{0}) = (\sum_{i=1}^{n} x_i^2)^{\frac{1}{2}} = (\textbf{x} \cdot \textbf{x})^{\frac{1}{2}} $$
이렇게 정의한 구조, 즉 metric $d$를 가지는 vector space $\mathbb{R}^n$을 우리는 'Euclidean n-space' 라고 부른다.
The Cauchy-Schwarz Inequality
Theorem 1. (The Cauchy-Schwarz Inequality) For any points $\mathbf{x} = (x_1, ..., x_n)$ and $\mathbf{y} = (y_1, ..., y_n)$ in $\mathbb{R}^n$, $$|\mathbf{x} \cdot \mathbf{y}| \leq ||\mathbf{x}|| ||\mathbf{y}||.$$
Proof. If either $\mathbf{x}$ or $\mathbf{y}$ is the origin, then the inequality is valid. Thus we may assume that both $\mathbf{x}$ and $\mathbf{y}$ have at least one non-zero coordinate.
For each $i = 1, ..., n,$ $$0 \leq (\frac{|x_i|}{||\mathbf{x}||} - \frac{|y_i|}{||\mathbf{y}||})^2$$ so $$\frac{2|x_iy_i|}{||\mathbf{x}||||\mathbf{y}||} \leq \frac{x^2_i}{||\mathbf{x}||^2} + \frac{y^2_i}{||\mathbf{y}||^2}.$$ Then $$\sum_{i=1}^{n} \frac{2|x_iy_i|}{||\mathbf{x}||||\mathbf{y}||} \leq \sum_{i=1}^{n} (\frac{x^2_i}{||\mathbf{x}||^2} + \frac{y^2_i}{||\mathbf{y}||^2}).$$ Splitting the right-hand side into two sums and factoring constant terms from both sides gives $$\frac{2}{||\mathbf{x}||||\mathbf{y}||} \sum_{i=1}^{n} |x_iy_i| \leq \frac{1}{||\mathbf{x}||^2} \sum_{i=1}^{n} x^2_i + \frac{1}{||\mathbf{y}||^2} \sum_{i=1}^{n} y^2_i = \frac{||\mathbf{x}||^2}{||\mathbf{x}||^2} + \frac{||\mathbf{y}||^2}{||\mathbf{y}||^2} = 2.$$ Thus $$\frac{1}{||\mathbf{x}||||\mathbf{y}||} \sum_{i=1}^{n} |x_iy_i| \leq 1$$ so $$|\mathbf{x} \cdot \mathbf{y}| = |\sum_{i=1}^{n} x_iy_i| \leq \sum_{i=1}^{n} |x_iy_i| \leq ||\mathbf{x}||||\mathbf{y}||.$$
The Minkowski Inequality
Theorem 2. (The Minkowski Inequality) For any points $\mathbf{x} = (x_1, ..., x_n)$ and $\mathbf{y} = (y_1, ..., y_n)$ in $\mathbb{R}^n,$ $$||\mathbf{x} + \mathbf{y}|| \leq ||\mathbf{x}|| + ||\mathbf{y}||.$$
Proof. By The Cauchy-Schwarz Inequality, $$||\mathbf{x} + \mathbf{y}||^2 = \sum_{i=1}^{n} (x_i + y_i)^2 = \sum_{i=1}^{n} (x^2_i + 2x_iy_i + y^2_i)^2 = \sum_{i=1}^{n} x^2_i + 2 \sum_{i=1}^{n} x_iy_i + \sum_{i=1}^{n} y^2_i \\= ||\mathbf{x}||^2 + 2\mathbf{x} \cdot \mathbf{y} + ||\mathbf{y}||^2 \leq ||\mathbf{x}||^2 + 2||\mathbf{x}||||\mathbf{y}|| + ||\mathbf{y}||^2 = (||\mathbf{x}|| + ||\mathbf{y}||)^2.$$ Hence $$||\mathbf{x} + \mathbf{y}|| \leq ||\mathbf{x}|| + ||\mathbf{y}||.$$
Proposition
Proposition. Suppose $\textbf{x}, \textbf{y}, \textbf{z} \in \mathbb{R}^n$, and $\alpha \in \mathbb{R}$. Then the following statements hold:
(a) $||\textbf{x}|| \geq 0$
(b) $||\textbf{x}|| = 0 \Longleftrightarrow \textbf{x} = \textbf{0}$
(c) $||\alpha \textbf{x}|| = |\alpha|||\textbf{x}||$
(d) $||\textbf{x} - \textbf{z}|| \leq ||\textbf{x} - \textbf{y}|| + ||\textbf{y} - \textbf{z}||$
Proof.
(a) Since each $x^2_i$ is nonnegative, it's clear.
(b) $||\textbf{x}|| = 0 \Longleftrightarrow x^2_i = 0 (i = 1, ... ,n) \Longleftrightarrow x_i = 0 (i = 1, ..., n) \Longleftrightarrow \textbf{x} = \textbf{0}$
(c) $||\alpha \textbf{x}|| = |(\alpha \textbf{x}) \cdot (\alpha \textbf{x})| = (\sum_{i=1}^{n} (\alpha x_i)^2)^{\frac{1}{2}} = |\alpha| (\sum_{i=1}^{n} x^2_i)^{\frac{1}{2}} = |\alpha|||\textbf{x}||$
(d) By The Minkowski Inequality, $||\mathbf{x} - \mathbf{z}|| = ||(\mathbf{x} - \mathbf{y}) + (\mathbf{y} - \mathbf{z})|| \leq ||\mathbf{x} - \mathbf{y}|| + ||\mathbf{y} - \mathbf{z}||$
Proposition (a), (b), (d)에 의해서 우리가 definitions 2에서 정의한 $d$는 metric이며, $(\mathbb{R}^n, d)$는 metric space이다.