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Proving Contractive

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  1. Jul 6, 2016 #1
    1. The problem statement, all variables and given/known data
    I worked on this question and I made it so far, and now I am stuck on how to finish it. Here is the problem and below I will explain what I attempted.

    upload_2016-7-6_8-41-4.png

    2. Relevant equations


    3. The attempt at a solution
    I know looking at the last part about using previous homework, I want to prove that the function is contractive. If I can prove that, I can use a previous assignment which says contractive functions have a fixed point. Here is what I have in my proof so far.
    upload_2016-7-6_8-43-44.png

    I am stuck now and not completely sure where to go next. Any advice is appreciated.
     
  2. jcsd
  3. Jul 6, 2016 #2

    fresh_42

    Staff: Mentor

    I don't understand what your last line in the picture means. Nevertheless, are you allowed to apply Banach's fixed-point theorem?
     
  4. Jul 6, 2016 #3
    Yes I could use that theorem. Would I just need to completely erase everything I did and start over?

    I was getting the last part from using the previous homework problems below.

    upload_2016-7-6_10-33-2.png
     
  5. Jul 6, 2016 #4

    fresh_42

    Staff: Mentor

    So you are already done. The other exercise (Lipschitz continuity for ##f## on the condition on ##f'##) showed you that ##b## is your Lipschitz constant. Since ##b<1, \; f## is a contraction and the fixed-point theorem applies.
     
  6. Jul 6, 2016 #5
    What other exercise are you talking about? I am lost on that statement. I understand the reasoning afterwards.
     
  7. Jul 6, 2016 #6
    I think you are looking at the proof I found in the book and asked about. I get it now.
     
  8. Jul 6, 2016 #7

    fresh_42

    Staff: Mentor

    Mean value theorem: ##|f(x) - f(y)| = f'(t) |x-y| < b |x-y|##, i.e. ##f## is a contraction because ##b<1##.
     
  9. Jul 6, 2016 #8
    Thanks I see it now. I forgot I could use that. Thanks.
     
  10. Jul 6, 2016 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Even easier: just use very elementary methods. If ##|f'(t)| \leq m## on ##R## (or an an interval ##[a,b]##), that means that ##-m \leq f'(t) \leq m##. Thus, for ##x_1 < x_2## we have
    [tex] f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) \, dt \leq \int_{x_1}^{x_2} m \, dt = m (x_2 - x_1) [/tex]
    and
    [tex] f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) \, dt \geq \int_{x_1}^{x_2} (-m) \, dt = -m (x_2 - x_1) [/tex]
    Thus, ##|f(x_2) - f(x_1)| \leq m |x_2 - x_1|##.
     
  11. Jul 6, 2016 #10
    That works too. Thanks.
     
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