Proof that a contractive function has a fixed point

In summary: I'm glad I could help you get it.In summary, the proof shows that if F is a contractive function over the interval [a,b], then there exists a unique point x in [a,b] such that F(x)=x. The proof relies on the fact that a Cauchy sequence in the real line converges, and uses the compactness of [a,b] to show that the sequence x_n converges to a point in [a,b], thus proving that x is also in [a,b].
  • #1
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Homework Statement


I must understand the proof that if [itex]F:[a,b] \to [a,b][/itex] and [itex]F[/itex] is contractive then there exist a unique [itex]x \in [a,b][/itex] such that [itex]F(x)=x[/itex].

Homework Equations


Definition of a contractive function: F is contractive over [a,b] if and only if there exist [itex]\lambda[/itex] such that [itex]0<\lambda <1[/itex] and [itex]|F(y)-F(x)| \leq \lambda |y-x| \forall x[/itex] and [itex]y \in [a,b][/itex]. That's from my memory.

The Attempt at a Solution


I'm almost done understanding the existence (I already have the uniqueness proof and I understand it).
I'm stuck at understanding the latest part.
Here is the -watered down- proof:
Let [itex]x_{n+1}=F(x_n)[/itex] for [itex]n \geq 1[/itex].
[itex]\exists \lambda \in (0,1)[/itex] such that [itex]|x_n-x_{n+1}|=|F(x_{n+1})-F(x_{n+2})| \leq \lambda |x_{n+1}-x_{n+2}| \leq ... \leq \lambda ^{n-1} |x_1-x_0|[/itex].
I can take [itex]x_n=x_0+S_n[/itex] where [itex]S_n=\sum _{j=1}^{\infty } (x_j-x_{j-1})[/itex].
If [itex]\{ S_n \}[/itex] converges, then so do [itex]\{ x_n \}[/itex].
But [itex]\{ S_n \}[/itex] does converge, since it's lesser than [itex]|x_1-x_0|\sum _{j=0} \lambda ^{j-1}[/itex] which is convergent.
Thus [itex]\lim _{n \to \infty} x_n=x[/itex].
Now I must show that [itex]x=F(x)[/itex].
[itex]x=\lim _{n \to \infty} x_n=\lim _{n \to \infty} F(x_{n-1})=F(\lim _{n \to \infty} x_{n-1}) =F(x)[/itex]. Thus [itex]x=F(x)[/itex]. Notice here that the proof used the fact that F is continuous without ever demonstrating it. So it seems that if F is contractive over an interval, it is continuous.
Now the proof wants to show that [itex]x \in [a,b][/itex].
And here comes the part that doesn't make any sense to me.
"Furthermore, since [itex]x\in [a,b][/itex], it follows that [itex]x_n \in [a,b] \forall n[/itex], for [itex]F([a,b])[/itex] is included in [itex][a,b][/itex].
Since [itex][a,b][/itex] is closed in [itex]\mathbb{R}[/itex], F is continuous and [itex]\lim _{n \to \infty} x_n =x[/itex], it follows that [itex]x\in [a,b][/itex]."
My problem is:
The first line assumes what it wants to prove and I don't see the point of the comment that follows.
For the second line, I do not understand the implication. Is there any theorem that justifies the implication? It's not at all intuitive to me.
I'd like an explanation on how to end the proof. I mean, how to show that [itex]x \in [a,b][/itex].
Thanks in advance.
 
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  • #2
The first line sets up a sequence for you, the point [itex]x_{n+1}[/itex] and [itex]x_{n}[/itex] are different points. The second line just shows that the distance between tow consecutive point is less than the distance between the first two points, so the points are getting closer together. The important thing to realize is that as the real line is complete, cauchy sequences converge.

Does that help at all?
 
  • #3
hunt_mat said:
The first line sets up a sequence for you, the point [itex]x_{n+1}[/itex] and [itex]x_{n}[/itex] are different points. The second line just shows that the distance between tow consecutive point is less than the distance between the first two points, so the points are getting closer together. The important thing to realize is that as the real line is complete, cauchy sequences converge.

Does that help at all?

Thanks for your help. Yeah I understood this part and I understand what you mean. A Cauchy sequence does converge in R.
My problem resides in that at the end of the proof we want to show that [itex]x \in [a,b][/itex]. And the proof states
Furthermore, since [itex]x \in[a,b][/itex], it follows that [itex]x_n \in[a,b] \forall n[/itex], for [itex]F([a,b])[/itex] is included in [itex][a,b][/itex].
Since [a,b] is closed in [itex]\mathbb{R}[/itex], F is continuous and [itex]\lim _{n \to \infty } x_n=x[/itex], it follows that [itex]x \in [a,b][/itex].
I still don't understand this part. It assumes what it wants to prove as true and at the end it states that it's true. Eh?
 
  • #4
Okay, as [a,b] is a compact set, then any sequence in [a,b] will converge to a point in [a,b].
 
  • #5
hunt_mat said:
Okay, as [a,b] is a compact set, then any sequence in [a,b] will converge to a point in [a,b].

Thanks a lot! Now I know how and understand to end the proof.
The sequence [itex]\{ x_n \}[/itex] is inside [a,b] because [itex]x_{n+1}=F(x_n)[/itex] and the image of F is [itex][a,b][/itex].
Since I've already proven that [itex]\{ x_n \}[/itex] converges, it does it in [a,b].

By the way, did you mean "Okay, as [a,b] is a compact set, then any converging sequence in [a,b] will converge to a point in [a,b]."? But I get what you mean.
 
  • #6
Yes, like that.
 
  • #7
hunt_mat said:
Yes, like that.

Thanks, problem solved. I now fully understand the proof. :biggrin:
 
  • #8
Responses like that are the reason why I come out and help.
 

1. What is a contractive function?

A contractive function is a function that maps elements from one set to another set, and the distance between the images of any two elements in the first set is always smaller than the distance between the original elements. In other words, a contractive function "contracts" the distance between points in the domain and their corresponding images in the range.

2. How do you prove that a contractive function has a fixed point?

To prove that a contractive function has a fixed point, we can use the Banach fixed point theorem. This theorem states that if a function is contractive on a complete metric space, then it has a unique fixed point in that space. Therefore, to prove that a contractive function has a fixed point, we need to show that the function is contractive and that the metric space is complete.

3. What is a fixed point of a function?

A fixed point of a function is a point in the domain that maps to itself under the function. In other words, when we apply the function to a fixed point, the output is equal to the input. For example, if we have a function f(x) = x^2, then the fixed points are 0 and 1, because f(0) = 0 and f(1) = 1.

4. Why is it important to prove that a contractive function has a fixed point?

Proving that a contractive function has a fixed point is important because it guarantees the existence of a solution to certain problems. For example, in optimization or numerical analysis, we often encounter situations where we need to find a point that satisfies a certain condition or equation. If we can prove that the function defining the problem is contractive, then we can use the Banach fixed point theorem to show that a solution exists.

5. Can a non-contractive function have a fixed point?

Yes, a non-contractive function can have a fixed point. However, the Banach fixed point theorem only guarantees the existence of a fixed point for contractive functions. For non-contractive functions, we would need to use other methods to prove the existence of a fixed point, such as the Brouwer fixed point theorem.

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