Math4501 Lecture 3

Review from last lecture

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 $g(\xi)=\xi$ .

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

Fixed point theorem:

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

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.

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

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

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 $x_0\in [a,b]$ .

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.

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

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

Then

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 $\xi$ .