絕對值與三角不等式 (Absolute value and the triangle inequality)

絕對值(Absolute value)

• $x \in \mathbb{R}$, the absolute value of $x$ is $\begin{vmatrix} x \end{vmatrix}$.

  • $\begin{vmatrix} x \end{vmatrix} = \left \lbrace \begin{array}{ll} x, & x \geq 0, \\ -x, & x \leq 0. \end{array} \right.$
• Theorem: $a \geq 0$, then $\begin{vmatrix} x \end{vmatrix} \leq a \ \Leftrightarrow \ -a \leq x \leq a$.

三角不等式 (Traingle inequality)

• $\forall x, y \in \mathbb{R} \Rightarrow \ \begin{vmatrix} x+y \end{vmatrix} \leq \begin{vmatrix} x \end{vmatrix} + \begin{vmatrix} y \end{vmatrix}$.
  • We have $-\begin{vmatrix} x \end{vmatrix} \leq x \leq \begin{vmatrix} x \end{vmatrix}$ and $- \begin{vmatrix} y \end{vmatrix} \leq y \leq \begin{vmatrix} y \end{vmatrix}$.
  • $\therefore - (| x| + |y| ) \leq x+y \leq |x| + |y|$. [1]
  • $\because | x + y| \leq x + y$. [2]
  • By [1][2], $| x+y| \leq |x| + |y|$ (QED).

• Corollary: $| a- b | \leq | a-c| + |c -b|$.

• Corollary: $| x_1 + x_2 + \cdots + x_n | \leq |x_1| + |x_2| + \cdots + |x_n|$.

Cauchy-Schwarz不等式

• $\lbrace a_k \in \mathbb{R}, k=1,2,\cdots,N \rbrace, \lbrace b_k \in \mathbb{R}, k=1,2,\cdots, N \rbrace$ then
• $\left( \sum_{k=1}^N a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^N b_k^2 \right)$

• matrix form: $(\mathbf{a} \cdot \mathbf{b})^2 \leq \begin{Vmatrix} a \end{Vmatrix}^2 \begin{Vmatrix} b \end{Vmatrix}^2$.

Lagrange identity

• $\lbrace a_k \in \mathbb{R}, k=1,2,\cdots,N \rbrace, \lbrace b_k \in \mathbb{R}, k=1,2,\cdots, N \rbrace$ then

• $\left( \sum_{k=1}^N a_k b_k \right)^2 = \left( \sum_{k=1}^N a_k^2 \right) \left( \sum_{k=1}^N b_k^2 \right) - \sum_{1 \leq k < j \leq N} (a_k b_j - a_j b_k)^2$.

• 可用Lagrange等式推出Cauchy-Schwarz inequality.

Minkowski's inequality

• $\lbrace a_k \in \mathbb{R}, k=1,2,\cdots,N \rbrace, \lbrace b_k \in \mathbb{R}, k=1,2,\cdots, N \rbrace$ then

• $\left( \sum_{k=1}^N (a_k + b_k)^2 \right)^{1/2} \leq \left( \sum_{k=1}^N a_k^2 \right)^{1/2} + \left( \sum_{k=1}^N b_k^2 \right)^{1/2}$.

• matrix form: $\begin{Vmatrix} \mathbf{a} + \mathbf{b} \end{Vmatrix} \leq \begin{Vmatrix} \mathbf{a} \end{Vmatrix} + \begin{Vmatrix} \mathbf{b} \end{Vmatrix}$.