Proof that a contractive function has a fixed point

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The discussion centers on understanding the proof that a contractive function F on the interval [a,b] has a unique fixed point x such that F(x) = x. A contractive function is defined by the existence of a constant λ in (0,1) that bounds the distance between F values. The proof involves showing that the sequence generated by iterating F converges, leveraging the completeness of the real numbers. The participants clarify that since [a,b] is compact and closed, any converging sequence within it will converge to a point in [a,b]. Ultimately, the proof is confirmed to be understood, highlighting the importance of the properties of compact sets in the context of fixed points.
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Homework Statement


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

Homework Equations


Definition of a contractive function: F is contractive over [a,b] if and only if there exist \lambda such that 0<\lambda <1 and |F(y)-F(x)| \leq \lambda |y-x| \forall x and y \in [a,b]. 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 x_{n+1}=F(x_n) for n \geq 1.
\exists \lambda \in (0,1) such that |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|.
I can take x_n=x_0+S_n where S_n=\sum _{j=1}^{\infty } (x_j-x_{j-1}).
If \{ S_n \} converges, then so do \{ x_n \}.
But \{ S_n \} does converge, since it's lesser than |x_1-x_0|\sum _{j=0} \lambda ^{j-1} which is convergent.
Thus \lim _{n \to \infty} x_n=x.
Now I must show that x=F(x).
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). Thus x=F(x). 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 x \in [a,b].
And here comes the part that doesn't make any sense to me.
"Furthermore, since x\in [a,b], it follows that x_n \in [a,b] \forall n, for F([a,b]) is included in [a,b].
Since [a,b] is closed in \mathbb{R}, F is continuous and \lim _{n \to \infty} x_n =x, it follows that x\in [a,b]."
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 x \in [a,b].
Thanks in advance.
 
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The first line sets up a sequence for you, the point x_{n+1} and x_{n} 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?
 
hunt_mat said:
The first line sets up a sequence for you, the point x_{n+1} and x_{n} 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 x \in [a,b]. And the proof states
Furthermore, since x \in[a,b], it follows that x_n \in[a,b] \forall n, for F([a,b]) is included in [a,b].
Since [a,b] is closed in \mathbb{R}, F is continuous and \lim _{n \to \infty } x_n=x, it follows that x \in [a,b].
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?
 
Okay, as [a,b] is a compact set, then any sequence in [a,b] will converge to a point in [a,b].
 
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 \{ x_n \} is inside [a,b] because x_{n+1}=F(x_n) and the image of F is [a,b].
Since I've already proven that \{ x_n \} 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.
 
Yes, like that.
 
hunt_mat said:
Yes, like that.

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

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