# Banach fixed point theorem

1. Feb 3, 2008

### sara_87

I really dont understand nothing from the Banach fixed point theorem, i know that it should satisfy:
[g(x)-g(y)]<K(x-y) for all x and y in[a,b]
but i dont even understand what thats supposed to mean?

any help will be appreciated.
thank you.

2. Feb 3, 2008

### CompuChip

The statement is:
Let g be a map (function) which satisfies
$$| g(x) - g(y) | < K |x - y|$$
for some real number 0 <= K < 1, for every point x and y
Then there exists one and only one $x_0$ such that $g(x_0) = x_0$ (that is: x0 is a "fixed point" of g).

Now, which part don't you understand? Is it the meaning or application of the theorem that confuses you? Is it the distances? The variables?

Basically, the condition says that g is a contraction mapping, that is: if you take any two points x and y, then their images under g will be closer together than the points themselves. Now the theorem states that if you have such a function g, then somewhere there is a point which doesn't move at all.

Last edited: Feb 3, 2008
3. Feb 3, 2008

### sara_87

if there was a question like:
a)use the banach fixed point iteration to solve the equation x=1/(2+x^2)
b)show that the assumptions of the banach fixed point theorem on the interval [0,1]

how would i go about solving it?

4. Feb 4, 2008

### HallsofIvy

Staff Emeritus
No, you don't know that! It should be $[g(x)- g(y)]\le k(x- y)$ where K< 1.
Applying g to two points in the set moves them closer together.

Draw a picture! Take a sheet of paper and draw some "set"- that is, some closed curve representing the original set. If you were to apply your contraction map to every point in that set what would happen? Since every point gets closer to every other point, the entire set is "contracted" to a smaller set inside the original set. Now apply your function to every point in THAT set. Again, it is "contracted"- you have a still smaller set. As you apply the function again and again, you get smaller and smaller sets, contracting to a single point. If you were to apply the function to THAT point, it has to go to itself- it is a "fixed point". If you pick x to be any point in the set and apply the function repeatedly, it must "contract" to that same point as all the points in the set- that sequence, x, f(x), f(f(x)), etc. converges to that fixed point.

Technical point which you may ignore: How do we know the "contractions" don't get smaller and smaller so that the set converges to a non-trivial subset and not to a single point? That's why we need "$\le k(x- y)$ with k< 1[/itex] rather than just "< 1". The contraction can never be less that 1-k.

Let F(x)= 1/(2+ x2). Saying that x= 1/(2+x2) is the same as saying F(x)= x, that is, that x is a fixed point. Choose some reasonable value of x, say x= 0.5. (I "cheated" I did a quick graph of 1/(2+ xx) and y= x to get an estimate of where they cross). Now F(x)= F(0.5)= 1/(2+ 0.25)= 1/2.25= .4444. F(.4444)= 1/(2+ 1.975)= 0.4551. F(0.4551)= 1/(2+ .207)= 0.4531. Keep doing that until you get sufficient accuracy.