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.

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