# Real Analysis

Extensions of Real Numbers

# Extended Real Numbers $$\overline{\mathbb{R}}$$

$\overline{\mathbb{R}} := \mathbb{R} \cup \{-\infty, +\infty\}.$

Order: $\forall x \in \mathbb{R}, -\infty < x < +\infty.$

Arithmetic: $x + (+\infty) := +\infty. \\ x + (-\infty) := -\infty. \\ \text{If } x > 0, x \cdot (+\infty) := +\infty. \\ \text{If } x < 0, x \cdot (+\infty) := -\infty. \\ \dots$

Every subset of $$\overline{\mathbb{R}}$$ has a supremum (possibly $$+\infty$$).

# Euclidean Space $$\mathbb{R}^k$$

$\mathbb{R}^k := \{(x_1, x_2, \dots, x_k) : x_i \in \mathbb{R}\}.$

Arithmetic: $\underbrace{(x_1, x_2, \dots, x_k)}_{\vec{x}} + \underbrace{(y_1, y_2, \dots, y_k)}_{\vec{y}} := (x_1 + y_1, x_1 + y_2, \dots, x_k + y_k). \\ \alpha (x_1, x_2, \dots, x_k) := (\alpha x_1, \alpha x_2, \dots, \alpha x_k)\text{, where } \alpha \in \mathbb{R}\text{, called a scalar}.$

Inner product, or dot product: $\vec{x} \cdot \vec{y} := \sum_{i = 1}^{k} x_iy_i.$

Norm, or length: $|\vec{x}| := \sqrt{\vec{x} \cdot \vec{x}}.$

# Complex Number Field $$\mathbb{C}$$

Complex numbers $$\mathbb{C} = \mathbb{R}^2$$ has a field structure.

$(a, b) + (c, d) := (a + c, b + d). \\ (a, b) \cdot (c, d) := (ac - bd, ad + bc).$

The additive identity is $$0^\star := (0, 0)$$. The multiplicative identity is $$1^\star := (1, 0)$$.

$$\mathbb{C}$$ extends $$\mathbb{R}$$. Consider subfield $$\{(a, 0) : a \in \mathbb{R}\}$$ which behaves just like $$\mathbb{R}$$, or is isomorphic to $$\mathbb{R}$$.

Note: $$(0, 1) + (0, 1) = (-1, 0)$$. So, let $$i = (0, 1)$$ then $$i^2 = -1$$, a real number.

So, write $$a + bi$$ for $$(a, b)$$.

Let $$z = a + bi$$. The real part of $$z$$ is $$\text{Re}(z) = a$$. The imaginary part of $$z$$ is $$\text{Im}(z) = b$$.

The conjugate of $$z$$ is $$\overline{z} = a - bi$$. $\overline{z + w} = \overline{z} + \overline{w}. \\ \overline{z \cdot w} = \overline{z} \cdot \overline{w}. \\ z + \overline{z} = 2\text{Re}(z). \\ z - \overline{z} = 2\text{Im}(z). \\ z \cdot \overline{z} = \text{Re}(z)^2 + \text{Im}(z)^2. \\$

Define length, or absolute value, of $$z$$ to be $$|z| = \sqrt{z \cdot \overline{z}}$$, which is the same as the length in $$\mathbb{R}^2$$.

In $$\mathbb{C}^k = \{(z_1, z_2, \dots, z_k) : z \in \mathbb{C}\}$$, the inner product is $$\langle \vec{x}, \vec{y} \rangle := \sum_{i = 1}^{k} x_i\overline{y_i}$$.

Notice that $$\overline{\left \langle \vec{a}, \vec{b} \right \rangle} = \left \langle \vec{b}, \vec{a} \right \rangle$$.

Proof. \begin{align} \overline{\left \langle \vec{a}, \vec{b} \right \rangle} & = \overline{\sum_{i} a_i\overline{b_i}} & \text{(by definition of } \left \langle \vec{a}, \vec{b} \right \rangle \text{)} \\ & = \sum_{i} \overline{a_i\overline{b_i}} & \text{(due to } \overline{z + w} = \overline{z} + \overline{w} \text{)} \\ & = \sum_{i} \overline{a_i}b_i & \text{(due to } \overline{z \cdot w} = \overline{z} \cdot \overline{w} \text{)} \\ & = \sum_{i} b_i\overline{a_i} & \text{(due to commutativity of multiplication)} \\ & = \left \langle \vec{b}, \vec{a} \right \rangle & \text{(by definition of } \left \langle \vec{b}, \vec{a} \right \rangle \text{)} \\ \end{align} QED.

## Properties of $$\mathbb{C}$$

$|z| \geq 0. \\ |\overline{z}| = |z|. \\ |zw| = |z||w|. \\ \text{Re}(z) \leq |z|. \\ |z + w| \leq |z| + |w|. \\$

Proof for $$|z + w| \leq |z| + |w|$$. \begin{align} |z + w|^2 & = (z + w)(\overline{z} + \overline{w}) \\ & = z \cdot \overline{z} + z \cdot \overline{w} + w \cdot \overline{z} + w \cdot \overline{w} \\ & = |z|^2 + 2\text{Re}(z \overline{w}) + |w|^2 \\ & \leq |z|^2 + 2|z||w| + |w|^2 = (|z| + |w|)^2. \end{align}

This yields the desired inequality. QED.

# Cauchy - Schwarz Inequality

Version for complex numbers: If $$a_1, \dots, a_n, b_1, \dots, b_n \in \mathbb{C}$$, then $\left|\sum_{i = 1}^{n} a_i\overline{b_i} \right|^2 \leq \sum_{i = 1}^{n} |a_i|^2 \cdot \sum_{i = 1}^{n} |b_i|^2.$

Version for real numbers: If $$a_1, \dots, a_n, b_1, \dots, b_n \in \mathbb{R}$$, then $\left|\sum_{i = 1}^{n} a_ib_i \right|^2 \leq \sum_{i = 1}^{n} a_i^2 \cdot \sum_{i = 1}^{n} b_i^2.$

Version for real vectors: Given $$\vec{x}, \vec{y} \in \mathbb{R}^k$$. Then $$|\vec{x} \cdot \vec{y}| \leq |\vec{x}||\vec{y}|$$.

Version for complex vectors: Given $$\vec{x}, \vec{y} \in \mathbb{C}^k$$. Then $$\left| \langle \vec{x}, \vec{y} \rangle \right|^2 \leq \langle \vec{x}, \vec{x} \rangle \langle \vec{y}, \vec{y} \rangle$$.

Proof. Let $$\vec{a}, \vec{b} \in \mathbb{C}^n$$, and $$y = \frac{\left \langle \vec{a}, \vec{b} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle}$$.

Consider $$\overline{y}$$. $\overline{y} = \overline{\left(\frac{\left \langle \vec{a}, \vec{b} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle} \right)} = \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle}}{\overline{\left \langle \vec{b}, \vec{b} \right \rangle}} = \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle}}{\overline{\sum_{i = 1}^{n} b_i\overline{b_i}}} = \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle}}{\sum_{i = 1}^{n} \overline{b_i\overline{b_i}}} = \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle}}{\sum_{i = 1}^{n} b_i\overline{b_i}} = \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle}}{\left \langle \vec{b}, \vec{b} \right \rangle}.$

Consider the following inequality. \begin{align} 0 \leq \left|\vec{a} - y\vec{b}\right|^2 & = \left \langle \vec{a} - y\vec{b}, \vec{a} - y\vec{b} \right \rangle \\ & = \sum_{i = 1}^{n} (a_i - yb_i)(\overline{a_i} - \overline{y}\overline{b_i}) \\ & = \left \langle \vec{a}, \vec{a} \right \rangle - \overline{y}\left \langle \vec{a}, \vec{b} \right \rangle - y\left \langle \vec{b}, \vec{a} \right \rangle + |y|^2 \left \langle \vec{b}, \vec{b} \right \rangle \\ & = \left \langle \vec{a}, \vec{a} \right \rangle - \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle} \left \langle \vec{a}, \vec{b} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle} - \frac{\left \langle \vec{a}, \vec{b} \right \rangle \left \langle \vec{b}, \vec{a} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle} + \left|\frac{\left \langle \vec{a}, \vec{b} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle}\right|^2 \left \langle \vec{b}, \vec{b} \right \rangle \\ & = \left \langle \vec{a}, \vec{a} \right \rangle - \frac{\overline{\left \langle \vec{a}, \vec{b} \right \rangle} \left \langle \vec{a}, \vec{b} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle} - \frac{\left \langle \vec{a}, \vec{b} \right \rangle \overline{\left \langle \vec{a}, \vec{b} \right \rangle}}{\left \langle \vec{b}, \vec{b} \right \rangle} + \frac{\left|\left \langle \vec{a}, \vec{b} \right \rangle \right|^2 \left \langle \vec{b}, \vec{b} \right \rangle}{\left| \left \langle \vec{b}, \vec{b} \right \rangle\right|^2} \\ & = \left \langle \vec{a}, \vec{a} \right \rangle - \frac{\left|\left \langle \vec{a}, \vec{b} \right \rangle \right|^2}{\left \langle \vec{b}, \vec{b} \right \rangle} - \frac{\left|\left \langle \vec{a}, \vec{b} \right \rangle \right|^2}{\left \langle \vec{b}, \vec{b} \right \rangle} + \frac{\left|\left \langle \vec{a}, \vec{b} \right \rangle \right|^2 \left \langle \vec{b}, \vec{b} \right \rangle}{\left \langle \vec{b}, \vec{b} \right \rangle \left \langle \vec{b}, \vec{b} \right \rangle} \\ & = \left \langle \vec{a}, \vec{a} \right \rangle - \frac{2\left|\left \langle \vec{a}, \vec{b} \right \rangle \right|^2}{\left \langle \vec{b}, \vec{b} \right \rangle} + \frac{\left|\left \langle \vec{a}, \vec{b} \right \rangle \right|^2}{\left \langle \vec{b}, \vec{b} \right \rangle} \\ & = \left \langle \vec{a}, \vec{a} \right \rangle - \frac{\left| \left \langle \vec{a}, \vec{b} \right \rangle \right|^2}{\left \langle \vec{b}, \vec{b} \right \rangle}. \end{align}

This yields the desired inequality after multiplying by $$\left \langle \vec{b}, \vec{b} \right \rangle$$. QED.