# Math4501 Lecture 3

## Review from last lecture

### Bisection method for finding root

P1. Let $f$ be a continuous function on $[a,b]\to \mathbb{R}$, find $\xi \in [a,b]$ such that $f(\xi)=0$.

P2. Let $g$ be a continuous function on $[a,b]\to \mathbb{R}$, find $\xi \in [a,b]$ such that $g(\xi)=\xi$.

#### Theorem 1:

solution to P1 exists if $f(a)f(b)<0$.

#### Theorem 2:

Fixed point theorem:

If solution to P2 exists, then $g:[a,b]\to [a,b]$.

#### Bijection method

Obtain two sequence $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$.

Initially, we set $a_0=a$ and $b_0=b$.

If $|a_n-b_n|<2^{-n}|a_0-b_0|$, then $a_n$ and $b_n$ are Cauchy sequence. So their limit exists.

$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=\xi$.

### Simple iteration method

#### Definition of Simple Iteration

Given $x_0\in [a,b]$, we define a sequence $(x_n)_{n=0}^\infty$ by

$$
x_{n+1}=g(x_n)
$$

If a simple iteration converges, the it converges to a fixed point of

<details>

<summary>Proof</summary>

Let $c\coloneqq \lim_{n\to\infty} x_n=g(c)$.

$g:[a,b]\to \mathbb{R}$ is continuous if and only if $g$ is continuous at $x_0\in [a,b]$.

</details>

#### Definition of Lipschitz continuous

A function $g:[a,b]\to \mathbb{R}$ is Lipschitz continuous if for all $x,y\in [a,b]$, there exists a constant $L>0$ such that

$$
|g(x)-g(y)|\leq L|x-y|
$$

for some $L>0$.

> [!NOTE]
> 
> Lipschitz continuous is a stronger condition than continuous.
>
> If a function is Lipschitz continuous, then it is continuous.
>
> However, the converse is not true.

#### Definition of contraction mapping

A function $g:[a,b]\to \mathbb{R}$ is a contraction mapping if it is Lipschitz continuous with $L<1$.

#### Theorem of simple iteration

Let $g:[a,b]\to [a,b]$ is a contraction mapping (Lipschitz continuous with $L<1$, with pointwise ontinuous $g$).

Then

- $g$ has a unique fixed point $\xi \in [a,b]$
- Simple iteration $x_{n+1}=g(x_n)$ converges to $\xi$ for any $x_0\in [a,b]$.

<details>

<summary>Proof</summary>

**Uniqueness**:

Suppose $\xi_1$ and $\xi_2$ are two fixed points of $g$.

Then $|x_1-x_2|=|g(\xi_1)-g(\xi_2)|\leq L|\xi_1-\xi_2|$.

Thus, $(1-L)|\xi_1-\xi_2|\leq 0$, which implies $|\xi_1-\xi_2|=0$.

A more general result:

Brouwer's fixed point theorem

**Convergence**:

Let $\xi\in [a,b]$ be the unique fixed point of $g$.

Then,

$$
\begin{aligned}
|x_n-\xi|&=|g(x_{n-1})-g(\xi)|\\
&\leq L|x_{n-1}-\xi|\\
&=L|g(x_{n-2})-g(\xi)|\\
&\leq L^2|x_{n-2}-\xi|\\
&\vdots\\
&\leq L^n|x_0-\xi|
$$

Thus, we can always find $N$ such that $L^N|x_0-\xi|<\epsilon$ for any $\epsilon>0$. Choose $N=\log(\frac{\epsilon}{|x_0-\xi|})/\log(L)$.

Therefore, $x_n$ converges to $\xi$.

</details>
