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Proof that a contractive function has a fixed point

  1. Jun 30, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    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].


    2. Relevant 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.


    3. 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.
     
  2. jcsd
  3. Jun 30, 2011 #2

    hunt_mat

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    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 realise is that as the real line is complete, cauchy sequences converge.

    Does that help at all?
     
  4. Jun 30, 2011 #3

    fluidistic

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    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
    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?
     
  5. Jun 30, 2011 #4

    hunt_mat

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    Okay, as [a,b] is a compact set, then any sequence in [a,b] will converge to a point in [a,b].
     
  6. Jun 30, 2011 #5

    fluidistic

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    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.
     
  7. Jun 30, 2011 #6

    hunt_mat

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    Yes, like that.
     
  8. Jun 30, 2011 #7

    fluidistic

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    Thanks, problem solved. I now fully understand the proof. :biggrin:
     
  9. Jun 30, 2011 #8

    hunt_mat

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    Responses like that are the reason why I come out and help.
     
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