What Does the Banach Fixed Point Theorem Mean?

F(x)= x is the same as solving the equation 0= x- F(x) which is the same as solving the equation 0= x- (1/(2+ x2). (That is the Banach fixed point theorem!)Now, to show that the assumptions of the Banach fixed point theorem are satisfied on [0,1] we need to show that F(x) is a contraction mapping on that interval. We need to show that, for all x and y in that interval, |F(x)- F(y)|< K|x- y| for some K< 1. We can actually find a K< 1 such that that is true for all x and y
  • #1
sara_87
763
0
I really don't 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 don't even understand what that's supposed to mean?

any help will be appreciated.
thank you.
 
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  • #2
The statement is:
Let g be a map (function) which satisfies
[tex]| g(x) - g(y) | < K |x - y|[/tex]
for some real number 0 <= K < 1, for every point x and y
Then there exists one and only one [itex]x_0[/itex] such that [itex]g(x_0) = x_0[/itex] (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.
 
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  • #3
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
sara_87 said:
I really don't 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]
No, you don't know that! It should be [itex][g(x)- g(y)]\le k(x- y)[/itex] where K< 1.
Applying g to two points in the set moves them closer together.

but i don't even understand what that's supposed to mean?

any help will be appreciated.
thank you.
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 "[itex]\le k(x- y)[/itex] with k< 1[/itex] rather than just "< 1". The contraction can never be less that 1-k.


sara_87 said:
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?

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.
 

Related to What Does the Banach Fixed Point Theorem Mean?

What is the Banach fixed point theorem?

The Banach fixed point theorem is a mathematical result that guarantees the existence and uniqueness of a fixed point for certain types of mappings in metric spaces. It states that if a complete metric space has a mapping that is a contraction, meaning it reduces distances between points, then there exists a unique fixed point for that mapping.

How is the Banach fixed point theorem used in real-world applications?

The Banach fixed point theorem has a wide range of applications in various fields such as economics, engineering, and physics. It is commonly used in optimization problems, where finding the fixed point of a function represents the optimal solution. It is also used in the study of dynamical systems and differential equations, where it helps to prove the existence and stability of solutions.

What are the conditions for the Banach fixed point theorem to hold?

There are two main conditions for the Banach fixed point theorem to hold: the metric space must be complete, meaning that every Cauchy sequence in the space converges to a point in the space, and the mapping must be a contraction, meaning that it reduces distances between points. Additionally, the mapping must be self-mapping, meaning it maps from the space onto itself.

What is the significance of the Banach fixed point theorem in mathematics?

The Banach fixed point theorem is a fundamental result in mathematics that has many important consequences. It provides a powerful tool for proving the existence and uniqueness of fixed points for various types of mappings. It has also been used to prove other important theorems in mathematics, such as the Brouwer fixed point theorem and the Kakutani fixed point theorem.

Can the Banach fixed point theorem be extended to infinite-dimensional spaces?

Yes, the Banach fixed point theorem can be extended to infinite-dimensional spaces, as long as the space is complete and the mapping is a contraction. This is known as the Banach-Caccioppoli fixed point theorem and has many applications in functional analysis and partial differential equations.

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